The Mole Concept
Senior 2 · Chemistry
The Mole Concept
Lesson 3 of 5
25 min
A farmer can buy nitrogen as urea, as ammonium nitrate or as ammonium sulfate. Which gives the most nitrogen for each kilogram carried home on a bicycle? The answer is a calculation of percentage composition.
The reverse calculation is just as useful. When chemists analyse an unknown substance, they measure the mass of each element it contains and use those masses to work out its formula. By the end of this lesson you will be able to calculate percentage composition and find empirical and molecular formulae.
The percentage by mass of an element in a compound is:
% of element = (number of atoms of the element × Ar) ÷ Mr of the compound × 100
| Fertiliser | Mr | Mass of N in the formula | % nitrogen |
|---|---|---|---|
| Urea, CO(NH2)2 | 60 | 2 × 14 = 28 | 28 ÷ 60 × 100 = 46.7% |
| Ammonium nitrate, NH4NO3 | 80 | 2 × 14 = 28 | 28 ÷ 80 × 100 = 35.0% |
| Ammonium sulfate, (NH4)2SO4 | 132 | 2 × 14 = 28 | 28 ÷ 132 × 100 = 21.2% |
Urea supplies the most nitrogen per kilogram. A 50 kg bag of urea contains 46.7 ÷ 100 × 50 = 23.3 kg of nitrogen.
What percentage of the mass of blue copper(II) sulfate crystals, CuSO4·5H2O, is water?
Solution. Mr = 249.5 and the water is 5 × 18 = 90. % water = 90 ÷ 249.5 × 100 = 36.1%.
The empirical formula is the simplest whole-number ratio of the atoms of each element in a compound. To find it:
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find its empirical formula.
| C | H | O | |
|---|---|---|---|
| Mass in 100 g | 40.0 | 6.7 | 53.3 |
| ÷ Ar | 40.0 ÷ 12 = 3.33 | 6.7 ÷ 1 = 6.7 | 53.3 ÷ 16 = 3.33 |
| ÷ smallest (3.33) | 1 | 2 | 1 |
The empirical formula is CH2O.
2.4 g of magnesium burns to give 4.0 g of magnesium oxide. Find the formula of the oxide.
Solution. Mass of oxygen that combined = 4.0 − 2.4 = 1.6 g. Moles: Mg = 2.4 ÷ 24 = 0.10; O = 1.6 ÷ 16 = 0.10. Ratio 1 : 1, so the formula is MgO.
An oxide of iron contains 70.0% iron. Find its empirical formula.
Solution. Oxygen = 100 − 70.0 = 30.0%. Moles: Fe = 70.0 ÷ 56 = 1.25; O = 30.0 ÷ 16 = 1.875. Dividing by 1.25 gives Fe : O = 1 : 1.5. That is not a rounding error; multiply both by 2 to get 2 : 3. The formula is Fe2O3.
The molecular formula gives the actual number of atoms of each element in one molecule. It is always a whole-number multiple of the empirical formula. To find it you also need the relative molecular mass:
multiplier = Mr ÷ empirical formula mass
A sugar has the empirical formula CH2O and Mr = 180. Find its molecular formula.
Solution. Empirical formula mass = 12 + 2 + 16 = 30. Multiplier = 180 ÷ 30 = 6. Molecular formula = C6H12O6 (glucose).
A hydrocarbon contains 85.7% carbon and 14.3% hydrogen, and its Mr is 56. Find its empirical and molecular formulae.
Solution. C: 85.7 ÷ 12 = 7.14; H: 14.3 ÷ 1 = 14.3. Ratio 1 : 2.00, so the empirical formula is CH2 (mass 14). Multiplier = 56 ÷ 14 = 4, so the molecular formula is C4H8.
Tip
Know your "hidden" fractions. After dividing by the smallest, a result ending in .5 means multiply everything by 2; .33 or .67 means multiply by 3; .25 or .75 means multiply by 4. Only round a result that is within about 0.1 of a whole number.
Use Ar: H 1, C 12, N 14, O 16, Mg 24, P 31, S 32, Ca 40. Try each item before you read the solution.
Calculate the percentage of calcium in calcium carbonate, CaCO3.
Solution: 40 ÷ 100 × 100 = 40.0%.
Calculate the percentage of nitrogen in DAP, (NH4)2HPO4 (Mr = 132).
Solution: 28 ÷ 132 × 100 = 21.2%.
A compound contains 52.2% carbon, 13.0% hydrogen and 34.8% oxygen. Find its empirical formula.
Solution: C: 52.2 ÷ 12 = 4.35; H: 13.0 ÷ 1 = 13.0; O: 34.8 ÷ 16 = 2.175. Divide by 2.175: C 2.00, H 5.98, O 1. The empirical formula is C2H6O.
A compound has the empirical formula CH and Mr = 78. What is its molecular formula?
Solution: Empirical formula mass = 13; 78 ÷ 13 = 6. Molecular formula: C6H6.
5.00 g of hydrated magnesium sulfate, MgSO4·xH2O, is heated until all the water is lost, leaving 2.44 g of anhydrous MgSO4 (M = 120 g/mol). Find x.
Solution: Mass of water = 5.00 − 2.44 = 2.56 g. Moles: MgSO4 = 2.44 ÷ 120 = 0.0203; H2O = 2.56 ÷ 18 = 0.142. Ratio 0.142 ÷ 0.0203 = 7.0, so x = 7: MgSO4·7H2O.
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