Compound of D = 4.76g
CO2 = 10.46g
H2O = 5.71g
Empirical formula = ?
First, you will find the grams of C and H by:
10.46g of CO2 = 1 mol of CO2 / 44.01 g/mol = 0.2376 mol of C x 12.01 g/mol = 2.8525g of C
5.71g of H2O = (1 mol of H2O / 18.02 g/mol) x 2 = 0.63374 mol of H x 1.01 g.mol = 0.64g of H
Then, subtract grams of C and H by the compound to find the O:
4.76 g of compound - 2.8535 g of C - 0.64 g of H = 1.2665 g of O
Find the mol of O by using the formula: mol = mass / Molar mass
1.2665 g of O / 16.00 g/mol = 0.079 mol of O
Finally, take the smallest number of those moles and divide for each.
C = 0.2376 / 0.079 = 3
H = 0.63374 . 0.079 = 8
O = 0.079 / 0.079 = 1
The Empirical formula is C3H8O
Hope this help!
