Question

At 300 k and 1 atm, 15 ml of a gaseous hydrocarbon requires 375 ml of air containing 20% o2 by volume for complete combustion. After combustion, the gases occupy 330 ml. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is

Answer 1

combustion of hydrocarbon is given by chemical equation ..
$C_xH_y+\left(x+\frac{y}{4}\right)O_2\rightarrow xCO_2+\frac{y}{2}H_2O$

here, it is clear that,
1vol of hydrocarbon reacts with (x + y/4) vol of O2 to form x vol of carbon dioxide.

so, 15ml hydrocarbon reacts with 15(x + y/4) ml of oxygen to form 15x ml of carbon dioxide.

volume of O2 = 20% of 375ml
= 20 × 375/100 = 75 ml

so, 15(x + y/4) = 75

x + y/4 = 5 .....(1)

a/c to question, final volume of gases = 330ml
initial volume of gases = 375ml + 15ml = 390ml
amount of water formed = 390ml - 330ml = 60ml

so, 15 × y/2 = 60 => y = 8

put it in equation (1), x = 3

so, hydrocarbon is $C_3H_8$

Answer 2

Explanation:

Chemical equation which shows combustion of a hydrocarbon is as follows.

    $C_{x}H_{y} + (x + (\frac{y}{4})O_{2} \rightarrow xCO_{2} + \frac{y}{2}H_{2}O$

Here, 1 mole of hydrocarbon reacts with $(\frac{y}{4})$ mole of $O_{2}$ to form x mole of $CO_{2}$.

Therefore, 15 ml of hydrocarbon reacts with 15$(\frac{x+y}{4})$ ml of oxygen to form 15x ml of $CO_{2}$.

It is given that volume of $O_{2}$ is 20% of 375 ml. Hence, total volume of oxygen will be calculated as follows.

              $20 \times \frac{375}{100}$ = 75 ml

this, also means that 15$(\frac{x+y}{4})$ = 75

                            x + $\frac{y}{4}$ = 5

so,                       x = $5 - \frac{y}{4}$     ........ (1)

As it is given that final volume of gases is 330 ml.

So, initial volume of gases is 375 ml + 15 ml = 390 ml

Hence, amount of water formed is calculated as follows.

                  Total volume - volume occupied by gases

                      = 390 ml - 330 ml

                      = 60 ml

Therefore, $15 \times \frac{y}{2}$ = 60

so,       y = 8

Now, putting the value of y in equation (1), we will get the value of x as equal to 3.

Thus, we can conclude that the molecular formula of the given hydrocarbon is $C_{3}H_{8}$.