Question #343ff

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Question #343ff

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Answer 1 · 2016-09-27T08:43:26

Let the molecular formula of HC be $C_{x} H_{y}$ $C_xH_y$

And the balanced equation of the combustion reaction of the HC in oxygen is

$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} (g) \to x C O_{2} (g) + \frac{y}{2} H_{2} O (l)$ $C_xH_y+(x+y/4)O_2(g)->xCO_2(g)+y/2H_2O(l)$

So by this equation the stochiometric ratio of $C O_{2}$ $CO_2$ and $H_{2} O$ $H_2O$ produced on combustion is
$= \frac{x m o l}{\frac{y}{2} m o l} = \frac{44 x g}{\frac{18}{2} y g} = \frac{44 x}{9 y}$ $=(xmol)/(y/2 mol)=(44xg)/(18/2yg)=(44x)/(9y)$
$Where 44 \frac{g}{mol} is the molar mass of C O_{2} and 18 \frac{g}{mol} is the molar mass of H_{2} O$ $"Where " 44g/"mol"" is the molar mass of "CO_2 and 18g/"mol" " is the molar mass of "H_2O$

But by the given data this ratio is

$\frac{321 m g}{64 m g} = \frac{321}{64} \approx 5$ $(321mg)/(64mg)=321/64~~5$

So equating these two we get

$\frac{44 x}{9 y} = 5$ $(44x)/(9y)=5$

$\Rightarrow \frac{x}{y} = \frac{45}{44} \approx \frac{1}{1}$ $=>x/y=45/44~~1/1$

Since the ratio of number of atoms C and H in the HC molecule is $1 : 1$ $1:1$ then we can easily say the Empirical formula of HC is $C H$ $color(red)(CH)$

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