Chapter 1 Foundations of Chemistry, Review and Introduction

Chemistry is the study of matter and the changes it undergoes. Matter is anything that has mass and

takes up space. A good understanding of chemical principles will enable you to better understand the

natural world around you independent of what your career goals might bee. Engineers design things

made up of matter. Therefore it is even more important for students of engineering disciplines to know

and understand key chemical and physical properties of the materials they work with. Not

understanding them can have consequences ranging anywhere from an upset customer to a

catastrophic event.

Mater come in all sorts of different forms, shapes, feels, looks and compositions creating the diverse

environment you live in including your very own body. To begin the study of chemistry we will try to

classify matter according to a) its physical state and b) its composition.

a) Classification of matter according to its physical state which can be gaseous, liquid or solid:

A gas has no fixed volume or shape. It assumes the volume and shape of the container. Gases

are highly compressible and can be expanded.

A liquid has a distinct volume independent of its container but has no specific shape. It assumes

the shape of the portion of the container it occupies.

A solid has both a definitive volume and a definite shape. Neither solids nor liquids are

appreciably compressible.

Plasma is another physical state of matter with properties very different from the state that

matter assumes under standard state conditions.

b) Classification of matter according to its composition:

All mater is composed of atoms which is why atoms are sometimes called the building blocks of

matter. There are fewer than 100 different kinds of atoms called elements in our world. Each

element is made up of the same kind of atoms. A compound is made up of two or more

different kinds of atoms. Lead is an element. It is made up of lead atoms only. Sodium chloride

(table salt) is a compound because it is made up of two different types of atoms.

A pure substance is matter that has distinct properties and a composition that does not vary

from sample to sample. A pure substance can either be en element or a compound.

An element cannot be decomposed in to simpler substances by chemical means.

A compound can be broken down into simpler substances by chemical means but not by

physical means. Water for example can be decomposed into the elements hydrogen and oxygen

using an electrolytic cell.

If two ore more elements react to give a compound, the properties of the elements are not

retained in the compound.

2Mg + O2 → 2MgO

Shiny metal colorless gas white solid

Compounds are represented by chemical formulas which show their elemental composition.

Every one knows the formula for water is H2O meaning the ratio of hydrogen atoms that

combine with oxygen atoms is 2:1. We will talk more about different types of formulas later.

Mixtures

Very few pure substances exist in our natural environment. Most matter occurs in form of

mixtures of elements and/or compounds. Air for example consists mainly of the elements

nitrogen, oxygen and argon which together make up more than 99 % of the atmosphere. Soil on

the other hand is a complex mixture of many compounds.

Unlike compounds, mixtures can be separated by physical means (distillation, crystallization,

filtration, floatation, etc.) Mixtures also differ from compounds in that the composition of the

mixture is variable e.g. home made bread is a different mixture every time!

The components of a mixture retain their own identity. Saltwater tastes salty but also has

properties similar to pure water.

This is in contrast to magnesium and oxygen forming magnesium oxide.

There are two types of mixtures:

- heterogeneous: a mixture that does not have the same composition, properties and

appearance throughout (rocks, soil).

- homogeneous: a mixture which is uniform throughout. Homogeneous mixtures are called

solutions (cup of tee).

Figure XXX Classification of matter

Matter

Is it uniform throughout?

NO

YES

YES

YES

NO

NO

Heterogeneous mixture

Homogeneous

Does it have a variable composition?

Pure substance Homogeneous mixture (solution)

Can it be separateinto simpler substances?

Element Compound

Dalton’s Atomic Theory

Some basic laws had been established by the year 1800. These included:

• Law of definite composition:

Every pure compound has a constant composition. (Berzelius)

• Law of conservation of mass

Matter can neither be created nor destroyed during a chemical change. (Lavoisier)

To account for those laws Dalton proposed:

1. All matter is composed of extremely small particles called atoms.

2. All atoms of a given element are alike, but atoms of one element differ from the atoms of any other

element.

3. Compounds are formed when atoms of different elements combine in fixed proportion.

4. A chemical reaction involves a rearrangement of atoms. No atoms are created or destroyed or

broken apart in a chemical reaction.

The law of constant composition is explained by the third statement. Water is always formed by

two atoms of hydrogen joining with one atom of oxygen. Its elemental composition never

changes, no matter were it was from or how it was synthesized – which is why we write its

formula as H2O.

According to the atomic theory chemical change involves a rearrangement of atoms – the same

atoms are combined in different ways to form the products. This explains why mass is conserved

during a chemical reaction.

The law of multiple proportions states that if two elements form more than a single compound, the

masses of one element that combines with a fixed mass of the other element are in the ratio of small

whole numbers.

Example: Consider the compound that can form between the elements hydrogen and oxygen. There is

of course water, H2O, but there is also hydrogen peroxide, H2O2. In hydrogen peroxide 16.000 g of

oxygen are combined with 1.008 g of hydrogen. In water 16.000 g of oxygen are combined with 2.016 g

of Hydrogen (exactly twice as much as in H2O2)

Atoms and Molecules

An atom can be defined as the smallest part of an element that can take part in chemical changes (an

oxygen atom is the smallest part of the element oxygen that is found in any substance that contains this

element.)

Usually individual atoms are not found in nature except the noble gases which occur in atomic form:

Helium, He used as an extreme coolant and in helium balloons

Neon, Ne used in neon lights and neon lasers

Argon, Ar used in welding and chemical industry as an inert gas

Krypton, Kr uses in automobile headlines (banned in a lot of European counties now because too

bright)

Xenon, Xe used in Xenon Ion Propulsion Engines (space missions)

A molecule can be defined as the smallest particle of a substance that still has the properties of that

substance. Several elements exist as diatomic molecules:

Hydrogen, H2 Used as a fuel and reactant and in small weather balloons

Oxygen, O2 vital to sustain life, essential to carry out combustion.

Nitrogen, N2 most abundant gas in the atmosphere, used as inert gas and to generate ammonia

Fluorine,F2 most reactive of all elements. Used as a reactant in the manufacture of Teflon and

refrigerants

Chlorine, Cl2 Used as a disinfectant and for bleaching and in the manufacture of PVC

Bromine, Br2 Important reactant.

Iodine, I2 essential trace element in the human body (thyroid)

Atomic Structure

Although Daltons atomic theory is still largely true 200 years after it was proposed it has a weak spot.

Atoms are not indestructible. Around the turn of the 19th Century experimental observations were

starting to accumulate that supported our modern view of atoms.

Discovery of the Electron

• Streams of negatively charged particles were found to emanate from cathode tubes. A cathode

tube is glass tube from which most of the air has been evacuated and which has two metallic

terminals, a cathode and an anode.

• The emitted particles (electrons) were independent of the nature of the cathode metal used

indicating that all metals possessed electrons.

• J. J. Thompson is credited with its discovery in 1897.

• He determined the charge/mass ratio of the electron to be 1.76 x 10-5 coulomb/g

Millikan’s Oil Drop Experiment

After the mass to charge ratio of the electron was known determining either its mass or its charge would

yield the other quantity.

Millikan used x-rays, a powerful form of radiation to produce ions (charged atoms or molecules) which

became attached to oil droplets produced by a mystifyer (commonly used in a barber shop). The fall of

the oil drops depended on the magnitude and the sign of the charge of the oil drop. Analyzing data from

a large number of experiments Millikan observed that the charge on an oil droplet was always an integer

multiple of the electric charge e.

Discovery of Radioactivity

The discovery and isolation of radioactive elements largely by Pierre and Marie Curie led the way to

even more important discoveries by Rutherford and Soddy. The chemical and physical properties of a

radioactive element changes as it undergoes radioactive decay suggesting a fundamental change of the

atom at the subatomic level, a process that we now call transmutation.

While studying radioactive decay three types of radiation were discovered by Ernest Rutherford

– particles

– particles

– rays

-particles were sharply deflected towards the positive plate of a charged plate capacitor and showed

properties identical to those of cathode rays.

-particles were deflected towards the negative plate of a charged capacitor but the degree of

deflection was very much smaller than that for -particles. We now know that -particles carry a charge

of +2e and have the mass of helium atoms. They are in fact helium nuclei [He]2+.

-rays (also called X-rays) were found to be a highly energetic form of electromagnetic radiation that

passes through an electric field un-deflected.

Discovery of the Nucleus

• Ernest Rutherford shot particles at a thin sheet of gold foil and observed the pattern of scatter

of the particles.

• Most particles went straight through the gold foil but a few (one in 20,000) particles were

deflected at large angles. This lead Rutherford to postulate that the atom had a nucleus.

• Rutherford postulated a very small, dense nucleus with the electrons around the outside of the

atom.

• Most of the volume of the atom is empty space

Class exercise

Which of the preceding experiments was in disagreement with Daltons postulate that atoms are small,

indestructible particles? Justify your answer.

How could Millikan deduce the charge on the oil droplet?

Locate 5 naturally occurring radioactive elements in the periodic table.

Summary:

Each of the roughly one hundred elements found in nature is composed of the same three

fundamental particles:

particle Charge* Mass (u)**

electron -1 0.0005486

proton +1 1.0073

neutron 0 1.0087

* the charge of -1 refers to the charge of an electron which is -1.602176462 Coulomb

** 1 u (formerly amu for atomic mass unit) = 1.6605 x 10-24 g = 1/12 of the mass of one 12C atom

Atoms have a small dens nucleus that holds almost the entire mass of the atom. Atoms are mostly

empty space.

The Atom

electronsprotons

neutrons

The Nucleus

(positive charge)(negative charge)

(neutral - no charge)

rnra

ra

rn≈ 100,000

Mass ≈ 1 u

Mass ≈ 1 u

Mass ≈ 0 u

Most mass

in the

nucleus

• The number of protons in the nucleus identifies the element (look up Z in periodic table)

• The number of protons = ATOMIC NUMBER (Z)

• The number of protons + the number of neutrons = MASS NUMBER (A)

All atoms of the same element have the same atomic number

For a helium atom (Z=2) we have two protons, two electrons and two neutrons. The roll of the neutrons

in the nucleus is partially to buffer the positive charge of the protons.

Symbolic Representation of an Element: EA

Z

The element symbol E consists either of a capital letter e.g. C or U or a capital letter followed by a small

letter e.g. Fe or Al. The mass number A appears in front of the element symbol as superscript. The

atomic number Z might be written in front of the element symbol as subscript but is often omitted since

it can be deduced unambiguously from the element symbol.

Isotopes

• While all atoms of a particular element have the same atomic number they may have differing

mass numbers. If they do, they are called isotopes of that element.

• Although the masses of the isotopes are different they behave similarly chemically.

• Most elements have isotopes, only a few are mono-isotopic.

• Some isotopes are radioactive and have important applications (radio carbon dating, diagnostic

and therapy).

Three isotopes of the element hydrogen exist

H1

1 (Hydrogen) H2

1 (Deuterium) H3

1 (Tritium)

Because the Isotopes of hydrogen are so important they have special names H1

1 is the most abundant

isotope of hydrogen with 99.985% of all natural occurring hydrogen consisting of this isotope

independent of its chemical form (e.g. water, oil, cellulose). H2

1 (deuterium) is an isotope of hydrogen

that is naturally occurring with 0.015 %.

Water made from deuterium and oxygen (D2O) has chemical properties very similar to ordinary water

(H2O). It will react with active metals to generate hydrogen, it will react with nonmetal oxides to form

acids. Because deuterium is much heavier than hydrogen the rate at which the reactions proceed might

be noticeably different. The mass also affects some physical properties.

Deuterium oxide which is also known as heavy water is used in nuclear reactors and as a solvent for

nuclear magnetic resonance spectroscopy.

Applications

Radiocarbon dating: High in our atmosphere neutrons from cosmic rays constantly bombard nitrogen at

a presumably steady rate* resulting in the formation of a radioactive carbon isotope according to the

equation bellow.

n1

0 + N14

7 → C14

6 + p1

1

It is worth reminding you here that this is not a chemical reaction but a nuclear reaction which produces

an entirely new element (transmutation). The 14C is eventually converted to CO2 and incorporated in the

carbon cycle. Plants assimilate it to carbohydrates and other fuels for higher organisms and a balance is

established in all living organisms between 12C and 14C that is about 1/1ppt (ppt = parts per trillion). This

balance is offset after the organism dies since the radioactive 14 C starts to decay but no new 14C is

incorporated. Even though the amount or 14 C in carbon containing samples such as wood is very small

radiation from the decay can be detected. The half life of the 14C isotope is 5770 years making it a

sensitive method to determine the age of prehistoric artefacts to approximately 10000 years. After that

the signal from the decaying 14 C grows so week that it becomes difficult to distinguish it from

background noise.

*) There is active debate as to how valid this assumption is. Neutrons arriving at our upper atmosphere

originate from the sun wind which might be affected by the cycles the sun is going through. There is in

fact a small discrepancy between dates obtained through radio carbon dating and tree ring counts.

Class exercise:

• Boron has two naturally occurring isotopes whose masses have been determined by experiment

B10

5 B11

5

10.012937 u 11.009305 u

• Determine the average atomic mass of naturally occurring boron if the lighter isotope is 20 %

and the heavier isotope is 80 % abundant.

Class exercise: Fill in the empty spaces.

Symbol 103Rh3+ _____ _____

Protons _____ 15 _____

Neutrons _____ 16 1

Electrons _____ 18 1

Net charge _____ _____ 0

Atomic weights

99.985% of all natural occurring hydrogen consists of the hydrogen isotope H1

1 which is composed of a

single proton and a single electron. Adding up the mass of a proton and an electron gives us the mass of

a hydrogen atom. If we try to do this for any other atom, though, we are bound to fail. For example

adding the mass of 5 protons, 5 neutron and 5 electrons give us a mass that is slightly greater than the

experimentally determined mass of the B10

5 isotope. The difference in mass is related to the nuclear

binding energy released in fusion according to Einstein’s famous equation E = mc2. You can feel a small

part of the nuclear binding energy released by the sun, which fusses hydrogen nuclei to helium, every

time you go out in to the sun. For chemists who need to know the atomic weight of all elements

precisely this means the masses of all elements even the mass of every isotope needs to be determined

by experiment. This is done in a mass spectrometer which can measure the mass and the abundance of

the different isotopes. All masses are compared to the mass of a 12C atom and reported as average

relative atomic masses of all naturally occurring isotopes of that element. Magnesium for example has a

relative atomic Mass of 24.305 which means it is 24.305/12.00000 = 2.025 times heavier that a 12C atom.

To use relative atomic masses in calculations we can do one of two things. We can express the mass of

one atom by adding the unit u (atomic mass unit) to the value of the relative atomic mass. E.g. 24.305 u

is the mass of one magnesium atom which equals 24.305 x 1.6605 x 10-24 g or we add the unit g/mol to

the relative atomic mass and get the mass of a mole of magnesium atoms in g (24.305 g/mol).

The mole concept

Since Compounds are formed when atoms of different elements combine in fixed proportion we can

carry out stoichiometric calculations based on the amount of a substance. It is very inconvenient if not

impossible to do a headcount of the participating atoms because of the vast quantity of atoms in even

the smallest speck of matter. By knowing the mass of an atom though we can count them by weighing.

The mole is an amount of substance, much like o dozen (12) or a gross (144). It is defined as the amount

of substance that contains as many entities as there are atoms in exactly 12 g of pure 12C. There are

6.023 × 1023 atoms in 12 g of 12C.

By knowing the average atomic mass of each element we can “count” atoms by weighing them. The

relative atomic mass of iron is 55.847. This means 1 mole of iron atoms has a mass of 55.847 g. We say

the molar mass of iron is 55.847 g/mol. Now we are ready to engage in stoichiometric calculations.

Stoichiometry refers to the amount of all compounds which react in a balanced chemical equation.

Let’s say we are interested in the chemistry of magnesium with oxygen. We have already seen a

balanced reaction equation for this reaction:

2Mg + O2 → 2MgO

How could you have come up with this yourself? We know the reactants, magnesium and oxygen, so

simply write down their chemical symbols remembering that oxygen occurs as O2 a diatomic molecule.

Writing O for oxygen would imply you are using atomic oxygen which has chemical properties very

different from that of molecular O2 or ozone O3 and would thus result in a serious error.

Coming up with the formula for the product is often more challenging? In this case it will be quite easy

as we will see a little later but if we knew nothing of the product as is often the case with new reactions,

we would have to determine the empirical formula of the product. This could simply be done by

weighing the mass of the magnesium metal that was allowed to react with an excessive amount of

oxygen and weighing the mass of the product formed.

Assume we started with 211.5 mg magnesium and obtained 350.1 mg product. We know that the

product only contains the elements oxygen and magnesium. The gain in mass must thus be due to

oxygen chemically bound (locked up) in the compound.

350.1 mg – 211.5 mg =138.6 mg = mass of oxygen atoms in the product.

To get to the stoichiometric ratio of the two elements and express an empirical formula we need to

convert mass of to mole:

0.2115 g Mg ×

= 8.702 x 10 -3 mol

0.1386 g O ×

= 8.663 x 10 -3 mol

Dividing both amounts by the smallest number of moles yields the mole ratio of the two elements:

8.702 x 10 -3 mol : 8.663 x 10 -3 mol = 1.004

8.663 x 10 -3 mol : 8.663 x 10 -3 mol = 1.000

Within experimental error the ratio between magnesium and oxygen atoms is 1:1 which is why we write

the formula MgO (Mg1O1 is not written!). The empirical formula expresses the smallest whole number

ratio of atoms in a compound (more about formulas later).

2Mg (s) + O2(g) → 2MgO(s)

A note on good practice: We should ensure that our equation is balanced and get used to the idea of

including the physical state of reactants and products in brackets behind their symbols. Steam for

example is more reactive then liquid water but liquid oxygen is much more dangerous than gaseous

oxygen. If elements exist in different allotropes (e.g. diamond and graphite, or white phosphorus versus

red, violet or black phosphorus) we also indicate this behind the symbol usually as subscript e.g.

Cgraphite(s)

There are two major kinds of stoichiometry problems you will encounter. They are:

a) You know the amount of one reactant

b) You know the amount of at least two reactants

to a) How much magnesium oxide will form if 5.73 g magnesium burn in excess oxygen?

1.) Convert mass of magnesium to moles of magnesium:

5.73 g Mg ×

= 2.36 x 10 -1 mol Mg

2.) Convert moles of magnesium to moles of magnesium oxide using the coefficients in the

balanced equation:

2.36 x 10 -1 mol Mg ×

= 2.36 x 10 -1 mol MgO

3.) Convert moles of MgO to mass MgO: To find the molar mass (or formula weight) of MgO

simply add the mass of all atoms in the formula (24.304 g/mol + 16.000 g/mol = 40.304 g/mol).

2.36 x 10 -1 mol MgO ×

= 9.50 g MgO

Keep more significant figures in intermediate calculations to avoid rounding errors.

to b) A sample of 1.51 g magnesium is ignited in a sealed 1L flask that contains 1.42 g oxygen

gas. How much magnesium oxide can be formed and how much of which reactant is left behind?

For this problem you first need to determine which reactant limits the amount of product

that can form.

1.) Convert the masses of the reactants to moles of reactant:

1.51 g Mg ×

= 6.21 x 10 -2 mol Mg

1.42 g O2 ×

= 4.44 x 10 -2 mol O2

2.) Determine which the limiting reactant is. At this point you might be persuaded to take a

shortcut but be careful. In the example above even though we have a greater number of

moles of magnesium we cannot simply assume that it is in excess. We need to consider the

mol ratio. Simply calculate the amount of product that could be formed in each case.

6.21 x 10 -2 mol Mg ×

= 6.21 x 10 -2 mol MgO (if enough O2 was available)

4.44 x 10 -2 mol O2 ×

= 8.88 x 10 -2 mol MgO (if enough Mg was available)

Magnesium, even though there is more of it in terms of mass and mol is the limiting

reactant. It limits the amount of MgO that can form. Some oxygen will be left over.

3.) Calculate the mass of product that can form:

6.21 x 10 -2 mol MgO ×

= 2.50 g MgO

2.50 g MgO can be formed. Some oxygen is left over. Since there are only two reactants

which masses we know and one product whose mass we know as well we can simply use

the law of conservation of mass to calculate the mass of oxygen left over.

The mass of the reactants is:

1.51 g Mg + 1.42 g O2 = 2.93 g reactants.

Since only 2.50 g of product is formed the difference must be O2 left over:

2.93 g – 2.50 g = 0.43 g

0.43 g oxygen are left un-reacted.

Had we mistakenly assumed that oxygen was the limiting reactant 8.88 x 10 -2 mol of MgO

should have been produced. This amount of MgO would weigh in at:

8.88 x 10 -2 mol MgO ×

= 3.58 g MgO

which is a mass far greater that the mass of reactants available (2.93 g). An error is not

always found so easily but it often pays off the check if your result makes sense.

The example above actually has practical applications. Magnesium is sometimes used as a getter

in tungsten light bulbs where it removes trace amounts of oxygen and nitrogen which greatly

enhances the lifetime of the light bulb (as long as you ensured that oxygen is the limiting

reactant).

Solution Stoichiometry

In aqueous solution the number of moles of any component dissolved in on litre of solution determines

the molar concentration of that component. This molar concentration is referred to as Molarity and

hugely simplifies stoichiometriy problems in solution over other concentrations such as g/L, %, molality

or ppm. A simple example shall illustrate this:

23.4 mL of a 1.200 M solution of NaOH was necessary to neutralize 10.00 mL of a sulphuric acid sample

of unknown concentration. What is the molarity of the sulfuric acid.

2NaOH + H2SO4 →Na2SO4 + 2H2O

since the molar concentration is defined as:

, or M =

the number of moles of sodium hydroxide can obtained as follows:

n NaOH = M × V = 1.200

× 0.0234 L = 0.02808 mol NaOH

At the equivalence point (neutral) the number of moles of sulfuric acid can be calculated using the

coefficients in the balanced chemical equation:

0.02808 mol NaOH ×

= 0.01404 mol H2SO4

The exact concentration of the sulfuric acid is thus:

M =

=

1.404

How would you prepare a 1 Molar solution. Simply dissolve one mole of the compound in enough water

to make 1 L of solution.

For diluting solutions you might find the equation ci × Vi = cf × Vf useful where:

ci = the initial concentration

Vi = the initial volume

cf = the final concentration

Vf = the final volume

The equation works for all concentrations (not just moles/L) and is based on the assumption that the

density of the solution is not changing (reasonable assumption for dilute aqueous solutions at constant

temperature)

Example: How much of a 6.5 M Ni2+(aq) stock solution is necessary to fill a 220 L nickel-plating bath if the

desired concentration of Ni2+(aq) is 0.10 M?

Vi =

=

= 3.38 L

% Yield

A side problem that may arise is to calculate the % yield of a reaction which is simply the amount of

product that you isolated from the reaction mixture divided by the amount of product that in theory

could have formed if everything went perfect.

% yield =

x 100 %

Example: 4.75 g metallic copper is precipitated from a solution containing 25.23 g CuSO4×5H2O by

adding excess iron metal. Calculate the % Yield.

Cu2+ (aq) + Fe(s) → Cu(s) + Fe2+ (net ionic equation for the redox reaction)

25.23 g (CuSO4×5H2O) ×

= 0.1010 mol CuSO4×5H2O

0.1010 mol (CuSO4×5H2O) ×

= 0.1010 mol Cu2+

0.1010 mol Cu2+ ×

= 0.1010 mol Cu (you might omit this explicit conversion in such

simple stoichiometry but it is good practice for more challenging problems)

0.1010 mol Cu2+ ×

= 6.42 g Cu

% yield =

x 100 % =

x 100 % = 74.0 %

Chemical Nomenclature

Positive Ions (Cations)

a) Cations formed from metal atoms have the same name as the metal, e.g. Na+ is the sodium ion

b) If a metal can form different cations, the positive charge is indicated by a Roman numeral in

parentheses following the name of the metal, e.g. Au+ is the gold(I) ion and Au3+ is the gold(III) ion.

c) Cations formed from nonmetal atoms have names that end in -ium, e.g. NH4+ is the ammonium ion.

Some examples of cations that you might encounter:

Charge Formula Name Formula Name

+1 H+ Hydrogen ion NH4+ Ammonium ion

+1 Li+ Lithium ion Cu+ Copper (I) ion

+1 Na+ Sodium ion

+1 K+ Potassium ion

+1 Rb+ Rubidium ion

+1 Cs+ Cesium ion

+1 Ag+ Silver ion

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

+2 Ca2+ Calcium ion Cu2+ Copper(II) ion

+2 Sr2+ Strontium ion Fe2+ Iron (II) ion

+2 Ba2+ Barium ion Mn2+ Manganese (II) ion

+2 Zn2+ Zinc ion Hg22+ Mercury(I) ion (dimer)

+2 Cd2+ Cadmium ion Hg2+ mercury (II) ion

Ni2+ Nickel (II) ion

Pb2+ Lead (II) ion

Sn2+ Tin (II) ion

3+ Al3+ Aluminum ion Cr3+ Chromium (III) ion

Fe3+ Iron(III) ion

PS: You will have a periodic table including the names of the elements on every exam

Chemical Nomenclature: Negative Ions (Anions)

a) The names of the monatomic anions are formed by replacing the ending of the name of the element

with -ide, e.g. O2- is the oxide ion.

b) Polyatomic anions containing oxygen (called oxyanions) have names ending in -ate or -ite, e.g. SO42- is

the sulfate ion and SO32- is the sulfite ion.

c) Anions derived by adding H+ to an oxyanion are named by adding the prefix hydrogen or dihydrogen,

e.g. HCO3- is the hydrogen carbonate ion

More on naming oxyanions:

ClO4- perchlorate ion (one more O atom than chlorate)

ClO3- chlorate

ClO2- chlorite ion (one less O atom than chlorate)

ClO- hypochlorite ion (one O atom less than chlorite)

Some examples of anions that you might encounter: Know the bolt ions by heart!

Charge Formula Name Formula Name

-1 H- Hydride ion CH3COO- Acetate ion

-1 F- Fluoride ion ClO3- Chlorate ion

-1 Cl- Chloride ion ClO4- Perchlorate ion

-1 Br- Bromide ion NO3- nitrate ion

-1 I- Iodide ion MnO4- Permanganate ion

-1 CN- Cyanide ion

-1 OH- Hydroxide ion

-2 O2- Oxide ion CO32- Carbonate ion

-2 O22- Peroxide ion CrO4

2- Chromate ion

-2 S2- Sulfide ion Cr2O72- Dichromate ion

SO42- Sulfate ion

-3 N3- Nitride ion PO43- Phosphate ion

Chemical Nomenclature Ionic Compounds:

Names of ionic compounds consist of the name of the cation followed by the name of the anion, e.g.

Cu(ClO4)2 is copper(II) perchlorate, and CaCO3 is calcium carbonate.

Chemical Nomenclature Name and Formulae of Acids:

1. Acids containing anions whose names end in -ide are named by changing the -ide ending to -ic, adding

the prefix hydro- to this anion name, and then following with the word acid (e.g. chloride becomes

hydrochloric acid)

2. Acids containing anions whose names end in -ate or -ite are named by changing the -ate ending to -ic

and the -ite ending to -ous and then adding the word acid. (e.g. sulfate becomes sulfuric acid, sulfite

becomes sulfurous acid)

Chemical Nomenclature: Binary Molecular Compounds

1. The name of the element farther to the left in the periodic table is written first (e.g. H2O not OH2)

2. If both elements are in the same group in the periodic table, the one having the higher atomic

number is written first ( e.g. SO2 not OS2)

3. The name of the second element is given an -ide ending (e.g. H2S is hydrogen sulfide)

4. Greek prefixes are used to indicate the number of atoms of each element.

Examples: N2O4 is dinitrogen tetroxide and P4S10 is tetraphosphorus decasulfide

Know the Greek prefixes up to 10: mono, di, tri, terta, penta, hexa, hepta, octa, nona, deca

Finally some acids you need to know:

Sulfuric acid H2SO4

Sulfurous acid H2SO3

Phosphoric acid H3PO4

Carbonic acid H2CO3

Nitric acid HNO3

Acetic acid CH3COOH

Perchloric acid HClO4

An important base that is not simply a hydroxide salt that you need to know is ammonia NH3.

Practice question:

1. Which of the following figures represents?

a) A pure element

b) A mixture of two elements

c) A pure compound

d) A mixture of an element and a compound

2. Does the following diagram represent a chemical or physical change? How do you know?

3. Define the following forms of matter as element, compound or mixture. If you indicate mixture

specify if homogeneous or heterogeneous:

a) Sand

b) Diamond

c) Gasoline

d) Sugar

4. Indicate which of the following changes represent chemical change or physical change:

a) Burning candle wax

b) Melting candle wax

c) Cutting wire

d) Painting your bike

5. Fill in the empty spaces:

Symbol 27Al3+ _____ 150Sm

# of protons _____ 42 _____

# of neutrons _____ 54 _____

# of electrons _____ _____ _____

Charge _____ 0 _____

6. Which atomic property is described by the following equation?

Cl + e- → Cl- + energy

7. Name 3 metals, 3 non-metals and 2 metalloids. Name three properties that characterize metals

and three properties that characterize non-metals.

8. Balance the following equation

C2H6O2 + O2 →CO2 + H2O

P4 + O2 → P4O10

Ca(OH)2 + HCl → CaCl2 + H2O

Fe3O4 + H2 → Fe + H2O

9. In a typical car battery lead (IV) oxide reacts with lead according to the following equation:

(During charging the battery the reveres reaction occurs!)

PbO2 + Pb + 2H2SO4 → 2PbSO4 + 2H2O

One way of checking the “livelihood” of a battery is to measure the density of the acid. At first

approximation battery acid consists only of sulfuric acid and water. Calculate the amount of lead

sulfate that formed and the amount of lead that reacted if the density of the acid changed from

1.2769 g/ml in the freshly charged battery to 1.1243 g/ml. Assume a total of 3 L of battery acid

which does not change considerably during the reaction. Use the table specific gravity of

aqueous sulfuric acid to solve the problem. Specific gravity refers to the density of a substance

at 20 C divided by the density of water at 4 C and is thus a dimensionless ratio.

% H2SO4 Sp. Gr. %H2SO4 Sp. Gr. %H2SO4 Sp.Gr. %H2SO4 Sp.Gr. %H2SO4 Sp.Gr.

1 1.0051 21 1.1471 41 1.3116 61 1.5091 81 1.7383

2 1.0118 22 1.1548 42 1.3205 62 1.52 82 1.7491

3 1.0184 23 1.1626 43 1.3294 63 1.531 83 1.7594

4 1.025 24 1.1704 44 1.3384 64 1.5421 84 1.7693

5 1.0317 25 1.1783 45 1.3476 65 1.5533 85 1.7786

6 1.0385 26 1.1862 46 1.3569 66 1.5646 86 1.7872

7 1.0453 27 1.1942 47 1.3663 67 1.576 87 1.7951

8 1.0522 28 1.2023 48 1.3758 68 1.5874 88 1.8022

9 1.0591 29 1.2104 49 1.3854 69 1.5989 89 1.8087

10 1.0661 30 1.2185 50 1.3951 70 1.6105 90 1.8144

11 1.0731 31 1.2267 51 1.4046 71 1.6221 91 1.8195

12 1.0802 32 1.2349 52 1.4148 72 1.6338 92 1.824

13 1.0874 33 1.2432 53 1.4248 73 1.6456 93 1.8279

14 1.0947 34 1.2515 54 1.435 74 1.6574 94 1.8312

15 1.102 35 1.2599 55 1.4453 75 1.6692 95 1.8337

16 1.1094 36 1.2684 56 1.4557 76 1.681 96 1.8355

17 1.1168 37 1.2769 57 1.4662 77 1.6927 97 1.8364

18 1.1243 38 1.2855 58 1.4768 78 1.7043 98 1.8361

19 1.1318 39 1.2941 59 1.4875 79 1.7158 99 1.8342

20 1.1394 40 1.3028 60 1.4983 80 1.7272 100 1.8305

10. Copper can be refined from copper salt solutions using iron according to the following equation

CuCl2 + Fe → Cu + FeCl2

What is the %yield of copper if 150 g Copper chloride reacted and 55 g Copper were obtained.

11. A main component of Portland cement that is made up of the elements calcium, silicon and

oxygen has the following analytical data: 34.50 % Ca, 24.18 % Si. Calculate the empirical formula

of the compound.

12. You have to determine the identity of a polymer used to insulate a copper wire in an older

industrial complex: Elemental analysis of the polymer gave 38.42 % C and 4.88 % H. Which of

the following polymers is very likely to be the material in question?

Polyethylene (PE)

C

C

nH

H H

H

Polyurethane (PU)

C

C

C

C

C

C

N

C

O

C

C

O

N

C

O

H

H

O

H

H

H

H

H H

H Hn

Polyvinylchloride (PVC)

C

C

nH

Cl H

H

13. How much of a 5 M stock solution of NaCl do you need to make 2 L of a 2 M NaCl solution.

14. Name the following ions: SO42-

Cl-

PO43-

CO32-

ClO3-

NO3-

15. Write names and formula of the parent acid for the ions above.

A brief note on standard state:

Standard state conditions: Many if not most properties of matter depend on temperature and usually

to a lesser extend on pressure. To be able to compare the properties of different substances they are

often measured and reported at a given temperature and pressure. Standard Temperature and Pressure

or STP refers to a specific temperature and pressure. For gases STP is 0 °C (273.15 K) and 1 atm (101.3

KPa). For electrochemistry and thermochemistry room temperature and 1 atm is usually associated with

standard conditions. Room Temperature nowadays means 25 C but might refer to 18 or 20 °C in older

literature.