Percentage yield from a reacting mass

Quantitative chemistry

GCSE Chemistry (8462) · 4.3.3.1 · Higher

ORIGINAL

Find the theoretical mass from the equation, then actual ÷ theoretical × 100 (chemistry only).

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Question

Calcium carbonate decomposes when heated:

\mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)}

. A student heats 25.0 g of calcium carbonate and obtains 11.2 g of calcium oxide. Calculate the percentage yield.

A_r

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

Every step worked, with the reasoning.

1
$M_r(\mathrm{CaCO_3}) = 40 + 12 + 3(16) = 100$ and $M_r(\mathrm{CaO}) = 40 + 16 = 56$
Both formula masses are needed for the reacting-mass step.
2
$n(\mathrm{CaCO_3}) = \dfrac{25.0}{100} = 0.250\ \text{mol}$
Amount in moles = mass ÷ $M_r$ .
3
$n(\mathrm{CaO}) = 0.250\ \text{mol}$ , so $\text{theoretical mass} = 0.250 \times 56 = 14.0\ \text{g}$
The equation's ratio is 1 : 1. This is the maximum mass of product if every bit reacted and nothing was lost.
4
$\text{percentage yield} = \dfrac{11.2}{14.0} \times 100 = 80.0\%$
Actual mass of product ÷ maximum theoretical mass × 100. Common error: dividing by the 25.0 g of calcium carbonate — a yield compares the product with the most of that same product you could make.

Answer: $80.0\%$ (3 s.f.)