Question

A carbon compound on analysis gave the following percentage composition 14.5 hydrogen and chlorine 64.46 and oxygen 19.2 find the empirical formula of the compound​

Answer 1

Answer:

the empirical formula of the compound H₂₄Cl₃O₂

Explanation:

Given

H = 14.5 %

Cl = 64.46 %

O = 19.2 %

empirical formula = ??

Solution

first divide each with respective atomic mass

H                        Cl                        O

14.5/1            64.46/35.5            19.2/16

14.5                     1.81                      1.2

divided by with lowest number

14.5/1.2               1.81/1.2                 1.2/1.2

12.0                        1.5                        1

multiply with suitable number to convert them into whole number

12.0 X 2            1.5  X 2                  1  X 2  

24                           3                          2

the empirical formula of the compound

H₂₄Cl₃O₂