Question

Question #910ee

Answer 1

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`