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 (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.
(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`
Percent of H in the compound `=(0.0767 g)/(0.9984 g)xx100`
Moles of carbon in the compound `=92.32/12.00`
Moles of hydrogen in the compound `=7.68/1`
∴ Ratio of carbon to hydrogen in the compound = 7.69: 7.68
= 1: 1
Hence, the empirical formula of the gas is CH.
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
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