During an analysis of 255mg of a molecule that contains only C,H and O, 561mgCO2 is produced along with 306mgH2O on combustion. If the molar mass of the compound is 180g/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 DC9H24O3 CxHyOz
On combustion we have : CxHyOz+(x+y4−z2)O2→xCO2+y2H2O
Moles of C are equal to moles of CO2 so : nCO2=56144=12.75×10−3 nC=nCO2=12.75×10−3∴wC=12×12.75=153mg
moles of H are equal to twice of moles of H2O so : nH=2×nH2O=2×30618×10−3=34×10−3∴wH=34mg
wC+wH=153+34=187mg
wO=wcompound−187=255−187=68mg
So moles of O : nO=6816=4.25×10−3=4.25mmol
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 =180g/mol and n =integer)
So we get n=3
Empirical formula =molecular formulan
so, Molecular formula =C9H24O3