Masses in Reactions — Chemistry (Grade 11)

Define and explain Masses in Reactions in your own words
Use key terms such as relative formula mass (Mr) accurately
Apply what you have learned to new examples and questions
Avoid the common mistakes learners make with this topic

This lesson focuses on Masses in Reactions: calculate how much product a reaction makes.

Definition: Masses in Reactions

Calculate how much product a reaction makes.

Key ideas

Real reactions are never perfect

Percentage yield compares what you actually make with the theoretical maximum: (actual ÷ theoretical) × 100. Yields fall short because of reversible reactions, side reactions and losses in filtering or transfer. Atom economy goes further, asking what fraction of the reactant atoms end up in the desired product — important for sustainable industry.

Use the equation's molar ratio

A balanced equation gives the ratio of moles that react: in 2Mg + O₂ → 2MgO, 2 moles of Mg make 2 moles of MgO, a 1:1 ratio. The method is always the same: convert the known mass to moles, use the ratio to find moles of the unknown, then convert back to mass. If a reactant runs out first, it is the limiting reactant and it alone sets the maximum product.

Key term — relative formula mass (Mr): The sum of the relative atomic masses in a formula, e.g. Mr of H₂O = 18.

Worked example: Masses in Reactions

A reaction should theoretically make 8.40 g of product but only 6.72 g is obtained. Calculate the percentage yield.

(6.72 ÷ 8.40) × 100 = 80.0%.

Answer: (6.72 ÷ 8.40) × 100 = 80.0%.

Common mistakes

Using the mass ratio instead of the mole ratio Grams do not react in the equation's ratio — moles do. Always convert masses to moles first, apply the ratio from the balanced equation, then convert back to grams.
Giving percentage yield above 100% without questioning it A yield over 100% means something is wrong: the product is probably wet or impure, or a mass was mismeasured. Check your results rather than reporting an impossible yield.

Practice

Explain why a chemist might prefer a reaction with 100% atom economy even if its percentage yield is lower.

What happens to the wasted atoms?

High atom economy means most reactant atoms end up in the desired product rather than waste, saving raw materials and reducing waste disposal — even if some product is lost in practice.

Calculate the number of moles in 10.0 g of sodium hydroxide, NaOH. (Ar: Na = 23.0, H = 1.0, O = 16.0)

First find Mr of NaOH.

Mr = 23.0 + 16.0 + 1.0 = 40.0. Moles = 10.0 ÷ 40.0 = 0.250 mol.

In the worked example, what mass of oxygen reacts with the 12.0 g of magnesium?

Use the 2:1 ratio of Mg to O₂.

Moles of Mg = 0.494; moles of O₂ = 0.494 ÷ 2 = 0.247 mol; mass = 0.247 × 32.0 = 7.90 g. (Check: 12.0 + 7.90 = 19.9 g of product.).

Calculate the concentration of a solution containing 0.20 mol of solute in 0.50 dm³ of solution.

Concentration = moles ÷ volume.

0.20 ÷ 0.50 = 0.40 mol/dm³.

Quick check

Masses in Reactions — quick check

Which of these best defines "relative formula mass (Mr)"?

The sum of the relative atomic masses in a formula, e.g. Mr of H₂O = 18.

5.84 g of NaCl is dissolved to make 1.00 dm³ of solution. What is its concentration? (Mr of NaCl = 58.4)

Moles = 5.84 ÷ 58.4 = 0.100 mol; concentration = 0.100 ÷ 1.00 = 0.100 mol/dm³.

Key takeaways

Masses in Reactions: calculate how much product a reaction makes.
Real reactions are never perfect: Percentage yield compares what you actually make with the theoretical maximum: (actual ÷ theoretical) × 100.
limiting reactant: The reactant that runs out first and therefore decides how much product forms.
Watch out for: using the mass ratio instead of the mole ratio

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