Calculate the number of moles in 250\ cm^3 of a solution of concentration 0.2\ mol per dm^3. [2 marks]

A reaction should give 50\ g of product but only 45\ g is collected. Calculate the percentage yield. [1 mark]

SEAB Original-style practice: Calcium carbonate decomposes when heated: CaCO_3 → CaO + CO_2. Calculate the mass of calcium oxide (CaO) formed when 50\ g of calcium carbonate decomposes completely. (A_r: Ca = 40, C = 12, O = 16.)

Step 1: relative formula masses. M_r(CaCO_3) = 40 + 12 + (3 × 16) = 100. M_r(CaO) = 40 + 16 = 56. Step 2: moles of calcium carbonate. n = 50100 = 0.5\ mol. Step 3: mole ratio. The equation is 1 : 1, so moles of CaO = 0.5\ mol. Step 4: mass of CaO. mass = n × M_r = 0.5 × 56 = 28\ g. What markers reward: correct M_r values, moles of reactant, the 1 : 1 ratio, and the final mass of 28\ g.

SingaporeChemistrySyllabus dot point

How do we use a balanced equation and the mole to work out the mass of product from the mass of reactant?

Use mole ratios from balanced equations to calculate reacting masses, work with concentration of solutions, and find a simple percentage yield

A focused answer to the N(A) Chemistry outcome on reacting-mass calculations. Using the mole ratio from a balanced equation to find masses, the meaning of concentration, and a simple percentage yield, kept to gentle numbers.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

The syllabus wants you to use the mole ratio from a balanced equation to calculate reacting masses, to understand the concentration of a solution, and to find a simple percentage yield. This is where the formula writing, balancing, and the mole all come together. At N(A) level the numbers stay friendly, so the marks come from following the right sequence of steps rather than from hard arithmetic.

The answer

The standard reacting-mass method

Almost every reacting-mass question follows the same four steps:

Work out the relative formula masses ( $M_r$ ) you need.
Change the given mass into moles using moles = mass divided by $M_r$ .
Use the mole ratio from the balanced equation to find the moles of the substance you want.
Change those moles back into a mass using mass = moles multiplied by $M_r$ .

The mole ratio in step 3 comes straight from the big numbers in the balanced equation.

Concentration of a solution

The concentration tells you how much solute is dissolved in a given volume. It is measured in moles per cubic decimetre. The relationship is:

\text{moles} = \text{concentration} \times \text{volume}

where the volume is in cubic decimetres ( $1\ \text{dm}^3 = 1000\ \text{cm}^3$ ). So always divide a volume in cubic centimetres by $1000$ first.

Percentage yield

In real experiments you rarely collect all of the product, so the percentage yield compares what you got to what you should have got:

\text{percentage yield} = \frac{\text{actual mass of product}}{\text{theoretical mass of product}} \times 100\%

The theoretical mass is the mass calculated from the equation; the actual mass is what was measured. Yield is below $100\%$ when some product is lost, some reactant does not react, or a side reaction occurs.

Mass of product from mass of reactant

Hydrogen reacts with oxygen to form water: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ . Calculate the mass of water formed from $8\ \text{g}$ of hydrogen. (Use H = 1, O = 16.)

Step 1: Find the relative formula masses

$M_r(\text{H}_2) = 2 \times 1 = 2$ . $M_r(\text{H}_2\text{O}) = (2 \times 1) + 16 = 18$ .

Step 2: Convert the given mass to moles

Moles of hydrogen:

n(\text{H}_2) = \frac{8}{2} = 4\ \text{mol}

Step 3: Use the mole ratio

The equation shows $2\text{H}_2$ gives $2\text{H}_2\text{O}$ , a $1 : 1$ ratio. So moles of water $= 4\ \text{mol}$ .

Step 4: Convert moles of water back to mass

\text{mass} = n \times M_r = 4 \times 18 = 72\ \text{g}

So $8\ \text{g}$ of hydrogen produces $72\ \text{g}$ of water.

Examples in context

Example 1. Scaling up a reaction in industry. A factory making lime from limestone uses the reacting-mass method to work out how much limestone to heat to get a target mass of lime. The mole ratio from the balanced equation lets them plan exactly how much raw material to buy, showing why these calculations matter beyond the classroom.

Example 2. Why a yield is rarely perfect. When a student prepares a salt and dries the crystals, some are always left stuck to the filter paper or the beaker. The percentage yield captures this loss, and comparing yields between attempts shows how careful technique improves how much product is recovered.

Try this

Q1. Calculate the number of moles in $250\ \text{cm}^3$ of a solution of concentration $0.2\ \text{mol per dm}^3$ . [2 marks]

Cue. Convert volume: $250 \div 1000 = 0.25\ \text{dm}^3$ ; moles $= 0.2 \times 0.25 = 0.05\ \text{mol}$ .

Q2. Magnesium reacts with oxygen: $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ . Calculate the mass of magnesium oxide formed from $0.2\ \text{mol}$ of magnesium ( $M_r$ of MgO = 40). [2 marks]

Cue. The ratio is $2 : 2$ , so $1 : 1$ ; moles of MgO $= 0.2$ ; mass $= 0.2 \times 40 = 8\ \text{g}$ .

Q3. A reaction should give $50\ \text{g}$ of product but only $45\ \text{g}$ is collected. Calculate the percentage yield. [1 mark]

Cue. Percentage yield $= (45 \div 50) \times 100 = 90\%$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksCalcium carbonate decomposes when heated:

\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2

. Calculate the mass of calcium oxide (

\text{CaO}

) formed when

50\ \text{g}

of calcium carbonate decomposes completely. (

A_r

: Ca = 40, C = 12, O = 16.)

Show worked answer →

Step 1: relative formula masses. $M_r(\text{CaCO}_3) = 40 + 12 + (3 \times 16) = 100$ . $M_r(\text{CaO}) = 40 + 16 = 56$ .

Step 2: moles of calcium carbonate.

$n = \dfrac{50}{100} = 0.5\ \text{mol}$ .

Step 3: mole ratio. The equation is $1 : 1$ , so moles of $\text{CaO} = 0.5\ \text{mol}$ .

Step 4: mass of $\text{CaO}$ .

$\text{mass} = n \times M_r = 0.5 \times 56 = 28\ \text{g}$ .

What markers reward: correct $M_r$ values, moles of reactant, the $1 : 1$ ratio, and the final mass of $28\ \text{g}$ .

Original4 marksIn an experiment,

0.5\ \text{mol}

of magnesium should produce

20\ \text{g}

of magnesium oxide. Only

17\ \text{g}

is actually obtained. (a) Calculate the percentage yield. (b) Suggest one reason the yield is less than

100\%

Show worked answer →

(a) Percentage yield = (actual mass divided by theoretical mass) $\times 100$ :

$\dfrac{17}{20} \times 100 = 85\%$ .

(b) Some product may have been lost during the experiment, for example some magnesium oxide escaped as smoke or some magnesium did not react.

What markers reward: using actual over theoretical times $100$ , the answer $85\%$ , and a sensible reason for loss.