Question

A particular compound contains, by mass, 41.4 % carbon, 3.47 % hydrogen, and 55.1 % oxygen. A 0.150-mol sample of this compound weighs 17.4 g. The molecular formula of this compound is Select one:

a. C3H3O3

b. C3H3O

c. CHO

d. C4H4O4

e. C5H5O5

I came up with C2H2O2 so I am not sure where I went wrong.  

Answer 1

step 1: find the empirical formula:
we have mass of each elements as:
C: 41.4 g
H: 3.47 g
O: 55.1 g


Divide by molar mass to get number of moles of each:
C: 41.4/12.01 = 3.4471
H: 3.47/1.008 = 3.4425
O: 55.1/16.0 = 3.4438


Divide by smallest to get simplest whole number ratio:
C: 3.4471/3.4425 = 1
H: 3.4425/3.4425 = 1
O: 3.4438/3.4425 = 1


So empirical formula is:
CHO

step 2: find the molar mass
use:
number of mol = mass / molar mass
0.150 = 17.4/MM
MM = 116 g/mol

step 3: find the molecular formula

Molar mass of CHO,
MM = 1*MM(C) + 1*MM(H) + 1*MM(O)
= 1*12.01 + 1*1.008 + 1*16.0
= 29.018 g/mol

Now we have:
Molar mass = 116.0 g/mol
Empirical formula mass = 29.018 g/mol
Multiplying factor = molar mass / empirical formula mass
= 116.0/29.018
= 4

Hence the molecular formula is : C4H4O4

Answer: d