Chemistry, asked by halimatunsaadiah00, 5 months ago

A certain compound, used as a welding fuel, contains carbon and hydrogen only. Burning
completely a small sample of the compound in oxygen gives 3.38 g of CO2, 0.692 g of water
and no other products. Calculate the empirical formula of the compound.

Answers

Answered by mkprasanna15
0

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

.

Answered by shaanullah
0

Answer:

Empirical formula of the compound  = CH

Now molecular formula  calculation

Empirical formula mass = 12 + 1 = 13 amu

Also molecular mass  = 26 g

Therefore n = molecular mass/empirical formula mass = 26/13 = 2

Explanation:

Since, 3.38 g of carbon dioxide are obtained, the mass of C present in the sample of welding fuel gas is

4412×3.38=0.92  g.

Since, 0.690 g of water are obtained, the mass of hydrogen present in the welding fuel gas sample is

182×0.690=0.077g.

The percentage of C is 0.92+0.0770.92×100=92.3 %.

The percentage of H is 0.92+0.0770.077×100=7.7%.

(i) The number of moles of carbon =1292.2=7.7.

The number of moles of hydrogen =17.7=7.7.

The mole ratio C:H=7.77.7=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 10.011.6×22.4=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 1326=2.

The molecular formula is 2(CH)=C2H2.

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