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Ledermann Unit 7: Moles, Empirical & Molecular Formulas
I. Empirical Formulas
A. The lowest whole-number ratio of atoms or moles of atoms for each element in a molecule
B. This may or may not be the same as molecular formula
If the molecular formula is C6H12O6, what is the empirical formula?
If the molecular formula is CH4, what is the empirical formula?
C. Calculating empirical formulas:
1. Find the mass of each element. If the percent composition is given, assume you have a 100-
gram sample. The percent then becomes your grams of each element.
2. Divide the mass of each element by its atomic mass to find the number of moles.
3. The number of moles of each element is then divided by the smallest number of moles to get the
ratio of the moles in the compound.
4. The ratios (whole numbers) are used as subscripts in writing the formula.
D. Example: A compound is 71.4% Calcium and 28.6% Oxygen. What is the empirical formula?
1. Assume a 100.0g sample. That means you have 71.5g Ca and 28.5g of O.
2. Find the number of moles.
71.5g Ca x 1 moles = 1.78 moles 28.5g O x 1 mole = 1.78
40.1g 16.0g
1.78 is the smallest number of moles, so divide each by 1.78. 1.78/1.78 = 1.00
The ratio is 1:1 so the subscripts are each 1 and the empirical formula is CaO.
V. Molecular Formulas
A. The actual whole-number ratio of the atoms is a compound.
B. It is the same as empirical formula or a whole-number ratio of it. (e.g. Double or triple all subscripts.)
C. Calculating a molecular formula:
1. Calculate the empirical formula. (It may be given to you.)
2. Find the formula mass of the empirical formula.
3. Divide the given molecular weight (formula mass) by the mass of the empirical formula.
4. Multiply the subscripts of the empirical formula by the answer in #3.
D. Example: If the empirical formula is C3H4O3 and the molecular weight is 176g, what is the molecular
formula? C: 3 x 12.0 = 36.0
H: 4 x 1.01 = 4.04
O: 3 x 16.0 = 48.0
88.0 grams
molecular weight = 176 = 2 so double the subscripts and the molecular formula is C6H8O6
mass of emp. formula 88
E. Example: A compound is 72.4% Fe and 27.6% O. Its molecular weight is 232. Find its empirical and
molecular formulas.
1. Assume a 100g sample. 72.4g Fe and 27.6g O.
2. 72.4g Fe x 1 mole = 1.29 moles 1.29 = 1 1:1.33 ratio, but you need whole
55.8g 1.29 numbers. Doubling gives you
2:2.66 (no help)
27.6g O x 1 mole = 1.73 1.73 = 1.33 If you triple it, you get 3:3.999
16.0g 1.29 which is 3:4.
3. Fe3O4 is the empirical formula. The formula mass of Fe3O4 is 232g.
4. Molecular weight = 232 = 1 Since it is equal to 1, the molecular formula
Empirical weight 232 and the empirical formula are the same!
