A $5.00 g$ sample of impure $CuSO X_{4} \cdot 5 H X_{2} O$ loses $1.44 g$ on strong heating (only the water of crystallisation is lost; the impurity is non-volatile and anhydrous). Find the percentage purity of the sample, to the nearest integer (Cu = 63.5, S = 32, O = 16, H = 1).

Numerical answer type, answer in %, tolerance 0.5.

Answer: 80 %

Water lost = 1.44/18 = 0.08 mol, and each mole of hydrate holds 5 mol water, so hydrate = 0.016 mol. Molar mass of hydrate = 249.5, so mass = 3.992 g. Purity = 3.992/5.00 x 100 = 79.84%, i.e. 80 to the nearest integer.

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A 5.00 g sample of impure CuSOX4​⋅5HX2​O loses 1.44 g on strong heating (only the water of crystallisation is lost; the impurity is non-volatile and anhydrous). Find the percentage purity of the sample, to the nearest integer (Cu = 63.5, S = 32, O = 16, H = 1).

Answer: 80 %

More questions on Empirical and molecular formula, percentage composition

A $5.00 g$ sample of impure $CuSO X_{4} \cdot 5 H X_{2} O$ loses $1.44 g$ on strong heating (only the water of crystallisation is lost; the impurity is non-volatile and anhydrous). Find the percentage purity of the sample, to the nearest integer (Cu = 63.5, S = 32, O = 16, H = 1).