Question

0.254g lead(ii)ethanoate, on adding excess K2CrO4 solution, gave 0.130g of lead(ii)chromate precipitate. what is the percentage composition of Pb in the organic salt? (O = 16, Cr = 52, Pb = 207).​

Answer 1

Answer:

32.8%

Explanation:

All of the Pb⁺² species precipitated as lead(II) cromate, PbCrO₄ (we know this as excess K₂CrO₄ was used).

First we convert 0.130 g of PbCrO₄ into moles, using its molar mass:

• 0.130 g ÷ 323 g/mol = 4.02x10⁻⁴ mol PbCrO₄

There's 1 Pb⁺² mol per PbCrO₄ mol, so in total 4.02x10⁻⁴ moles of Pb⁺² were in the ethanoate sample.

We convert those 4.02x10⁻⁴ moles of Pb into grams:

• 4.02x10⁻⁴ mol * 207 g/mol = 0.083 g Pb

Finally we calculate the percentage composition of Pb:

• 0.083 g Pb / 0.254 g salt * 100% = 32.8%