Question #a4626

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Answer 1 · 2016-10-05T23:45:33

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

So 0.88g $CO_2$ will contain

$(0.88xx12)/44g=0.24g" "C$

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

So 0.36g $H_2O$ will contain

$(2xx0.36)/18g=0.04g" " H$

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

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

Hence 1 mole or 132g of the sample will contain

$C->(0.24xx132)/0.44g=72g=(72g)/(12g/"mol")=6" mol"$

$H->(0.04xx132)/0.44g=12g=(12g)/(1g/"mol")=12" mol"$

$O->(0.16xx132)/0.44g=48g=(48g)/(16g/"mol")=3" mol"$

Hence the molecular formula of the sample is- $" "C_6H_12O_3$