5.72 grams of hydrated sodium carbonate is heated strongly. The white residue left over is 2.12 g. Calculate:

(i) the formula of hydrated sodium carbonate

(ii) the percentage of carbon in the compound.

[Na = 23, C = 12, O = 16, H = 1]

(i) Hydrated sodium carbonate = Na₂CO₃.xH₂O

Mass of hydrated sodium carbonate = 5.72g

Mass of white residue = 2.12g

Mass of x water = 5.72 - 2.12 = 3.6 g

Molecular wt. of sodium carbonate Na₂CO₃ = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106

Molecular wt. of H₂O = 2(1) + 16 = 18

$\dfrac{\text{Mass of water}}{\text{Mass of anhydrous sodium carbonate}}$ = $\dfrac{18\text{x}}{106}$ = $\dfrac{3.6}{2.12}$

∴ x = $\dfrac{3.6 \times 106}{2.12 \times 18}$ = $\dfrac{381.6}{38.16}$ = 10

Hence, formula of hydrated sodium carbonate = Na₂CO₃.10H₂O

(ii) Molecular wt. of hydrated sodium carbonate Na₂CO₃.10H₂O = 2(23) + 12 + 3(16) + 10(18) = 46 + 12 + 48 + 180 = 286g

286 g of sodium carbonate contains 12 g of carbon

100 g of sodium carbonate will contain $\dfrac{12}{286}$ x 100 = 4.19 g of carbon.

Hence, percentage of carbon in the compound = 4.19%