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 litre (Measured at STP) of this welding gas is found weigh `11.6 g`. Calculate
(i) empirical formula,
(ii) molar mass of the gas, and
(iii) molecular formula.

Answer

Empirical formula CH , molar mass `26.0 g mol ^(-1)` . Molecular formula `C_(2) H_(2)`

Explanation

(i) 1 mole (44g) of` CO_(2)` contains 12 g of carbon .
`therefore `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 .
`therefore ` 0.0690 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
`therefore ` Percent of C in the compound `=(0.9217 g)/(0.9984 g_ xx100`
=92.32 %
percent of H the copmpound `=(0.0767g)/(0.9984 g)xx100`
=7.68 %
Moles of carbon in the compound `=(92.32)/(12.00)`
Moles of hydrogen in the compound `=(7.68)/(1)`
= 7.68
`therefore ` 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 10.0 L of the gas (at S.T.P ) = 11.6 g
`therefore `Weight of 22.4 L of gas at STP `+(11.6 g)/(10.0g)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
`therefore` Molecular formula of gas `=(CH)_(n)`
`=C_(2)H_(2)`