Empirical formulas
Definitions & Key takeaways
An empirical formula is a specific type of chemical formula that specifies the ratios of the atoms of each element in a compound. It is obtained by measuring the relative masses of the atoms present in a molecule and dividing them by the number of atoms of each element. The subscripts in an empirical formula indicate the number of atoms of each element in a molecule, not their relative masses.
For example, if carbon dioxide is analyzed and found to have one carbon atom for every two oxygen atoms, then the empirical formula for carbon dioxide would be CO2 . If it were found to have one carbon atom for every four oxygen atoms, then the empirical formula would be C2O4. In both cases, there are twice as many oxygen atoms as carbon atoms.
An empirical formula is a chemical formula that gives the smallest whole number ratio of atoms in a compound. As an example, let’s take the molecule ethene, which has two carbons and four hydrogens, so the molecular formula is C2H4.
But notice that both subscripts are divisible by two. So to figure out the empirical formula for ethene, we have to find the smallest whole-number ratio that describes the relative number of different atoms.
So we divide both the two and the four by two, to get the empirical formula CH2. If we want to get the molecular formula from the empirical formula, we need to do the opposite of this: we would multiply all of the subscripts by two.
This is because, for ethene, the empirical formula is different than the molecular formula. For an unknown compound, we would need to determine the common factor of the subscripts (two in the case of ethene) using an experimental technique such as mass spectrometry.Sometimes the empirical formula and the molecular formula are the same.
A common example is water, which is H2O. That means that there are two hydrogens for every one oxygen.
Now, let’s suppose we have a mystery compound that is 84.1% carbon and 15.9% hydrogen and we want to figure out the empirical formula for our mystery compound.A useful trick is to assume that we start with exactly 100 grams of the mystery compound.
That means we have 84.1 grams of carbon and 15.9 grams of hydrogen. Our first step is to convert the masses into moles.
We can look up the molar mass for carbon and hydrogen on the Periodic Table. The molar mass of carbon is 12.0 grams per mole and the molar mass of hydrogen is 1.01 grams per mole.We take 84.1 grams of carbon and divide it by 12.0 grams per mole.
The grams cancel out of the division, and after we divide the numbers we are left with around 7 moles of carbon. We next take 15.9 grams of hydrogen, and divide this number by by 1.01 grams per mole.
This tells us that we have 15.7 moles of hydrogen.We now have the relative number of moles of carbon and hydrogen in our unknown sample.
We next try to divide all of of our numbers by the smaller number, in order to figure out the ratio of these two elements in our sample.
So we divide 7 moles of carbon by 7 moles to get 1.0, and we divide 15.7 moles of hydrogen by 7 moles to get 2.24.At this point, we might be tempted to write that the empirical formula of our unknown compound is C1H2.24.
But the empirical formula has to have whole numbers as the subscripts. So we need to figure out the smallest whole number to multiply all of our subscripts by, so that all of them are whole numbers.
2.24 times two is 4.48, which is not a whole number. 2.24 times three is 6.72, which is still not very close to a whole number.
But 2.24 times four is 8.96, which is extremely close to a whole number - 9. So we multiply both of our subscripts by four, in order to produce the empirical formula C4H9, which is the smallest whole-number ratio of different elements in our sample.Now let’s say we use a mass spectrometer which tells us the molar mass of the mystery compound is 114 grams per mole.
Next, we need to use the empirical formula to find the molecular formula.We start by figuring out the total molar mass of our empirical formula.
There are four carbons, each with a mass of 12 grams per mole, and 9 hydrogens, each with a mass of 1 gram per mole. So the molar mass of our empirical formula is 4 times 12 grams per mole, plus 9 times 1 grams, which is 57 grams per mole.If we take the molar mass reported by the mass spectrometer and divide it by the total molar mass of our empirical formula, that’s 114 grams per mole divided by 57 grams per mole, which equals two.
This means that to get the molecular formula, we should multiply every subscript by two, which gives us the molecular formula C8H18.
This molecular formula tells us the true structure of the molecule. C8H18 is a molecule called octane.
Drawing out the eight carbon atoms and eighteen hydrogen atoms, we can see how these different atoms covalently bond together to form a molecule.Now, it’s also possible to do the reverse - to start with the empirical formula and then calculate the percent composition.
As an example, let’s calculate the percent of chlorine by mass in three separate compounds. The first compound is sodium chloride.Whenever we go from an empirical formula to percent composition, we should assume that we have one mole of our compound.
So the mass of one mole of sodium chloride is 23 grams plus 35.5 grams, or 58.5 grams. To figure out the percent of chlorine by mass, we can just divide the 35.5 grams of chlorine by our 58.5 grams of sodium chloride and then multiply by 100 to get 60.7% chlorine by mass.
Doing the same calculation for sodium tells us that sodium chloride is 39.3% sodium by mass. Or, since the total should add to 100%, we could find the 39.3% by subtracting 60.7 from 100%.
Let's do the same calculation for potassium chloride. One mole of potassium chloride gives us one mole of potassium cations and one mole of chloride anions.
Potassium has a molar mass of 39.1 grams per mole, and chlorine has a molar mass of 35.5 grams per mole. Adding these masses together, the total mass of one mole of potassium chloride is 74.6 grams.
So once again, we can find the percent composition by dividing each constituent element’s mass by the total mass. So 35.5 divided by 74.6 gives us 0.476.
We then multiply by 100 to get a percent composition of 47.6% chlorine. Doing the same calculation for potassium tells us that potassium chloride is 52.4% potassium by mass.47.6% is much lower than the percent chlorine composition in sodium chloride, which was 60.7%.
Now, if we divide the mass of chlorine by the total mass, which is 71.0 grams divided by 95.3, then we get 0.745. Multiply by 100 to get a percentage.
##SummaryAlright, as a quick recap...empirical formulas tell us the smallest whole-number ratio of different types of atoms or ions in a compound.
For covalent compounds, empirical formulas may or may not be identical to the molecular formula, depending on the actual structure of the molecule.
Empirical formulas are helpful in figuring out the chemical structure of unknown compounds when given analysis results showing the percentage composition of the compound.
We can also use empirical formulas to compare the mass percentages of elements in different compounds.
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