Question

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 1

 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.