Moles, Molar Mass and Empirical Formula Calculations
Empirical and molecular formula calculations build on the mole concept, covering how to calculate a compound's empirical formula from experimental mass or percentage composition data, find its molecular formula once its relative molecular mass (Mr) is known, use the molar gas volume (24 dm3 per mole at room temperature and pressure) to convert between the volume and moles of a gas, and calculate concentrations in mol/dm3 to solve titration calculations.
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Method
- To find an empirical formula from the masses of each element, divide each element's mass by its Ar to find moles of atoms, then divide every value by the smallest of these moles to get a simplest ratio, multiplying every value by a small whole number (2 or 3) if needed to clear any fractions and reach whole numbers.
- If percentage composition is given instead of masses, treat each percentage as if it were a mass in grams (since the total is 100), then follow the same method as for masses.
- To find a molecular formula from an empirical formula, divide the given relative molecular mass (Mr) by the empirical formula mass, then multiply every subscript in the empirical formula by that whole number.
- Learn the molar gas volume rule: at room temperature and pressure (rtp), one mole of any gas occupies 24 dm3 (24,000 cm3); use moles = volume (dm3) / 24 to convert a gas volume to moles, or volume = moles x 24 to convert moles to a volume.
- Learn the concentration formula: concentration (mol/dm3) = moles / volume (dm3); remember to convert a volume given in cm3 into dm3 by dividing by 1000 before using this formula.
- For a titration calculation, use the volume and concentration of the solution you know to calculate its moles, use the mole ratio from the balanced equation to find the moles of the unknown solution, then divide by its volume (in dm3) to find its concentration, or divide moles by concentration to find its volume.
Worked example
A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its relative molecular mass is 42. Determine its molecular formula. (Ar: C = 12, H = 1)
- Treat the percentages as masses: 85.7 g carbon, 14.3 g hydrogen.
- Divide by Ar to find moles of atoms: carbon = 85.7/12 = 7.14; hydrogen = 14.3/1 = 14.3.
- Divide both values by the smaller value (7.14): carbon = 1; hydrogen = 14.3/7.14 = 2.00.
- Write the empirical formula: CH2, with empirical formula mass = 12 + (1 x 2) = 14.
- Divide the given Mr by the empirical formula mass: 42/14 = 3.
- Multiply every subscript in CH2 by 3: molecular formula = C3H6. Final answer: C3H6.
Practice questions
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Q1Calculate the moles of atoms in 4.8 g of magnesium. (Ar: Mg = 24)Show answer
Answer: 0.2 mol (4.8/24)
Q2Calculate the volume, in dm3, occupied by 0.5 mol of oxygen gas at rtp.Show answer
Answer: 12 dm3 (0.5 x 24)
Q3Calculate the number of moles of hydrogen gas in 6 dm3 of hydrogen at rtp.Show answer
Answer: 0.25 mol (6/24)
Q4Calculate the concentration, in mol/dm3, of a solution containing 0.2 mol of solute in 500 cm3 of solution.Show answer
Answer: 0.4 mol/dm3 (500 cm3 = 0.5 dm3; 0.2/0.5 = 0.4)
Q5A compound contains 2.4 g of carbon and 0.4 g of hydrogen only. Determine its empirical formula. (Ar: C = 12, H = 1)Show answer
Answer: CH2 (moles C = 2.4/12 = 0.2; moles H = 0.4/1 = 0.4; dividing by 0.2 gives C = 1, H = 2)
Q6Calculate the number of moles of hydrochloric acid in 25 cm3 of a 2.0 mol/dm3 solution.Show answer
Answer: 0.05 mol (25 cm3 = 0.025 dm3; 0.025 x 2.0 = 0.05)
Q7An oxide of nitrogen has the empirical formula NO2 and a relative molecular mass of 92. Determine its molecular formula. (Ar: N = 14, O = 16)Show answer
Answer: N2O4 (empirical formula mass = 14 + 32 = 46; 92/46 = 2; multiply every subscript in NO2 by 2)
Exam-style questions
Written in the style of a IGCSE Science exam paper, with a full mark scheme.
A sample of gas has a volume of 3.6 dm3 at rtp. Calculate the number of moles of gas present, and hence the mass of the gas if its molar mass is 32 g/mol.
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25.0 cm3 of a sodium hydroxide solution of unknown concentration is exactly neutralised by 15.0 cm3 of 0.100 mol/dm3 hydrochloric acid, according to the equation NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l). Calculate the concentration, in mol/dm3, of the sodium hydroxide solution.
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A compound used as a fertiliser contains only nitrogen, hydrogen and oxygen. When 8.0 g of the compound is analysed, it is found to contain 2.8 g of nitrogen and 0.4 g of hydrogen, with the remainder being oxygen. The relative molecular mass of the compound is 80. (a) Calculate the mass of oxygen in the sample. (b) Determine the empirical formula of the compound. (c) Determine its molecular formula. (Ar: N = 14, H = 1, O = 16)
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Free printable worksheet
Want more practice on paper? Download the moles, molar mass and empirical formula calculations worksheet pack - 6 pages of exam-style questions with a full mark scheme. One email opens every download in this browser for 14 days - no account, no card. Print it for personal and classroom use.
This topic is chapter 19 of IGCSE Science Workbook, the whole course as one free printable PDF.
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