Question

1.34 A welding fuel gas contains carbon and hydrogen only. Burning a small sama
in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other
volume of 10.0 L (measured at STP) of this welding gas is found to weigh the
Calculate (1) empirical formula, (ii) molar mass of the gas. and (iii) molen
formula.

Answer 1

Since, 3.38 g of carbon dioxide are obtained, the mass of C present in the sample of welding fuel gas is 12/44 × 3.38 = 0.92 g.Since, 0.690 g of water ...

Answer 2

(i) percentage of C can be calculated as follows:

CO2 = C

i.e 44 parts of CO2= 12 parts of C

OR

44g of CO2 = 12 g of C

Therefore according to question

3.38 g of CO2 contains C = 12/44 * 3.38 = 0.921 g

18 g of water contains hydrogen = 2g

Therefore 0.690 g of water contains hydrogen = 2/18 * 0.690 = 0.0767g

0.690 g of water will contain hydrogen

= 0.0767 g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:

= 0.9217 g + 0.0767 g = 0.9984 g

Now percentage of carbon = weight of carbon/weight of compound * 100

=0.921 /0.998 * 100= 92.32 %

Also percentage of hydrogen = weight of hydrogen/weight of compound *100

=0.0766/0.998 * 100 = 7.68 %

(ii) Given,

Weight of 10.0L of the gas (at S.T.P) = 11.6 g

Weight of 22.4 L of gas at STP

= 25.984 g

≈ 26 g

Hence, the molar mass of the gas is 26 g.

(iii) empirical formula

Element Percentage Atomic mass Atomic ratio Simplest ratio Simplest whole no ratio

C 92.32 12 92.32/12 = 7.69 7.69/7.65 = 1.00 1

H 7.65 1 7.65/1 = 7.65 7.65/7.65 = 1 1

Empirical formula of the compound = CH

Now molecular formula calculation

Empirical formula mass = 12 + 1 = 13 amu

Also molecular mass = 26 g (calculated in previous step)

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