Molecular weight of CaCO3 = 40 + 12 + 3(16) = 40 + 12 + 48 = 100 g
100 g of CaCO3 contains 12 g of carbon.
Given, purity is 55% therefore,
55% of 12 g = 55\over100 × 12
= 6.6%
Hence, percentage of carbon is 6.6%.
Question
Calculate the percentage of carbon in a 55% pure sample of carbon carbonate. [Ca = 40, C = 12, O = 16]
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