Question

A compound found in the athletes urine had a percent composition of 80.8% carbon, 8.97% hydrogen and 10.3% oxygen. What is the Empirical formula of the compound? What is the molecular formula of the compound if the molar mass of the compound is 312g/mol?​

Answer 1

C21H2802

Step-by-step explanation:

C=12g/mol

H=1g/mol

O=16g/mol

Part (C) of compound-80.18%

(0.8018 x 312)/12=21

Part (H) of compound-8.97%

(0.897 x 312)/1=28

Part (O) of compund-10.3%

(0.103 x 312)/16 = 2

Therefore the emp. formula is C12H28O2