Moles, Formulae and Equations
The mole is the chemist's counting unit: one mole of any substance contains 6.02 x 10^23 particles, the Avogadro constant, and has a mass in grams equal to its relative formula mass (Mr). Moles = mass / Mr is the relationship that connects a balanced equation's mole ratios to masses measured in grams, making calculations from reacting masses possible.
Method
- Convert to moles first, always. Grams cannot be compared between different substances, but moles can, because the balancing numbers are mole ratios.
- Use moles = mass / Mr, and remember the rearrangements: mass = moles x Mr, and Mr = mass / moles.
- Read the mole ratio straight from the balanced equation, for example in N2 + 3H2 gives 2NH3 the ratio of nitrogen to ammonia is 1 to 2, so 0.5 mol of nitrogen produces 1.0 mol of ammonia.
- To find a formula or balance an equation from masses (Higher), convert each mass to moles, divide all the values by the smallest, and round to the nearest simple whole-number ratio.
- To find the limiting reactant (Higher), calculate the moles of each reactant, divide each by its balancing number, and the smallest value identifies the limiting reactant. Base the product calculation on that reactant only.
- Check the answer is sensible: a product mass should be of the same order as the reactant mass, and if it is out by a factor of exactly 2 or 3 the mole ratio has usually been missed.
Worked example
6.0 g of carbon reacts with 32.0 g of oxygen: C + O2 gives CO2. Determine the limiting reactant and calculate the mass of carbon dioxide formed. Relative atomic masses: C = 12, O = 16.
- Moles of carbon = 6.0 / 12 = 0.50 mol.
- Mr of oxygen, O2 = 2 x 16 = 32, so moles of oxygen = 32.0 / 32 = 1.0 mol.
- Divide each by its balancing number: carbon 0.50 / 1 = 0.50; oxygen 1.0 / 1 = 1.0. The smaller value is carbon, so carbon is the limiting reactant and oxygen is in excess.
- Use the limiting reactant for the product: the ratio of C to CO2 is 1 to 1, so 0.50 mol of carbon dioxide is formed.
- Mr of carbon dioxide = 12 + (2 x 16) = 44, so mass = 0.50 x 44 = 22 g.
- Final answer: carbon is limiting and 22 g of carbon dioxide is formed, with 0.50 mol (16 g) of oxygen left unreacted.
Practice questions
Try each question, then tap to reveal the answer.
Q1State the value of the Avogadro constant and what it counts.Show answer
Answer: 6.02 x 10^23, the number of particles (atoms, molecules or ions) in one mole of a substance.
Q2Calculate the number of moles in 88 g of carbon dioxide. (Mr = 44)Show answer
Answer: 88 / 44 = 2.0 mol.
Q3Calculate the mass of 0.25 mol of sodium hydroxide, NaOH. (Na = 23, O = 16, H = 1)Show answer
Answer: Mr = 40, so mass = 0.25 x 40 = 10 g.
Q4In the equation 2H2 + O2 gives 2H2O, how many moles of water are formed from 3 mol of hydrogen?Show answer
Answer: The ratio of H2 to H2O is 2 to 2, that is 1 to 1, so 3 mol of water.
Q5Define the limiting reactant.Show answer
Answer: The reactant that is completely used up in the reaction, which therefore determines the maximum amount of product that can form.
Q6Why is the amount of product directly proportional to the amount of limiting reactant?Show answer
Answer: Once the limiting reactant is used up the reaction stops, so every additional mole of it would produce a fixed additional amount of product, while adding more of the reactant in excess changes nothing.
Q7How many atoms are there in 0.5 mol of helium?Show answer
Answer: 0.5 x 6.02 x 10^23 = 3.01 x 10^23 atoms.
Exam-style questions
Written in the style of a GCSE Science exam paper, with a full mark scheme.
Calculate the mass of ammonia produced when 28 g of nitrogen reacts completely with excess hydrogen. N2 + 3H2 gives 2NH3. Relative atomic masses: N = 14, H = 1.
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A student reacts 4.0 g of magnesium with 100 cubic centimetres of hydrochloric acid containing 0.10 mol of HCl. Mg + 2HCl gives MgCl2 + H2. Determine which reactant is limiting, and explain what this means for the volume of hydrogen produced. (Ar of Mg = 24)
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This topic is chapter 20 of GCSE Chemistry Workbook, the whole course as one free printable PDF.
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