Question #a7489

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Question #a7489

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Answer 1 · 2016-10-03T09:20:36

Percentage Composition of constituent elements in a compound

$C \to 74.0 %$ $C->74.0%$

$H \to 7.4 %$ $H->7.4%$

$N \to 8.6 %$ $N->8.6%$

$O \to 10.0 %$ $O->10.0%$

Considering their atomic masses as

$C \to 12 g/mol$ $C->12" g/mol"$

$H \to 1 g/mol$ $H->1" g/mol"$

$N \to 14 g/mol$ $N->14" g/mol"$

$O \to 16 g/mol$ $O->16" g/mol"$

Now dividing each percentage composition by the respective atomic mass and then taking the ratio we get the ratio of number of atoms of constituent elements in the compound as

$C : H : N : O = \frac{74}{12} : \frac{7.4}{1} : \frac{8.6}{14} : \frac{10}{16}$ $C:H:N:O=74/12:7.4/1:8.6/14:10/16$

$= 6.16 : 7.4 : 0.614 : 0.625$ $=6.16:7.4:0.614:0.625$

$= \frac{6.16}{0.614} : \frac{7.4}{0.614} : \frac{0.614}{0.614} : \frac{0.625}{0.614}$ $=6.16/0.614:7.4/0.614:0.614/0.614:0.625/0.614$

$= 10 : 12 : 1 : 1$ $=10:12:1:1$

Hence the empirical formula is

$C_{10} H_{12} N O$ $C_10H_12NO$

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