A welding fuel gas contains carbonn & hydrogen only. Burning a smal sample of it in oxygen gives 3.38g `CO_(2)`, 0.690g of `H_(2)O` and no other products. A volume of 10.0 L (at STP) of this welding gas is found to weigh 11.6g. Calculate (1) empirical formula,
(2) molar mass of the gas.

Explanation

(1) Mass of carbon in 44g of `CO_(2)=12g`
`therefore`Mass of C in 3.38g of `CO_(2)=(12)/(44)xx3.38g=0.9218g`
Similarly, 18g `H_(2)O-=2g` hydrogen
Therefore, mass of hydrogen in 0.690 of `H_(2)O`
`=(2//18)xx0.690g=0.0767g`
`therefore`Total mass of fuel gas`=(0.9218+0.0767)=0.9985g`
`therefore`% amount of C in the gas`=(0.9218)/(0.9985)xx100" "=92.32`
`therefore`% amount of H in the gas `=(0.0767)/(0.9985)xx100=7.68`.
Determination of the empirical formula of the fuel gas

Therefore, empirical formula of the fuel gas=CH
(2) As given in the question,
Mass of 10 L gas at STP=11.6g
`therefore` Mass of 22.4 L gas at `STP=(11.6xx22.4)/(10)=22.984g`
`therefore` Molecular mass=25.984`~~26`.