Finding the molecular formula given combustion data


> 
> $\pu{150 cm^3}$ of oxygen were added to $\pu{20 cm^3}$ of a gaseous hydrocarbon D.
> After explosion and cooling, the gaseous mixture occupied $\pu{130 cm^3}$  
> 
> and, after absorption by potassium hydroxide, $\pu{90 cm^3}$ of oxygen remained.
> 
> 
> 


$$\ce{C\_xH\_y + (x + \frac{y}{4}) O2 -> x CO2 + y/2 H2O}$$


$\ce{C\_xH\_y}$: $\pu{20 cm^3}$


Why is $(x+y/4) \ce{O2} = \pu{150 cm^3} - \pu{90 cm^3}$  

and $\ce{x CO2} = \pu{130 cm^3} - \pu{90 cm^3}$ ?


I genuinely don't know why.why do we subtract 90 from both sides and not just the product? Why are we subtracting 90 from the $\ce {CO2}$




* $\pu{20 cm^3}$ is the volume of the hydrocarbon.
* $\pu{130 cm^3} - \pu{90 cm^3} = \pu{40 cm^3}$  

is the volume of $\ce{CO2}$ absorbed in the hydroxide, therefore $x=\frac{\pu{40 cm^3}}{\pu{20 cm^3}} = 2$.
* $\pu{150 cm^3} - \pu{90 cm^3} = \pu{60 cm^3}$ is volume of spent $\ce{O2}$.
* Therefore $x + \frac{y}{4} = \frac{\pu{60 cm^3}}{\pu{20 cm^3}} = 3 \implies y = 4 \implies \ce{C\_xH\_y} = \ce{C2H4}$.


 1

If starting from $\pu{150 mL}$ $\ce{O2}$, only a part of it is used to finally obtain a residue of $\pu{90 mL}$ $\ce{O2}$, it means that $150 - 90 = \pu{60 mL}$ $\ce{O2}$ has been consumed. Now added $\ce{KOH}$ after the combustion reacts with $\ce{CO2}$ and reduces the final gas volume from $\pu{130 mL}$ to $\pu{90 mL}$. This indicates that $130 - 90 = \pu{40 mL}$ is the volume of $\ce{CO2}$ produced. Let's rewrite these results:


$\ce{
 \pu{20 mL} C\_xH\_y + \pu{60 mL} O2 -> \pu{40 mL} CO2 + $a$ H2O
}$


The number of moles is proportional to the volumes. So that this equation can be also rewritten so, after division by $10$:


$\ce{2 C\_xH\_y + 6 O2 -> 4 CO2 + $a'$ H2O}$


To get $\ce{4 C}$ atoms at right, we need: $x = 2$ in the initial hydrocarbon. The equation can be rewritten:


$\ce{2 C2H\_y + 6 O2 -> 4 CO2 + $a'$ H2O}$.


And now $4$ of the $\ce{6 O2}$ are required to make $\ce{4 CO2}$. The remaining $\ce{2 O2}$ (or $\ce{4 O}$) are required to burn the $2y$ $\ce{}$ atoms and form $\ce{4 H2O}$. So $a' = 4$ and $y = 4$. And the equation becomes:


$\ce{2 C2H4 + 6 O2 -> 4 CO2 + 4 H2O}$.


This can even be more simplified :


$\ce{C2H4 + 3 O2 -> 2 CO2 + 2 H2O}$.


 1