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Moles, reacting masses and yield
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The mole lets chemists count particles by weighing. With a balanced equation, you can work out the mass of product to expect.
The mole
The mole (mol) is the unit for the amount of a substance.
One mole of a substance has a mass in grams equal to its A_r or M_r.
moles = mass ÷ M_r, so mass = moles × M_r
One mole of a substance has a mass in grams equal to its A_r or M_r.
moles = mass ÷ M_r, so mass = moles × M_r
Reacting masses
1. Work out the moles of the substance you know.
2. Use the ratio from the balanced equation.
3. Convert moles back to mass.
Example: CaCO3 → CaO + CO2. 50 g CaCO3 (M_r 100) = 0.5 mol. Ratio 1 : 1, so 0.5 mol CaO (M_r 56) = 28 g.
2. Use the ratio from the balanced equation.
3. Convert moles back to mass.
Example: CaCO3 → CaO + CO2. 50 g CaCO3 (M_r 100) = 0.5 mol. Ratio 1 : 1, so 0.5 mol CaO (M_r 56) = 28 g.
Percentage yield
The theoretical yield is the mass you calculate. The actual yield is the mass you get.
percentage yield = actual yield ÷ theoretical yield × 100
Yields are below 100% because of incomplete reactions, losses during separation, and side reactions.
percentage yield = actual yield ÷ theoretical yield × 100
Yields are below 100% because of incomplete reactions, losses during separation, and side reactions.
2Mg + O2 → 2MgO. What mass of magnesium oxide forms from 12 g of magnesium? (A_r: Mg = 24, O = 16)
- Moles of Mg = 12 ÷ 24 = 0.5 mol
- Ratio Mg : MgO = 2 : 2 = 1 : 1, so 0.5 mol MgO
- M_r of MgO = 40. Mass = 0.5 × 40
Answer: 20 g
moles = mass ÷ M_r. Use the equation ratio to go from the moles of one substance to another, then mass = moles × M_r. Percentage yield = actual ÷ theoretical × 100.
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