Question #910ee

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Question #910ee

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Answer 1 · 2016-10-04T02:08:15

Given 0.44g of sample on combustion produces 0.88g $C O_{2}$ $CO_2$
We know molar mass of $C O_{2} = 44 g/mol$ $CO_2=44" g/mol"$ which contains 12g carbon.

So 0.88g $C O_{2}$ $CO_2$ will contain

$\frac{0.88 \times 12}{44} g = 0.24 g C$ $(0.88xx12)/44g=0.24g" "C$

Similarly molar mass of $H_{2} O = 18 g/mol$ $H_2O=18" g/mol"$ which contains 2g hydrogen.

So 0.36g $H_{2} O$ $H_2O$ will contain

$\frac{2 \times 0.36}{18} g = 0.04 g H$ $(2xx0.36)/18g=0.04g" " H$

So the remaing amount in 0.88g of sample $(0.88 - 0.24 - 0.04) g = 0.16 g$ $(0.88-0.24-0.04)g=0.16g$ must be Oxygen.

Given molar mass of sample $132 g/mol$ $132" g/mol"$

Hence 1 mole or 132g of the sample will contain

$C \to \frac{0.24 \times 132}{0.44} g = 72 g = \frac{72 g}{12 \frac{g}{mol}} = 6 mol$ $C->(0.24xx132)/0.44g=72g=(72g)/(12g/"mol")=6" mol"$

$H \to \frac{0.04 \times 132}{0.44} g = 12 g = \frac{12 g}{1 \frac{g}{mol}} = 12 mol$ $H->(0.04xx132)/0.44g=12g=(12g)/(1g/"mol")=12" mol"$

$O \to \frac{0.16 \times 132}{0.44} g = 48 g = \frac{48 g}{16 \frac{g}{mol}} = 3 mol$ $O->(0.16xx132)/0.44g=48g=(48g)/(16g/"mol")=3" mol"$

Hence the molecular formula of the sample is- $C_{6} H_{12} O_{3}$ $" "C_6H_12O_3$

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Question #910ee

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