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Empirical Formula

An organic co...

Question

An organic compound on analysis gave $C = 5.45$ % , $H = 9.1$ % by mass. Its empirical formula is:

A

C H O_{2}

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B

C H_{2} O

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C

C_{2} H_{4} O

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D

C_{3} H_{4} O

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Solution

The correct option is C $C_{2} H_{4} O$
Of :- % age of carbon $= 54.5 %$
% age of Hydrogen $= 9.1 %$
% age of oxygen $= 100 - (54.5 + 9.1)$
$= 36.4 %$

Element % age mass Molar mass n Ratio
O 36.4 16 $\frac{36.4}{16} = 2.27$ $\frac{2.27}{2.27} = 1$
C 54.5 12 $\frac{54.5}{12} = 4.54$
$\frac{4.54}{2.27} = 2$
H 9.1 1 $\frac{9.1}{1} = 9.1$ $\frac{9.1}{2.27} = 4$
$∴$ Formula $= C_{2} H_{4} O$

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