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What is the empirical formula of a compound that is 54.5% C, 9.2% H, and 36.3% O by mass?

-C₂HO
-C₄H₈O₂
-C5H₉O₂
-C₂H₄O
-C₆HO₄

1 Answer

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Final answer:

The empirical formula of the compound is C₂H₄O.

Step-by-step explanation:

The empirical formula of a compound can be determined using the percent composition by mass of each element in the compound. To find the empirical formula, you need to convert the mass percentages to moles, then divide each mole value by the lowest mole value obtained. In this case, for a compound that is 54.5% C, 9.2% H, and 36.3% O by mass, we can assume we have 100g of the compound.

First, convert the mass percentages to grams: 54.5g C, 9.2g H, 36.3g O.

Next, calculate the number of moles for each element by dividing the grams by the atomic masses: moles of C = 54.5g / 12.01g/mol = 4.54 mol, moles of H = 9.2g / 1.01g/mol = 9.11 mol, moles of O = 36.3g / 16.00g/mol = 2.27 mol.

Finally, divide each mole value by the lowest mole value to obtain the simplest, whole number ratio: C = 4.54 mol / 2.27 mol ≈ 2, H = 9.11 mol / 2.27 mol ≈ 4, O = 2.27 mol / 2.27 mol = 1.

Therefore, the empirical formula of the compound is C₂H₄O.

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