Calculations With Moles and Formulas

There are a number of calculations (other than mole-gram conversions) that are associated with moles and molecular masses. We will look at several of them in this section. The first deals with determining the molecular mass of a chemical when you do not know its molecular formula. That is what you will be doing in your lab work for this lesson. The next two deal with formulas and composition. One considers determining the percent composition from the formula or from the masses of each element in the sample and the other considers determining the empirical formula from the composition. The last deals with using the composition and/or empirical formula along with the molecular mass to determine the molecular formula.

Molar Mass By Experiment

As you have seen, it is relatively straightforward to find the formula mass of a substance if you know its formula; you learned how to do this in the previous section, by adding up the atomic masses of all the atoms in the formula. This value then can give you the molar mass by simply changing the units from amu to g/mol.

In some cases, it is also possible to calculate the molar mass of a substance without knowing the formula using data gathered by experiment. The core idea here is this: because of the Law of Constant Composition, the molar mass for a substance holds true for any sample of that substance. The molar mass of CO₂, for example, is 44.01 g/mol whether you have a tiny sample of CO₂ or a huge one. This means that for any sample of a substance, if you are able to determine both the mass of the sample and the number of moles in it, dividing the mass value by the moles value will give the molar mass.

It is usually easy to determine the mass of a sample: you just put it on a balance. However, it can be more difficult to determine the number of moles. After all, you don't know the formula of the substance. In this week's lab, you will get around this issue by using Avogadro's law and measuring the masses of several gaseous substances. Details on Avogadro's law are in your lab workbook.

Once again, the core idea here is this: dividing the mass of a sample by the number of moles in that sample always gives the molar mass.

Percent Composition From Formulas or Masses

Atoms explain why compounds have fixed composition. It's because the atoms combine with one another in specific fixed ratios. Formulas give us the composition of the compounds in terms of the ratio of atoms. In addition, atomic masses give us the relative masses of the atoms. We can combine all that information and use it to calculate the composition in mass percent for a compound, starting with the formula.

Percent Composition From Formulas

Let's start with a simple case, KF, the formula for potassium fluoride. We want to know what percent of the mass of KF is made up of potassium, and what percent is made up of fluorine.

The first step is to find the atomic masses for potassium and fluorine, which are 39.1 grams for potassium and 19.0 grams for fluorine. Next, you have to figure the formula mass for the compound KF by adding the masses together, and that is 58.1 g.

Now let's take a look at what that really represents. One mole of KF contains one mole of potassium and one mole of fluorine. That means 58.1 grams of KF contains 39.1 grams of potassium and 19.0 grams of fluorine. So we have the individual masses of the elements that are contained in the compound and the total mass of one mole of that compound. From there you calculate the percentage composition the same way you calculate any percentage. Take the mass of the element and divide by the total mass of the compound and convert it into percent by multiplying by 100. The percentage of potassium in this compound is calculated out to be 67.3% and the percent of fluorine is calculated to be 32.7%.

As you can see, the idea here is to take the mass of a single element and divide it by the mass of all the elements in the compound. In this case, the method is to get those masses from atomic masses on the periodic table.

As a second example, we can determine the percent composition of a compound with subscripts, like Li₂O. The mass of two lithium atoms is 2*6.941 amu = 13.882 amu, and the mass of one oxygen atom is 15.999 amu. Note that this time we are using amu for the masses, where above we used grams. There is no issue using either unit here as long as you are consistent.

Adding these masses together gives a formula mass of 29.881 amu. To determine the percent lithium we divide 13.882/29.881 and multiply by 100% to get a value we will round to 46.46% Li. The corresponding calculation for oxygen gives 53.54% O.

Percent Composition From Lab Data

In the above examples, we got our element masses and formula mass from the periodic table and used them to find the percent composition of compounds. However, if you do not know the formula of a compound, you can still find its percent composition if you know the masses of the elements that make it up.

For example, say you have a sample of table salt with a mass of 11.7 g. You take that sample and electrolyze it, decomposing it entirely, leaving you with 4.6 g of sodium metal and 7.1 g of chlorine gas (DON'T TRY THIS AT HOME ... there are at least three ways doing this experiment on your own could kill you!).

These data give us element masses (4.6 g Na and 7.1 g Cl) and a compound mass. We don't know our compound formula, but we can still divide our element masses by the compound mass (multiplying by 100% at the end) to get the percent composition of our compound.

Empirical Formula From Composition

Because you can calculate the composition given the formula, you should be able to work the other way around and get formulas from their compositions. That's our next topic, determining the empirical formula of a compound, starting with some information about its composition.

Here is an overview of the process: First change the masses or mass percentages to moles. Then calculate the mole ratio. Express the mole ratio in whole numbers. Next, switch your thinking from moles to atoms and remember that the mole ratio is equal to the atom ratio, which you use to write the formula.

This approach will necessarily give you the simplest ratio of atoms in the formula, but not necessarily the actual number of atoms in a molecule. Other information must be available to determine molecular formulas. We will address that issue just a little bit later.

Example 1

We are given information about how much carbon and how much hydrogen there is in a particular sample of methane. Specifically 6.0 grams of methane contains 4.5 grams of carbon and 1.5 grams of hydrogen. What's involved in determining the empirical formula is to eventually get the ratio of atoms within the compound. As mentioned before, the way we go about doing this is to start with mass and change it to moles. Once we get the mole ratio, then the mole ratio will be the same as the atom ratio and from that we can get the empirical formula.

Step 1: Change Masses to Moles

Starting with 4.5 grams of carbon and 1.5 grams of hydrogen, we need to find out how many moles there are of each element. We take the 4.5 grams of carbon and multiply by the conversion factor that changes from grams to moles by using the atomic weight and we get 0.375 moles of carbon. (Notice that I carried an extra digit here rather than rounding off right away. This is because I'm not finished calculating with that number yet. If I rounded off now, that would change the number and it would change the results of the following calculations. So I won't round off yet, I will wait until the calculations are done and then round off.) We do the same with hydrogen and get 1.488 moles.

Step 2: Calculate The Mole Ratio

To get the mole ratio, divide the each mole amount by the smaller amount. In this case, we divide both by carbon simply because there is more hydrogen (by moles) than carbon. When we divide 1.5 by 0.375, it comes out to be 4.0; and when we divide 0.375 by itself we get 1.0. As a mole ratio that should be interpreted as 4.0 mol H to 1.0 mol C.

Step 3: Express Ratio in Whole Numbers

In this case, we already have a whole-number ratio of 4:1, so this step is easy. In later examples, it will be more complicated.

Step 4: Write Formula

We have calculated a ratio of moles. This ratio will match the ratio of atoms in the empirical formula, so in our empirical formula we should have a ratio of four hydrogen atoms to one carbon atom. This is expressed as the formula CH₄.

Example 2

Here we're given the weights of iron and oxygen that combine with one another to form a particular compound. Specifically, 8.65 grams of iron combines with 3.72 grams of oxygen. Let's find the empirical formula of the compound they make.

Step 1: Change Masses to Moles

Converting these masses to moles should be getting easier for you now. Turning 8.65 g Fe into moles gives 0.155 mol Fe, and the same calculation for oxygen gives 0.233 mol O. Check in with your instructor if you do not get these values.

Step 2: Calculate the Mole Ratio

Because there's more oxygen, I will divide both by iron. 0.233 divided by 0.155 comes out to be 1.50. Notice that in this case it does not come out to be a nice whole number ... the interpretation of this is that for every one mole of iron there are 1.50 moles of oxygen. But we want a ratio of atoms, and there's no such thing as 1.50 atoms of oxygen.

Step 3: Express Ratio in Whole Numbers

You need to realize that 1.50 is the same as 1 and 1/2; and if you convert that into an improper fraction, you come up with 3/2 or 3 to 2 ratio. (Another technique is to think about finding the smallest whole number possible to use as a common multiple to obtain whole numbers. We know that 1.50 x 2 = 3; if we multiply the O ratio by 2 then we must do the same thing to the Fe ratio. That gives us a 3 to 2 ratio again.) So our 1.50:1 ratio has become a 3:2 ratio, and that's a ratio we can work with.

Step 4: Write Formula

With a 3:2 ratio of oxygen to iron, we will write a formula of Fe₂O₃. If you flip the order of iron and oxygen, it will not be marked wrong for now. Later, when we learn more about formulas, you will learn that this formula is never written as O₃Fe₂.

Example 3

The composition of a compound is 40% sulfur and 60% oxygen by mass. What is its empirical formula?

Step 1: Change Masses to Moles

In this third example we start with a percent composition instead of masses. The simplest way to deal with this is to assume you have a 100 gram sample, 40 percent of that is sulfur, 60 percent is oxygen. Thus you have 40 g of sulfur and 60 g of oxygen. (It doesn't really matter whether you assume a 100g sample or any other size sample, because the ratio will remain the same. Choosing 100 grams is convenient and makes the math easy.)

Changing these masses to moles gives 1.25 moles of sulfur and 3.75 moles of oxygen.

Step 2: Calculate the Mole Ratio

Because there's more oxygen, I will divide both by sulfur. 3.75 divided by 1.25 comes out to be 3.00.

Step 3: Express Ratio in Whole Numbers

Once again we have a nice round number right off the bat, for a ratio of three oxygen to one sulfur.

Step 4: Write Formula

The empirical formula for this compound would be SO₃. As with the above example, you will learn later in this course that it is improper to write this formula as O₃S, but for now that answer would not be marked wrong.

Example 4

Pure formaldehyde consists of 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen. What is its empirical formula?

Step 1: Change Masses to Moles

The new aspect in this fourth example is that we are dealing with three elements. The starting part of the calculation is the same. Since we're dealing with percents again, assume a 100 gram sample. Thus we have 40.0 grams of carbon, 6.7 grams of hydrogen, 53.3 grams of oxygen. The atomic weight of each of those elements is used to calculate moles. For C, H, and O respectively, there are 3.33 moles, 6.7 moles, and 3.33 moles.

Step 2: Calculate the Mole Ratio

Now, we're faced with what to do with a ratio of three elements rather than just two. Luckily, we can use the same method that we used in the previous examples. Simply divide each of those numbers by the smallest number of moles that is there; and if you do that, you come up with the simple relationship of 1.00 mole C, 2.01 mole H and 1.00 mole O.

Step 3: Express Ratio in Whole Numbers

We have a tiny bit of rounding to do here, but it should be obvious that 2.01 moles of H rounds to 2.00, for a 1:2:1 ratio of C:H:O.

Sometimes, when your mole ratios look like decimals, it is appropriate to round them to whole numbers (as in this case) in other times (Example 2 above) it is not appropriate to round, and instead you look for an integer ratio that matches your value.

A good rule of thumb is to look for numbers that end in the following: X.25, X.33, X.50. X.67 and X.75. These values represent small integer ratios. For example, if you found a ratio of 1.68:1 for two elements, it would be best to assume that is really a 5:3 ratio (since 5/3 is 1.67). However, if you found a ratio of 1.98:1, that should be rounded to be a 2:1 ratio.

Step 4: Write Formula

A 1:2:1 ratio of C:H:O gives us an empirical formula for formaldehyde of CH₂O.

Example 5

Determine the empirical formula of a compound that is 29.0% sodium, 40.5% sulfur, and 30.4% oxygen by mass.

Step 1: Change Masses to Moles

This example starts with percentage data and deals with more than two elements, but also has a mole ratio that doesn't come out nice and simple. To begin, we will convert all three of our assumed masses (29.0 g Na, 40.5 g S, and 30.4 g O) to moles: 1.26 moles each of Na and S, and 1.90 moles of O.

Step 2: Calculate the Mole Ratio

Dividing each of our mole values by 1.26 (the smallest) gives us a ratio of 1:1:1.51.

Step 3: Express Ratio in Whole Numbers

Here it would not be appropriate to round 1.51 to 2 ... those numbers are quite far apart! Instead, we once again recognize 1.51 as essentially a 3:2 ratio (or 3:2:2 in this case, since there are three elements).

Step 4: Write Formula

A 3:2:2 ratio of these elements should be expressed as the empirical formula Na₂S₂O₃.

Molecular Formulas From Data

The last topic for this lesson is determining the molecular formula. The empirical formula represents the simplest ratio of atoms of different elements contained in a particular compound. With molecular compounds we generally want to know more. We want to know the actual number of each kind of atom that is contained in the molecule.

For example hydrogen peroxide has a 1:1 ratio of hydrogen to oxygen. Thus its empirical formula is HO. But molecules of hydrogen peroxide contain four atoms, two each of hydrogen and oxygen. Thus, hydrogen peroxide has a molecular formula of H₂O₂. Note: this cannot be determined just by looking at the empirical formula.

Another example is the compound benzene. It has a molecular formula of C₆H₆. There are actually 12 atoms in the molecule. But the empirical formula only gives you CH, a 1-to-1 ratio. Importantly, this is the same empirical formula as you would get if your molecular formula were, say C₄H₄ or C₁₀H₁₀. Empirical formulas contain less information than molecular formulas, and you cannot deduce a molecular formula solely by using the empirical formula.

The way you can go about determining the molecular formula involves the fact that the molecular weight can generally be determined independently of the formula. In this week's lab activity, you will be doing just that.

In homework and tests for this class, you will be given (or asked to find) the empirical formula of a compound, along with its molecular mass. Using these two pieces of data, you can find the molecular formula. To do so, find the empirical formula mass of the compound. For example, if you know hydrogen peroxide has the empirical formula HO, then the empirical formula mass of that compound is about 17 g/mol (technically closer to 17.007, but we will not get bogged down in the thousandths place).

To finish up the lesson, you should do the practice problems in your lab workbook related to finding empirical and molecular formulas.