A compound is 54.53% C, 9.15% H, and 36.32% O by mass. What is its empirical formula?


#1

A compound is 54.53% C, 9.15% H, and 36.32% O by mass. What is its empirical formula?
The molecular mass of the compound is 132 amu. What is the molecular formula?

Answer:

First - you need the empirical formula.
So, assume you have 100 g of the compound.
If so, you’ll have 54.53 gram of C, 9.15 g of H and 36.32 g of O. Find the number of moles of each.

54.53 g C (1 mole C / 12.01 g C) = 4.540
9.15 g H (1 mole H / 1.008 g H) = 9.077
36.32 g O (1 mole O / 15.9994 g O) = 2.270

Take the smallest number found and divide the others by it to get the empirical formula.

4.540/2.270 = 2.
9.077/2.270 = 4.
2.270/2.270 =1.

So, that gives you the empirical formula of C2H4O.
Find the weight of this compound. C = 12, H = 1, O = 16. So, C2H4O is 44 amu.
132/44 = 3.
So, 3 (C2 H4 O) = C6H12O3 = molecular formula.