Empirical and Molecular Formulas
Empirical Formulas:
Definition – the lowest, whole number, ratio of atoms in a molecule
Chemical Formulas – (1) tells the types of elements
(2) shows the number of atoms
(3) indicates the ratio of atoms or moles of atoms
In the past we have used the number of atoms of each element and their gram atomic masses to determine the mass of a compound in one mole. Now it is necessary to use the mass and gram atomic mass to determine a compound’s empirical formula.
For example, in the compound BaCl2 we know that the ratio of barium (Ba) to chlorine (Cl) is 1:2. This could be the ratio of atoms or the ratio of moles of atoms. Similarly, Na2SO4 would have a ratio of 2:1:4. From this, if you know the ratio of moles of atoms in a compound you could determine the empirical formula. Let’s say that a compound has a ratio of one mole of sodium to one mole of nitrogen to three moles of oxygen. Then the empirical formula would be NaNO3.
Practice problem:
1. Determine the empirical formula using the following ratios.
a. two bariums to three nitrogens ______
b. one aluminum to three acetates ______
c. one potassium to one half oxygen ______
In an experiment you will not be able to directly measure the moles of each element. Rather
you will probablly the measure the mass (grams), volume (liters) or atoms of each element. From there all that you need to do is convert any one of the above to moles using your chemistry road map and then fins the ratio of moles. For example, if you have a compound that contains 18.25 grams of sodium, 12.70 grams of sulfur, and 19.05 grams of oxygen, you can determine empirical formula by converting the grams to moles using gram atomic mass.
Na 18.25 g Na 1.0 mole Na = 0.7938 mole Na
22.99 g Na
S 12.70 g S 1.0 mole S = 0.3960 mole S
32.07 g S
O 19.05 g O 1.0 mole O = 1.191 mole O
16.00 g O
Now that you have found the number of moles of each element, divide each mole by the
smallest number of moles. This will give you the lowest whole number ratio.
0.7938 mole = 2.005
0.3960 mole ratio = 2:1 :3
0.3960 mole = 1.000 empirical formula = Na2SO4
0.3960 mole
1.191 mole = 3.008
0.3960 mole
Practice problem:
2. A 90.0 gram sample of a colorless gas is removed from a coal fired power plant and found to contain 45.0 grams of sulfur and 45.0 grams of oxygen (O). Determine the empirical formula for this gas.
Often instead of knowing the mass of each element in a compound you may know the percent by mass of each element. Even if you know the percent by mass you can still determine the empirical formula. The easiest way to do this is to assume that there are 100 grams of the compound. Then the percentages tell you the number of grams of each element. For example, a compound containing 20.23% aluminum and 79.77% chlorine by mass would have its empirical formula determined in the following manner.
Al – 20.23% = 20.23 g Al 1 mole = 0.7493 mole Al
26.98 g Al
Cl – 79.77% = 79.77 g Cl 1 mole = 2.250 moles Cl
35.45 g Cl
0.7493 mole = 1.000 Al
0.7493 mole ratio = 1:3
2.250 mole = 3.003 Cl empirical formula = AlCl3
0.7493 mole
Practice problem:
3. An organic compound is found to contain 69.73% carbon, 11.70% hydrogen, and 18.57% oxygen by mass. What is the empirical formula of this organic compound?
A final word about empirical formulas is that sometimes when you find the ratio of moles of each
element in the compound, one or two of the elements may not be whole numbers. In order to handle such a situation you need to get them all to whole numbers. This is not done by just rounding the numbers since you would not have the same ratio. Rather, find a number to multiply each element by so that they are whole numbers. For example, if you had a compound that contained phosphorus and oxygen and after calculating the ratio of moles you came up with 1:2.5 P:O. What would you multiply each number by to keep the same ratio and have whole number? Why you would multiply both by 2 of course. This would make the ratio 2:5 which is the same as 1:2.5 so that the empirical formula for this compound would be P2O5.
Practice problem:
4. A mineral is unearthed by a geologist and analyzed to find that it contains 1.13 grams of potassium,1.51 grams of chromium, and 1.62 grams of oxygen. Determine the empirical formula of the compound.
Molecular Formulas:
Definition: Shows the actual number of atoms in a chemical.
In our earlier discussion of chemical formulas you saw that the molecular formula can be reduced down to give the empirical formula for the compound. For example the molecular formula for benzene is C6H6 and its empirical formula is CH. This was done by dividing by six (6). One thing that can be seen is that the molecular formula is six times larger than the empirical formula for benzene. How many times bigger do you think the molar mass for benzene’s molecular formula is than its empirical formulas mass?
78.06 g/mol = 6
13.01 g/mol
As you can see it is six times bigger. This relationship between the molar mass of the empirical formula and molecular formula can be used to determine the molecular formula of a compound if you know the empirical formula and the molar mass of the substance.
For example, a compound is known to have a molar mass of 391.5 g/mol. The analysis of a 310.8 gram sample revealed that the sample contains only boron and iodine. The sample is found to contain 302.2 grams of iodine. The molecular formula can be found by first finding the empirical formula.
Mass of Boron = 310.8g – 302.2g = 8.6g
B 8.6g B 1.00 mole B = 0.80 mole B
10.81g B
I 302.2g I 1.00 mole I = 2.381 mole I
126.90g I
0.80 mole 2.381 mole ratio = 1:3
------= 1 B ------= 3.0 I
0.80 mole 0.80 mole empirical formula = BI3
To determine the molecular formula you now determine how many times bigger the molar mass is than the empirical formulas gram formula mass.
BI3 1(10.81) + 3(126.90) = 391.51 g/mole
Molar mass 391.5 g/mol
------= ------= 1.00
Empirical formula mass 391.51 g/mol
Molecular formula = 1(BI3) = BI3
In this case the molecular formula and the empirical formula are the same. But, as you saw with benzene this is not always the case.
Practice Problem:
5. Tooth enamel, or hydroxyapatite, has a molar mass of 1004.64 g/mole. Its composition is 39.89% calcium, 18.50% phosphorous, 41.41% oxygen, and 0.20% hydrogen by mass. Determine the molecular formula for hydroxyapatite.
6. During an experiment 105.33 mg of lysine, an essential amino acid, is burned with an excess amount of oxygen gas. Lysine is known to contain carbon, hydrogen, nitrogen, and oxygen atoms. When the lysine is burned the reaction produces 190.22 mg of carbon dioxide gas, 90.90 mg of water vapor, and 20.18 mg of nitrogen gas. What is the molecular formula of lysine?