Chem 1: Chapter 7: Chemical Formulas and Chemical Compounds
Part 2
Determining an empirical formula
If the composition of each part of a compound is known, the simplest formula (called
empirical formula) can be found. Remember that subscripts must be whole numbers.
Simplest ratio between the atoms in a compound: (whole numbers)
Stress that mole ratios are indicated by the subscripts since 1 mole = NA of atoms
Composition of a compound could be in % of each element in the compound or the grams
of each element in the compound. Remember that the % sum will be 100 %.
A compound is composed of 78.1 % Boron and 21.9 % hydrogen. What is the empirical
formula?
Because atoms in a compound are always in a small whole number ratio, convert each
percent to moles. Convert the % to g by using a 100 g sample of the compound. Use
your formula of g of the atom x 1 mole/ mole mass of the atom = # of moles.
General format
78.1 % B = 78.1 g B x 1 mole B/10.81 g B = 7.22 moles of B
21.9 % H = 21.9 g of H x 1 mole H/1.01 g H = 21.9 moles of H
This is not a whole number ratio so the next step is to divide each mole value by the
smallest # of moles. This will give a whole # ratio.
7.22 moles of B / 7.22 moles = 1.00 mole B
21.7 moles of H / 7.22 moles = 3.01 moles of H
The formula is BH3
Due to rounding off, slight liberties are taken with the ratio.
If gram masses are given, the same process is needed.
A compound of P and O has a mass of 10.15 g. The phosphorus content is 4.43 g. What
is the EF?
( 10.15 g of compound – 4.43 g of P will yield 5.72 of O)
Convert g to moles
4.43 g P x 1 mole P/30.97 g of P = 0.143 moles of P
5.72 g O x 1 mole O/ 16.00 g of O = 0.357 moles of O
Mole ratio will be 0.143 moles of P / 0.142 moles = 1 mole of P
0.357 moles of O / 0.143 moles = 2.50 moles of O
This is not a whole number ratio. Both mole values can be X 2 and will yield a whole
number ration of 2/5. The EF is P205
Example:
77.7% Fe and 22.3 % oxygen. Determine the EF
77.7 g of Fe x 1 mole Fe/55.8 g = 1.39 moles Fe
22.3 g of O x 1 mole O/16.0 g = 1.39 moles O
The number of moles in each is equal so the ratio is 1-1
FeO (Iron II oxide)
Example 2
3.50 g of Fe combine with 1.50 g of O to form a compound. What is the EF?
3.50 g of Fe x 1 mole Fe/55.8 g = 0.0627 moles Fe
1.50 g of O x 1 mole O/16.0 g = 0.0938
Divide by the smallest number of moles (0.0627 moles)
Ratio comes out to be 1 Fe:1.5 )
If the ratios are not whole numbers, determine the multiplier needed and convert to whole
numbers.
Multiplier of 2 so the ratio is 2Fe : 3 O
Fe2O3 (Iron III oxide)
If the compound is a hydrate: Keep H20 together as a unit
A compound is composed of the following atoms and molecules. Determine the
empirical formula
Na 16.1 %
C 4.2 %
O 16.8 %
H2O 62.9 %
Convert to moles, find the mole ratio
It will be 2-1-3-10 mole ratio to yield the following formula
Na2CO3 ● 10 H2O A raised dot is placed between the compound and the moles of water.