Revision notes for AQA GCSE Chemistry Amounts of substances in equations (HT only). Open the guide for explanations and worked examples. Written against the AQA GCSE Chemistry (8462) specification, so the content matches what's examinable rather than general Chemistry background.
Amounts of substances in equations (HT only)
What you'll learn
- How a balanced symbol equation works like a recipe in moles.
- How to convert between mass and amount of substance using MrM_rMr.
- How to calculate masses of reactants or products from one known mass.
- The common Higher Tier trap: using mole ratios as if they were mass ratios.
Why balanced equations matter
In a chemical reaction, atoms are rearranged. No atoms are created or destroyed, so the equation must be balanced.
Definition
Balanced symbol equation
A balanced symbol equation uses chemical formulae and numbers in front of formulae so that there are the same numbers of each type of atom on both sides of the reaction arrow.
For example:
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
This says magnesium reacts with hydrochloric acid to form magnesium chloride and hydrogen.
The small state symbols mean:
- (s) = solid
- (l) = liquid
- (g) = gas
- (aq) = aqueous, dissolved in water
Definition
Coefficient
A coefficient is the big number placed in front of a formula in a balanced equation. In 2HCl, the coefficient is 2. If there is no number written, the coefficient is 1.
Key Idea
Equations are recipes in moles
The coefficients in a balanced equation give the mole ratio of the substances reacting and being made.
Example
Interpreting a balanced equation
For the equation:
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
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The coefficients are 1 for Mg, 2 for HCl, 1 for MgCl₂ and 1 for H₂.
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So the mole ratio is:
Mg : HCl : MgCl₂ : H₂ = 1 : 2 : 1 : 1
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If 0.30 mol of Mg reacts completely, the amount of HCl needed is twice as much: 0.30×2=0.60 mol0.30 \times 2 = 0.60 \text{ mol}0.30×2=0.60 mol. The amount of H₂ made is the same as Mg: 0.30 mol.
Common Mistake
Coefficient vs subscript
The 2 in 2HCl means two moles of HCl. The 2 in H₂ is part of the formula and means each hydrogen molecule contains two hydrogen atoms. Do not treat these as the same thing.
Amount of substance and moles
Definition
Amount of substance
Amount of substance, symbol nnn, tells you how many particles you have. It is measured in moles, symbol mol.
A mole is a counting unit, like a dozen — but much bigger. One mole contains about 6.02×10236.02 \times 10^{23}6.02×1023 particles.
In this topic, you usually do not need to count particles directly. Instead, you use moles as the link between masses and equations.
Relative formula mass, MrM_rMr
Definition
Relative formula mass
The relative formula mass, MrM_rMr, is found by adding the relative atomic masses, ArA_rAr, of all the atoms in a formula.
For example, magnesium oxide is MgO.
- ArA_rAr of Mg = 24
- ArA_rAr of O = 16
- So MrM_rMr of MgO = 40
At GCSE, the numerical value of MrM_rMr tells you the mass of one mole in grams. So one mole of MgO has a mass of 40 g.
The mass–moles equations
The key equation is:
n=mMrn = \frac{m}{M_r}n=Mrmwhere:
- nnn = amount of substance in mol
- mmm = mass in g
- MrM_rMr = relative formula mass
You can rearrange it to find mass:
m=nMrm = nM_rm=nMrExample
Converting mass to moles
Calculate the amount of substance in 10.0 g of magnesium oxide, MgO.
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Work out the relative formula mass: Mr=24+16=40M_r = 24 + 16 = 40Mr=24+16=40.
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Substitute into the equation: n=10.040n = \frac{10.0}{40}n=4010.0.
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Calculate the amount: n=0.250 moln = 0.250 \text{ mol}n=0.250 mol.
Tip
Keep the units consistent
Use grams for mass in these GCSE reacting-mass calculations. If a mass is given in kilograms, convert it to grams first.
The mole-ratio bridge
A balanced equation does not directly compare masses. It compares moles.
So the overall route is:
mass of known substance → moles of known substance → mole ratio → moles of wanted substance → mass of wanted substance
A useful ratio formula is:
nwanted=nknown×coefficient of wantedcoefficient of knownn_{\text{wanted}} = n_{\text{known}} \times \frac{\text{coefficient of wanted}}{\text{coefficient of known}}nwanted=nknown×coefficient of knowncoefficient of wantedThis is often the most important Higher Tier step.
Common Mistake
Using the mole ratio as a mass ratio
If the equation says 1 mol of Mg makes 1 mol of H₂, that does not mean 1 g of Mg makes 1 g of H₂. Different substances have different MrM_rMr values, so you must convert through moles.
Calculating the mass of a product
Use this method when you are given the mass of a reactant and asked for the mass of a product.
Example
Calculating mass of hydrogen produced
Calculate the mass of hydrogen produced when 6.0 g of magnesium reacts completely with excess hydrochloric acid.
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
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Convert the known mass of Mg into moles. Magnesium has Mr=24M_r = 24Mr=24, so n=6.024=0.25 moln = \frac{6.0}{24} = 0.25 \text{ mol}n=246.0=0.25 mol.
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Use the mole ratio from the equation. Mg : H₂ is 1 : 1, so 0.25 mol of Mg produces 0.25 mol of H₂.
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Convert moles of H₂ into mass. Hydrogen gas is H₂, so Mr=1+1=2M_r = 1 + 1 = 2Mr=1+1=2. Therefore m=0.25×2=0.50 gm = 0.25 \times 2 = 0.50 \text{ g}m=0.25×2=0.50 g.
Definition
Excess
A reactant is in excess if there is more of it than needed. It will not run out, so the calculation is based on the other reactant.
Calculating the mass of a reactant needed
You can also work backwards from the mass of a product to the mass of a reactant.
The method is the same: convert the known mass into moles, use the mole ratio, then convert back into mass.
Example
Calculating mass of oxygen needed
Calculate the mass of oxygen needed to make 20.0 g of magnesium oxide.
2Mg(s) + O₂(g) → 2MgO(s)
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Convert the known mass of MgO into moles. Magnesium oxide has Mr=24+16=40M_r = 24 + 16 = 40Mr=24+16=40, so n=20.040=0.500 moln = \frac{20.0}{40} = 0.500 \text{ mol}n=4020.0=0.500 mol.
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Use the mole ratio. O₂ : MgO is 1 : 2, so the amount of O₂ needed is 0.500×12=0.250 mol0.500 \times \frac{1}{2} = 0.250 \text{ mol}0.500×21=0.250 mol.
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Convert moles of O₂ into mass. Oxygen gas is O₂, so Mr=16+16=32M_r = 16 + 16 = 32Mr=16+16=32. Therefore m=0.250×32=8.00 gm = 0.250 \times 32 = 8.00 \text{ g}m=0.250×32=8.00 g.
A reliable step-by-step method
When you see a reacting-mass question, use this structure:
- Write or check the balanced symbol equation.
- Work out the MrM_rMr values for the known and wanted substances.
- Convert the known mass into moles using n=mMrn = \frac{m}{M_r}n=Mrm.
- Use the coefficients in the balanced equation to find the moles of the wanted substance.
- Convert the wanted moles into mass using m=nMrm = nM_rm=nMr.
Common Mistake
Pure substances
This method assumes the substances are pure and the reaction goes as described by the equation. If a question mentions percentage purity, yield or an impurity, there is an extra step.
Exam technique
In the exam
- Always use the balanced equation; the big numbers in front give the mole ratio.
- Convert grams to moles before using the ratio — never compare masses directly.
- Put the ratio in the direction you need: wanted coefficient divided by known coefficient.
- Carry units through your working, especially g and mol.
- If a reactant is described as excess, base your calculation on the other reactant.
Self review
Check yourself
- What does the 2 in 2HCl tell you, and why is it different from the 2 in H₂?
- In 2Mg(s) + O₂(g) → 2MgO(s), how many moles of O₂ are needed to make 0.60 mol of MgO?
- What two conversions happen on either side of the mole ratio in a reacting-mass calculation?
Use of amount of substance in relation to masses of pure substances
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