Question #87de9

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Question #87de9

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Answer 1 · 2016-10-01T11:39:02

$C_{2} H_{4} O$ $color(red)(C_2H_4O)$

Explanation:

Atomic mass of $H = 1 g/mol$ $H=1" g/mol"$
Atomic mass of $C = 12 g/mol$ $C=12" g/mol"$
Atomic mass of $O = 16 g/mol$ $O=16" g/mol"$
Molar mass of $C O_{2} = 44 g/mol$ $CO_2=44" g/mol"$
Molar mass of $H_{2} O = 18 g/mol$ $H_2O=18" g/mol"$

Mass of C in 7.59g $C O_{2} = \frac{7.59 g}{44 g/mol} \times 12 g/mol = 2.07 g$ $CO_2=(7.59g)/(44"g/mol")xx12"g/mol"=2.07g$

Mass of H in 3.11g $H_{2} O = \frac{3.11 g}{18 g/mol} \times 2 g/mol = 0.345 g$ $H_2O=(3.11g)/(18"g/mol")xx2"g/mol"=0.345g$

These two amounts of C and H have been obtained from 3.8g compound containing C ,H and O.

So the remaining (3.80-2.07-0.35)g=1.38g is the mass of Oxygen present in 3.8 g compound.

Hence dividing these masses of elements with respective atomic masses of elements and taking their ratio we get the ratio of number of atoms of C, H and O in the molecule of the compound as below

$C : H : O = \frac{2.07}{12} : \frac{0.345}{1} : \frac{1.38}{16}$ $C:H:O=2.07/12:0.345/1:1.38/16$
$= 0.1725 : 0.345 : 0.08625$ $=0.1725:0.345:0.08625$

$= (\frac{0.1725}{0.08625}) : (\frac{0.345}{0.08625}) : (\frac{0.08625}{0.08625})$ $=(0.1725/0.08625):(0.345/0.08625):(0.08625/0.08625)$
$= 2 : 4 : 1$ $=2:4:1$

Hence empirical formula of the compound is $C_{2} H_{4} O$ $color(red)(C_2H_4O)$

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Question #87de9

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