Elements

Quantitative Chemistry1Activity:Relative atomic massRelative molecular massRelative formula massAvogadros numberElementCompoundMoleculeMoleIsotopeDiatomic

Print out the keywords worksheet and add the following words, once you know the meanings fill them in.Chemical Formulas and the Mole conceptElementsCompoundsMole concept and Avogadro's constantIsotopesFormulas of compoundsEmpirical formulaMolecular formulaStructural formula

ElementsAll substances are made up of one or more __________.An element ___________be broken down by any chemical process into ___________substances.There are just over _____ known elements.The smallest part of an element is called an __________.

elementscannotatomsimpler 100 Hand out IB periodic table/ data book that they must bring each lesson

4Molecules and compoundsSome substances are made up of a single element although there may be more than one atom of the element in a particle of the substance.For example, Oxygen is diatomic, that is, a molecule of oxygen containing two oxygen atoms.A compound contains more than one element.For example; a molecule of water contains two hydrogen atoms and one oxygen atom.Water is a compound not an element because it can be broken down chemically in to its constituent elements: hydrogen and oxygen

Mole concept and Avogadros constantA single atom of an element has an extremely small mass. For example, an atom of carbon12 has a mass of 1.993 x 10-23 g.This is far too small to weighInstead of weighing just one atom we can weigh a mole of atoms. 1 mole contains 6.02 x 10 23 particles6.02 x 10 23 atoms of carbon12 = ?This number is known as Avogadros constant(NA or L)1 mole of Carbon = 12g

Mole concept and Avogadros constantChemists measure amounts of substances in moles.A mole is the amount that contains L particles of that substance.The mass of one mole of any substance is known as the molar mass and has the symbol M. For example, Hydrogen atoms have one 12th the mass of carbon12 atoms so one mole of hydrogen atoms contains 6.02 x 10 23 hydrogen atoms and has a mass of 1.01 g.For diatomic molecules e.g. H2 there are 6.02 x 1023 molecules of hydrogen and therefore 12.04 x 1023 atoms.

Try some molar calculations involving Avogadros const.

7Relative Atomic massThe actual atomic mass of an individual atom of an element is so small that we use a relative atomic mass as seen on the periodic table.Since hydrogen is the lightest atom (it has an actual mass of 8.275 x 10-25) we say that hydrogen has a relative atomic mass of 1. Experiments have shown that an atom of carbon weighs 12 times as much as an atom of hydrogen. So the relative atomic mass of carbon is 12. Relative atomic mass can be abbreviated to Ar or RAM and is found on the periodic table. It has no units.ActivityComplete the worksheet IB 1.1 Using Avogadros NumberThere are 20 questions to completeShow all workingIsotopes and RMM/RFMIn reality elements are made up of a mixture of isotopes.The relative atomic mass of an element Ar is the weighted mean of all the naturally occurring isotopes of the element relative to carbon12. This explains why the relative atomic masses given for elements on the periodic table are not whole numbersThe units of molar mass are g mol-1 (this means grams per mole) but relative molar masses Mr have no units. For molecules relative molecular mass is used (RMM). For example the Relative Molecule Mass of glucose, C6H12O6 = (6 x 12.01) + (12 x 1.01) + (6 x 16.00) = 180.18

Recall what an isotope is/ look it up in text bookRMM is the Relative Molecular Mass for covalent molecules / RFM is the relative formula mass for any compounds it is safer to use RFM rather than RMM if youre not sure what type of compound it is.mol-1 means per

10Calculating RFMThis is very simple but there are one or two rules to remember:Numbers in subscript only refer to the number directly in front of it. E.g. SO4 there is one Sulphur atom and 4 Oxygen atoms RFM = 96 in totalEverything inside a bracket is multiplied by the small number directly outside of it on the right hand side. E.g. (NH4)2SO4 So the RFM of NH4 must be multiplied by 2 RFM = 132 in totalWhen a . is placed into a formula it means another compound is attached to the formula and must be included in the RFM e.g. CuSO4.5H2O means there are 5 water molecules added to this compound so you must include them in the total RFM = 249.5 in total

ActivityComplete the worksheet IB 1.2 - Calculation of the Molar Mass of CompoundsThere are 60 questions to completeDont forget to refer to the previously stated rulesShow all workingUsing Molar massesAs we have seen 1g of Hydrogen contains many more atoms than 1g of carbon.Calculate the number of atoms in 1g of hydrogen and 1g of carbon.Suggest why we cannot use mass as a means of keeping a fair test during an experiment i.e. The reaction of substances with oxygen which would give out more energy 1g of hydrogen or 1g of carbon?To keep a test fair we use molar quantities.1g of hydrogen = 6 x 1023 atoms1g of carbon = 0.5 x 1023Not fair because there are far more hydrogen atoms to react than carbon atoms therefore will have more reactions and give out more energy.13Calculating molar quantitiesAs we have seen a mole is a fixed number of particles, 6 x 1023, we can find the molar mass of an element from the periodic table (mass number). The molar mass of a compound is just the sum of all the elements in the formula in their correct proportions. E.g. 1 mole of Na = 23g1 mole of NaOH = 23 + 16 + 1 = 40g1 mole of Na2SO4 = (2 x 23) + 32 + (4 x 16) = 142g

Finding the number of molesIt is very rare to use exactly one mole of a substance in a reaction. So it is important to be able to find the number of moles of elements and compounds in different reactions.We can do this using the following equation triangle.You must learn this!!!!!!MassNumber of molesRAM or RFMxExample 1;Find the number of moles in 46g of sodium

MassNumber of molesRAM or RFMxNumber of moles=Mass________RAM or RFM46232Example 2;Find the mass of 0.2 moles of sodium

MassNumber of molesRAM or RFMxNumber of moles=MassRAM or RFMx0.2234.6 gActivityComplete the worksheet IB 1.3 Molar calculations and IB 1.4 Molar calculations (II)There are 60 questions to completeDont forget to refer to the previously stated rulesShow all workingFormulas of compoundsCompounds can be described by different chemical formulas:EmpiricalMolecularStructural

Empirical FormulaLiterally this is the formula obtained by experiment.This shows the simplest whole number ratio of atoms of each element in a particle of the substanceIt can be obtained by either knowing the mass of each element in the compound or from the percentage composition by mass of the compound.The percentage composition can be converted directly into mass by assuming 100g of the compound are taken.

Formula TriangleYou can use a formula triangle to help you rearrange the equation:MassNumber of molesRAM or RFMxNumber of moles=Mass________RAM or RFM

Practice questions on molesExampleA compound contains 40.00% carbon, 6.73% hydrogen and 53.27% oxygen by mass, determine the empirical formula. Amount / molratioCHO40.00/12.01 = 3.3316.73/1.01 = 6.6653.27/16.00 =3.3321Empirical formula is therefore CH2O

22Worked exampleAn organic compound contains 60.00% carbon and 4.46% hydrogen by mass. Calculate the empirical formula of the compound.As the percentages do not add up to 100% it can be assumed that the compound also contains 35.54% oxygen by mass.Assume that there are 100g of compound, and calculate the amount of each element by dividing the mass of the element by its atomic mass ArFind the simplest ratio by dividing through by the smallest amount and then converting any decimals to give the simplest whole number ratioWrite the empirical formula, using subscripts for the numbers in the ratio.The empirical formula of the compound isC9H8O4

Practice questionsActivity;Read through the examples 16, 17 and 18 on pages 31 and 32 in Calculations in AS/A Level Chemistry Jim ClarkComplete problems 15 and 16 part a only.Answers are at the back of the book page 279Complete worksheet 1.5 Formulae and Percentage compositionMolecular formulaFor molecules this is much more useful as it shows the actual number of atoms of each element in a molecule of the substance.It can be obtained from the empirical formula if the molar mass of the compound is known.Methanal CH2O (Mr = 30), ethanoic acid C2H4O2 (Mr = 60) and glucose C6H12O6 (Mr = 180) are different substances with different molecular formulas but all with the same empirical formula CH2O. Note that the subscripts are used to show the number of atoms of each element in the compound.

Activity;Using your answers from part a (from last lesson)complete problems 15 and 16 part b only.Answers are at the back of the book page 279Complete worksheet; Worksheet IB1.6 Section B-Molecular FormulaeStructural formulaThis shows the arrangement of atoms and bonds within a molecule and is particularly useful in organic chemistry.The three different formulas can be illustrated using ethene:

CH2EmpiricalC2H4MolecularCH2CH2StructuralStructural formulaTwo compounds with the same molecular formula but different structural formulas are known as isomers.For example, both 1,1-dichloroethane and 1,2-dichloroethane have the same empirical formula, CH2Cl, and the same molecular formula, C2H4Cl2, but their structural formulas are different. In 1,1-dichloroethane both the chlorine atoms are bonded to the same carbon atom, whereas in 1,2-dichloroethane each of the two carbon atoms has one chlorine atom bonded to it

ClCCHHHClHHCCHHHClCl1,1-dichloroethane1,2-dichloroethaneStructural formulasThe use of structural formulas is particularly important in organic chemistry.Compounds with the same molecular formula but with different structural formulas are called structural isomers.Structural isomers often have very different physical and chemical properties.

For example, methoxymethane, CH3OCH3, and ethanol, C2H5OH, both have the same molecular formula, C2H6O but, unlike methoxymethane, ethanol is completely miscible with water and is generally much more reactive chemically.HCCHHHHHOHCCHHHHHOmethoxymethaneethanolQuick QuestionsWhat is the empirical formula of the following compounds? You could try to name them too.A liquid containing 2.0g of hydrogen, 32.1 g sulphur, and 64g oxygen.A white solid containing 0.9g beryllium, 3.2g oxygen, and 0.2g hydrogen.A white solid containing 0.234g magnesium and 0.710g chlorine.3.888g magnesium ribbon was burnt completely in air and 6.488g of magnesium oxide were produced.How many moles of magnesium and of oxygen are present in 6.488g of magnesium oxide?What is the empirical formula of magnesium oxide?More to followJust a few more!What are the empirical formulae of the following molecules?Cyclohexane C6H12Dichloroethene C2H2Cl2Benzene C6H6Mr for ethane-1,2-diol is 62.0. It is composed of carbon, hydrogen and oxygen in the ratio by moles of 1:3:1. What is its molecular formula?Answers1H2SO4 sulphuric acidBeO2H2 which is Be(OH)2, beryllium hydroxideMgCl2 magnesium chloride10.16 moles of eachMgO1CH2CHClCHC2H6O2Working out formulae for ionic compoundsYou cant write equations until you can write formulae.People tend to remember the formulae for common covalent substances like water or carbon dioxide or methane, and will rarely need to work them out.That is not true of ionic compounds. You need to know the symbols and charges of the common ions and how to combine them into a formula.

The need for equal numbers of pluses and minusesIons are atoms or groups of atoms which carry electrical charges, either positive or negative.Compounds are electrically neutral.In an ionic compound there must therefore be the right number of each sort of ion so that the total positive charge is exactly the same as the total negative charge.Obviously, then, if you are going to work out the formula, you need to know the charges on the ions.Some charges you may remember!ChargeGroup 1 metalsGroup 2 metalsGroup 3 metalsGroup 5 non-metalsGroup 6 non-metalsGroup 7 non-metals+3+1-1+2-2-3Can you put the correct charges into the table?Cases where the name tells you the charge on an ionA name like lead(II)oxide tells you that the charge on the lead (a metal) is +2.Iron(III)chloride contains 3+ iron ion.Copper(II) sulphate contains a Cu2+ ionNotice that all metals form positive ions.Ions that need to be learnt:

Positive ionsNegative ionsZincNitrateSilverHydroxideHydrogenHydrogencarbonateammoniumCarbonateLeadSulphatePhosphateZn2+Ag+H+NH4+NO3-OH-HCO3-CO32-SO42-PO43-Pb2+Confusing endingsDo not confuse ions like sulphate with sulphide.A name like sodium sulphide means that it contains sodium and sulphur only.Once you have an ate ending it means that there is something else there as well often, but not always, oxygenTry naming the following compounds:Mg3N2Mg(NO3)2CaC2CaCO3ActivityFill in the worksheet; Worksheet IB1.6 Important ions for IB calculations how many can you complete without looking them up?Click here for answers.Working out the formula of an ionic compoundExample 1.To find the formula for sodium oxide, first find the charges on the ionsSodium is in Group 1, so the ion is Na+Oxygen is in Group 6, so the ion is O2-To have equal numbers of positive and negative charges, you would need 2 sodium ions for each oxide ionThe formula is therefore Na2OWorking out the formula of an ionic compoundExample 2.To find the formula for barium nitrate, first find the charges on the ionsBarium is in Group 2, so the ion is Ba2+Nitrate ions are NO3-To have equal numbers of positive and negative charges, you would need 2 nitrate ions for each barium ionThe formula is therefore Ba(NO3)2Notice the brackets around the nitrate group. Brackets must be written if you have more than one of these complex ions (ions containing more than one type of atom).Working out the formula of an ionic compoundExample 3.To find the formula for iron(III)sulphate, first find the charges on the ionsIron(III) tells us that the ion is Fe3+Sulphate ions are SO42-To have equal numbers of positive and negative charges, you would need 2 iron (III) ions for every 3 sulphate ionsThe formula is therefore Fe2(SO4)3ActivityComplete the worksheet; Worksheet IB1.7 Writing formulae from namesFor answers click hereChemical Reactions and equationsProperties of chemical reactionsChemical equationsState symbolsOne way or reversible reactionsIonic equations

Properties of chemical reactionsOnce the correct formulas of all the reactants and products are known, it is possible to write a chemical equation to describe a reaction taking place.New substances are formed.Bonds in the reactants are broken and bonds in the products are formed, resulting in an energy change between the reacting system and the surroundings.There is a fixed relationship between the number of particles of reactants and products, resulting in no overall change in mass.

Chemical EquationsIn a chemical equation the reactants are written on the left hand side, and the products are written on the right hand side. As there is no overall change in mass, the total amount of each element must be the same on the two side of the equation.For example, consider the reaction between zinc metal and hydrochloric acid to produce zinc chloride and hydrogen gas.The correct formulas for all the reacting species and products are first written down.

Zn+HClZnCl2+H2REACTANTSPRODUCTSChemical EquationsThe equation is then balanced by adding the correct coefficients

When the correct coefficients are in place, the reaction is said to be stoichiometrically balanced.The stoichiometry tells us that in this reaction two moles of hydrochloric acid react with one mole of zinc to form one mole of zinc chloride and one mole of hydrogen.

Zn+HClZnCl2+H22REACTANTSPRODUCTSState symbolsThe physical state that the reactants and products are in can affect both the rate of the reaction and the overall energy change so it is good practice to include the state symbols in the equation.

(s)solid(l)liquid(g)gas(aq)In aqueous solutionZn+2HClZnCl2+H2REACTANTSPRODUCTS(s)(aq)(aq)(g)ActivityComplete the worksheet; Worksheet IB1.8 Balancing equationsFor answers click hereOne way or reversibleA single arrow is used if the reaction goes to completion. Sometimes the reaction conditions are written on the arrow:

Reversible arrows are used for reactions where both the reactants and products are present in the equilibrium mixture:

C2H4(g)H2(g)C2H6(g)+Ni Catalyst, 180oCIonic equationsIonic compounds are completely dissociated in solution so it is sometimes better to use ionic equations to describe their reactions. For example, when silver nitrate solution is added to sodium chloride solution a precipitate of silver chloride is formed.

Na +(aq) and NO3 (aq) are spectator ions and do not take part in the reaction. So the ionic equation becomes:

AgNO3(aq)+NaCl(aq)NaNO3(aq)AgCl(s)+Ag+(aq)+Cl-(aq)AgCl(s)ActivityCarry out Experiment 1.1 Precipitation reactionsPrecipitation ReactionsTo each of the following solutions add NaOH solution drop-wise to see a precipitate formMetal compoundMetal ionProduct (ObservationMgCl2(aq)Mg2+Mg(OH)2(s)White solidFeSO4(aq)Fe2+Fe(OH)2(s)Dirty green solidCaCl2(aq)Ca2+Ca(OH)2(s)White solidFeCl3(aq)Fe3+Fe(OH)3(s)Rusty brown solidCu(NO3)2(aq)Cu2+Cu(OH)2(s)Blue solidIonic equationsMgCl2(aq)2NaOH(aq)Mg(OH)2(s)2NaCl(aq)FeSO4(aq)2NaOH(aq)Fe(OH)2(s)Na2SO4(aq)CaCl2(aq)2NaOH(aq)Ca(OH)2(s)2NaCl(aq)FeCl3(aq)3NaOH(aq)Fe(OH)3(s)3NaCl(aq)Cu(NO3)2(aq)2NaOH(aq)Cu(OH)2(s)2NaNO3(aq)++++++++++Mg2+(aq)2OH-(aq)2OH-(aq)2OH-(aq)3OH-(aq)2OH-(aq)Fe2+(aq)Ca2+(aq)Fe3+(aq)Cu2+(aq)Measurements and calculationsError and uncertaintyMeasurements of molar quantitiesSolidsSolutionsLiquidsGasesWorked examples

Error and uncertaintyIn the laboratory moles can conveniently be measured using either mass or volume depending on the substances involved.But with any measurement there is always uncertainty and error.

Systematic errorsQuantitative chemistry involves measurement.A measurement is a method which some quantity or property of a substance is compared with a known standard.If the instrument used to take the measurements has been calibrated wrongly or if the person using it consistently misreads it then the measurements will always differ by the same amount.Such an error is known as a systematic error.An example might always be reading a pipette from the sides of the meniscus rather than from the middle of the meniscus.Random uncertaintiesRandom uncertainties occur if there is an equal probability of the reading being too high or too low from one measurement to the next.These might include variations in the volume of glassware due to temperature fluctuations or the decision on exactly when an indicator changes colour during an acid base titration.Precision and accuracy.Precision refers to how close several experimental measurements of the same quantity are to eachother.Accuracy refers to how close the readings are to the true value. This may the standard value,SolidsSolids are normally measured by weighing to obtain the massMasses are measured in grams and kilograms remember 1.000 kg = 1000gWhen weighing a substance the mass should be recorded to show the accuracy of the balance.For example, exactly 16 g of a substance would be recorded as 16.00 g on a balance weighing to + or 0.01 g but as 16.000 g on a balance weighing to + or 0.001 g.

LiquidsPure liquids may be weighed or the volume recorded.The density of the liquid = mass / volume and is usually expressed in g cm-3 or kg dm-3In the laboratory, volume is usually measured using different apparatus depending on how precisely the volume is requiredFor very approximate volumes a beaker or conical flask can be used.Measuring cylinders are more precise, but still have a large amount of uncertainty.For fixed volumes, volumetric flasks or pipettes are used for precise measurements, and for variable volumes a burette or graduated pipette is used.

Measuring LiquidsThe uncertainty associated with the burette, pipette or volumetric flask can vary. In schools, grade B equipment is usually used, but more expensive grade A (analytical) equipment can be accurate to 0.01cm3.

SolutionsVolume is usually used for solutions.

1.000 litre = 1.000dm3 = 1000cm3

Concentration is the amount of solute (dissolved substance) in a known volume of solution (solute plus solvent). It is expressed either in g dm-3 or more usually in mol dm-3We will work out how to calculate concentrations and carry out volumetric calculations later in the topic.We can use this formula..molesconcentrationVol in dm3xconcentrationmolesVol in dm3=You can use a formula triangle to help you rearrange the equation:Students complete exercisesDo you still remember it????You can use a formula triangle to help you rearrange the equation:MassNumber of molesRAM or RFMxNumber of moles=Mass________RAM or RFM

Practice questions on molesConcentrations of solutionsConcentrations can be measured in:g dm-3mol dm-3And you have to be able to convert between themYou have already practiced converting moles into grams and vice versa.When you are doing the conversions in concentration sums, the only thing that is different is the amount of substance you are talking about happens to be dissolved in 1dm3 of solution. This does not affect the sum in any way.

Calculations involving solutionsA solution of sodium hydroxide, NaOH, had a concentration of 4g dm-3. What is its concentration in mol dm-3? (H = 1; O = 16; Na = 23)1 mole of NaOH weighs 40g4g is 4/40 moles = 0.1 mol4 g dm-3 is therefore 0.1 mol dm-3

Calculations involving solutionsWhat is the concentration of a 0.0500 mol dm-3 solution of sodium carbonate, Na2CO3, in g dm-3? (C = 12; O = 16; Na = 23)1 mol Na2CO3 weighs 106g0.0500 mol weighs 0.0500 x 106g = 5.30gSo 0.0500 mol dm-3 is therefore 5.30g dm-3

Calculations involving concentrationsRecall ionic charges and names of ionic formulae.Know how to solve problems involving concentration and amount of solute.Basic calculations from equations involving solutions.Example:What mass of barium sulphate would be produced by adding excess barium chloride to 20.0cm3 of copper(II)sulphate solution of concentration 0.100mol dm-3 (O = 16, S = 32, Ba = 137) BaCl2(aq) + CuSO4(aq) BaSO4(aq) + CuCl2(aq)From this equation we can see that 1 mole of CuSO4 gives 1 mole BaSO4With the information given you can only work out the number of moles of copper(II)sulphate that you are starting with.

Number of moles of CuSO4=20---------1000x0.100Volume of CuSO4 in QnConverts cm3 to dm3Concentration of CuSO4 given in qn.= 0.00200 molGoing back to the question!!If 1 mole of CuSO4 gives 1 mole of BaSO4 therefore 0.00200mol of CuSO4 will give 0.00200mol of BaSO4RFM of 1 mol of BaSO4 = 233Therefore 0.00200mol of BaSO4

0.00200 x233== 0.466g of BaSO4Next example:What is the maximum mass of calcium carbonate which will react with 25cm3 of 2.00mol dm-3 hydrochloric acid? (C = 12, O = 16, Ca = 40)Write the balanced symbol equation.Calculate the number of moles of hydrochloric acid.Work out the ratio of moles that HCl reacts with CaCO3Calculate the number of moles of Calcium chlorideWork out the RFM of calcium carbonateWhat is the maximum mass which could react with HCl?Here it is..CaCO3 + 2HCl CaCl2 + H2O + CO2Number of moles of HCl = 25.0 / 1000 x 2.00 = 0.0500 mol1 mol CaCO3 reacts with 2 mole HCl= 0.0250mol1 mol CaCO3 weighs Therefore 0.0250 mol weighs 0.0250 x 100 = So the max mass of CaCO3 which you could react with this amount of hydrochloric acid is 2.50g

100g2.50gHere it is..CaCO3 + 2HCl CaCl2 + H2O + CO2Number of moles of HCl = 25.0 / 1000 x 2.00 = 0.0500 mol1 mol CaCO3 reacts with 2 mole HCl0.5 mol CaCO3 reacts with 1 mole HCl= 0.0250mol1 mol CaCO3 weighs Therefore 0.0250 mol weighs 0.0250 x 100 = 2.50g100gSo the max mass of CaCO3 which you could react with this amount of hydrochloric acid is 2.50gLast one.What is the minimum volume of 0.500 mol dm-3 sulphuric acid needed to react with 0.240g of magnesium? (Mg = 24)Write a balanced symbol equation.Work out the ratio which magnesium reacts with sulphuric acid.Calculate the number of moles of magnesiumCalculate the number of moles of sulphuric acidCalculate the volume of sulphuric acid needed.Here it isMg + H2SO4 MgSO4 + H21 mole of Mg reacts with 1 mol of H2SO4Number of moles of Mg =0.240 / 24=0.0100 molSo number of moles of H2SO4 = 0.0100 molNext we can use this formula..molesconcentrationVol in dm3xSo the volume in dm3=Moles / Concentration=0.0100 / 0.500=0.0200 dm3Which is what in cm3Students complete exercisesCalculations involving concentrationsKnow how to solve problems involving concentration and amount of solute.Know how to solve simple titration equationsSimple volumetric calculations25.0cm3 of 0.100 mol dm-3 NaOH solution required 23.5cm3 of dilute hydrochloric acid for neutralisation. Calculate the concentration of the hydrochloric acid.Write the balanced symbol equationWork out the ratio of reactants and productsWhat volumes of substances are known?What concentrations are known?What are you trying to calculate?NaOH + HCl NaCl + H2Oratio1111:::volumes25cm323.5cm3concentrations0.100M?????The only thing we can do with the numbers given is to calculate the number of moles of NaOH..Moles of NaOH=Vol (in dm3) x concentration=25/1000 x 0.100=0.00250 molNaOH + HCl NaCl + H2Oratio1111:::volumes25cm323.5cm3concentrations0.100M?????No. of moles0.00250 mol0.00250 molSo now we can calculate the concentration of HCl using the formula:concentration=Mole / volume in dm3=0.0025 / (23.5/1000)=0.106 mol dm3What else do we now know?Mass and Gaseous volume relationshipsCalculate theoretical yields from chemical equations.Determine the limiting reactant and the reactant in excess when quantities of reacting substances are givenSolve problems involving theoretical, experimental and percentage yield.Molar volume of a gasAvogadros law states that:equal volumes of different gases at the same temperature and pressure will contain the same number of molesFrom this it follows that one mole of any gas will occupy the same volume at the same temperature and pressure.This is known as the molar volume of gas.At 273K (0oC) and 1.013 x 105 Pa (1atm) pressure this volume is 2.24 x 10-2m3 (22.4 dm3 or 22 400cm3)Calculations from equationsWrite down the correct formulae for all the reactants and productsBalance the equation to obtain the correct ratio of the reactants to productsIf the amounts of all reactants are known work out which are in excess and which one is the limiting reagent. By knowing the limiting reagent the maximum yield of any of the products can be determined.Work out the number of moles of the substance required.Convert the number of moles into the mass or volume.Express the answer to the correct number of significant figures and include the appropriate units.ExampleCalculate the volume of hydrogen gas evolved at 273K and 1 atm pressure when 0.623g of magnesium reacts with 27.3 cm3 of 1.25 moldm-3 hydrochloric acid.

Equation and amountsMg(s) + 2HCl(aq) H2(g) + MgCl2(aq)RAM for Mg = 24.31 Amount of Mg present = mass / RAM= 0.623 / 24.31= 2.56 x 10-2molAmount of HCl present= vol / 1000 x concn= 27.3 / 1000 x 1.25= 3.41 x 10-2 mol Excess and limiting Mg(s) + 2HCl(aq) H2(g) + MgCl2(aq)Ratio:1: 2 : 1: 1Amounts: 2.56 x 10-2mol 3.41 x 10-2 mol So 2 x 2.56 x 10-2 mol of HCl will react completely with the Mg present = 5.12 x 10-2 molSo what is in excess?MagnesiumSo the Hydrochloric acid is the limiting reagent

Theoretical yield of HydrogenThe maximum no. of moles of hydrogen that can be produced = 3.41 x 10-2 / 2= 1.705 x 10-2 molSo the volume of the hydrogen at 273K and 1atm= mol x 22.4= (1.705 x 10-2) x 22.4= 0.382 dm3To convert this to cm3 we need to multiply by 1000= 382cm3

GasesMass or volume may be used for gases.Normally it is easier to measure the volume of a gas. However, as well as the amount of gas present, the volume of a gas also depends on the pressure and the temperature.The physical behaviour of all gases is governed by the ideal gas equation:P V = n R Twhere: p represents the pressure. V represents the volume measured in m3N represents the amount of gas in moles = mass of gas/ MrT represents the absolute temperature measured in kelvin K.R represents the gas constant.GasesThe units of the gas constant R can be derived from the equation.pxV=N m-2xm3=JnxT=mol KSo the SI units of R must be J K-1 mol -1R has the value 8.314 J K -1 mol -1, and is one of the best known constants in science.Ideal GasesThe gas equation pV = nRT is only true for an ideal gas.Unlike ideal gases, real gases such as oxygen or hydrogen do not obey the law equation exactly.This is because there are still some weak attractive forces between the molecules in the gas, and the molecules themselves occupy some space even though most of the volume of a gas is empty space.However, for practical purposes we can use this ideal gas equation to describe the behaviour of real gases.Avogadro's constant and molar volume of a gas.Consider two gases, A and B:

andIf the gases occupy equal volumes, and the temperature and pressure are the same, then

and therefore

Calculations involving gasesFind the volume occupied by 2.20g of carbon dioxide, CO2, at 298K and a pressure of 100kPa. ( C = 12, O = 16, R = 8.31J K-1 mol-1It helps to rearrange the ideal gas equation before you start: pV = nRTpV = mass (g) / mass of 1 mole (g) x RTSo; V = mass (g) / mass of 1 mole (g) x RT /p

Calculations involving gases.Substitute in the numbers. Remember to check the units.Everything is ok except for the pressure.This has to be entered as 100 000Pa.The mass of one mole of CO2 is 44g.V = mass (g) / mass of 1 mole (g) x RT / pSo; V = 2.20 (g) / 44 (g) x (8.31 x 298) / 100 0001.24 x 10 -3 m3

Students need to practice questions on gas lawMole Calculationsnumber of moles of material in a given mass of that material.mass of material in a given number of moles of that material.concentrations of solutions.volume of a given number of moles of a gasnumber of moles of gas in a given volume of that gasvolume of a given mass of a gasmass of a given volume of gasmolar mass of a gas from the mass and the volume data for the gas.Mole calculations form the basis of many of the calculations that you will meet in IB, they include calculating: