40 ml of a hydrocarbon undergoes combustion in 260ml of oxygen and gives 160ml of CO2 . If all volumes are measured under similar conditions of temperature and pressure, the formula of the hydrocarbon is?

40 ml of a hydrocarbon undergoes combustion in 260ml of oxygen and gives 160ml of ${\mathrm{CO}}_{2}$ . If all volumes are measured under similar conditions of temperature and pressure, the formula of the hydrocarbon is?

1. A

${\mathrm{C}}_{3}{\mathrm{H}}_{8}$

2. B

${\mathrm{C}}_{4}{\mathrm{H}}_{8}$

3. C

${\mathrm{C}}_{6}{\mathrm{H}}_{14}$

4. D

${\mathrm{C}}_{4}{\mathrm{H}}_{10}$

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Solution:

At constant temperature and pressure (at STP), volume is proportional to moles.

Hence, the combustion of 1 mole of hydrocarbon will require 6.5 moles (260/40​=6.5)  of oxygen and will produce 4 moles (160/​40=4) of carbon dioxide.

i,e x=4
Thus, the molecular formula contains 4 carbon atoms.

x+y/4​=6.5

y=10

Thus, the molecular formula contains 10 hydrogen atoms.

Out of 6.5 moles of oxygen, 4 moles will combine with carbon to form 4 moles of carbon dioxide.
The remaining 2.5 moles of oxygen will combine with 10 moles of hydrogen atoms (from the hydrocarbon) to form 5 moles of water.

The hydrocarbon is C4​H10 and the combustion reaction is C4​H10​+6.5O2​→4CO2​+5H2​O.

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