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Question

During an analysis of 255 mg of a molecule that contains only C, H and O, 561 mg CO2 is produced along with 306 mg H2O on combustion. If the molar mass of the compound is 180 g/mol, what is the molecular formula of the compound ?

A
C16H4O9
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B
C10H21O9
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C
C5H21O3
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D
C9H24O3
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Solution

The correct option is D C9H24O3
CxHyOz
On combustion we have :
CxHyOz+(x+y4z2)O2xCO2+y2H2O

Moles of C are equal to moles of CO2 so :
nCO2=56144=12.75×103
nC=nCO2=12.75×103wC=12×12.75=153 mg

moles of H are equal to twice of moles of H2O so :
nH=2×nH2O=2×30618×103=34×103wH=34 mg

wC+wH=153+34=187 mg

wO=wcompound187=255187=68 mg

So moles of O :
nO=6816=4.25×103=4.25 mmol
therefore the ratio C:H:O=12.75:34:4.25
=3:8:1

Therefore Empirical formula : C3H8O
Empirical mass =60
Empirical mass =molecular massn

(molar mass of compound =180 g/mol and n =integer)

So we get n=3

Empirical formula =molecular formulan
so, Molecular formula =C9H24O3

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