2 Activity:Print out the keywords worksheet and add the following words, once you know the meanings fill them in.Relative atomic massMoleculeRelative molecular massMoleRelative formula massIsotopeAvogadro’s numberDiatomicElementCompound
3 Chemical Formulas and the Mole concept ElementsCompoundsMole concept and Avogadro's constantIsotopesFormulas of compoundsEmpirical formulaMolecular formulaStructural formula
4 Elements All 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 __________.Hand out IB periodic table/ data book that they must bring each lessoncannot100elementsatomsimpler
5 Molecules and compounds Some 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 it’s constituent elements: hydrogen and oxygen
6 Mole concept and Avogadro’s constant A single atom of an element has an extremely small mass. For example, an atom of carbon–12 has a mass of x 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 particles6.02 x atoms of carbon–12 = ?This number is known as Avogadro’s constant(NA or L)1 mole of Carbon = 12g
7 Mole concept and Avogadro’s constant Chemists 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 carbon–12 atoms so one mole of hydrogen atoms contains x hydrogen atoms and has a mass of 1.01 g.For diatomic molecules e.g. H2 there are 6.02 x molecules of hydrogen and therefore x 1023 atoms.Try some molar calculations involving Avogadro’s const.
8 Relative 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 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.
9 Activity Complete the worksheet IB 1.1 – Using Avogadro’s Number There are 20 questions to completeShow all working
10 Isotopes 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 carbon–12.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) =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 you’re not sure what type of compound it is.mol-1 means per
11 Calculating 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 = in total
12 ActivityComplete the worksheet IB Calculation of the Molar Mass of CompoundsThere are 60 questions to completeDon’t forget to refer to the previously stated rulesShow all working
13 Using 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.
14 Calculating molar quantities As 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 = = 40g1 mole of Na2SO4 = (2 x 23) (4 x 16) = 142g
15 Finding the number of moles Mass÷Number of molesRAM or RFMxIt 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!!!!!!
16 Example 1; ÷ x Find the number of moles in 46g of sodium Mass RAM or RFMxMass46Number of moles________=223RAM or RFM
17 Example 2; ÷ x Find the mass of 0.2 moles of sodium Mass Number of molesRAM or RFMxNumber of molesRAM or RFM4.6 g0.2x23Mass=
18 ActivityComplete the worksheet IB 1.3– Molar calculations and IB 1.4 Molar calculations (II)There are 60 questions to completeDon’t forget to refer to the previously stated rulesShow all working
19 Formula’s of compounds Compounds can be described by different chemical formulas:EmpiricalMolecularStructural
20 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.
21 Formula TriangleYou can use a formula triangle to help you rearrange the equation:Mass÷Number of molesRAM or RFMxMassNumber of moles________=Practice questions on molesRAM or RFM
22 ExampleA 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 =6.66253.27/16.00 =3.331Empirical formula is therefore – CH2O
23 Worked 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 isPractice questionsC9H8O4
24 Activity;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 composition
25 Molecular 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.
26 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 Formulae
27 Structural 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:CH2C2H4CH2CH2EmpiricalMolecularStructural
28 Structural 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 itClCHHCCl1,1-dichloroethane1,2-dichloroethane
29 Structural 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.HCOFor 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.methoxymethaneHCOethanol
30 Quick 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 follow…
31 Just a few more!What are the empirical formulae of the following molecules?Cyclohexane C6H12Dichloroethene C2H2Cl2Benzene C6H6Mr for ethane-1,2-diol is It is composed of carbon, hydrogen and oxygen in the ratio by moles of 1:3:1. What is its molecular formula?
32 Answers 1 C2H6O2 H2SO4 sulphuric acid BeO2H2 which is Be(OH)2, beryllium hydroxideMgCl2 magnesium chloride0.16 moles of eachMgOCH2CHClCHC2H6O2
33 Working out formulae for ionic compounds You can’t 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.
34 The need for equal numbers of “pluses” and “minuses” Ions 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.
35 Some charges you may remember! Can you put the correct charges into the table?ChargeGroup 1 metalsGroup 2 metalsGroup 3 metalsGroup 5 non-metalsGroup 6 non-metalsGroup 7 non-metals+3+1-1+2-2-3
36 Cases where the name tells you the charge on an ion A 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+NO3-Ag+OH-H+HCO3-NH4+CO32-Pb2+SO42-PO43-
37 Confusing endings Do not confuse ions like sulphate with sulphide. Try naming the following compounds:A name like sodium sulphide means that it contains sodium and sulphur only.Mg3N2Mg(NO3)2CaC2CaCO3Once you have an “ate” ending it means that there is something else there as well – often, but not always, oxygen
38 ActivityFill in the worksheet; Worksheet IB1.6 Important ions for IB calculations how many can you complete without looking them up?Click here for answers.
39 Working out the formula of an ionic compound Example 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 Na2O
40 Working out the formula of an ionic compound Example 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).
41 Working out the formula of an ionic compound Example 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)3
42 ActivityComplete the worksheet; Worksheet IB1.7 Writing formulae from namesFor answers click here
43 Chemical Reactions and equations Properties of chemical reactionsChemical equationsState symbolsOne way or reversible reactionsIonic equations
44 Properties of chemical reactions Once 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.
45 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+HCl→ZnCl2+H2REACTANTSPRODUCTS
46 Chemical EquationsThe equation is then balanced by adding the correct coefficientsWhen 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+2HCl→ZnCl2+H2REACTANTSPRODUCTS
47 State 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(aq)In aqueous solution(l)liquid(g)gasZn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)REACTANTSPRODUCTS
48 Activity Complete the worksheet; Worksheet IB1.8 Balancing equations For answers click here
49 One 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:Ni Catalyst, 180oCC2H4(g)+H2(g)C2H6(g)
50 Ionic 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)
51 ActivityCarry out Experiment 1.1 Precipitation reactions
52 Precipitation Reactions Metal 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)FeCl3(aq)Fe3+Fe(OH)3(s)Rusty brown solidCu(NO3)2(aq)Cu2+Cu(OH)2(s)Blue solidTo each of the following solutions add NaOH solution drop-wise to see a precipitate form
54 Measurements and calculations Error and uncertaintyMeasurements of molar quantitiesSolidsSolutionsLiquidsGasesWorked examples
55 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.
56 Systematic errors Quantitative 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.
57 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.
58 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,
59 Solids Solids are normally measured by weighing to obtain the mass Masses are measured in grams and kilograms remember 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 g on a balance weighing to + or – 0.01 g but as g on a balance weighing to + or – g.
60 Liquids Pure 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.
61 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.
62 Solutions Volume is usually used for solutions. 1.000 litre = dm3 = 1000cm3Concentration 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.
63 We can use this formula….. You can use a formula triangle to help you rearrange the equation:moles÷molesconcentration=Students complete exercisesVol in dm3xconcentrationVol in dm3
64 Do you still remember it???? You can use a formula triangle to help you rearrange the equation:Mass÷MassNumber of moles________=Practice questions on molesNumber of molesRAM or RFMRAM or RFMx
65 Concentrations of solutions Concentrations 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.
66 Calculations involving solutions A 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
67 Calculations involving solutions What is the concentration of a mol dm-3 solution of sodium carbonate, Na2CO3, in g dm-3?(C = 12; O = 16; Na = 23)1 mol Na2CO3 weighs 106gmol weighs x 106g = 5.30gSo mol dm-3 is therefore 5.30g dm-3
68 Calculations involving concentrations Recall ionic charges and names of ionic formulae.Know how to solve problems involving concentration and amount of solute.
69 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.
70 Volume of CuSO4 in Qn20Number of moles of CuSO40.100=xConcentration of CuSO4 given in qn.1000Converts cm3 to dm3= mol
71 Going back to the question!! If 1 mole of CuSO4 gives 1 mole of BaSO4therefore mol of CuSO4 will give mol of BaSO4RFM of 1 mol of BaSO4 = 233Therefore mol of BaSO4=x233= g of BaSO4
72 Next 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?
73 1 mol CaCO3 reacts with 2 mole HCl Here it is…..CaCO3 + 2HCl → CaCl2 + H2O + CO2Number of moles of HCl = 25.0 / 1000 x 2.00= mol1 mol CaCO3 reacts with 2 mole HCl= mol1 mol CaCO3 weighsTherefore mol weighs x 100 =So the max mass of CaCO3 which you could react with this amount of hydrochloric acid is 2.50g100g2.50g
74 Here it is….. CaCO3 + 2HCl → CaCl2 + H2O + CO2 Number of moles of HCl = 25.0 / 1000 x 2.00= mol1 mol CaCO3 reacts with 2 mole HCl0.5 mol CaCO3 reacts with 1 mole HCl= mol1 mol CaCO3 weighs100gTherefore mol weighs x 100 =2.50gSo the max mass of CaCO3 which you could react with this amount of hydrochloric acid is 2.50g
75 Last one….What is the minimum volume of 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.
76 Here it is… Mg + H2SO4 → MgSO4 + H2 1 mole of Mg reacts with 1 mol of H2SO4Number of moles of Mg0.240 / 24=mol=So number of moles of H2SO4 =mol
77 Next we can use this formula….. So the volume in dm3moles=Moles / Concentration/ 0.500÷=dm3=Students complete exercisesxconcentrationVol in dm3Which is what in cm3
78 Calculations involving concentrations Know how to solve problems involving concentration and amount of solute.Know how to solve simple titration equations
79 Simple volumetric calculations 25.0cm3 of 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?
80 NaOH + HCl → NaCl + H2O 1 : 1 : 1 : 1 ratio 25cm3 23.5cm3 volumes ?????concentrations0.100MThe 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=mol
81 NaOH + HCl → NaCl + H2O 1 : 1 : 1 : 1 ratio 25cm3 23.5cm3 volumes What else do we now know??????concentrations0.100MNo. of molesmolmolSo now we can calculate the concentration of HCl using the formula:concentration=Mole / volume in dm3=/ (23.5/1000)=0.106 mol dm3
82 Mass and Gaseous volume relationships Calculate 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.
83 Molar volume of a gas Avogadro’s law states that: ‘equal volumes of different gases at the same temperature and pressure will contain the same number of moles’From 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 x 105 Pa (1atm) pressure this volume is 2.24 x 10-2m3 (22.4 dm3 or cm3)
84 Calculations from equations Write 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.
85 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.
86 Equation and amounts Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq) RAM for Mg = 24.31Amount of Mg present = mass / RAM= / 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
87 Excess and limitingMg(s) + 2HCl(aq) → H2(g) + MgCl2(aq) Ratio: 1 : 2 : 1 : 1 Amounts: 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 mol So what is in excess? Magnesium So the Hydrochloric acid is the limiting reagent
88 Theoretical yield of Hydrogen The maximum no. of moles of hydrogen that can be produced = 3.41 x 10-2 / 2= 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= dm3To convert this to cm3 we need to multiply by 1000= 382cm3
89 Gases Mass 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:where: p represents the pressure.P V = n R TV 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.
90 GasesThe units of the gas constant R can be derived from the equation.pxV=N m-2xm3=Jmol KnxT=So the SI units of R must be J K-1 mol -1R has the value J K -1 mol -1, and is one of the best known constants in science.
91 Ideal Gases The 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.
92 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, thenand therefore
93 Calculations involving gases Find 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
94 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 Pa.The mass of one mole of CO2 is 44g.V = mass (g) / mass of 1 mole (g) x RT / pStudents need to practice questions on gas lawSo; V = 2.20 (g) / 44 (g) x (8.31 x 298) /1.24 x m3
95 Mole CalculationsMole calculations form the basis of many of the calculations that you will meet in IB, they include calculating:number 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.