Question

Question #5c3b5

Answer 1

n = 10. The relative atomic mass of A is 23 u.

This is a disguised empirical formula problem.

Assume 100 g of A₂CO₃•nH₂O. Then we have 16 g of A and 63 g of H₂O.

Mass of CO₃ = (100 – 16 – 63) g = 21 g CO₃

Moles of CO₃ = 21 g CO₃ × `(1 mol CO₃)/(60.01 g CO₃)` = 0.35 mol CO₃

Moles of H₂O = 63 g H₂O × `(1 mol H₂O)/(18.02 g H₂O)` = 3.5 mol H₂O

Moles of A = 2 × moles of CO₃ = 2 × 0.35 mol = 0.70 mol A

Molar ratios A:CO₃:H₂O = 0.70:0.35:3.5 = 2.0:1:10 ≈ 2:1:10

The empirical formula is A₂CO₃•10H₂O.

16 g A = 0.70 mol A

∴ Atomic mass of A = `(16 g)/(0.70 mol)` = 23 g/mol.

Relative atomic mass of A = 23 u.

(A must be Na, and the formula is Na₂CO₃•10H₂O).