I need help with this chemistry problem. A certain compound can be used as a rocket fuel. A 12.8 g sample of this fuel is found to contain 11.2 g of nitrogen and 1.60 g of hydrogen. The molar mass of fuel is the same as that of oxygen gas. Determine the empirical formula and the molecular formula of this compound.
Answers
Answer:
Since, 3.38 g of carbon dioxide are obtained, the mass of C present in the sample of welding fuel gas is
44
12
×3.38=0.92 g.
Since, 0.690 g of water are obtained, the mass of hydrogen present in the welding fuel gas sample is
18
2
×0.690=0.077g.
The percentage of C is
0.92+0.077
0.92
×100=92.3 %.
The percentage of H is
0.92+0.077
0.077
×100=7.7%.
(i) The number of moles of carbon =
12
92.2
=7.7.
The number of moles of hydrogen =
1
7.7
=7.7.
The mole ratio C:H=
7.7
7.7
=1:1.
Hence, the empirical formula of the welding fuel gas is CH.
(ii) The weight of 10.0 L of gas at N.T.P is 11.6 g.
22.4 L pf gas at N.T. P will weigh
10.0
11.6
×22.4=26g/mol
This is the molar mass.
(iii) Empirical formula mass is 12+1=13.
Molecular mass is 26.
The ratio, n of the molecular mass to empirical formula mass is
13
26
=2.
The molecular formula is 2(CH)=C
2
H
2
.