Reacting Masses

Section 1 of 7The key idea

Equations give ratios

• Equations give ratios The numbers in front of formulae in a balanced equation tell you the ratio of moles of each substance.
• Multiplying each coefficient by the relative formula mass converts that ratio into a ratio of masses.

Scaling

• Scaling Reacting masses are always proportional, so doubling the mass of a reactant doubles the mass of product, whatever unit of mass you use.

Section 2 of 7Mass-to-mass method

1

Write the balanced symbol equation for the reaction.

2

Work out the M_r of the two substances involved: the one you are given and the one you want.

3

Multiply each M_r by its coefficient to get the reacting masses in the equation.

4

Divide to find the mass of wanted substance made from 1 g (or 1 tonne) of the given substance.

5

Multiply by the actual mass in the question and give the answer to a sensible number of significant figures with units.

Section 3 of 7Worked example

Stage	Magnesium burning, 6.0 g of Mg
Equation	2Mg + O2 → 2MgO
Mr values	Mg = 24; MgO = 40
Reacting masses	2 × 24 = 48 g Mg gives 2 × 40 = 80 g MgO
Mass from 1 g	80 ÷ 48 = 1.67 g MgO per 1 g Mg
Answer	6.0 × 1.67 = 10.0 g of magnesium oxide

Section 4 of 7Using moles instead (Higher level approach)

1Moles of known

• Moles of known moles = mass ÷ M_r.

2Use the ratio

• Use the ratio Multiply or divide by the ratio from the equation to get the moles of the wanted substance.

3Convert to mass

• Convert to mass mass = moles × M_r.

4Check

• Check The mass of product should be sensible when compared with the mass of reactant used.

Section 5 of 7Percentage composition and yield

Related calculations

Percentage by mass

• Percentage by mass percentage of an element = (A_r × number of atoms) ÷ M_r × 100.

Formula from masses (HT only)

• Formula from masses (HT only) Divide each mass by its A_r and simplify to the lowest whole-number ratio to find the simplest formula.

Percentage yield

• Percentage yield percentage yield = actual mass obtained ÷ maximum theoretical mass × 100.
  • It is always less than 100% in practice because of losses.

Section 6 of 7Key terms

Term	Meaning
Reacting mass	The mass of a substance in a balanced equation, found from its coefficient multiplied by its Mr.
Coefficient	The number in front of a formula in an equation.
Theoretical yield	The maximum mass of product that could form from the reactants.
Percentage yield	The actual yield expressed as a percentage of the theoretical yield.

Section 7 of 7Exam technique

1

Forgetting to multiply M_r by the coefficient from the equation, which gives the wrong ratio.

2

Using an unbalanced equation to work out the ratio.

3

Mixing units in the same calculation, such as grams for one substance and tonnes for another.

4

Dividing the wrong way round, giving a product mass larger than is possible from the reactant used.