An organic compound contains C 66.7%, H11.1 % and cl 22.2% by mass, what is it's empirical formula
Question
An organic compound contains C 66.7%, H 11.1% and Cl 22.2% by mass, what is its empirical formula?
Solution
To find the empirical formula, follow these steps:
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Assume that you have 100g of the compound. This means you have 66.7g of Carbon (C), 11.1g of Hydrogen (H), and 22.2g of Chlorine (Cl).
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Convert the mass of each element to moles.
- For Carbon: 66.7g / 12.01g/mol = 5.56 mol
- For Hydrogen: 11.1g / 1.01g/mol = 11 mol
- For Chlorine: 22.2g / 35.45g/mol = 0.63 mol
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Divide each mole quantity by the smallest number of moles calculated (which is 0.63 in this case) to get the ratio of atoms.
- For Carbon: 5.56 / 0.63 = 8.8 ≈ 9
- For Hydrogen: 11 / 0.63 = 17.5 ≈ 18
- For Chlorine: 0.63 / 0.63 = 1
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The empirical formula of the compound is therefore C9H18Cl.
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