Calculate the number of atoms of each kind in 5.3 grams of sodium carbonate.

Molecular mass of Na₂CO₃ = 2Na + C + 3O = (2 x 23) + 12 + (3 x 16) = 46 + 12 + 48 = 106 g

(i) 106 g of Na₂CO₃ has = 2 × 6.022 × 10²³ atoms of Na

∴ 5.3 g of Na₂CO₃ will have = $\dfrac{2 \times 6.022 \times 10^{23} \times 5.3 }{106}$ = 6.022 × 10²² atoms of Na

(ii) 106 g of Na₂CO₃ has = 6.022 × 10²³ atoms of carbon

∴ 5.3 g of Na₂CO₃ will have = $\dfrac{6.022 \times 10^{23} \times 5.3 }{106}$ = 3.01 × 10²² atoms of carbon

(iii) 106 g of Na₂CO₃ has 3 x 6.022 × 10²³ atoms of oxygen

∴ 5.3 g of Na₂CO₃ will have = $\dfrac{3 \times 6.022 \times 10^{23} \times 5.3 }{106}$ = 9.03 × 10²² atoms of oxygen