Acompound 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 ?

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

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