Lesson 2: Associated Calculations

Let's start with a simple case, KF, the formula for potassium fluoride. The first step is to find the atomic weights for potassium and fluorine, which are 39.1 grams for potassium and 19.0 grams for fluorine. Next, you have to figure the formula weight for the compound KF by adding the weights together, and that is 58.1g.

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 weights of the elements that are contained in the compound and the total weight of one mole of that compound.

From there you calculate the percentage composition the same way you calculate any percentage. Take the weight of the element and divide by the total weight 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%.

In this case, one mole of Fe₂O₃ has a total formula weight of 159.6. That one mole of Fe₂O₃ contains two moles of iron, which weighs 111.6 grams. It also contains three moles of oxygen, which weighs 48.0 grams.

So again, by taking into account what those subscripts are, we can figure out the weight of iron and the weight of oxygen in that particular compound and then use those numbers to calculate that there is 69.944% iron and 30.06% oxygen in that compound.

11.7 g NaCl

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.

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..

In this first example, we are starting with mass data and will get a simple (x:1) atom ratio. 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.

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.)

Next, do the same thing with hydrogen, 1.5 grams times the conversion factor of one mole over 1.0 grams, which is the atomic weight for hydrogen, and we get 1.5 moles of hydrogen. 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. Since we want to interpret this as an atom ratio and since atoms come in whole numbers, the ratio is expressed as a nice even 4 to 1 ratio. That tells us there are four hydrogen atoms to every one carbon atom. So the empirical formula is CH₄. At this point don't worry about what order you put the hydrogen and carbon; we will deal with that later in the course. For now you could write H₄C, but the formula is generally written as CH₄.

As an aside let me point out that these rules are not terribly rigid as long as you keep in mind what you are doing and how you are doing it. For example, what if you set up the mole ratio with the smaller number on top. You don't get 4:1, you get 0.25 instead. But 0.25 = 1/4 and if you keep track of the units or the order of the ratio, you still get one mole of carbon for four moles of hydrogen. Most students find it easier, though, to set up the mole ratio with the larger number on the top.

A number of complicating factors can arise. The next few examples take these into account one by one. In this second example, we still start with mass data, but we will not get a simple (x:1) atom ratio.

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. If you follow through the calculations, you can see that the amount of iron is changed from grams of iron to moles of iron, 0.155 moles. Then the amount of oxygen is changed from grams to moles. The sample contains 0.233 moles of oxygen. Next, we need to get the mole ratio of oxygen to iron. 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. But to get a formula, we are going to need a whole number ratio of oxygen atoms to iron atoms.

We cannot work with 1.50 atoms. We need whole numbers of oxygen and iron. So, here you need to be able to change from decimal fractions into whole number fractions. You need to realize that 1.50 is the same as 1 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.) If there are 3 moles of oxygen to 2 moles of iron, then there are 3 oxygen atoms to 2 iron atoms. So, the empirical formula for the compound is Fe₂O₃. If you were to write O₃Fe₂, that is fine for now. We'll get into proper formula writing later on. Again the numbers in the ratio could have been inverted. If we put 0.155 moles Fe over 0.233 moles O, we would get 0.665 and if you recognize that as the same as 2/3, then you still get 2 moles Fe to 3 moles O, which gives you the same formula Fe₂O₃.

In this third example we start with a percent composition instead of weights. 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.) Change those weights into moles by using atomic weights as is shown, then take the mole ratio of oxygen to sulfur, 3.75 to 1.25. That comes out to be a 3-to-1 ratio, 3 oxygen's to every one sulfur, so the empirical formula is SO₃. That was very much like the first example, except that in this one we started with percentages rather than weights.

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. 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. In whole numbers that is a 1-to-2-to-1 ratio of carbon to hydrogen to oxygen and the empirical formula is CH₂O. Since these calculated values are very close to whole numbers, they can be rounded off. You cannot always do that, as is shown in the next example.

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. Take a moment to follow through the calculations until you get down to determining the mole ratio. Now, look at the mole ratio of sodium to sulfur to oxygen. It is 1.26-to-1.26-to-1.90. Again, we divide by the smallest because we're dealing with three elements. This gives us a new ratio of 1.00 to 1.01 to 1.51. The 1.00 is as close to 1 as you can get using three digits so that presents no problem. However, the 1.51 is not close enough to either 1 or 2 to be rounded off that much. But it is close enough to 1.5 to be treated as 1-1/2. Now, we have a 1 to 1 to 1-1/2 ratio, we can get that into a whole number ratio by multiplying each number by 2 in order to clear the fraction. We use 2 because it is the denominator in one-half. After doing that we come up with a 2 to 2 to 3 ratio and that gives us the empirical formula Na₂S₂O₃. In other cases the fractional mole amount might be 1-1/3 or 2-1/4. In any case that you have to deal with a fractional mole amount multiply everything by the denominator of the fraction.

The principle of the process is this. Start with the empirical formula (EF) and use whatever multiple of it gives the correct molecular weight (MW). In the tables shown here you can see that H₂O and H₂O₂ are the molecular formulas (MF) that match the molecular weights.

One example of how to do this is shown here (and in example 26 in your workbook). The empirical formula gives the formula weight of 15.0. The molecular weight is 30.0 g/mole which is just twice the empirical formula weight. You can figure that out by dividing the molecular weight by the empirical formula weight. In this case you get that it's two times heavier. That means that the molecular formula must contain twice as many atoms as the empirical formula but still in the same ratio. That means twice as many of each kind of atom. In this case the ratio is 1-to-3. Twice that many would be 2-to-6, so the molecular formula is C₂H₆.