A sample of a gaseous hydrocarbon occupying 1.12 litres at NTP when completely burnt in air produced 2.2 g of `CO_(2)` and 1.8 g of `H_(2)O`. Calculate the weight of the compound taken and the volume of `O_(2)` at NTP required for its burning. Find the molecular formula of the hydrocarbon.

Explanation

Mole of hydrocarbon = `(1.12)/(22.4)=(1)/(20)` (Rule 3, Chapter 1)
Mole of `CO_(2)=(2.2)/(44)=(1)/(20)`. (Rule 1, Chapter 1)
Mole of `H_(2)O=(1.8)/(18)=(1)/(10)` (Rule 1, Chapter 1)
Thus,
`C_(x)H_(y)` (Suppose) + `O_(2)rarrCO_(2)+H_(2)O`
`{:(C_(x)H_(y)" (suppose)"+O_(2)rarrCO_(2)+H_(2)O),((1)/(20)"mole "(1)/(20)"mole"(1)/(10)"mole"):}`
Applying POAC for C and H atoms we get respectively,
`x xx` moles of `C_(x)H_(y)=1xx` moles of `CO_(2)`
`x xx(1)/(20)=1xx(1)/(20),x=1`
and `yxx` moles of `C_(x)H_(y)=2xx` moles of `H_(2)O`
Thus, formula of the hydrocarbon is `CH_(4)`.
Wt. of `CH_(4)` = moles of `CH_(4)xx` mol. wt.
`=(1)/(20)xx16=0.8g`.
Now, again applying POAC for O atoms,
`2xx` moles of `O_(2)=2xx` moles of `CO_(2)+1xx` moles of `H_(2)O`
`2xx("volume of "O_(2)" (NTP) in lit.")/(22.4)=2xx(1)/(20)+(1)/(10)=(1)/(5)` .
`therefore` vol. of `O_(2)` at NTP = 2.24 litres.