Question

A compound is 52.14% C, 13.13% H, and 34.73% O. What is the empirical formula of the compound?

Answer 1

% H = 100 - ( 52.14 + 34.73 )=13.13 % 

assume 100 g of this compound 
mass H = 13.13 g 
moles H = 13.13 g / 1.008 g/mol=13 

mass C = 52.14 g 
moles C = 52.14 g/ / 12.011 g/mol=4 

mass O = 34.73 g 
moles O = 34.73 g/ 15.999 g/mol=2 

the empirical formula is C4H13O2

Answer 2

Answer: The empirical formula is $C_2H_6O$.

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 52.14 g

Mass of H = 13.13 g

Mass of O = 34.73 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{52.14g}{12g/mole}=4.35moles$

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{13.13g}{1g/mole}=13.13moles$

Moles of O =$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{34.73g}{16g/mole}=2.17moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =$\frac{4.35}{2.17}=2$

For H =$\frac{13.13}{2.17}=6$

For O =$\frac{2.17}{2.17}=1$

The ratio of  C: H: O = 2: 6: 1

Hence the empirical formula is $C_2H_6O$.