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Catalytic Transfer Hydrogenation

180 g of a hy...

Question

180 g of a hydrocarbon reacts fully with excess oxygen and produces 264 g of $C O_{2}$ and 108 g of $H_{2} O$ . If the ratio of molar mass of the hydrocarbon to the empirical formula mass is 6 then find the molecular formula of given hydrocarbon.

C H_{2} O

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C_{6} H_{12} O_{6}

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C_{12} H_{24} O_{12}

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C_{42} H_{84} O_{42}

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Solution

The correct option is B $C_{6} H_{12} O_{6}$
Mass of carbon produced $= \frac{12}{44} \times 264 = 72 g$
Number of moles of carbon $= \frac{72}{12} = 6 m o l$
Mass of hydrogen produced: $= \frac{2}{18} \times 108 = 12 g$
Number of moles of hydrogen $(H)$ $= \frac{12}{1} = 12$
Mass of oxygen in 180 g of sample $180 - 72 - 12 = 96 g$
Number of moles of oxygen atoms $= \frac{96}{16} = 6 m o l$
So, ratio of moles in 180 g of sample:
$n_{C} : n_{H} : n_{O} = 1 : 2 : 1$
Emprical formula of the given hydrocaron $\to C H_{2} O$
Molecular formula of the given hydrocarbon: $\to n \times e m p r i c a l f o r m u l a$
Where, n is ratio of the molar mass to the empirical formula mass which is 6, molecular formula of the given hydrocarbon is $C_{6} H_{12} O_{6}$ .

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Catalytic Transfer Hydrogenation

Standard XII Chemistry

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180 g of a hydrocarbon reacts fully with excess oxygen and produces 264 g of CO2 and 108 g of H2O. If the ratio of molar mass of the hydrocarbon to the empirical formula mass is 6 then find the molecular formula of given hydrocarbon.

180 g of a hydrocarbon reacts fully with excess oxygen and produces 264 g of $C O_{2}$ and 108 g of $H_{2} O$ . If the ratio of molar mass of the hydrocarbon to the empirical formula mass is 6 then find the molecular formula of given hydrocarbon.