Question

a compound contains 6.7% hydrogen, 39.9% carbon and rest is oxygen. its molar mass is 60 what are its empirical and molecular formula​

Answer 1

Explanation:

In this question first of all we have to find the percentage of oxygen which can be easily found by simply subtracting percentage of hydrogen & carbon from 100 and then the atomic mass of these elements are taken & the % is divided by their atomic masses & then the least answer is considered & by which all the results are divided & it is rounded off to get integers which gives empirical formula & then the mass of the empirical formula is calculated and is divided by molecular mass which is the value of N & value of N is multiplied to the empirical formula to get answer as acetic acid

Answer 2

The empirical and molecular formula is CH2O and CH3COOH respectively.

Given:

A compound contains 6.7% hydrogen, 39.9% carbon and the rest is oxygen and its molar mass is 60.

To Find:

The empirical and molecular formula.

Solution:

To find the empirical and molecular formula we will follow the following steps:

As we know,

The molecular weight of carbon, oxygen and hydrogen is 12,16 and 1 respectively.

Now,

$number \: of \: moles \: of \: carbon = \frac{39.9}{12} = 3.32$

$number \: of \: moles \: of \: hydrogen \: = \frac{6.7}{1} = 6.7 \:$

Oxygen percentage = 100-39.9-6.7 = 53.4

$number \: of \: moles \: of \: oxygen = \frac{53.4}{16} = 3.34$

Diving the number of moles with the smallest containing moles we get carbons, oxygen and hydrogen in the empirical formula.

$Number of carbon = \frac{3.34}{3.34} = 1$

$Number of hydrogen = \frac{6.7}{3.34} = 2$

$Number of oxygen = \frac{3.34}{3.34} = 1$

So, the empirical formula = CH2O

The molecular mass of CH2O = 12+2+16 = 30 grams.

Also,

$multiplication \: factor \: for \: molecular \: formula \: = \frac{molecular \: mass}{empirical \: mass} \: = \frac{60}{30} = 2$

Molecular formula = 2 × empirical formula = 2 × CH2O = C2H4O2 = CH3COOH.

Henceforth, the empirical and molecular formula is CH2O and CH3COOH respectively.