Question

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5ml of gas containing only carbon and hydrogen were mixed with an excess of oxygen and the mixture exploded by means of an electric sparks . After the explosion the volume of the remained gas was 25 ml . On adding a concentrated solution of potassium hydroxide , the volume further diminished to 15ml of the residual gas being pure oxygen . All volumes has been reduced to NTP . Calculate the molecular formula of the hydrogen gas?

Class - 11th
Chapter - Some basic concepts of chemistry



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Answer 1

$\mathfrak \purple{Question}$

> 5ml of gas containing only carbon and hydrogen were mixed with an excess of oxygen and the mixture exploded by means of an electric sparks . After the explosion the volume of the remained gas was 25 ml . On adding a concentrated solution of potassium hydroxide , the volume further diminished to 15ml of the residual gas being pure oxygen . All volumes has been reduced to NTP . Calculate the molecular formula of the hydrogen gas?

$\mathfrak \purple{Answer}$

Volume of oxygen = 30 ml

Volume of unused oxygen = 15 ml

Volume of carbon-dioxide after explosion = 25 ml

Volume of carbon -dioxide before explosion = 15 ml

So , the volume of carbon-dioxide produced = 25-15 = 10 ml

Volume of hydrocarbon =15 ml

Combustion: CxHy +(x+y/4) O² ⟶ xCo² +y/4 H²O

So, volume of carbon dioxide produced = 5x

Equating ,5x = 10

⟶ x = 2 &

5( x + y/4) = 15

⟶ y = 4

Molecular formula = C2H4

Answer 2

$\\ \bold{ \bigstar \: \large \underline{ \underline{AnsWer :- }}}$

The general equation of combustion of hydrocarbons are as follows :

$\\ \: \: \: \: \sf \: \: C _x H_y + \bigg( \frac{4x + y}{4} \bigg) O_2 \longrightarrow \: xCO_2 + \frac{y}{2} H_2O$

$\\ \sf \: \: 1mL \: \: \: \: \: \: \bigg( \frac{4x + y}{4} \bigg)mL \: \: \: \: \: \: x \: mL \: \: \: \: \: \: \: \frac{y}{2} mL$

$\\ \sf \: \: 5mL \: \: \: \: \: \: 5\bigg( \frac{4x + y}{4} \bigg)mL \: \: \: \: \: \:5 x \: mL \: \: \: \: \: \: \: \frac{5y}{2} mL$

[Since volume of hydrogen taken = 5mL]

The value of oxygen added = 30mL

Volume of oxygen remained = 15mL

Volume of oxygen used = 30-15 = 15mL

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: = \: 5\bigg( \frac{4x + y}{4} \bigg)$

$\\ \sf \: \: \: \frac{4x + y}{4} = 3 \: \: \: ...(1)$

Volume of $\sf CO_2$ produced,

$\\ \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 25 - 15 = 10ml$

$\\ \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 5x$

$\\ \sf \: \: Thus, \: \: \: \: \: \: \: \: \: 5x = 10mL$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{10 }{5}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = 2$

Put the value of x in equation (1)

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \therefore \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{4 \times 2 + y}{4} = 3$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \therefore \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{8 + y}{4} = 3$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \therefore \: \: \: \: \: \: \: \: \: \: \: \: \: \: y= 12 - 8$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \therefore \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \frak{ \pink{y= 4}}}$

Thus , The molecular formula of hydrogen atom is $\bold{C_2H_4}$.