posted by ABCD .
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.
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.