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Elemental Composition of Pure Substances

AP Chemistry· difficulty 4/5

A compound is 40.0% C, 6.7% H, 53.3% O by mass. Its empirical formula is

  • A

    CH2O\text{CH}_2\text{O}

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  • B

    CHO\text{CHO}

  • C

    CH3OH\text{CH}_3\text{OH}

  • D

    C2H4O2\text{C}_2\text{H}_4\text{O}_2

Explanation

Per 100 g: 40.0/12 = 3.33, 6.7/1 = 6.7, 53.3/16 = 3.33 mol. Ratio 1:2:1 → CH2_2O. (Could be glucose's empirical formula.)

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