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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

- empirical formula,
- molar mass of the gas, and
- molecular formula.

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#### Solution

**(i) **1 mole (44 g) of CO_{2} contains 12 g of carbon.

∴ 3.38 g of CO_{2} 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}

= C_{2}H_{2}

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