Question

A compound contains 72.62 g carbon , 9.885 g hydrogen , and 69.50 g chlorine . If the molar mass is 153 g/mol , what is the molecular formula ?

Answer 1

carbon=72.62/12=6.05
Hydrogen:9.885/1=9.885
chlorine=69.50/35.4=1.96

6.05/1.96=3.09 , 9.885/1.96=5.04, 1.96/1.96=1

C3H5CL1
(3+5+1)N=153
9n=153
N=153/9
N=17
C=3x17
H=5x17
Cl=1x17
C51H85CL17

Answer 2

C₆H₁₀Cl₂

Step-by-step explanation:

Step 1. Calculate the empirical formula

a. Calculate the moles of each element

Moles of C = 72.62 g C × (1 mol C/12.01 g C) = 6.0466 mol C

Moles of H = 9.885 g H × (1 mol H/1.008 g H) = 9.8065 mol H

Moles of Cl = 69.50 g Cl × (1 mol Cl/35.45 g Cl) = 1.9605 mol O

b. Calculate the molar ratio of each element

Divide each number by the smallest number of moles and round off to an integer

C:H:Cl = 6.0466:9.8065:1.9605

C:H:Cl = 3.084:5.002:1

C:H:Cl ≈ 3:5:1

c. Write the empirical formula

EF = C₃H₅Cl

Step 2. Calculate the molecular formula

EF Mass = 76.52 u

MF mass = 153 u

MF = (EF)_n

n = MF Mass/EF Mass

n = 153 u/76.52 u

n = 1.999 ≈ 2

MF = (C₃H₅Cl)₂

MF = C₆H₁₀Cl₂