The sum of mass percents of Oxygen and Carbon is 100 - 39.95 = 60.05%.
Let x be the percentage of Carbon. Therefore, the percentage of Oxygen equals 4x.
x + 4x = 60.05;
x = 12.01
So the compound contains 12.01% of Carbon and 4 x 12.01 = 48.04% of Oxygen by mass.
Considering 100 g of the compound, there is 39.95 g of Ca, 12.01 g of C, and 48.04 g of O. To determine the empirical formula, the mole ratio of these elements should be calculated.
n (Ca) = 39.95g / 40.08g/mol ≈ 1 mol;
n (C) = 12.01g / 12.01g/mol = 1 mol;
n (O) = 48.04g / 16.00g/mol ≈ 3 mol.
Therefore the empirical formula is CaCO3.
Answer: CaCO3
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