Question

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate

1. empirical formula,
2. molar mass of the gas, and
3. molecular formula.

Answer 1

(i) 1 mole (44 g) of CO₂ contains 12 g of carbon.

∴ 3.38 g of CO₂ will contain carbon `=(12  "g")/(44  "g")xx3.38  "g"`

= 0.9217 g

18 g of water contains 2 g of hydrogen.

∴ 0.690 g of water will contain hydrogen `=(2  "g")/(18  "g")xx0.690`

= 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

∴ Percent of C in the compound `(0.9217  g)/(0.9984  g)xx100`

= 92.32%

Percent of H in the compound `=(0.0767  "g")/(0.9984  "g")xx100`

= 7.68%

Moles of carbon in the compound `=92.32/12.00`

= 7.69

Moles of hydrogen in the compound `=7.68/1`

= 7.68

∴ Ratio of carbon to hydrogen in the compound = 7.69 : 7.68

= 1 : 1

Hence, the empirical formula of the gas is CH.

(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 `=(11.6  "g")/(10.0  "L")xx22.4  "L"`

= 25.984 g

≈ 26 g

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

(iii) Empirical formula mass of CH = 12 + 1 = 13 g

`"n" ="molar mass of gas"/"Empirical formula mass of gas"`

`=(26  "g")/(13  "g")`

n = 2

∴ Molecular formula of gas = (CH)_n

= C₂H₂