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A sample of a compound contains 38.74% of carbon, 9.63% of hydrogen, and 51.63% of oxygen. Its molar mass is 62 g/mol. Determine the empirical and molecular formula.

empirical formula ________
molecular formula __________"

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Final answer:

The empirical formula of the compound is CH₃O and the molecular formula is C₂H₅O₂.

Step-by-step explanation:

To determine the empirical formula, we need to find the simplest whole number ratio of the atoms in the compound. The percentages given can be used to find the moles of each element in a 100g sample.

Carbon: 38.74g * (1 mol/12.01 g/mol) = 3.227 mol

Hydrogen: 9.63g * (1 mol/1.01 g/mol) = 9.534 mol

Oxygen: 51.63g * (1 mol/16.00 g/mol) = 3.227 mol

Dividing each of the mole values by the smallest mole value, we get the ratio of C:H:O as 1:3:1. Therefore, the empirical formula is CH₃O.

To find the molecular formula, we need to know the molar mass of the compound. If the molar mass is given as 62 g/mol, we can divide the molar mass by the empirical formula mass to find the multiplying factor. The empirical formula mass of CH₃O is 12.01+1.01+16.00 = 29.02 g/mol. The multiplying factor is 62 g/mol divided by 29.02 g/mol, which is approximately 2.137. Multiplying the subscripts in the empirical formula by the multiplying factor gives us the molecular formula, which is C₂H₅O₂.

User Tim Andersen
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