Determine the empirical and molecular formula for a compound made from 54.09% calcium, 2.72% hydrogen, and 43.18% oxygen with a total molar mass of 222.3 g.

Answer 1

Here is how you do a problem such as this.
Take 100 g sample. That gives you
54.09 g Ca
2.72 g H
43.18 g O.
Convert grams to moles.
54.09/40.1 = about 1.35 moles Ca.
2.72/1 = 2.72 moles H.
43.18/16 = about 2.7 moles O.

Now divide by the smallest number.
1.35/1.25 = 1.00 moles Ca
2.72/1.25 = 2.01 moles H = 2.00 (rounding to whole number)
2.7/1.35 = 2.00 moles O
Empirical formula is CaO2H2
Empirical formula mass is (40.1 + 2*1 + 2*16) = 74
To find the molecular formula, we divide the molar mass by the empirical mass or 222.3/74 = 3.0 (and round to whole number BUT this one comes out whole.). So the molecular formula is
(CaO2H2)₃ or we can write it as
Ca3O6H6 which doesn't make any sense to me but this probably is just a problem someone made up.