The compound chloral hydrate, known in detective stories as knockout drops, is composed of 14.52% C, 1.83% H, 64.30% Cl, and 19.35% O by mass and has a molar mass of 165.4 g/mol. Part A What is the empirical formula of this substance

C₂H₃Cl₃O₂

Explanation:

Given parameters:

Percentage composition:

14.52% C

1.83% H

64.30% Cl

19.35% O

Molar mass = 165.4 g/mo

Unknown:

Empirical formula = ?

Solution:

The empirical formula of a compound is the simplest formula of a compound. To solve this problem, let us follow the process below:

                                 C                     H                  Cl                  O

%composition        14.52                1.83              64.3              19.35

Molar mass               12                     1                   17                  16      

number of

moles                  14.52/12               1.83/1           64.3/3.5         19.35/16    

                               1.21                     1.83               1.81                1.21

Divide by

the smallest       1.21/1.21              1.83/1.21           1.81/1.21          1.21/1.21

                                1                        1.51                    1.51                 1

multiply

by 2                         2                          3                       3                   2

 Empirical formula   C₂H₃Cl₃O₂

i think its 100

i hope this !