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Formulae, Equations and Amount of Substance

1. Formulae2. Formulae of ionic compounds3. Relative molecular mass4. Chemical equations5. Further chemical equations6. Ionic equations7. The mole8. Calculating masses and Mr values9. Calculating formulae10. Mass equations11. Calculations on gas volumes12. The volumes of gases involved in reactions13. Quantitative chemistry with solutions14. Finding an unknown concentration by titration15. Key words and ideas16. Questions17. Marking schemes

1 FormulaeIntroduction

Every pure chemical has a formula. Chemical reactions are often described using balanced chemical equations of various types. In order to find out what is the correct formula of a substance, or what is the correct equation to describe a chemical reaction, measurements of the amounts of various substances involved can be made. This booklet is intended to provide an independent learning scheme to help you understand how formulae and equations are decided upon, and how to handle the mathematical methods that are inevitable when measurements have been made. The booklet includes worked examples to show how the methods work and a variety of questions so you can practice the methods for yourself.

The Ar values used are those on the OCR Data Sheet for Chemistry.

Compounds are made of different atoms in fixed proportions

A substance made of one type of atom is called an element. There are currently 115 known elements. However, most of the substances on this planet are made of more than one type of atom, chemically combined together. Such substances are called compounds. ‘Chemically combined’ means that two (or more) different types of atom are bonded (held) to each other by forces called chemical bonds. The different types of atom a compound is made of are always present in set, fixed, proportions. For example, water is a compound of hydrogen atoms and oxygen atoms, and every drop of water contains twice as many hydrogen atoms as oxygen atoms. This is because water is made of small groups of atoms called molecules. One water molecule consists of two hydrogen atoms and one oxygen atom, chemically bonded together. Since every water molecule is made thus, any drop of water (which will contain many billions of molecules) contains twice as many hydrogen atoms as oxygen atoms.

The molecular formula of a compound tells you about the atoms in its moleculesThe molecular formula of a compound tells you two things:

• It tells you which elements the compound is made of. • It tells you how many atoms of each element are in one molecule of the compound.

The molecular formula of water is H2O. The 2 means that one water molecule contains two hydrogen atoms. The lack of a number after the O means that a water molecule contains one oxygen atom.

Question 1What is the molecular formula of each of the compounds whose molecules are drawn below? (These drawings are called displayed formulae.)

a. b. c. d.

e. C=O f. g.

2 © Cambridge University Press 2005 Formulae, equations and amount of substance

An empirical formula tells you the simplest ratio of the numbers of each type of atom in a compoundChemists have more than one way to express the formula of a compound. Instead of using a molecular formula, a chemist may describe a compound using an empirical formula. The empirical formula of a compound gives the information about the elements it contains in their simplest ratios. We can take the compound hydrogen peroxide as an example, the molecules of which are H–O–O–H. Its molecular formula is H2O2 but its empirical formula is HO, since the ratio of hydrogen atoms to oxygen atoms is 1:1. For most, but not all, compounds the molecular formula and the empirical formula are the same. For water, for example, both are H2O, and for carbon dioxide both are CO2.

Question 2Copy and complete the table below.

Name of Molecule (displayed Molecular formula Empirical formulacompound formula)

EthaneH H

C2H6 CH3

H HC C

H H

EtheneH H

C C

H H

PropaneH H H

H HC C C

H H H

Propene H H

C C

H C H

H H

Benzene H

H C HC C

H HC C

C

H

Phenol H

H C O – HC C

H HC C

C

H

Lactic acid H HO

H CC CO H

H OH

[12]

Formulae, equations and amount of substance © Cambridge University Press 2005 3

Many non-metal elements are molecular

The empirical formula of an element is simple. A sample of any element contains one type of atom only, so there is no ratio to calculate. For example, the empirical formula of oxygen is O, sulphur is S and magnesium is Mg.

In a sample of a metal the atoms are packed as individual atoms in a vast array called a lattice, so for a metal the molecular formula is also that of the single atom, e.g. sodium’s is Na.

Many non-metal elements are different, as their atoms form small groups (molecules) held together by chemical bonds. Oxygen atoms bond together to form molecules consisting of two atoms so the molecular formula of oxygen is O2.

Question 3Copy and complete the table below.

Name of Molecule (displayed Molecular formula Empirical formulaelement formula)

OxygenO O

O2 O

NitrogenN N

FluorineF F

HydrogenH H

Sulphur S SS S

S SS S

Phosphorus P

PP P

IodineI I

[12]

Not all non-metal elements have a structure with the atoms in small groups. Both carbon and silicon have their atoms bonded in a lattice, so their formulae are just given as C and Si. The gases helium, neon, argon, krypton, xenon and radon consist of totally independent single atoms not bonded to any other, so their formulae are also given as He, Ne, etc.

The substances whose molecules are drawn on these pages all have a simple molecular structure. The bonds that hold the atoms together within the molecules are covalent bonds. These pages include examples of single, double and triple covalent bonds. The next section discusses compounds held together by a very different type of chemical bond – the ionic bond.

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2 Formulae of ionic compoundsThe particles in some compounds are ions

The compounds whose formulae you have studied on these pages up to now are all covalent compounds with simple molecular structures. Many compounds, however, have their components held together by ionic bonds. Such compounds usually consist of a metal element combined with one or more non-metal elements:

• The metals are present as positive ions – atoms that have lost one or more electrons. • The non-metals are present as negative ions – atoms or small groups of atoms that

have gained one or more electrons.

The positive and negative ions in a sample of the compound do not form a structure consisting of small, distinct groups. The structures formed are vast arrays of ions called giant lattices. The constituent ions are present in fixed proportions, so an ionic compound also has a formula.

Ions can be positive or negative, simple or compoundMost positive ions are simple ions – this means that they consist of single atoms that have lost one or more electrons. Examples of simple ions that you will have come across are the sodium ion, Na+, and the copper ion, Cu2+. The most important exception is the ammonium ion. Each ammonium ion consists of four hydrogen atoms bonded covalently to one central nitrogen atom, and since this group of five atoms has always lost one electron from its constituent atoms it has a single positive charge – a ‘one plus’ charge. The ammonium ion is a compound ion, and is shown as NH4

+. The table below includes many of the positive ions whose names and charges you need to know.

Name Symbol Charge Name Symbol Charge

Lithium Li+ 1+ Zinc Zn2+ 2+

Sodium Na+ 1+ Copper(I) Cu+ 1+

Potassium K+ 1+ Copper(II) Cu2+ 2+

Magnesium Mg2+ 2+ Cobalt(II) Co2+ 2+

Calcium Ca2+ 2+ Iron(II) Fe2+ 2+

Barium Ba2+ 2+ Iron(III) Fe3+ 3+

Aluminium Al3+ 3+ Tin(II) Sn2+ 2+

Hydrogen H+ 1+ Lead(II) Pb2+ 2+

Ammonium NH4+ 1+ Manganese(II) Mn2+ 2+

Silver Ag+ 1+ Chromium(III) Cr3+ 3+

Assignment 1Learn the table of positive ions above.

• Copy the table onto a blank sheet of paper and fill in the ‘Name’ columns only, using the table.

• Cover this page and complete the ‘Symbol’ and ‘Charge’ columns from memory. Repeat until you get it right every time.

• See if you can still complete the assignment successfully in an hour’s time, and again tomorrow. Put it in your diary to be redone once a week between now and your Module 1 exam. Use a copy of the Periodic Table to help you at first.

• Note that the Group I elements all form simple ions with a 1+ charge, and the Group II elements all form simple ions with a 2+ charge. You could also include rubidium, caesium, beryllium and strontium in the assignment.

Negative ions can also be simple, consisting of single atoms that have gained one or more electrons. However, many important negative ions are compound ions, consisting of small groups of covalently bonded atoms that have gained one or more electrons. There are many different compound negative ions, most of which contain oxygen atoms and some of which contain metal atoms, so metals in compounds are not found exclusively as positive ions. The table below includes many of the negative ions whose charges you need to know.

Name Symbol Charge Name Symbol Charge

Fluoride F– 1– Sulphate(VI) SO42– 2–

Chloride Cl– 1– Sulphate(IV) SO32– 2–

Bromide Br– 1– Thiosulphate S2O32– 2–

Iodide I– 1– Nitrate(V) NO3– 1–

Oxide O2– 2– Nitrate(III) NO2– 1–

Sulphide S2– 2– Manganate(VII) MnO4– 1–

Hydroxide OH– 1– Manganate(VI) MnO42– 2–

Silicate SiO32– 2– Dichromate(VI) Cr2O7

2– 2–

Carbonate CO32– 2– Chromate(VI) CrO4

2– 2–

Hydrogencarbonate HCO3– 1– Phosphate(V) PO4

3– 3–

Assignment 2Learn the table of negative ions above, using the method given in Assignment 1. Note that the Group VII elements form simple ions with a 1– charge and the Group VI elements form simple ions with a 2– charge. Note also how the names change, for example from ‘chlorine’ for the element to ‘chloride’ for the ion.

It is essential that you know the charges on these ions and the formulae of the compound ions in order to be able to work out the formulae of ionic compounds.

The amount of plus charge must equal the amount of minus charge in one formula unit of an ionic compoundAll ionic compounds consist of positive ions and negative ions arranged in a lattice. The name of the compound gives the positive ion first, e.g. zinc oxide, ammonium carbonate, iron(III) nitrate. Every ionic compound has a formula; the small group of particles represented by the formula is called the formula unit of the compound. For example the formula of common salt (sodium chloride) is NaCl, therefore one formula unit of common salt consists of one sodium ion and one chloride ion.

The formula of an ionic compound tells you the relative amounts of the two types of ion present in one formula unit of the compound. These relative amounts are such that the amount of plus charge and the amount of minus charge in one formula unit are always the same.

Worked examples1 Zinc oxide The zinc ion is Zn2+. The oxide ion is O2–. One Zn2+ ion has two units

of plus charge, one O2– ion has two units of minus charge; these amounts are the same. Therefore for every one Zn2+ ion in the zinc oxide formula unit there is one O2– ion. The formula is ZnO.

2 Ammonium carbonate The ammonium ion is NH4+, the carbonate ion is CO3

2–. One NH4

+ ion has one unit of plus charge, one CO32– ion has two units of minus

charge. Therefore for every one CO32– ion in the ammonium carbonate formula unit

there are two NH4+ ions. The formula is (NH4)2CO3.

Note the use of brackets in the formula of ammonium carbonate. This is necessary on occasions with compound ions. (NH4)2 means two of the NH4 units. NH42 would cause obvious confusion.

3 Iron(III) nitrate(V) The iron(III) ion is Fe3+, the nitrate(V) ion is NO3–. One Fe3+ ion has

three units of plus charge, one NO3– ion has one unit of minus charge. Therefore for

every one Fe3+ ion in the formula unit there are three NO3– ions. The formula is

Fe(NO3)3.

These three examples are summarised in this table.

Compound Positive ion Negative ion Number of each Formula ofpresent present to balance charges compound

Zinc oxide Zn2+ O2– 1 Zn2+ and ZnO

1 O2–

Ammonium NH4+ CO3

2– 2 NH4

+ and (NH4)2CO3

carbonate 1 CO32–

Iron(III) nitrate(V) Fe3+ NO3– 1

Fe3+ and Fe(NO3)3

3 NO3–

Question 1What is the formula of each of the following compounds?

a sodium chloride

b lithium oxide

c lead(II) nitrate(V)

d ammonium dichromate(VI)

e potassium manganate(VI)

f potassium manganate(VII)

g aluminium oxide

h sodium phosphate(V)

i copper(I) chloride

j copper(II) hydroxide

k iron(III) sulphate(VI)

l sodium sulphate(IV)

m calcium carbonate

n calcium hydrogencarbonate [14]


3 Relative molecular massThe mass of a molecule relative to carbon-12 is its Mr

In order to be able to work out the masses of reactants and products involved in chemical reactions, you must be able to calculate the Mr of a compound. For a compound with a simple molecular structure this means the mass of one molecule of the compound, measured relative to 1/12-th of the mass of one atom of carbon-12. For a compound with a giant lattice structure this means the mass of one formula unit – the atoms represented in the formula – of the compound relative to 1/12-th of the mass of one atom of carbon-12. To calculate the Mr of a compound you add together the Ar values of the atoms it contains in the amounts shown in the formula. The Ar values required for all the examples on these pages are found on the Periodic Table. You can find a copy of the Periodic Table on page 191 of Chemistry 1.

Worked examples1 What is the Mr of ammonia?

i Write down its formula. As this compound has a simple molecular structure its formula needs to be learnt. It is NH3.

ii Add together the Ar values of the atoms involved. N is 14.0, H is 1.0. Since there is one N and three H in the molecular formula, the Mr is 14.0 + (3 1.0). The answer = 17.0

2 What is the Mr of calcium hydroxide?

i Write down its formula. As this is an ionic compound you can work it out using the technique from the previous pages. The calcium ion is Ca2+. The hydroxide ion is OH–. Therefore for every one calcium ion in the structure there are two hydroxide ions. The formula is Ca(OH)2.

ii Add together the Ar values of the atoms involved. Ca is 40.1, O is 16.0, H is 1.0. Since there is one Ca, two O and two H in one formula unit, the Mr is 40.1 + (2 16.0) + (2 1.0). The answer = 74.1

Question 1What is the Mr of the following compounds?

They all have simple molecular structures. Refer to pages 2 and 3 for some help with formulae.

a Water e Phenol

b Benzene f Phosphorus pentachloride

c Propene g Lactic acid

d Sulphuric(VI) acid [7]

Question 2The elements listed in Question 3, Section 1 (Formulae), all have simple molecular structures. This means that, as well as having an Ar, each of these elements also has an Mr.

What is the Mr of the following elements?

a Hydrogen e Phosphorus

b Nitrogen f Sulphur

c Oxygen g Iodine

d Fluorine [7]

Question 3Chlorine and bromine have simple molecular structures, with molecular formulae Cl2 and Br2.

a What is the Mr of chlorine?

b What is the Mr of bromine? [2]

Question 4Work out the formula and the Mr for each of the following compounds:

a Sodium thiosulphate f Ammonium sulphate(VI)

b Zinc sulphide g Copper(II) chromate(VI)

c Potassium dichromate(VI) h Sodium sulphate(VI)

d Aluminium nitrate(V) i Silver oxide

e Sodium silicate j Lead(II) bromide [20]

Compounds with water of crystallisation

Many ionic compounds in solid crystal form have water molecules included in their lattices. Such crystals will have been produced by taking an aqueous solution of the compound (aqueous = dissolved in water) and allowing the water to evaporate slowly. Not all the water evaporates, however, some of it remains as part of the solid crystals. This remaining water is called water of crystallisation. The amount of water is not random; a particular ionic compound will retain an exact number of water molecules associated with each formula unit in its lattice. One of the best known such compounds is copper(II) sulphate(VI). Its formula is CuSO4, but in the lattice structure there are five water molecules associated with each CuSO4 formula unit. Its formula is therefore written CuSO4.5H2O. The use of the ‘dot’ before the 5H2O is a convention to remind you that the water molecules are water of crystallisation.

If copper(II) sulphate(VI) solution is evaporated to dryness by gentle heating, blue crystals are formed. If these blue crystals are heated further the water of crystallisation can be driven off, leaving a very pale blue powder called anhydrous copper(II) sulphate(VI). Anhydrous means ‘without water’. Copper(II) sulphate(VI) can therefore exist in two forms, a hydrated (with water) form and an anhydrous form. You should note that:

• The hydrated forms have a glassy appearance and look much more ‘crystalline’ than the anhydrous forms.

• If we are dealing with a coloured compound, as in the case of copper(II) sulphate(VI), the hydrated form is usually much more strongly coloured.

The water of crystallisation must be included in the Mr

If you wish to calculate the Mr of CuSO4.5H2O, the mass of the five water molecules must be included. Its Mr is therefore 63.5 + 32.1 + (4 16.0) + 5(1.0 + 1.0 + 16.0). The answer = 249.6

Question 5a What is the Mr of the anhydrous form of sodium carbonate, Na2CO3?

b What is the Mr of the hydrated form, Na2CO3.10H2O? [2]

Assignment 1Look again at your answers to Question 4. Ask your teacher for a bottle of each of these chemicals. Read the labels carefully. Do any of these substances contain water of crystallisation? For any that do, work out a correct value of the Mr.


4 Chemical equationsAll the elements in the reactants appear in the products

A chemical reaction involves a permanent change in which the substances there at the beginning (the reactants) turn into the substances there at the end (the products). For example silver nitrate and sodium chloride react together. The equation is:

AgNO3 + NaCl → AgCl + NaNO3

Silver nitrate and sodium chloride are the reactants, silver chloride and sodium nitrate are the products. Note that the reactants and products contain the same five elements – silver, nitrogen, oxygen, sodium and chlorine. This is the case in every chemical reaction, whatever the elements happen to be. In a chemical reaction every element in the reactants must appear in the products and every element in the products must appear in the reactants. In reactions elements change what they are combined with but they are not created or destroyed.

Reactions take place on an atomic level

Chemical reactions take place between molecules or formula units of the substances involved. The equation tells you two things:

• Firstly, it tells you what the molecules consist of. For this to be true their formulae must be written correctly in the equation. The equation above tells you the formula of each reactant and product molecule.

• Secondly, it tells you how many molecules of each substance are involved. The equation above says that one molecule of each reactant reacts to give one molecule of each product. Working out the number of molecules is called balancing the equation.

A balanced equation has the same numbers of atoms on both sides

When reactant substances react and product substances are produced the total number of atoms of each element in the reactants must be the same as the total number of atoms of each element in the products. By looking at what the reactant and product molecules consist of it is usually possible to deduce how many of each are involved.

Worked examples1 What is the equation for the reaction between nitrogen and hydrogen, producing

ammonia?

i Write an equation showing the formula of each substance: N2 + H2 → NH3

After reaching this point in any equation do not change any formulae. Once they are correct leave them correct!

ii Look at the equation carefully and decide how many molecules of nitrogen will react with how many molecules of hydrogen and how many molecules of ammonia are produced. Start by looking at the formula of ammonia. It contains three times as many hydrogen atoms as nitrogen atoms, so, since nitrogen molecules and hydrogen molecules both contain two atoms, one nitrogen molecule must react with three hydrogen molecules. These four molecules consist of exactly enough atoms to make two ammonia molecules, so the equation is:

N2 + 3H2 → 2NH3

This equation has the same eight atoms on the reactant side as it does on the product side. It is said to be balanced. It is a correct description of what happens when ammonia is formed.

2 What is the equation for the reaction between phosphorus and oxygen, producing phosphorus oxide (P4O10)? i Write an equation showing the formula of each substance:

P4 + O2 → P4O10

ii Balance the equation. There are already four atoms of phosphorus on both sides, so it is said to be balanced for phosphorus. You can explain the origin of the ten oxygen atoms in the product molecule by stating that five oxygen molecules react with the phosphorus molecule:

P4 + 5O2 → P4O10

Question 1The reactions in Worked examples 1 and 2 involve the making of two compounds from their component elements. Such reactions are called synthesis reactions or syntheses.

Write balanced chemical equations for the following syntheses. The formulae of the covalent products (e.g. silicon dioxide) are given but you should be able to work out the formulae of the ionic products (e.g. sodium oxide).

a sodium + oxygen → sodium oxide

b aluminium + oxygen → aluminium oxide

c silicon + oxygen → silicon dioxide (SiO2)

d magnesium + chlorine → magnesium chloride

e aluminium + chlorine → aluminium chloride (Al2Cl6)

f silicon + chlorine → silicon tetrachloride (SiCl4)

g phosphorus + chlorine → phosphorus pentachloride (PCl5) [14]

Some reactions have only one reactant

Some compounds are unstable. They are liable to break down when they are heated, either into the elements they are made of, or into simpler compounds, or into a combination of elements and compounds. Such reactions are called decompositions and only one reactant is involved.

Worked examples3 Magnesium nitrate(V) decomposes to give magnesium oxide, nitrogen dioxide and

oxygen when heated. Write an equation for this reaction.

i Write an equation showing the formula of each substance: Mg(NO3)2 → MgO + NO2 + O2

ii Balance the equation. Where one of the substances involved is an element (oxygen in this case) it is usually easiest to leave it until last. The equation is already balanced for magnesium, so look at nitrogen first. There are two nitrogen atoms on the left so the equation will be balanced for nitrogen if you suppose two nitrogen dioxide molecules are produced:

Mg(NO3)2 → MgO + 2NO2 + O2

iii This only leaves the oxygen. The equation will balance (with six atoms of oxygen on each side) if you state that half an oxygen molecule is produced:

→ 1Mg(NO3)2 MgO + 2NO2 + –O2

2

However ‘half an oxygen molecule’ does not really exist. This can be put right by doubling everything:

2Mg(NO3)2 → 2MgO + 4NO2 + O2

Question 2Write an equation for each of the following decomposition reactions:

a Calcium nitrate(V) decomposing to give calcium oxide, nitrogen dioxide and oxygen.

b Lithium nitrate(V) decomposing to give lithium oxide, nitrogen dioxide and oxygen.

c Magnesium carbonate decomposing to give magnesium oxide and carbon dioxide.

d Magnesium hydroxide decomposing to give magnesium oxide and water.

e Calcium hydrogencarbonate decomposing to give calcium carbonate, water andcarbon dioxide. [10]

Question 3Why can you not be asked to write an equation for the decomposition of an element? [1]

Question 4The explosion of TNT (C7H5N3O6) is actually a very rapid decomposition reaction.Write a balanced equation for the decomposition of TNT into water, carbonmonoxide, nitrogen and carbon. [2]


5 Further chemical equationsEquations can be used to illustrate acidic and basic behaviour

If you are explaining the acidic or basic nature of a substance the best way is often to use an equation to describe one of its reactions:

• The acidic nature of a substance can be shown by an equation for its reaction with water to form a solution containing hydrogen ions, or its ability to react with an alkali, e.g. sodium hydroxide.

• The basic nature of a substance can be shown by an equation for its reaction with water to form a solution containing hydroxide ions, or its ability to react with an acid, e.g. hydrochloric acid.

Worked examples1 Show by equations how calcium oxide behaves as a base in its reactions a

with water to give calcium hydroxide and b with hydrochloric acid to give calcium chloride and water.

a i Write an equation showing the formula of each substance: CaO + H2O → Ca(OH)2

ii Balance the equation. You can see that it is already balanced, i.e. one formula unit of calcium oxide reacts with one molecule of water to give one formula unit of calcium hydroxide. Leave the equation as it is.

b i Write an equation with the formula of each substance: CaO + HCl → H2O + CaCl2

ii Balance the equation. It is already balanced for oxygen and calcium. There are two hydrogen atoms and two chlorine atoms in the products but only one of each in the reactants, so two formula units of hydrochloric acid must react with each formula unit of calcium oxide:

CaO + 2HCl → H2O + CaCl2

2 Show by an equation how phosphorus(V) oxide (P4O10) behaves as an acid in its reaction with water to give phosphoric(V) acid (H3PO4).

i Write an equation showing the formula of each substance: P4O10 + H2O → H3PO4

ii Balance the equation. This is a harder example. It is not obvious where to start. (At times you will need to have the confidence to start to balance the equation, and then be prepared to start again if your first attempt doesn’t work.) Start with the phosphorus. The reactants have four phosphorus atoms and the product only one, so if we assume that four product molecules are produced the equation will be balanced for phosphorus:

P4O10 + H2O → 4H3PO4

The product now contains more hydrogen and oxygen than the reactants, but these elements will balance too if we state that six water molecules react with each phosphorus(V) oxide molecule:

P4O10 + 6H2O → 4H3PO4

Question 1Write balanced chemical equations for the following reactions that illustrate basic behaviour:

a copper(II) oxide + sulphuric(VI) acid → copper(II) sulphate(VI) + water

b aluminium oxide + sulphuric(VI) acid → aluminium sulphate(VI) + water

c magnesium oxide + water → magnesium hydroxide

d lead(II) oxide + nitric(V) acid → lead(II) nitrate(V) + water [8]

Question 2Write balanced chemical equations for the following reactions that illustrate acidic behaviour. Some of the formulae are given in brackets to help you.

a sulphur trioxide (SO3) + water → sulphuric(VI) acid b carbon dioxide + water → carbonic acid (H2CO3)

c silicon dioxide + sodium hydroxide → sodium silicate + water

d lead(II) oxide + sodium hydroxide + water → sodium plumbate(II) (Na2Pb(OH)4)

e nitrogen dioxide + water → nitric(V) acid + nitrogen monoxide [10]

You have probably noticed in Questions 1d and 2d that you have written equations which show that lead(II) oxide can behave as a base and as an acid. Substances that behave like this are said to be amphoteric.

Rapid reactions involving oxygen are called combustion reactions

Many organic substances will burn if mixed with oxygen or air. A little heat (either a flame or a spark) is usually all that is needed to start the reaction. Such reactions are called burning or combustion reactions. If the organic substance is allowed sufficient oxygen it will burn completely. This means that the products will be carbon dioxide and water, provided the starting material consisted of no elements other than carbon, hydrogen and oxygen.

Worked examples3 Propane is a hydrocarbon; its formula is C3H8. Write a balanced equation for its

complete combustion in oxygen.

i Write an equation showing the formula of each substance: C3H8 + O2 → CO2 + H2O

ii Balance the equation for carbon and hydrogen. As oxygen is present as an element it should be left until last. Every carbon atom in each propane molecule will combine with sufficient oxygen to give one carbon dioxide molecule, therefore three carbon dioxide molecules form from each propane molecule. Every two hydrogen atoms in each propane molecule will combine with sufficient oxygen to give one water molecule, therefore four water molecules form from each propane molecule:

C3H8 + O2 → 3CO2 + 4H2O

iii Balance the equation for oxygen. There are ten oxygen atoms in the products, therefore ten oxygen atoms must appear in the reactants:

C3H8 + 5O2 → 3CO2 + 4H2O

4 Butanone is a compound of hydrogen, carbon and oxygen; its formula is C4H8O. Write a balanced equation for its complete combustion in oxygen.

i Write an equation showing the formula of each substance: C4H8O + O2 → CO2 + H2O

ii Balance the equation for carbon and hydrogen. The products will contain four carbon dioxide molecules and four water molecules per molecule of butanone that burns:

C4H8O + O2 → 4CO2 + 4H2O

iii Balance the equation for oxygen. There are twelve oxygen atoms in the products, therefore twelve oxygen atoms must appear in the reactants. Butanone contains one oxygen atom per molecule, therefore eleven more are needed:

1 →C

4H8O + 5–O2 4CO2 + 4H2O

2

iv If the whole equation is doubled, the somewhat misleading ‘half’ will disappear: 2C4H8O + 11O2 → 8CO2 + 8H2O

Question 3Write a balanced chemical equation for the complete combustion of the following compounds. Their formulae are written after their names.

a Methane, CH4

b Pentane, C5H12

c Benzene, C6H6

d Propene, C3H6

e Ethanoic acid, C2H4O2

f Glucose, C6H12O6

g Propanone, C3H6O

h Phenol, C6H6O [8]


6 Ionic equationsIonic equations are useful for describing acid/base reactions

Ionic equations deal only with the particular ions that undergo change during a chemical reaction. The other ions that are present in the reaction mixture, but which do not undergo change, are omitted from the ionic equation. This means that ionic equations cannot tell the whole story of a chemical reaction. Instead they focus attention where it is most relevant.

Take, for example, calcium oxide behaving as a base. The equations for calcium oxide reacting with water and with hydrochloric acid are shown in Worked example 1, Section 5 (Further chemical equations).

The first of these equations looked like this:

CaO + H2O → Ca(OH)2

If ionic charges are added it looks like this:

Ca2+O2– + H2O → Ca2+(OH)–2

The Ca2+ ion has not changed. It is the O2– ion and the H2O molecule which have reacted, giving two OH– ions. An ionic equation for this reactions looks like this:

O2– + H2O → 2OH–

The ionic equation cannot possibly tell the whole story, since O2– cannot exist on its own as a reactant and neither can OH– as a product, but it does concentrate on the species that have undergone change.

The second of the calcium oxide equations looked like this:

CaO + 2HCl → H2O + CaCl2

If ionic charges are added it looks like this:

Ca2+O2– + 2H+Cl– → H2O + Ca2+Cl–2

The Ca2+ ion has not changed. Neither have the two Cl– ions. It is the O2– ion and the two H+ ions which have reacted, giving an H2O molecule. An ionic equation for this reaction looks like this:

O2– + 2H+ → H2O

Worked examples1 Write an ionic equation for the reaction between calcium hydroxide and

sulphuric(VI) acid.

i Start with a balanced chemical equation for the reaction: Ca(OH)2 + H2SO4 → 2H2O + CaSO4

ii Add the ionic charges:

Ca2+(OH)–2 + H+2SO4

2– → 2H2O + Ca2+SO42–

iii Write the ionic equation by omitting all ions that are the same in the products as they are in the reactants:

2OH– + 2H+ → 2H2O

or OH– + H+ → H2O

2 Write an ionic equation for the reaction between nitric(V) acid and sodium hydrogencarbonate.

i Start with a balanced chemical equation for the reaction: HNO3 + NaHCO3 → NaNO3 + H2O + CO2

ii Add the ionic charges:

H+NO3– + Na+HCO3

– → Na+NO3– + H2O + CO2

iii Write the ionic equation by omitting all ions that are the same in the products as they are in the reactants:

H+ + HCO3– → H2O + CO2

Ionic equations are particularly useful when used to describe acid/base reactions, as in the worked examples above, since they focus attention on the acidic parts of the acids, the H+ ions, and the basic parts of the bases, which in these examples are the OH– ions and the HCO3

– ions. When you have answered the questions below you will see how similar the ionic equations are from apparently different acid/base reactions.

Question 1Write ionic equations for the following acid/base reactions.

a Sodium hydroxide reacting with hydrochloric acid.

b Calcium carbonate reacting with nitric(V) acid.

c Potassium hydrogencarbonate reacting with sulphuric(VI) acid.

d Magnesium carbonate reacting with sulphuric(VI) acid.

e Calcium hydrogencarbonate reacting with hydrochloric acid.

f Ammonium hydroxide reacting with nitric(V) acid. [18]

Question 2Write ionic equations for the reactions in Question 1, Section 5 (Furtherchemical equations). [12]

Ionic equations are useful for describing redox reactions

In certain reactions the changes that take place are accompanied by electrons moving from one species to another. Such reactions are called redox reactions. In describing such a reaction an ionic equation draws attention to the changes in ionic charges caused by the gain and loss of electrons.

Remember:• If a species loses electrons in a reaction it is oxidised.

Oxidation Is Loss of electrons – OIL • If a species gains electrons in a reaction it is reduced.

Reduction Is Gain of electrons – RIG

For example, when magnesium reacts with sulphuric(VI) acid the balanced chemical equation is:

Mg + H2SO4 → H2 + MgSO4

With ionic charges added it looks like this:

Mg + H+2SO4

2– → H2 + Mg2+SO42–

The ionic equation looks like this:

Mg + 2H+ → Mg2+ + H2

It is clear from the ionic equation that the magnesium atoms are losing two electrons each, therefore they are oxidised. The hydrogen ions gain one electron each, therefore they are reduced.

Note that this ionic equation is balanced for charge. There is a total of 2+ charge on each side of the arrow. Ionic equations must always be balanced for charge.

Question 3Write ionic equations for the following redox reactions.

a Zinc reacting with hydrochloric acid.

b Iron reacting with hydrochloric acid. (One of the products is an iron(II) compound.)

c Aluminium reacting with sulphuric(VI) acid.

d Copper(II) sulphate(VI) solution reacting with iron.

e Zinc oxide reacting with magnesium.

f Copper reacting with silver nitrate(V) solution. [18]

Question 4Identify which substance is oxidised and which substance is reduced in each partof Question 3. [12]

Question 5When potassium manganate(VII) reacts with sulphuric(VI) acid and iron(II) sulphate(VI), the products are manganese(II) sulphate(VI), iron(III) sulphate(VI), potassium sulphate(VI) and water. Each potassium manganate(VII) formula unit reacts with five iron(II) sulphate(VI) formula units.

a Produce a balanced chemical equation for this reaction.

b Produce an ionic equation for this reaction. [5]


7 The moleYou can find out a lot from taking measurements

Although much important information can be gained from making observations during a chemical reaction, some of the most telling results are obtained by making measurements:

• If you measure how much of each element a compound is made of you can calculate its formula.

• If you know how much of a substance is dissolved in a certain volume of water you can calculate the concentration of the solution.

• This solution might then be able to react with another solution and the concentration of the second solution can be calculated from measurements taken during the reaction.

The key to all these calculations is the concept of the mole.

You have to calculate the number of atoms involved

Suppose you discover that 6.35g of copper react with 1.60g of oxygen and produce 7.95g of copper oxide. What is the formula of the copper oxide produced? You should not just look at the masses, note that almost four times as many grams of copper are involved as oxygen, and write down the formula Cu4O. A correct formula tells you how many atoms of each element are in one formula unit (or one molecule) of a compound. The formula Cu4O would only be correct if oxygen atoms and copper atoms had the same mass as each other. However, you know they don’t – a copper atom has average mass of 63.5 and an oxygen atom has average mass of 16.0.

One mole of every substance contains the same number of particlesSince a copper atom has an average mass of 63.5 and an oxygen atom has an average mass of 16.0, it follows that 63.5g of copper contains the same number of atoms as 16.0g of oxygen. This amount is approximately 6 1023 atoms and is called a mole. 63.5g of copper contains 1 mole of copper atoms (6 1023 of them). 16.0g of oxygen contains one mole of oxygen atoms (6 1023 of them). Since the Ar values for all elements are relative to the mass of one atom of carbon-12, the definition of 1 mole is the amount of a substance that contains the same number of particles as 12.0g of carbon-12. Just as a dozen of anything consists of the same number – twelve – so a mole of anything consists of the same number – 6 1023. The word ‘amount’ in the definition above is important. To a chemist ‘amount’ means ‘number of particles’ and is always measured in moles.

The mass of an element that contains 1 mole of atoms is its Ar in grams. For example, the Ar of iron is 55.8, so 55.8g of iron contains 1 mole of atoms. The mass of a compound that contains 1 mole of molecules or formula units is its Mr in grams. For example, the Mr of ammonia is 17.0, so 17.0g of ammonia contains 1 mole of molecules.

Question 1What is the mass in grams of one mole of the following?

a Zinc atoms f Sulphur molecules

b Lead atoms g Copper(II) nitrate(V) formula units

c Hydrogen atoms h Water molecules

d Hydrogen molecules i Sodium chloride formula units

e Sulphur atoms [9]

Number of moles = mass of sample

mass of 1 mole

When you are given data about a reaction in terms of the masses of substances in grams, in order to make sense of them you must convert the masses into amounts. (The amount of a substance is the number of moles of that substance.) This is done by dividing the mass of the substance by the mass of 1 mole of the substance.

Worked exampleWhat is the amount of calcium carbonate in 83.0g of calcium carbonate?

i Write down the formula of calcium carbonate. To do this you must remember the formulae and charges of the ions involved. Calcium is Ca2+, carbonate is CO3

2–, so calcium carbonate is CaCO3.

ii Find the Mr of calcium carbonate. This is 40.1 + 12.0 + (3 16.0). Answer = 100.1. 1 mole of calcium carbonate therefore weighs 100.1g.

iii Divide the mass of the sample by the mass of 1 mole of calcium carbonate. This is 83.0g/100.1g. The answer = 0.829 moles. 83.0g of calcium carbonate therefore contains 0.829 moles of CaCO3 formula units.

Note: When you divide 83.0 by 100.1 on a calculator it gives an answer of 0.8291708. You should give your answer to the accuracy of the least accurate piece of data in the question. In this question the Ar values and the 83.0g are all given to three significant figures (3sf). The answer should therefore be given to 3sf, so it should be given as 0.829 moles.

You can now solve the copper oxide problem above. 6.35g of copper react with 1.60g of oxygen:

• The number of moles of copper atoms in 6.35g of copper is 6.35g/63.5g =0.100 moles of copper atoms.

• The number of moles of oxygen atoms in 1.60g of oxygen is 1.60g/16.0g =0.100 moles of oxygen atoms.

• Therefore 0.100 moles of copper atoms reacted with 0.100 moles of oxygen atoms. The ratio of amount of copper atoms to amount of oxygen atoms is 1:1. The formula of the product is therefore CuO.

Question 2a How many moles of atoms are there in the following?

i 4.6g of zinc iii 0.156g of calcium v 0.31g of lead

ii 79g of oxygen iv 109.6g of sodium vi 5.3g of hydrogen

b Which of these samples contains the greatest number of atoms?

c Which of these samples contains the smallest number of atoms? [8]

Question 3How many moles of molecules are there in the following?

a 9.0g of water c 56.3g of carbon monoxide

b 0.088g of carbon dioxide d 0.0465g of ammonia [4]

Question 4How many moles of formula units are there in the following?

a 1.00g of calcium carbonate c 74.63g of zinc chloride

b 26.0g of copper(II) nitrate(V) d 0.163g of aluminium oxide [4]


8 Calculating masses and Mr valuesN = m can be rearranged to give two other equations

Mr

You have learnt that the amount of a substance in a sample (measured in moles) can be calculated by using the equation:

mass of sample number of moles = mass of 1 mole

The following abbreviations are used:

N for number of moles

m for mass of sample

Mr for mass of 1 mole

So the abbreviated equation is:

N = m/Mr

This can be rearranged into two other equations:

m = N Mr and Mr = m/N

Assignment 1• Learn the meanings of the symbols N, m and Mr. • Learn the three equations

N = m/Mr m = N Mr Mr = m/N

and test yourself on them.

How to calculate what mass of sample you needIf you need a particular amount of a substance (i.e. a particular number of moles of it) and you know the Ar or Mr of the substance, the mass you should weigh out is calculated by using:

m = N Mr

Worked examples1 What mass of sample should be weighed out if you need:

a 0.0500 moles of anhydrous sodium carbonate? b 0.0300 moles of sulphur atoms?

c 0.00500 moles of sulphur molecules?

a The formula of anhydrous sodium carbonate is Na2CO3. Its Mr is therefore (23.0 2) + 12.0 + (16.0 3) = 106.0, so the mass of 1 mole of anhydrous sodium carbonate is 106.0g. The equation to calculate mass of sample is m = N Mr N is the number of moles needed =

0.0500 Mr is the mass of 1 mole = 106.0g

• Substituting: m = 0.0500 106.0 • Solving: m = 5.30g

b The Ar of sulphur is 32.1 • Substituting: m = 0.0300 32.1 • Solving: m = 0.963g

c The molecular formula of sulphur is S8, so the Mr of sulphur is 32.1 8 = 256.8 (Note how important it is to specify whether sulphur atoms or sulphur molecules are meant.)

• Substituting in the equation: m = 0.00500 256.8 • Solving: m = 1.284g (which is 1.28g to 3sf)

Question 1What mass of sample should you weigh out if you need:

a 3.00 moles of zinc sulphate(V)?

b 0.500 moles of sodium hydroxide?

c 0.0200 moles of sulphuric(VI) acid?

d 0.330 moles of oxygen atoms?

e 0.330 moles of oxygen molecules?

f 0.00100 moles of glucose? [12]

How to calculate an Ar or an Mr valueIf you need to know the Ar or Mr of a substance but don’t know its exact formula, you can calculate it. To do this you need to know how many moles of it are contained in a sample of known mass. The equation used is Mr = m/N. This technique is very useful for working out the Mr of a gas and for finding the number of molecules of water of crystallisation in one formula unit of certain ionic compounds.

Worked examples2 0.0350 moles of ethyne gas has a mass of 0.910g. Calculate the Mr of ethyne.

Mr = m/N

• Substituting: Mr = 0.910g/0.0350 moles • Solving: Mr = 26.0 So the Mr of ethyne is 26.0

3 How many molecules of water of crystallisation are there in one formula unit of copper(II) sulphate(VI), if 0.0250 moles of hydrated copper(II) sulphate(VI) has a mass of 6.25g?

i Mr = m/N

• Substituting: Mr = 6.25g/0.0250 moles • Solving: Mr = 250.0 So the Mr of hydrated copper(II) sulphate(VI) is 250

ii The formula of hydrated copper(II) sulphate(VI) is CuSO4.xH2O, where x is the number of molecules of water of crystallisation per formula unit. You know the Mr is 250. The copper(II) sulphate(VI) contributes 63.5 + 32.1 + (16.0 4) = 159.6 to this. The water of crystallisation must contribute 250.0 – 159.6 = 90.4

Since xH2O contributes 90.4, and the Mr of H2O is 18.0, x is 5. The formula of hydrated copper(II) sulphate(VI) is therefore CuSO4.5H2O Note that the numbers don’t work out exactly. 5 18.0 is of course 90.0, not 90.4. There are two reasons for this:

• Firstly, the Ar values we are using are rounded off to the nearest 0.1. • Secondly, the mass of sample and its amount given in this question were

given to 3sf so some rounding off was involved here too.

You will often have to look for the answer that best fits your figures. In this case 90.4 is obviously very close to 5 18.0.

Question 2a Calculate the Mr of the following gases:

i 0.0100 moles of hydrogen sulphide gas, which has a mass of 0.340g.

ii 0.0620 moles of nitrous oxide gas, which has a mass of 2.73g. iii 0.00500 moles of hydrazine gas, which has a mass of 0.160g.

b Suggest a molecular formula for each gas. (Hydrazine is a compound ofnitrogen and hydrogen.) [6]

Question 3Calculate the number of molecules of water of crystallisation in one formula unit of the following compounds:

a FeCl2.xH2O (0.0200 moles has a mass of 4.70g)

b CaSO4.xH2O (0.0310 moles has a mass of 5.33g)

c MgSO4.xH2O (0.00200 moles has a mass of 0.493g) [6]


9 Calculating formulaeCalculating empirical formulae from percentage composition data

If a compound is analysed, the relative proportions of its constituent elements can be found. These relative proportions are usually quoted as percentages. Such data is called percentage composition data and can be used to calculate the empirical formula of the compound.

Worked examples1 A compound of carbon, hydrogen and oxygen is analysed and is found to consist

of 37.5% carbon, 12.5% hydrogen and 50.0% oxygen by mass. Calculate its empirical formula.

i The data gives you the proportions by mass, but the formula tells you the proportions by numbers of atoms. The first step is therefore to convert the data from mass to numbers of atoms. This is done using the equation N = m/Mr. Each percentage must be divided by the Ar of that element: • for carbon: 37.5/12.0 = 3.125 • for hydrogen: 12.5/1.0 = 12.5 • for oxygen: 50.0/16.0 = 3.125

ii You now know that the proportions by number of atoms are 3.125 carbon to 12.5 hydrogen to 3.125 oxygen. To get the empirical formula, these proportions must be reduced to their simplest whole number ratio. To find this ratio divide each number by the smallest number, in this case 3.125:

• for carbon: 3.125/3.125 = 1 • for hydrogen: 12.5/3.125 = 4 • for oxygen: 3.125/3.125 = 1

The ratio is therefore 1:4:1 and the empirical formula is CH4O. This calculation is repeated in the table below.

Carbon Hydrogen Oxygen

Mass in 100g of compound 37.5g 12.5g 50g

i Dividing by the Ar of the 37.5/12.0 = 3.125 12.5/1.0 = 12.5 50/16.0 = 3.125element

ii Dividing by the lowest 3.125/3.125 = 1 12.5/3.125 = 4 3.125/3.125 = 1number

Therefore, the empirical formula of the compound is CH4O.

2 In the example above, the second step produced a whole number ratio straight away. This is not the case in this next example.

A compound of carbon and hydrogen is analysed and is found to consist of 81.8% carbon and 18.2% hydrogen by mass. Calculate its empirical formula.

i Convert the data from percentage by mass to proportions by numbers of atoms:

• for carbon: 81.8/12.0 = 6.82 • for hydrogen: 18.2/1.0 = 18.2

ii Divide by the smaller number:

• for carbon: 6.82/6.82 = 1 • for hydrogen: 18.2/6.82 = 2.67

iii This ratio should not be approximated to 1:3. The figures haven’t yet given a whole number ratio. If step ii doesn’t produce a whole number ratio the figures must all be multiplied by the same number until a whole number ratio emerges. To decide which number to multiply by, examine the data. Here we have 1 and 2.67. You need to notice that 2.67 is 8/3, so multiplying by 3 will produce a whole number:

• for carbon: 1 3 = 3 • for hydrogen: 2.67 3 = 8.01

The ratio is therefore 3:8 and the empirical formula is C3H8.

Question 1The following sets of data give the percentage composition by mass of three different oxides of sodium. Calculate the empirical formula of each oxide of sodium.

a 74.2% sodium, 25.8% oxygen

b 41.8% sodium, 58.2% oxygen

c 59.0% sodium, 41.0% oxygen [6]

Question 2The following sets of data give the percentage composition by mass of four different oxides of nitrogen. Calculate the empirical formula of each oxide of nitrogen.

a 36.8% nitrogen, 63.2% oxygen

b 46.7% nitrogen, 53.3% oxygen

c 25.9% nitrogen, 74.1% oxygen

d 63.6% nitrogen, 36.4% oxygen [8]

Calculating empirical formulae from composition data given in grams

Another way for the analytical data to be expressed is in grams. The technique used to calculate the empirical formula is identical.

Worked example3 3.30g of a compound consisting of carbon, hydrogen and oxygen only were analysed

and found to contain 1.29g of carbon and 0.29g of hydrogen. Calculate its empirical formula.

i Find out how much oxygen the sample contained: 3.30 – 1.29 – 0.29 = 1.72g of oxygen

ii Convert the masses into amounts of atoms, using N = m/Mr

• for carbon: 1.29/12.0 = 0.11 • for hydrogen: 0.29/1.0 = 0.29 • for oxygen: 1.72/16.0 = 0.11

iii Divide by the smallest number:

• for carbon: 0.11/0.11 = 1 • for hydrogen: 0.29/0.11 = 2.64 • for oxygen: 0.11/0.11 = 1

iv Find the simplest whole number ratio. This is 3:8:3, so the empirical formula is C3H8O3.

Question 3Use the following sets of data to calculate the empirical formulae of the compounds involved.

a Octane is a hydrocarbon. 5.00g of octane contains 4.21g of carbon.

b 6.45g of aluminium sulphide contains 2.32g of aluminium.

c Orthoclase, a mineral in feldspar, is a compound of potassium, aluminium, silicon and oxygen. 20.00g of orthoclase contains 2.81g of potassium, 1.94g of aluminium and 6.04g of silicon. [6]

Question 4Phosphorus and sulphur form several different compounds. The following sets of data give the mass of phosphorus contained in 1.000g of each of these three compounds. Calculate the empirical formula of each compound.

a 1.000g of compound A contains 0.356g of phosphorus.

b 1.000g of compound B contains 0.564g of phosphorus.

c 1.000 g of compound C contains 0.437 g of phosphorus. [6]


10 Mass equationsThe method explained in Sections 10, 11 and 12 calculates reacting amounts by the method of working in constant ratios, having first converted a balanced chemical equation into a ‘mass equation’. It is not anticipated that this will replace the usual method of calculating the number of moles of reactant, calculating the number of moles of product, and then calculating the mass of product. The method explained in Sections 10, 11 and 12 is offered as an alternative which some students may find easier to remember.

A mass equation enables the reacting mass of each substance to be foundHaving written a balanced chemical equation, if you write the Mr value of each substance underneath its formula you have a mass equation – a very useful device. For example, for the reaction of sulphuric(VI) acid and magnesium oxide, the balanced chemical equation is:

H2SO4 + MgO → MgSO4 + H2O

The mass equation is 98.1 + 40.3 → 120.4 + 18.0

If the mass equation is written in grams it reads:

98.1g + 40.3g → 120.4g + 18.0g

This means 98.1g of sulphuric(VI) acid reacts with 40.3g of magnesium oxide and produces 120.4g of magnesium sulphate(VI) and 18.0g of water. A mass equation can be used to work out the masses of reactants and products involved in reactions.

Worked examples1 What mass of calcium hydroxide is produced when 112.2g of calcium oxide reacts

with water, and what mass of water is used up in this reaction?

i Write a balanced chemical equation and a mass equation underneath it (in grams):

CaO + H2O → Ca(OH)2

56.1g + 18.0 g → 74.1 g

ii The question says 112.2 g of calcium oxide reacts. Notice that this is twice the mass of calcium oxide in the mass equation. Rewrite the mass equation with each mass multiplied by 2:

112.2 g + 36.0g → 148.2g

iii The answers can be read straight from this version of the mass equation. 112.2 g of calcium oxide will produce 148.2 g of calcium hydroxide, and 36.0 g of water are used up.

2 What mass of calcium chloride forms when 5.61g of calcium oxide reacts with excess hydrochloric acid? What mass of acid will be used up in the reaction?


CaO + 2HCl → CaCl2 + H2O

56.1g + 73.0g → 111.1g + 18.0g

Note that twice the Mr of hydrochloric acid is needed (2 36.5 = 73) since the balanced chemical equation involves two molecules of HCl.

ii The question says 5.61g of calcium oxide reacts. This is one-tenth the mass in the mass equation. Rewrite the mass equation with every mass divided by 10:

5.61g + 7.30g → 11.11g + 1.80g

iii The answers can be read straight from this version of the mass equation. 11.11g of calcium chloride are produced and 7.30g of hydrochloric acid are used up.

Question 1a How much calcium chloride forms when the following masses of calcium oxide

react with excess hydrochloric acid?

i 28.05g iv 11.22g

ii 2.805g v 18.7tonnes

iii 56.1kg b How much hydrochloric acid would react in each case?

c How much water would be produced in each case? [15]

Question 2a Write a balanced chemical equation and a mass equation for the reaction between

copper(II) oxide and nitric(V) acid.

b What mass of copper(II) nitrate(V) will be produced when the following masses of copper(II) oxide react with excess nitric(V) acid?

i 7.95g ii 15.9g iii 159kg [5]

You can calculate the mass of reactants needed

This method can be used just as effectively to calculate the mass of reactants required to make a particular mass of product.

Worked example3 If you need to know what masses of phosphorus(V) oxide and water are required to

make 98.0g of phosphoric(V) acid, you start with the balanced chemical equation and mass equation in grams as before:

P4O10 + 6H2O → 4H3PO4

284.0g + 108.0g → 392.0g

98.0g of phosphoric(V) acid is one-quarter the mass in the mass equation (98.0/392 = 0.25). The mass of phosphorus(V) oxide required is 284.0/4 = 71.0g. The mass of water required is 108.0/4 = 27.0g

Question 3What masses of carbon dioxide and water must react in order to produce the following masses of carbonic acid (H2CO3)?

a 310g b 6.20kg c 0.155g [8]

Question 4What masses of zinc oxide and sulphuric(VI) acid must react in order to produce the following masses of zinc sulphate(VI)?

a 3.00g b 17.2g c 0.130g [8]


11 Calculations on gas volumesOne mole of any gas occupies 24 000 cm3 at r.t.p.If you are given data telling you that a certain amount of gas has a particular mass, it is possible to calculate the Mr of the gas using the equation:

Mr = m

N

Finding N for a gas is particularly easy. 1 mole of any gas occupies a volume of 24dm3 (24000cm3) at a temperature of 298K and a pressure of 1 atmosphere. These conditions are called room temperature and pressure, abbreviated to r.t.p. To find N for a sample of gas just measure its volume in cm3 at r.t.p. and divide by 24000.

Worked example0.100g of a gaseous oxide of carbon has volume 86.4cm3 at r.t.p. Calculate its Mr and hence deduce whether it is carbon monoxide or carbon dioxide.

i Calculate the number of moles (N) of gas in the sample.

86.4cm3/24000cm3 = 0.00360 molesii Calculate Mr using Mr = m/N

• Substituting: Mr = 0.100/0.00360

• Solving: Mr = 27.8

The oxide is therefore carbon monoxide. Although the Mr of carbon monoxide is 28.0, 27.8 is sufficiently close – the Mr of CO2 is 44.0.

Question 1The sets of data below refer to three of the noble gases. Calculate the Ar in each case and hence identify each gas.

a 0.200g of X occupies 120cm3 at r.t.p.

b 0.500g of Y occupies 3000cm3 at r.t.p.

c 0.370g of Z occupies 106cm3 at r.t.p. [3]

Question 2Find the Mr of each gaseous compound below and use the additional information to identify its molecular formula.

a 0.100g of compound P occupies 80.0cm3 at r.t.p. P is a compound of carbon, hydrogen and oxygen only.

b 0.100g of compound Q also occupies 80.0cm3 at r.t.p. Q is a hydrocarbon with empirical formula CH3.

c 0.261g of a compound R occupies 195cm3 at r.t.p. R is the compound of a Group IV element and hydrogen. Each molecule of R contains one atom of the Group IV element and four atoms of hydrogen. [6]

Total: / 9 Score: %

12 The volumes of gases involved in reactionsThe volumes of gases involved in reactions can be calculatedFor any gas 1 mole of molecules always occupies the same volume at r.t.p., namely 24dm3. This enables you to calculate the volume of any gas produced (or used up) in a reaction. The value of 24dm3 in the questions that follow should be taken as 24000cm3. Therefore the masses (in g) and the Ar values are the least accurate pieces of data, so the answers can be given to 3sf.

Worked example1 What volumes of nitrogen dioxide and oxygen, measured under normal conditions,

will be produced when 14.83g of magnesium nitrate(V) is decomposed?


2Mg(NO3)2 → 2MgO + 4NO2 + O2

296.6g → 80.6g + 184.0g +32.0gii Include the volumes of the gases. The Mr of nitrogen dioxide is 46.0, so

46.0g of nitrogen dioxide is 1 mole of molecules and has a volume of 24dm3. 184.0g of nitrogen dioxide is 4 moles and has a volume of 96 dm3 at r.t.p. 32.0g of oxygen is 1 mole of molecules and has a volume of 24dm3, so:

2Mg(NO3)2 → 2MgO + 4NO2 + O2

296.6g → 80.6g + 184.0g + 32.0g

96dm3 24dm3

iii The problem says 14.83g of magnesium nitrate(V) decomposes: 14.83/296.6 = 0.05.

The volume of nitrogen dioxide that forms is 96dm3 0.05 = 4.80dm3.

The volume of oxygen that forms is 24dm3 0.05 = 1.20dm3.

Question 1When calcium carbonate is heated strongly it decomposes to give calcium oxide and carbon dioxide.

a Write a balanced chemical equation for this reaction.

b Write a mass equation (in grams) and include the volume of carbon dioxide measured at r.t.p.

c What volume of carbon dioxide, measured at r.t.p., will form when the following masses of calcium carbonate are decomposed?

i 1.00g ii 25.0g iii 40.0g [6]

Question 2When ammonia gas and hydrogen chloride gas mix together they react to form a solid called ammonium chloride.

a Write a balanced chemical equation for this reaction.

b Write a mass equation (in grams) and include the volumes of ammonia and hydrogen chloride measured at r.t.p.

c What volumes of ammonia and hydrogen chloride gases must react at r.t.p. in order to produce the following masses of ammonium chloride?

i 10.7g ii 214g iii 1.00g [10]

The combustion of hydrocarbons involves gaseous reactants and productsWhen a hydrocarbon burns it reacts with oxygen gas, one of the products of complete combustion is carbon dioxide gas, and the hydrocarbon itself may be a gas. It is therefore possible to calculate the volume of these substances that will take part in a reaction.

Worked example2 What volume of oxygen reacts with 3.00dm3 of propane gas when it undergoes

complete combustion? What volume of carbon dioxide will be produced? (All volumes are measured at r.t.p.)

i Write a balanced chemical equation and a mass equation including gas volumes:

C3H8 + 5O2 → 3CO2 + 4H2O

44.0g + 160.0g → 132.0g + 72.0g

24 dm3 120dm3 72dm3

ii The question says 3.00dm3 of propane reacted. 3/24 = 0.125.

The volume of oxygen that will react is therefore 120dm3 0.125 = 15.0dm3 The

volume of carbon dioxide that will be produced is 72dm3 0.125 = 9.00dm3

Question 3Methane gas (CH4) burns completely in oxygen producing carbon dioxide and water.

a Write a balanced chemical equation and a mass equation for this reaction, including gas volumes at r.t.p. (Ignore the volume of water, which will be a liquid at r.t.p.)

b What volumes of methane and oxygen must be reacting if 16.0dm3 of carbondioxide are produced under normal conditions? [7]

Question 4Butane (C4H10) is a gas at r.t.p.

a Write a balanced chemical equation and a mass equation including gas volumes for the complete combustion of butane. (Ignore the volume of water, which will be a liquid at r.t.p.)

b What volume of oxygen would be required for the complete combustion of36.0dm3 of butane and what volume of carbon dioxide would be produced? [7]


13 Quantitative chemistry with solutionsTitration involves measuring the volume of reacting solutions

The preceding sections have covered two main aspects of quantitative chemistry. Quantitative means that the quantity of something has been measured – so far these quantities have been either masses or gas volumes. Some of the most important quantitative chemistry involves careful measurement of the volumes involved when two solutions react. This technique is called titration.

Titration results can enable you to calculate the Mr of a substance, to work out how many molecules of one substance react with one molecule of another substance, or to calculate the amount of one substance contained in a mixture.

amount (in moles)

Concentration =volume (in dm3)

Concentration is the key concept in titration work. The concentration of a solution tells you how many moles of solute are dissolved in each dm3 of solution. The solvent involved is almost invariably water. The concentration of a solution is calculated by dividing the number of moles of solute by the total volume of solution produced:

C = N/V

The units of concentration are therefore moles per dm3, which can be abbreviated to mol/dm3 or moldm–3. We are going to use moldm–3.

Worked examples1 0.0500 moles of sulphuric(VI) acid are dissolved in 500cm3 of solution. Calculate the

concentration of sulphuric(VI) acid.

i Firstly the 500cm3 must be converted to dm3. 1dm3 is 1000cm3, so 500cm3 is 500/1000 = 0.500dm3.

ii Using C = N/V

• Substituting: C = 0.0500/0.500

• Solving: C = 0.100moldm–3

2 3.00g of sodium hydroxide are dissolved in 250cm3 of solution. Calculate the concentration of sodium hydroxide. i Firstly, calculate the number of moles of sodium hydroxide. Since the formula is

NaOH, its Mr is 23.0 + 16.0 + 1.0 = 40.0

Using N = m/Mr, the number of moles is 3.00/40.0 = 0.0750 moles

ii Secondly, convert 250cm3 to dm3. 250/1000 = 0.250dm3 iii Using C = N/V

• Substituting: C = 0.0750/0.250

• Solving: C = 0.300moldm–3

Question 1a Calculate the concentration of the following solutions:

i 0.840 moles of solute dissolved in 5.00dm3 of solution. ii 0.360 moles of solute dissolved in 300cm3 of solution.

iii 0.0200 moles of solute dissolved in 25.0cm3 of solution.

iv 24.0 moles of solute dissolved in 30.0dm3 of solution. b i Which of the solutions above is the most concentrated?

ii Which of the solutions above is the most dilute (least concentrated)?

iii Which of the solutions above have the same concentration? [7]

Question 2Calculate the concentration of the following solutions:

a 0.760g of potassium nitrate(V) dissolved in 80.0cm3 of solution.

b 12.0g of CuSO4.5H2O dissolved in 100cm3 of solution.

c 54.0g of glucose dissolved in 300cm3 of solution.

d 120g of nitric(V) acid dissolved in 200cm3 of solution. [12]

C = N

can be rearranged to give two more equationsV

If you have a solution of known concentration and a sample containing a particular number of moles is required, the volume that should be measured out can be calculated using this rearranged version of the equation:

NV = C

This equation demonstrates one reason why solutions of known concentration are so useful. If a small amount of substance is needed it is much easier to measure a reasonable volume of dilute solution than it is to measure a tiny mass of solid. If the substance is harmful, e.g. sodium hydroxide, the dilute solution will also be much safer to handle than the pure substance.

If a particular volume of a solution of known concentration is involved in a reaction, the number of moles involved can be calculated using this version of the equation:

N = C V

Assignment 1a Learn the meanings of the symbols N, C and V.

b Learn the three equations:

N = C V C = N/V V = N/C

c Learn the units of concentration, mol dm–3.

d Learn how to interconvert volumes in cm3 and dm3.e Test yourself.

f Test yourself again in 24 hours. Put it in your diary to repeat this exercise once a week between now and your Module 1 exam. These equations are always needed.

g Check you can still do the assignment in Section 8 ‘Calculating masses and Mr values’.

Worked examples3 What volume of 0.200moldm–3 hydrochloric acid should be measured out if

0.0250 moles are required? i The question asks you to calculate a volume, so the equation used is:

V = N/C

ii Substituting: V = 0.0250/0.200

iii Solving: V = 0.125dm3 or125 cm3

4 If 37.2cm3 of 0.100 moldm–3 sodium hydroxide were involved in a reaction, how many moles of sodium hydroxide reacted?

i Convert 37.2cm3 to dm3 (i.e. divide by 1000). The answer is 0.0372dm3. ii The question asks you to calculate an amount, so the equation used is:

N = C V

iii Substituting: N = 0.100 0.0372

iv Solving: N = 0.00372 moles

Question 3How many moles of solute do these samples contain?

a 60.0cm3 of a 2.00moldm–3 solution.

b 200cm3 of a 0.200moldm–3 solution.

c 0.500dm3 of a 1.00moldm–3 solution.

d 12.6cm3 of a 0.250moldm–3 solution. [4]

Question 4What volume of a solution of concentration 0.0800moldm–3 should be measured out to give the following amounts? Give your answers in both cm3 and dm3.a 0.000100 moles

b 0.000320 moles

c 0.0400 moles [6]


14 Finding an unknown concentration by titrationA solution of known concentration is titrated against a solution of unknown concentrationThe concentration of a solution of an acid of unknown concentration can be found by titrating the acid against an alkali of known concentration. In the first example below, the acid is hydrochloric acid and the alkali is sodium hydroxide solution of concentration 0.100moldm–3.

Experimental method

i 25.0cm3 of the sodium hydroxide solution are measured accurately using a graduated pipette and put into a conical flask.

ii Two drops of phenolphthalein indicator are added to the sodium hydroxide solution, which turns pink.

iii A burette is filled with the hydrochloric acid. A burette is a graduated glass tube that can be read to ±0.05cm3.

iv The acid is added to the sodium hydroxide solution until the indicator goes colourless. The aim of the titration is to find the exact amount of acid that does this. At this point all the sodium hydroxide solution has been neutralised.

v Repeat the titration until you have three results that are very nearly exactly the same.

As an example of a set of results to use to explain the calculation, we will say that 25.0cm3 of the sodium hydroxide was neutralised in three titrations by 17.20cm3, 17.25cm3 and 17.15cm3 of the hydrochloric acid. The average of these three (very close) results is 17.20cm3.

The concentration of the acid is calculated from the results

With any titration calculation first you need a balanced chemical equation:

NaOH + HCl → NaCl + H2O

This says that one formula unit of sodium hydroxide reacts with one formula unit of hydrochloric acid. At the instant when the indicator changes colour, when the alkali is just neutralised, the number of moles of acid added is equal to the number of moles of alkali originally in the flask:

number of moles of alkali = number of moles of acid

• Number of moles can be calculated using N = C V

so: C V (for NaOH) = number of moles of alkali

• Substituting: 0.100 25.0/1000 = 0.0025 moles of

NaOH number of moles of alkali = number of moles of acid

The number of moles of acid at neutralisation = 0.0025 moles

C V (for HCl) = number of moles of acid • Substituting: ? 17.2/1000 = 0.0025 moles of HCl • Solving: ? = 0.145 mol dm–3

In this calculation ‘?’ is the unknown concentration of hydrochloric acid.

Worked example23.9cm3 of sulphuric(VI) acid of unknown concentration are needed to neutralise 25.0cm3 of 0.500moldm–3 sodium hydroxide. Calculate the concentration of the acid.

i Write an equation for the reaction:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

ii In this case, two formula units of sodium hydroxide react with one of sulphuric(VI) acid. The number of moles of alkali originally in the flask is equal to twice the number of moles of acid that were added.

C V (for NaOH) = number of moles of alkali

• Substituting: 0.500 25.0/1000 = 0.0125 moles of NaOH

number of moles of alkali = 2 number of moles of acid

The number of moles of acid at neutralisation = 0.00625 moles

C V (for H2SO4) = number of moles of acid

• Substituting: ? 23.9/1000 = 0.00625 moles of H2SO4 • Solving: ? = 0.262moldm–3 In this calculation ‘?’ is the unknown concentration of sulphuric(VI) acid.

Question 1You have four different hydrochloric acid solutions of unknown concentrations. They are each separately titrated against 25.0cm3 of 0.100moldm–3 sodium hydroxide solution. Calculate the concentration of each hydrochloric acid solution, given that the average volume of each acid needed to neutralise the alkali is as follows:

a 3.80cm3

b 43.00cm3

c 12.55cm3

d 26.30cm3 [6]

Question 225.0cm3 of 0.200moldm–3 potassium hydroxide solution are just neutralised by 16.60cm3 of nitric(V) acid of unknown concentration.a Describe how this result would have been obtained in the lab. [5]

b Write a balanced chemical equation for the reaction involved. [1]

c Calculate the concentration of the nitric(V) acid. [3]

Question 325.0cm3 of barium hydroxide solution of unknown concentration are just neutralised by 19.95cm3 of 0.250mol dm–3 hydrochloric acid.a Write a balanced chemical equation for the reaction involved. [1]

b Calculate the concentration of the barium hydroxide. [3]


15 Key words and ideasAmphoteric Able to neutralise both acid and alkali. Anhydrous Without water of crystallisation.

Ar The mass of one atom of an element relative to 1/12-th the mass of one atom of carbon-12.

Balanced When the atoms on the two sides of the equation are the same, both in number and type.

Chemical equation A description of a chemical reaction which tells us the amount and formula of each substance involved.

Combustion Rapid reaction with oxygen. The technical chemical term for burning.

Compound ion An ion consisting of a small group of atoms held together by covalent bonds and having gained or lost one or more electrons.

Concentration The number of moles of solute dissolved in each dm3 of solution. Decomposition The breaking down of a compound into simpler substances.

Empirical formula The formula of a compound giving its constituent elements in their simplest ratio.

Formula unit The group of ions specified by the formula of an ionic compound. Giant lattice A vast array of particles in a regular pattern.

Hydrated With water of crystallisation.

Ionic equation An equation for a chemical reaction involving only those species that change. The species involved are predominately ionic.

Mass equation A chemical equation including the masses of each reactant and product molecule.

Mole The amount of a substance containing the same number of atoms, molecules or formula units as 12.0g of carbon-12.

Molecular formula The formula of an element or compound giving the number of each type of atom found in one molecule of the substance.

Mr The mass of one molecule or one formula unit of a compound relative to 1/12-th the mass of one atom of carbon-12.

r.t.p. Room temperature and pressure. A temperature of 298K and a pressure of 1 atmosphere (101kPa).

Redox reaction A chemical reaction in which electrons are gained and lost by the species involved.

Simple ion An ion consisting of a single atom that has gained or lost one or more electrons.

Simple molecular Substance whose atoms are bonded covalently into discrete groups called molecules.

Synthesis The making of a compound from its component elements.

Titration Measuring the amounts of two solutions which react together. Titration is always done to a high degree of accuracy involving precise volume measurement.

Water of crystallisation Water molecules included in a lattice structure. The number of water molecules present per formula unit of the compound is fixed.

16 QuestionsThe questions that follow are mostly more difficult than standard AS level Module 1 questions. Questions 3, 4 and 5 are particularly difficult. Attempt them if directed to by your teacher, and expect to need several goes at them. Go back to these questions towards the end of Year 13 as you prepare for your final modular exams, which will also involve calculations.

1 a 1.78g of a hydrocarbon were completely burnt in oxygen. The products consisted of 5.59g of carbon dioxide and 2.29g of water.

i Calculate the mass of carbon in the carbon dioxide formed. [1]

ii Calculate the mass of hydrogen in the water formed. [1]

iii Calculate the empirical formula of the hydrocarbon. [2]

b The Mr of the hydrocarbon is 42. What is its molecular formula? [1]

2 Q is a compound of carbon, hydrogen and oxygen only. The Mr of Q is 74.0. When 0.675g of Q is burnt in excess oxygen the products consist of 1.605g of carbon dioxide and 0.821g of water.

a Calculate the mass of carbon in 1.605g of carbon dioxide. [1]

b Calculate the mass of hydrogen in 0.821g of water. [1]

c Calculate the mass of oxygen in the 0.675g of Q. [1]

d Calculate the empirical formula of Q. [2]

e Calculate the molecular formula of Q. [1]3 R is the hydroxide compound of a Group II metal.

• R contains water of crystallisation. • The formula of R can be written M(OH)2.xH2O, where M is the Group II metal

and x is the number of molecules of water of crystallisation per M(OH)2 formula unit.

• 2.000g of R were heated strongly to drive off the water of crystallisation. • 1.086g of water were driven off. • The product of heating R, an anhydrous compound M(OH)2, was then heated

further to give 0.779g of an oxide, MO, and 0.135g of water.

a Calculate the number of moles of water driven off by the first heating. [2]

b Calculate the number of moles of water driven off by the second heating. [2]

c By comparing your answers to parts a and b, calculate the value of x in theformula M(OH)2.xH2O. [2]

d Calculate the Ar of metal M and hence identify it. [2]

4 In order to estimate the percentage of calcium carbonate in a sample of marble, the following was done.

i The marble chips were crushed and 2.14g of the powder were weighed out.

ii This powder was put into 50.0cm3 of hydrochloric acid of concentration 1.00moldm–3. The acid was in excess.

iii When the reaction was finished the resulting solution was titrated against 0.100moldm–3 sodium hydroxide solution. 98.8cm3 of the alkali were needed to neutralise the unreacted acid.

a How many moles of alkali were used in part iii? [1]

b How many moles of acid were left over from part ii? [1]

c How many moles of acid reacted in part ii? [2]

d How many moles of calcium carbonate reacted in part ii? [1]

e What mass of calcium carbonate is this? [2]

f Calculate the purity of the marble, stating any assumptions you make. [2]

5 In order to find the percentage of iron in a sample of steel wool, 3.00g of it were reacted with an excess of hot dilute sulphuric(VI) acid, giving a solution of iron(II) sulphate(VI). More dilute sulphuric(VI) acid was then added to make the volume of acidified iron(II) sulphate(VI) solution exactly 250cm3. 25.0cm3 portions of this solution were pipetted out and titrated against 0.100moldm–3 KMnO4. 10.65cm3 of the KMnO4 solution was necessary to fully oxidise the iron(II) sulphate(VI) in each 25.0cm3 portion to iron(III) sulphate(VI). You need to know that one KMnO4 formula unit can oxidise five iron(II) sulphate(VI) formula units.

a Calculate the number of moles of KMnO4 in 10.65cm3. [1]b Calculate the number of moles of iron(II) sulphate(VI) in each 25.0cm3

portion. [1]

c Calculate the mass of iron in each 25.0cm3 portion. The Ar of iron is 55.8. [1]d Calculate the total mass of iron in the steel wool. [1]

e Calculate the percentage of iron in the steel wool. [1]


17 Marking schemesThe Ar values used are those on the OCR Data Sheet for Chemistry.

1 Formulae1 a CO2 [1]; b NH3 [1]; c SO2 [1]; d PCl5 [1]; e CO [1]; f N2H4 [1]; g SO3 [1]

2 Ethene, C2H4 [1], CH2 [1]; Propane, C3H8 [1], C3H8 [1];Propene, C3H6 [1], CH2 [1]; Benzene, C6H6 [1], CH [1];Phenol, C6H6O [1], C6H6O [1]; Lactic acid, C3H6O3 [1], CH2O [1]

3 Nitrogen, N2 [1], N [1];

Sulphur, S8 [1], S [1];

Fluorine, F2 [1], F [1]; Phosphorus, P4 [1], P [1];

Hydrogen, H2 [1], H [1]; Iodine, I2 [1], I [1]

Total = 31

2 Formulae of ionic compounds1 a NaCl [1]; b Li2O [1]; c Pb(NO3)2 [1]; d (NH4)2Cr2O7 [1]; e K2MnO4 [1]; f KMnO4 [1];

g Al2O3 [1]; h Na3PO4 [1]; i CuCl [1]; j Cu(OH)2 [1]; k Fe2(SO4)3 [1]; l Na2SO3 [1];m CaCO3 [1]; n Ca(HCO3)2 [1]Total = 14

3 Relative molecular mass1 a 18.0 [1]; b 78.0 [1]; c 42.0 [1]; d 98.1 [1]; e 94.0 [1]; f 208.5 [1]; g 90.0 [1]

2 a 2.0 [1]; b 28.0 [1]; c 32.0 [1]; d 38.0 [1]; e 124.0 [1]; f 256.8 [1]; g 254.0 [1]

3 a 71.0 [1]; b 159.8 [1]

4 a Na2S2O3 [1], 158.2 [1]; b ZnS [1], 97.5 [1]; c K2Cr2O7 [1], 294.2 [1];d Al(NO3)3 [1], 213.0 [1]; e Na2SiO3 [1], 122.1 [1]; f (NH4)2SO4 [1], 132.1 [1];g CuCrO4 [1], 179.5 [1]; h Na2SO4 [1], 142.1 [1]; i Ag2O [1], 232.0 [1];j PbBr2 [1], 366.8 [1]

5 a 106.0 [1]; b 286.0 [1]

Total = 38

4 Chemical equations1 a 4Na + O2 → 2Na2O; b 4Al + 3O2 → 2Al2O3; c Si + O2 → SiO2;

d Mg + Cl2 → MgCl2; e 2Al + 3Cl2 → Al2Cl6; f Si + 2Cl2 → SiCl4;g P4 + 10Cl2 → 4PCl5[1] for all formulae correct, [1] for balancing each equation

2 a 2Ca(NO3)2 → 2CaO + 4NO2 + O2; b 4LiNO3 → 2Li2O + 4NO2 + O2;c MgCO3 → MgO + CO2; d Mg(OH)2 → MgO + H2O;e Ca(HCO3)2 → CaCO3 + H2O + CO2[1] for all formulae correct, [1] for balancing each equation

3 By definition, an element cannot be decomposed [1]

4 2C7H5N3O6 → 3N2 + 5H2O + 7CO + 7C [1] for balancing N and H, [1] for balancing C and O

Total = 27

5 Further chemical equations1 a CuO + H2SO4 → CuSO4 + H2O; b Al2O3 + 3H2SO4 → Al2(SO4)3 + 3H2O;

c MgO + H2O → Mg(OH)2; d PbO + 2HNO3 → Pb(NO3)2 + H2O[1] for all formulae correct, [1] for balancing, for each equation

2 a SO3 + H2O → H2SO4; b CO2 + H2O → H2CO3; c SiO2 + 2NaOH → Na2SiO3 + H2O;d PbO + 2NaOH + H2O → Na2Pb(OH)4; e 3NO2 + H2O → 2HNO3 + NO[1] for all formulae correct, [1] for balancing, for each equation

3 a CH4 + 2O2 → CO2 + 2H2O; b C5H12 + 8O2 → 5CO2 + 6H2O; c 2C6H6 + 15O2 → 12CO2 + 6H2O; d 2C3H6 + 9O2 → 6CO2 + 6H2O;e C2H4O2 + 2O2 → 2CO2 + 2H2O; f C6H12O6 + 6O2 → 6CO2 + 6H2O;g C3H6O + 4O2 → 3CO2 + 3H2O; h C6H6O + 7O2 → 6CO2 + 3H2O[1] for balancing, for each equation

Total = 26

6 Ionic equations1 a OH– + H+ → H2O; b CO3

2– + 2H+ → H2O + CO2; c HCO3– + H+ → H2O + CO2;

d CO32– + 2H+ → H2O + CO2; e HCO3

– + H+ → H2O + CO2; f OH– + H+ → H2O[1] for including relevant species only, [1] for all charges correct, [1] for balancing,for each equation

2 a O2– + 2H+ → H2O; b O2– + 2H+ → H2O; c O2– + H2O → 2OH–; d O2– + 2H+ → H2O[1] for including relevant species only, [1] for all charges correct, [1] for balancing,for each equation;

3 a Zn + 2H+ → Zn2+ + H2; b Fe + 2H+ → Fe2+ + H2; c 2Al + 6H+ → 2Al3+ + 3H2;d Cu2+ + Fe → Cu + Fe2+; e Zn2+ + Mg → Zn + Mg2+; f 2Ag+ + Cu → 2Ag + Cu2+

[1] for including relevant species only, [1] for all charges correct, [1] for balancing,for each equation

4 a Zn + 2H + → Zn2+ + H2; b Fe + 2H + → Fe2+ + H2; c 2Al + 6H + → 2Al3+ + 3H2;d Cu 2+ + Fe → Cu + Fe2+; e Zn 2+ + Mg → Zn + Mg2+; f 2Ag + + Cu → 2Ag + Cu2+

[1] for each oxidised substance (bold), [1] for each reduced substance (underlined) 5 a 2KMnO4 + 8H2SO4 + 10FeSO4 → 2MnSO4 + 5Fe2(SO4)3 + K2SO4 + 8H2O

[1] for all formulae correct, [1] for balancing; b 2MnO4

– + 16H+ + 10Fe2+ → 2Mn2+ + 10Fe3+ + 8H2O or MnO4

– + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O[1] for including relevant species only, [1] for all charges correct, [1] for balancing

Total = 65

7 The mole1 a 65.4g [1]; b 207g [1]; c 1.0g [1]; d 2.0g [1]; e 32.1g [1]; f 256.8g [1];

g 187.5 g [1]; h 18.0 g [1]; i 58.5 g [1]

2 a i 0.070mol (to 2sf) [1]; ii 4.9mol (to 2sf) [1]; iii 0.00389mol (to 3sf) [1];iv 4.77 mol (to 3sf) [1]; v 0.0015mol (to 2sf) [1]; vi 5.3mol (to 2sf) [1];

b vi [1]; c v [1]

3 a 0.50mol (to 2sf) [1]; b 0.0020mol (to 2sf) [1]; c 2.01mol (to 3sf) [1];d 0.00274 mol (to 3sf) [1]

4 a 0.00999mol (to 3sf) [1]; b 0.139mol (to 3sf) [1]; c 0.547mol (to 3sf) [1];d 0.00160mol (to 3sf) [1]

Total = 25

8 Calculating masses and Mr values

1a 3.00 161.5 = 484.5g; b 0.500 40.0 = 20.0g; c 0.0200 98.1 = 1.96g;d 0.330 16.0 = 5.28g; e 0.330 32.0 = 10.6g; f 0.00100 180.0 = 0.180g[1] for use of the correct Mr, [1] for the answer, for each substance

2 a i 0.340/0.0100 = 34.0 [1]; ii 2.73/0.0620 = 44.0 [1]; iii 0.160/0.00500 = 32.0 [1];b i H2S [1]; ii N2O [1]; iii N2H4 [1]

3 a Mr = 235, x = 6; b Mr = 172, x = 2; c Mr = 246.5, x = 7[1] for use of the correct Mr, [1] for the answer, for each substanceTotal = 24

9 Calculating formulae1 a Na2O; b NaO2; c NaO

[1] for dividing data by Ar values, [1] for each correct answer2 a N2O3; b NO; c N2O5; d N2O

[1] for dividing data by Ar values, [1] for each correct answer3 a C4H9; b Al2S3; c KAlSi3O8

[1] for dividing data by Ar values, [1] for each correct answer4 a P4S7; b P4S3; c P4S5

[1] for dividing data by Ar values, [1] for each correct answerTotal = 26

10 Mass equations1 a (CaCl2) b (HCl) c (H2O)

i 55.5g 36.5g 9.00g

ii 5.55g 3.65g 0.900g

iii 111kg 73.0kg 18.0kg

iv 22.2g 14.6g 3.60g

v 37.0 tonnes 24.3 tonnes 6.00 tonnes

[1] for each correct mass

2 a CuO + 2HNO3 → Cu(NO3)2 + H2O [1], 79.5 + 126 → 187.5 + 18.0 [1]; b i 18.7g [1]; ii 37.5g [1]; iii 375kg [1]

3 CO2 + H2O → H2CO3 [1], 44.0 + 18.0 → 62.0 [1];

a 220g CO2 [1] and 90.0g H2O [1];

b 4.40kg [1] and 1.80kg [1];

c 0.110 g [1] and 0.0450g [1]

4 ZnO + H2SO4 → ZnSO4 + H2O [1], 81.4 + 98.1 → 161.5 + 18.0

[1]; a 1.51g ZnO [1] and 1.82g H2SO4 [1];

b 8.67g [1] and 10.4g [1];

c 0.0655g [1] and 0.0790g

[1] Total = 36

11 Calculations on gas volumes

1 a Ar is 40.0, therefore argon [1]; b Ar is 4.00, therefore helium [1];c Ar is 83.8, therefore krypton [1]

2 a Mr is 30.0 [1], therefore CH2O (methanal) [1]; b Mr is 30.0 [1], therefore C2H6 (ethane) [1]; c Mr is 32.1 [1], therefore SiH4 (silane) [1]

Total = 9

12 The volumes of gases involved in reactions1 a CaCO3 → CaO + CO2 [1]; b 100.1g → 56.1g + 44.0g (24 dm3) [2];

c i 240cm3 [1]; ii 5.99dm3 [1]; iii 9.59dm3 [1]2 a NH3 + HCl → NH4Cl [1]; b 17.0g (24 dm3) + 36.5g (24 dm3) → 53.5g [3];

c i 4.80dm3 (NH3) [1] and 4.80dm3 (HCl) [1]; ii 96.0dm3 [1] and 96.0dm3 [1];iii 449cm3 [1] and 449cm3 [1]

3 a CH4 + 2O2 → CO2 + 2H2O [1], 16.0g (24 dm3) + 64.0g (48 dm3) → 44.0g (24dm3) + 36.0

g [4]; b 16dm3 methane [1], 32dm3 O2 [1]4 a 2C4H10 + 13O2 → 8CO2 + 10H2O [1],

116g (48dm3) + 416g (312dm3) → 352g (192dm3) + 180g [4]; b 234dm3 O2 [1], 144dm3 CO2 [1]Total = 30

13 Quantitative chemistry with solutions1 a i 0.168moldm–3 [1]; ii 1.20moldm–3 [1]; iii 0.800moldm–3 [1];

iv 0.800moldm–3 [1]; b i solution ii [1]; ii solution i [1]; iii solutions iii and iv [1]2 a Mr = 101.1 [1], moles solute = 0.00752mol [1], concentration = 0.0940moldm–3 [1]; b

Mr = 249.6 [1], moles solute = 0.0481mol [1], concentration = 0.481moldm–3 [1]; c Mr = 180 [1], moles solute = 0.300mol [1], concentration = 1.00moldm–3 [1]; d Mr = 63.0 [1], moles solute = 1.90mol [1], concentration = 9.52moldm–3 [1]

3 a 0.120mol [1]; b 0.0400mol [1]; c 0.500mol [1]; d 0.00315mol [1]

4 a 0.00125dm3 [1], which is1.25cm3 [1]; b 0.00400dm3 [1], which is 4.00cm3 [1];c 0.500dm3 [1], which is 500cm3 [1]Total = 29

14 Finding an unknown concentration by titration1 moles of alkali = moles of acid [2]; a 0.658moldm–3 [1]; b 0.0581moldm–3 [1];

c 0.199moldm–3 [1]; d 0.0951moldm–3 [1]

2 a 25.0cm3 of the potassium hydroxide solution are measured using a graduated pipette [1], phenolphthalein indicator is added to the alkali [1], a burette is filled with the nitric acid [1], the acid is added to the alkali until the indicator goes colourless [1], three consistent results are obtained [1];

b KOH + HNO3 → KNO3 + H2O [1];c moles of alkali = 0.00500mol [1], moles of acid = 0.00500mol

[1], concentration of acid = 0.301mol dm–3 [1]3 a Ba(OH)2 + 2HCl → BaCl2 + 2H2O [1];

bmoles of acid = 0.00499mol [1], moles of alkali = 0.00249mol [1], concentration of alkali = 0.0997moldm–3 [1]

Total = 19

16 Questions1 a i 5.59g 12.0/44.0 = 1.52g [1]; ii 2.29g 2.0/18.0 = 0.254g [1];

iii 1.52/12.0 = 0.127 C, 0.254/1 = 0.254 H [1], CH2 [1]; b C3H6 [1]

2a 1.605g 12.0/44.0 = 0.438g [1]; b 0.821g 2.0/18.0 = 0.0912g [1];c 0.675g – (0.438g + 0.0912g) = 0.146g [1];d 0.438/12.0 = 0.0365 C, 0.0912/1.0 = 0.0912 H, 0.146/16 = 0.00912 O [1], C4H10O [1];e C4H10O [1]

3 a 1.086g/18.0 = 0.0603mol [2]; b 0.135g/18.0 = 0.00750mol [2];c 0.0603/0.00750 = 8, if one H2O per molecule was driven off by the second heating,

in the reaction M(OH)2 → MO + H2O [1]then 8 were driven off by the first heating so x = 8 [1];

d 0.779g of MO is 0.00750 mol (same as moles of water lost in second heating),so Mr is 103.9 [1], M is therefore Sr [1] since Mr of SrO is 103.6 (or other method)

4 a N = V C so 0.0988 0.1 = 0.00988mol [1]; b 1 mole HCl reacts with 1 mole NaOH so 0.00988mol [1]; c moles at the start were 0.0500 1.00 = 0.0500 [1],

since 0.00988 mol were left 0.0401 mol had reacted [1];d equation is 2HCl + CaCO3 so moles CaCO3 were 0.0201 mol [1]; e 0.0201 100.1 = 2.01g [2]; f 2.01/2.14 = 93.9% [1],

assumption is that none of the impurities have acid/base nature [1]

5 a 0.01065 0.100 = 0.001065mol [1]; b 0.001065mol 5 = 0.005325mol [1];c 0.005325mol 55.8g = 0.297g [1]; d 0.297g 10 = 2.97g [1];e 2.97/3.00 = 99.0% [1]

Total = 33

· web view0.0500 moles of sulphuric(vi) acid are dissolved in 500cm3 of solution. calculate...

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