Composition Knowing either the empirical or molecular formula of a compound, can help to determine its composition in more detail. The opposite is also true. Knowing the composition of a substance can help you to determine its formula. There are four different types of composition problems that you might come across: 1. Problems where you will be given the formula of the substance and asked to calculate the percentage by mass of each element in the substance. 2. Problems where you will be given the percentage composition and asked to calculate the formula. 3. Problems where you will be given the products of a chemical reaction and asked to calculate the formulaof one of the reactants. These are often referred to as combustion analysis problems. 4. Problems where you will be asked to find number of moles of waters of crystallisation. The following worked examples will show you how to do each of these. Percentage composition and empirical formula Example 1: Calculating the percentage by mass of elements in a compound Question Calculate the percentage that each element contributes to the overall mass of sulphuric acid (H2SO4). Answer Calculate the molar masses Hydrogen=2 1,01=2,02 g mol 1 Sulphur=32,1 g mol 1 Oxygen=4 16,0=64,0 g mol 1 Use the calculations in the previous step to calculate the molecular mass of sulphuric acid. Mass=2,02 g mol 1+32,1 g mol 1+64,0 g mol 1=98,12 g mol 1
Use the equation Percentage by mass=atomic massmolecular mass of H2SO4 100% Hydrogen (1) 2,02 g mol 198,12 g mol 1 100%=2,0587% Sulphur 32,1 g mol 198,12 g mol 1 100%=32,7150% (2) Oxygen 64,0 g mol 198,12 g mol 1 100%=65,2263% (3) (You should check at the end that these percentages add up to 100%!) In other words, in one molecule of sulphuric acid, hydrogen makes up 2,06% of the mass of the compound, sulphur makes up 32,71% and oxygen makes up 65,23%. Example 2: Determining the empirical formula of a compound Question A compound contains 52,2% carbon (C), 13,0% hydrogen (H) and 34,8% oxygen (O). Determine its empirical formula. Answer Give the masses Carbon = 52,2 g, hydrogen = 13,0 g and oxygen = 34,8 g Calculate the number of moles (4) Therefore:
n(carbon)=52,2 g12,0 g mol 1=4,35 mol (5) n(hydrogen)=13,0 g1,01 g mol 1=12,871 mol (6) n(oxygen)=34,8 g16,0 g mol 1=2,175 mol (7) Find the smallest number of moles Use the ratios of molar numbers calculated above to find the empirical formula. units in empirical formula=moles of this elementsmallest number of moles In this case, the smallest number of moles is 2,175. Therefore: Carbon 4,352,175=2 (8) Hydrogen 12,8712,175=6 (9) Oxygen 2,1752,175=1 (10) Therefore the empirical formula of this substance is: C2H6O. Example 3: Determining the formula of a compound Question 207 g of lead combines with oxygen to form 239 g of a lead oxide. Use this information to work out the formula of the lead oxide (Relative atomic masses: Pb=207,2 u and O=16,0 u).
Answer Find the mass of oxygen 239 g 207 g=32 g (11) Find the moles of oxygen (12) Lead n=207 g207,2 g mol 1=1 mol (13) Oxygen n=32 g16,0 g mol 1=2 mol (14) Find the mole ratio The mole ratio of Pb:O in the product is 1:2, which means that for every atom of lead, there will be two atoms of oxygen. The formula of the compound is PbO. Example 4: Empirical and molecular formula Question Vinegar, which is used in our homes, is a dilute form of acetic acid. A sample of acetic acid has the following percentage composition: 39,9% carbon, 6,7% hydrogen and 53,4% oxygen. 1. Determine the empirical formula of acetic acid. 2. Determine the molecular formula of acetic acid if the molar mass of acetic acid is 60,06 g mol 1. Answer Find the mass In 100 g of acetic acid, there is 39,9 g C, 6,7 g H and 53,4 g O
Find the moles ncnhno===39,9 g12,0 g mol 1=3,325 mol6,7 g1,01 g mol 1=6,6337 mol53,4 g16,0 g mol 1=3,3375 mol (15) Find the empirical formula C H O 3,325 6,6337 3,3375 1 2 1 Empirical formula is CH2O Table 1 Find the molecular formula The molar mass of acetic acid using the empirical formula is 30,02 g mol 1. However the question gives the molar mass as 60,06 g mol 1. Therefore the actual number of moles of each element must be double what it is in the empirical formula (60,0630,02=2). The molecular formula is therefore C2H4O2 or CH3COOH Example 5: Waters of crystallisation Question Aluminium trichloride (AlCl3) is an ionic substance that forms crystals in the solid phase. Water molecules may be trapped inside the crystal lattice. We represent this as: AlCl3.nH2O. Carine heated some aluminium trichloride crystals until all the water had evaporated and found that the mass after heating was 2,8 g. The mass before heating was 5 g. What is the number of moles of water molecules in the aluminium trichloride before heating? Answer Find the number of water molecules
We first need to find n, the number of water molecules that are present in the crystal. To do this we first note that the mass of water lost is 5 g 2,8 g=2,2 g. Find the mass ratio The mass ratio is: AlCl3 H2O 2,8 2,2 Find the mole ratio Table 2 To work out the mole ratio we divide the mass ratio by the molecular mass of each species: AlCl3 H2O 2,8 g133,35 g mol 1 2,2 g18,02 g mol 1 0,02099... 0,12... Table 3 Next we convert the ratio to whole numbers by dividing both sides by the smaller amount: AlCl3 H2O 0,020 997 375 0,122 086 57 0,0210,021 0,1220,021 1 6 Table 4
The mole ratio of aluminium trichloride to water is: 1:6 Write the final answer And now we know that there are 6 moles of water molecules in the crystal. The formula is AlCl3.6H2O. We can perform experiments to determine the composition of substances. For example, blue copper sulphate (CuSO4) crystals contain water. On heating the waters of crystallisation evaporate and the blue crystals become white. By weighing the starting and ending products, we can determine the amount of water that is in copper sulphate. Another example is reducing copper oxide to copper. Exercise 1: Moles and empirical formulae Problem 1: Calcium chloride is produced as the product of a chemical reaction. 1. What is the formula of calcium chloride? 2. What is the percentage mass of each of the elements in a molecule of calcium chloride? 3. If the sample contains 5 g of calcium chloride, what is the mass of calcium in the sample? 4. How many moles of calcium chloride are in the sample? Practise more questions like this Answer 1: a) CaCl2 b) The percentage by mass is the atomic mass of the element divided by the molecular mass of the compound: Ca: % mass=40,08110,98 100=36,15%
Cl: % mass=(2(35,45))110,98 100=63,89% If we add these up we get 100%: 36,15+63,89=100 c) Calcium makes up 36,15% of the calcium chloride, so the mass of calcium in the sample must be 36,15% of 5g: m=5 36,15100=1,808 g d) The number of moles is: n=1,80840,08 n=0,045 mol (Notice that this is the same as the number of moles in 5g of calcium chloride: n=5110,9=0,045 mol) Problem 2: 13 g of zinc combines with 6,4 g of sulphur. 1. What is the empirical formula of zinc sulphide? 2. What mass of zinc sulphide will be produced?
3. What is the percentage mass of each of the elements in zinc sulphide? 4. The molar mass of zinc sulphide is found to be 97,44 g mol 1. Determine the molecular formula of zinc sulphide. Practise more questions like this Answer 2: a) We work out the number of moles of each reactant: Zinc: n=1365,38 n=0,20 mol Sulphur: n=6,432,06 n=0,20 mol Zinc and sulphur combine in a 1:1 ratio, i.e. there 1 atom of zinc reacts with 1 atom of sulphur. The empirical formula of zinc sulphide is therefore: ZnS
b) The mass of zinc sulphide produced is: m=nm m=(0,20)(97,44) m=19,45 g c) Zinc: 65,3897,44 100=67% Sulphur: 32,0697,44 100=33% d) The molecular mass from the empirical formula is the same as the molecular mass of the molecular formula and so the molecular formula is: ZnS Problem 3: A calcium mineral consisted of 29,4% calcium, 23,5% sulphur and 47,1% oxygen by mass. Calculate the empirical formula of the mineral. Practise more questions like this Answer 3: In 100g of the mineral there will be 29,4 g calcium, 23,5 g sulphur and 47,1 g oxygen. We can calculate the number of moles for each of these:
Calcium: n=29,440,08 n=0,73 mol Sulphur: n=23,532,06 n=0,73 mol Oxygen: n=47,116,00 n=2,95 mol The smallest number is 0,73 so we divide all the values by 0,73. This gives 1 for calcium and sulphur and 2,950,73=4 for oxygen. So the molar ratio of the mineral is: 1:1:4 and the empirical formula is:
CaSO4 Problem 4: A chlorinated hydrocarbon compound was analysed and found to consist of 24,24% carbon, 4,04% hydrogen and 71,72% chlorine. From another experiment the molecular mass was found to be 99 g mol 1. Deduce the empirical and molecular formula. Practise more questions like this Answer 4: In 100g of the mineral there will be 24,24 g carbon, 4,04 g hydrogen and 71,72 g chlorine. We can calculate the number of moles for each of these: Carbon: n=24,2412,011 n=2 mol Hydrogen: n=4,041,01 n=4 mol Chlorine: n=71,7235,45
n=2 mol The smallest number is 2 so we divide all the values by 2. This gives 1 for carbon and chlorine and 2 for hydrogen. So the molar ratio of the mineral is: 1:2:1 and the empirical formula is: CH2Cl The molecular mass is: 12,011+1,01+35,45=48,471 g mol 1 The molecular mass of the compound is found to 99 g mol 1 The empirical formula must be multiplied by: 9948,471=2 The molecular formula is therefore: C2H4Cl2 Problem 5: Magnesium sulphate has the formula MgSO4.nH2O. A sample containing 5,0 g of magnesium sulphate was heated until all the water had evaporated. The final mass was found to be 2,6 g. How many water molecules were in the original sample? Practise more questions like this Answer 5: We first need to find n, the number of water molecules that are present in the crystal. To do this we first note that the mass of water lost is 5 2,6=2,4g. The mass ratio is: 2,6:2,4 To work out the mole ratio we divide the mass ratio by the molecular mass of each species:
2,6120,4:2,418,02=0.021594684:0.13318535 1:6 So the number of water molecules is 6 and the formula is: MgSO4 6H2O