Question

Cumene, a hydrocarbon, is 89.94% carbon, and its molar mass is 120.2 g/mol. What are the a) empirical and b) molecular formulas for Cumene? Show your work

Answer 1

a)
we have mass of each elements as:
C: 89.94 g
H: 100 - 89.94 = 10.06 g


Divide by molar mass to get number of moles of each:
C: 89.94/12.01 = 7.489
H: 10.06/1.008 = 9.98


Divide by smallest:
C: 7.489/7.489 = 1
H: 9.98/7.489 = 1.33 = 4/3

Multiply by 3 to get simplest whole number ratio:
C: 1*3 = 3
H: 4/3*4 = 4
So empirical formula is:C3H4
Answer: C3H4

b)

Molar mass of C3H4,
MM = 3*MM(C) + 4*MM(H)
= 3*12.01 + 4*1.008
= 40.062 g/mol

Now we have:
Molar mass = 120.2 g/mol
Empirical formula mass = 40.062 g/mol
Multiplying factor = molar mass / empirical formula mass
= 120.2/40.062
= 3

So molecular formula is:C9H12
Answer: C9H12