Question

an 11.5g sample of ethonal 6.00g of Carbon 1.51g of hydrogen what is the percent composition of each element

Answer 1

C : 52.2%, H : 13.1%, O:34.7%

Further explanation  

The empirical formula is the smallest comparison of atoms of compound forming elements.  

A molecular formula is a formula that shows the number of atomic elements that make up a compound.  

(empirical formula) n = molecular formula  

11.5 g of Ethanol-C₂H₅OH contains 6.00 g of Carbon and 1.51 g of Hydrogen.

mass of Oxygen :

$\tt 11.5-(6+1.51)=4~g$

Percent composition :

• C

$\tt \dfrac{6}{11.5}\times 100\%=52.2\%$

• H

$\tt \dfrac{1.51}{11.5}\times 100\%=13.1\%$

• O

$\tt \dfrac{4}{11.5}\times 100\5=34.7\%$