A compound contains 40.0% carbon, 6.67% hydrogen and 53.33% oxygen by mass. Find its empirical formula. (C = 12, H = 1, O = 16)
ACHO
BC$_2$H$_4$O
CCH$_2$OCORRECT
DCH$_3$O
JUPEBChemistryStoichiometry and Volumetric Analysis
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Why the answer is C, and why the others tempt you.
Divide each percentage by the atomic mass: C $= \dfrac{40}{12} = 3.33$, H $= \dfrac{6.67}{1} = 6.67$, O $= \dfrac{53.33}{16} = 3.33$. Dividing through by 3.33 gives the ratio 1 : 2 : 1, so the empirical formula is CH$_2$O.
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