Empirical Formulas 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 weights or weight 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. The last step is to write the formula using the atom ratio. This approach will necessarily give you the simplest ratio of atoms in the formula, and not necessarily the actual number of atoms in a molecule. Other information must be available to determine molecular formulas. We will address that issue later in this lesson. |
Weights or Weight Percentages |
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| Moles | |
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| Mole Ratio | |
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| Atom Ratio | |
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| Express as Formula |
Five examples follow, each with an additional step or degree of complexity. (These are also shown in Example 18 in your workbook.) Contact your instructor if these are not sufficient. Then try the Quick Quiz on Determining Empirical Formulas before moving on.
First Example
| 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.) |
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| 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. |
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Second Example
| 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. |
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| 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. 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. | |||||||||||||||||||||||||||
Third Example
| In this third example we start with a percentage composition instead of weights. Simply 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.) 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. |
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Fourth Example
| 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. |
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Fifth 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. |
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If you have any questions about any of those examples, please check with the instructor. Make sure any problems and questions that you have are taken care of. If you need or want more practice on figuring out empirical formulas, you can find more exercises in your text or you can ask the instructor for some extra practice exercises (Ex. 19-alt.).
Comment on Rounding Off
Students often ask how much they can round off these kind of calculations. How far away from a whole number can we get and still round it off? That depends on how accurate the data was. If you want to round 1.05 to 1, you are saying that the data was off by 5%. But if you want to round 1.25 to 1, you are saying that the data was off by 25%. That's not very good data on which to base an empirical formula. Besides, it might have been off 25% the other way. At this point in your chemistry studies, I would say that if you have to round off more than 5% to get a whole number, you should not round off. Under other circumstances a different cut-off would be appropriate.
Practice with Determining Empirical Formulas from Composition
Try your hand at working the problems in Example 19. After you've done that, check your answers below then continue. (If, after completing these, you would like to try some additional problems of this type, you will find several in a "Practice Problems" page in the Wrap-Up for this lesson. Click here to go there now.)
Answers
The formulas for the compounds in Example 19 are SO2, CH3, C2H6O, and N2H8Cr2O7.
E-mail instructor: Eden Francis
Clackamas Community College
©1998, 2002 Clackamas Community College, Hal Bender