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0.262 g of a substance gave, on combusti...

0.262 g of a substance gave, on combustion 0.361 g of `CO_(2)` and 0.147 g of `H_(2)O`. What is the empirical formula of the substance

A

`CH_(2)O`

B

`C_(3)H_(6)O`

C

`C_(3)H_(6)O_(2)`

D

`C_(2)H_(6)O_(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
% of carbon
`=(12)/(44)xx("Wt. of "CO_(2))/("Wt. of organic compound")xx100`
`=(12)/(44)xx(0.361)/(0.262)xx100=37.57`
% of hydrogen
`=(2)/(18)xx("wt. of "H_(2)O)/("wt. of organic compound")xx100`
`=(2)/(18)xx(0.147)/(0.262)xx100=6.23`
% of oxygen = `100-(37.57+6.23)=56.2`
`C=(37.57)/(12)=3.13,H=(6.23)/(1)=6.23,O=(56.2)/(16)=3.51`
Empirical formula = `CH_(2)O`
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