Question #c839f

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Answer 1 · 2017-01-26T17:27:04

and 0.9397g of water. Determine the molecular formula for beta carotene. The molar mass is 536.96

Explanation:

Let the molecular formula of Beta carotene be $C_{x} H_{y}$ $C_xH_y$ and the balanced equation of its combustion in excess oxygen is as follows.

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

It is given that $1 g$ $1g$ beta carotene produces $3.278 g C O_{2}$ $3.278g" "CO_2$ and $0.9397 g H_{2} O$ $0.9397g" "H_2O$

Molar mass of BC is $536.96 g /mol$ $536.96g"/mol"$

So $1 m o l or 536.96 g$ $1" "mol or 536.96g$ BC will produce $3.278 \times 536.96 \approx 1760.15 g C O_{2}$ $3.278xx536.96~~1760.15g" "CO_2$ and $0.9397 \times 536.96 \approx 504.58 g H_{2} O$ $0.9397xx536.96~~504.58g" "H_2O$

Molar mass of $C O_{2} = 44 g /mol$ $CO_2=44g"/mol"$

So number of moles of $C O_{2}$ $CO_2$ produced $= \frac{1760.15 g}{44 g /mol} \approx 40 m o l$ $=(1760.15g)/(44g"/mol")~~40mol$

Hence number of C-atoms in one molecule of BC is $= 40$ $=40$ as one molecule of $C O_{2}$ $CO_2$ contains one atom of Carbon.

Again Molar mass of $H_{2} I = 18 g /mol$ $H_2I=18g"/mol"$

So number of moles of $H_{2} O$ $H_2O$ produced $= \frac{504.58 g}{18 g /mol} \approx 28 m o l$ $=(504.58g)/(18g"/mol")~~28mol$

So one mole BC produces $28 m o l$ $28mol$ water which contain $56$ $56$ mol atom Hydrogen as one molecule of water contains two atoms of Hydrogen.Hence number of Hydrogen atom present in one molecule of BC is $56$ $56$ .

Now we can say that molecular formula of BC is $C_{40} H_{56}$ $C_40H_56$

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Question #c839f

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