Question

A compound contains 13.13%of hydrogen 52.14%of carbon and 34.73%of oxygen It's molar mass is 46.66 what are its empirical and molecular formula
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Answer 1

Answer:

$C_{2}H_{6}O$ ethanol or dimethyl ether

Explanation:

Given that,

H=13.13%  

C= 52.14%

O= 34.73%

No.of moles of H= 13.13/1= 13.13  

No. of moles of C= 52.14/12= 4.345

No.of moles of O= 34.73/16= 2.170

Therefore,the ratio of the atoms of H,C and O

= 13.13 : 4.345 : 2.170 = 2 : 6 : 1

So,the empirical formula is $C_{2}H_{6}O$

Let the Molecular formula is ()n

Then, molar mass = (2×12+1×6+16)n

= 46n

As the question, 46n = 46.07  or,n= 1

So,the molecular formula is ($C_{2}H_{6}O$)1 = $C_{2}H_{6}O$ i.e.the compound is either ethanol or dimethyl ether

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