compound contains onlyC, H, and N. Combustion of 70 mg of the compound produces 132 mg CO2
and 54 mg H2O. What is empirical formula of compound?
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Answer:
C
x
H
y
O
z
+O
2
⟶xCO
2
+
z
y
H
2
O
moles of CO
2
=0.364m mol
It contains 0.364m mol of C.
∴ Mass of C=0.364×12=4.368 mg$$
moles of H
2
O=
18
4.37
=0.243m mol
It contains 2×0.243m mol of H
∴ Mass of H=0.486 mg
Mass of O=10.68−(4.368+0.486)=5.826 mg
∴ Moles of O=0.364
∴ Ratio C:H:O=0.364:0.486:0.364=1:1.34:1=3:4:3
∴ Empirical formula is C
3
H
4
O
3
,
Empirical formula mass =88=2× Molar mass.
∴ Molecular formula is C
6
H
8
O
6
.
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