This chemistry grades 9-10 lesson on reacting masses notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on reacting masses and check yourself with the quiz.
Quick answer: Reacting masses come from a balanced equation. Convert the known mass to moles with n = m ÷ M, use the mole ratio to find the moles of the other substance, then convert back to grams. The limiting reactant sets the product, and percentage yield compares actual with theoretical mass.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: HS-PS1-7, GCSE C3
Reacting masses notes: the key ideas
How do you use a balanced equation to find reacting masses?
A balanced equation is a recipe written in moles. The big numbers in front of the formulae give the mole ratio in which substances react and form. Chemists never compare masses directly, because a mole of one substance has a different mass from a mole of another. So they convert the mass to moles, use the ratio, then convert back to a mass.
n = m ÷ M
n = amount of substance (mol), m = mass (g), M = molar mass (g/mol), numerically equal to the relative formula mass.
Aarav burns 6.0 g of magnesium ribbon: 2Mg + O₂ → 2MgO. Calculate the mass of magnesium oxide formed (Ar: Mg = 24, O = 16).
n(Mg) = 6.0 ÷ 24 = 0.25 mol. The ratio Mg : MgO is 2 : 2, so n(MgO) = 0.25 mol. M(MgO) = 24 + 16 = 40 g/mol, so m = 0.25 × 40 = 10 g.
What is a limiting reactant?
The limiting reactant is the reactant that is completely used up first. It decides how much product can form. A reactant that is left over at the end is in excess.
To find it, work out the moles of each reactant, then divide by its coefficient in the equation. The smallest answer belongs to the limiting reactant. A fire needs fuel and oxygen in the same way: whichever runs out first stops the burning.
Wanjiru adds 6.5 g of zinc to 50 cm³ of 1.0 mol/dm³ hydrochloric acid: Zn + 2HCl → ZnCl₂ + H₂. Find the mass of hydrogen (Ar: Zn = 65, H = 1).
n(Zn) = 6.5 ÷ 65 = 0.10 mol. n(HCl) = 1.0 × 0.050 = 0.050 mol. Zinc would need 2 × 0.10 = 0.20 mol of HCl, so the acid is limiting. n(H₂) = 0.050 ÷ 2 = 0.025 mol, and m = 0.025 × 2 = 0.050 g.
How do you calculate percentage yield?
The theoretical yield is the mass of product the equation predicts. The actual yield is what you really collect. Some product is lost on filter paper, some reactant forms other substances, and some reactions do not go to completion, so the actual yield is almost always smaller.
percentage yield = (actual yield ÷ theoretical yield) × 100
Both yields must use the same unit, usually grams.
Heating 25.0 g of calcium carbonate: CaCO₃ → CaO + CO₂. Chidi collects 11.2 g of calcium oxide (Ar: Ca = 40, C = 12, O = 16).
n(CaCO₃) = 25.0 ÷ 100 = 0.250 mol, so n(CaO) = 0.250 mol and the theoretical yield is 0.250 × 56 = 14.0 g. Percentage yield = 11.2 ÷ 14.0 × 100 = 80%.
Do not divide the actual yield by the mass of the reactant. Compare the actual mass of product only with the theoretical mass of the same product.
How do you calculate concentration in mol/dm³?
Concentration tells you how many moles are dissolved in each cubic decimetre (1 dm³ = 1000 cm³) of solution. A solution that is twice as concentrated holds twice as many particles in the same volume.
c = n ÷ V and n = c × V
c = concentration (mol/dm³), n = amount (mol), V = volume (dm³ = cm³ ÷ 1000). In g/dm³: concentration = mass ÷ volume.
| Quantity | Unit | How to get it |
|---|---|---|
| Amount | mol | m ÷ M |
| Volume | dm³ | cm³ ÷ 1000 |
| Concentration | mol/dm³ | n ÷ V |
| Mass concentration | g/dm³ | c × M |
Kofi dissolves 4.0 g of sodium hydroxide in water and makes the volume up to 250 cm³ (Ar: Na = 23, O = 16, H = 1).
n = 4.0 ÷ 40 = 0.10 mol. V = 0.250 dm³. c = 0.10 ÷ 0.250 = 0.40 mol/dm³, which is 0.40 × 40 = 16 g/dm³.
Write the amount in moles under every formula in the equation before you calculate anything. Then you can see the ratio, the limiting reactant and the target at a glance.
- n = m ÷ M, with M in g/mol
- The coefficients give the mole ratio, not the mass ratio
- Limiting reactant: smallest (moles ÷ coefficient)
- Percentage yield = actual ÷ theoretical × 100
- c = n ÷ V, with V in dm³ (cm³ ÷ 1000)
Using equations to find masses and yields: frequently asked questions
What is a mole ratio?
A mole ratio is the proportion of the coefficients in a balanced equation. In 2Mg + O₂ → 2MgO, magnesium, oxygen and magnesium oxide react and form in the ratio 2 : 1 : 2.
How do I find the limiting reactant?
Convert each reactant to moles and divide by its coefficient. The reactant with the smallest value runs out first and is limiting. All product calculations then use its moles.
Why can percentage yield never be above 100%?
The theoretical yield is the most product the reactants can form. If you measure more than 100%, the product is probably wet, impure or contaminated by another substance.
How do I convert cm³ to dm³?
Divide by 1000. For example, 25.0 cm³ is 0.0250 dm³. Concentration in mol/dm³ is only correct when the volume is in dm³.
How do I change g/dm³ into mol/dm³?
Divide the mass concentration by the molar mass. A solution of 16 g/dm³ sodium hydroxide, with M = 40 g/mol, is 0.40 mol/dm³.
