Moles from mass, then number of molecules

Quantitative chemistry

GCSE Chemistry (8462) · 4.3.2.1 · Higher

ORIGINAL

Mass ÷ relative formula mass gives moles; moles × the Avogadro constant gives particles. Round only at the end.

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Question

Calculate the amount, in moles, of carbon dioxide in 13.2 g of

\mathrm{CO_2}

, and the number of molecules this contains.

A_r

: C = 12, O = 16. The Avogadro constant is

6.02 \times 10^{23}

per mole. Give both answers to 3 significant figures.

Every step worked, with the reasoning.

1
$M_r(\mathrm{CO_2}) = 12 + (2 \times 16) = 44$
Moles need the mass of one mole, so find $M_r$ first.
2
$n = \dfrac{\text{mass}}{M_r} = \dfrac{13.2}{44}$
Amount in moles = mass in grams ÷ $M_r$ .
3
$n = 0.300\ \text{mol}$
The division gives exactly 0.3; written as 0.300 to give 3 significant figures, matching the 13.2 g.
4
$\text{molecules} = 0.300 \times 6.02 \times 10^{23} = 1.806 \times 10^{23}$
Each mole contains the Avogadro constant's number of particles. Common error: multiplying the mass (13.2) by the Avogadro constant instead of the amount in moles.
5
$1.806 \times 10^{23} \approx 1.81 \times 10^{23}$
Round once, at the end, to 3 significant figures.

Answer: $0.300\ \text{mol}$ and $1.81 \times 10^{23}$ molecules (3 s.f.)