Question

A compound found to possess C=40%H=6.7%0=53.3% its molecular mass is 60 find the molecular formula of the compound.

Answer 1

C ==> 40/12 = 3.33 moles C
H ==> 6.7/1= 6.7 moles H
O ==> 53.3/16 = 3.33 moles O

Now divide through the least no. of moles to get simple integral ratio
C = >3.33/3.33 = 1
H = >6.7/3.33 = 2
O =>3.33/3.33 = 1
Empirical formula is CH2O
The molar mass of this EF is 12 + 2 + 16 = 30 g/mol
The molar mass of molecule 60 g/mol.

That means, there are two EFs in one MF. 
Or simply the Molecular formula = 2 x CH2O = C2H4O2

Answer 2

Answer:

Explanation:

Secondary SchoolChemistry 8+4 pts

A compound found to possess C=40%H=6.7%0=53.3% its molecular mass is 60 find the molecular formula of the compound.

Report by Indelixmeghaishera 29.07.2016

Answers

anandawasthi2002

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pratichi

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C ==> 40/12 = 3.33 moles C

H ==> 6.7/1= 6.7 moles H

O ==> 53.3/16 = 3.33 moles O

Now divide through the least no. of moles to get simple integral ratio

C = >3.33/3.33 = 1

H = >6.7/3.33 = 2

O =>3.33/3.33 = 1

Empirical formula is CH2O

The molar mass of this EF is 12 + 2 + 16 = 30 g/mol

The molar mass of molecule 60 g/mol.

That means, there are two EFs in one MF.

Or simply the Molecular formula = 2 x CH2O = C2H4O2