Lesson 19: Writing Empirical Formulas
(See pages 242-245)
In the last lesson, we completed a large amount of compound formula writing as well as calculation of percentages
and amounts of ingredients required to prepare various chemical compounds. In this lesson, we will investigate the
procedures followed in taking a report or analysis of an unknown compound and writing the chemical formula that we
suspect that unknown chemical compound to be. In other words, we will take clues about an unknown chemical and
figure out its identity based upon those clues.
There is an entire field of study known as quantitative analysis where students and chemists learn and develop new
ways to analyze unknown chemical compounds that have been produced in the laboratory. Most of these unknown
chemical compounds are products and byproducts of chemical reactions intended to create or produce a specific
product. For example, in the process of developing an improved fabric coating that could repel water from all-weather
jackets while allowing body moisture to evaporate, chemists may produce several potential products and byproducts
along the way. Each of these products must be analyzed for its components so that those people using the product
understand it properties and capabilities. In addition, all of the byproducts must be analyzed for their components to
insure that they are handled and disposed of, if necessary, in the proper ways.
In this lesson we take a previously prepared analysis of an unknown chemical compound and with some relatively
simple calculations, arrive at a chemical formula for that unknown compound. (This is how Chemistry is done folks!) In
the title of this chapter, the term empirical formula is used. In this context, empirical formula means a formula indicating
the most reduced whole-number ratio of ingredients found in the compound. Realize that the empirical formula is not
necessarily the correct formula for the compound. WHY??? Look at these chemical formulas for sodium chloride below:
Na350Cl350
(Ratio: 350:350)
Na200Cl200
(Ratio: 200:200)
Na75Cl75
(Ratio: 75:75)
NaCl
(Ratio: 1:1)
Although all of these chemical formulas represent the very same compound (sodium chloride), the ratio of sodium
ions to chloride ion in the last formula is in the most reduced form. It is the chemical formula in which the components
are written in the most reduced, whole number ratio, that makes up an empirical formula for a compound.
You might ask, "How do I know when I've reached the empirical formula for a particular compound?" The answer is
that when you can no longer reduce the ratio (or fraction relationship) between the two or more constituents within the
compound using only whole numbers (1,2,3, etc.), you have reached the empirical formula.
Below are some chemical formulas for some compounds. Circle the ones that are empirical formulas.
H2O
H4O2
K3PO4
K6(PO4)2
Cr(NO2)3
Note that all of the chemical formulas we have presented in previous chapters have been written as empirical
formulas. Now that you can recognize an empirical formula, let's continue with how to go about finding and writing the
empirical formula of a chemical compound based upon an analysis of an unknown compound you have received.
Suppose you have received the following analysis for an unknown chemical compound you have just produced.
The analysis for your compound is:
20% calcium
80% bromine
Based upon the information received in the analysis, you can tell that you have a compound which contains calcium
and bromine. To find the correct empirical formula for this compound you must first find the number of moles of each
ingredient present and then reduce that ratio of moles to the nearest whole number ratio. Remember - the suffixes in a
chemical formula like C6H12O6 represent number of atoms or number of moles, not mass (grams).
To begin finding the empirical formula for this unknown chemical compound, assume that you have 100 grams of the
compound. (Assuming you have 100 grams of the compound makes the math calculation somewhat easier!) Based upon
the analysis, you might then say:
20% of the 100 gram sample represents the calcium ingredient, which is the same as saying . .
20% x 100 grams = grams of calcium present
0.20 x 100 grams = grams of calcium present
20 grams = grams of calcium present.
And for the bromine ingredient, we could say:
80% of the 100 gram sample represents the bromine ingredient, which is the same as saying . .
80% x 100 grams = grams of bromine present
0.80 x 100 grams = grams of bromine present
80 grams = grams of bromine present.
Therefore, in our 100 gram sample, we have 20 grams of calcium and 80 grams of bromine. Recall that our goal is find
the number of moles present of each ingredient. We now have the number of grams present of each ingredient, so we
need to convert those grams of ingredients into moles of ingredients.
Recall from the last chapter that the atomic mass value for each element, found on the periodic table of elements,
tells the number of grams required to make one mole of that particular element. For example, the atomic mass of
carbon is 12 grams. This means that one mole of carbon would have a mass of 12 grams or, yet another way to say it is,
that it would require 12 grams of carbon to make one mole of carbon. Therefore, to convert grams to moles we just
need to compare the number of grams we have in our sample to the number of grams it takes to make one mole of that
element. Let's return to our example of calcium and bromine.
In regard to calcium, we have 20 grams present in our sample. By referring to the periodic table of elements, we can
see that calcium has an atomic mass of 40 grams (to 2 s.f.). Based upon this information we can build a ratio which says:
40 grams of Ca: 1 mole of Ca
as
20 grams of Ca: (x) moles of Ca
or
40 grams of Ca
1 mole Ca
20 grams of Ca
x moles Ca
Since these ratios are considered equal, we can solve for x by cross-multiplying.
40 grams of Ca
1 mole Ca
20 grams of Ca
x moles Ca
40x = 20 x 1
40x = 20
x = 20/40
x = 0.50 moles Ca
Another way (I think a better way) to do the same calculation above is:
I have 20 grams Ca x 1 mole Ca
40 grams Ca
Start with what you
have (20 grams Ca)
= 0.50 mole Ca
Answer
Convert units of what you have
(grams) to what you want (moles)
Remember 1 mole Ca
= 40 grams Ca
Based upon these calculations, we can say that in our 100 gram sample of unknown compound, we have 20 grams of
calcium which represents 0.5 moles of calcium. In our definition of empirical formula, we stated that the empirical
formula must be composed of ingredients in a whole number ratio. Obviously, the moles of calcium we just derived is
not a whole number (0.50 is not a whole number). We will not stop at this point to adjust the moles of calcium value to a
whole number. Instead, let us determine the number of moles of bromine present. After we have calculated that
amount, we will then adjust both values to a whole number ratio.
To find the number of moles of bromine present, we will follow the same procedure as we did with finding the moles
of calcium present. Find bromine on the periodic table of elements. Note that the atomic mass of bromine is 80 grams.
This means that one mole of bromine will have a mass of approximately 80 grams.
From our earlier calculations, we found that in our 100 gram sample of the unknown compound, we had 80 grams of
bromine present (80% of the sample was bromine). It may appear obvious that we, therefore, have one mole of bromine
present in our unknown compound. For completeness of our example, let's examine the calculations to arrive at that
value.
I have 80 grams Br x
Start with what you
have (80 grams Br)
1 mole Br
= 1.0 mole Br
80 grams Br
Answer
Convert units of what you have
(grams) to what you want (moles)
Based upon these calculations, we can indeed say that in our 100 grams sample of unknown compound, we have 80
grams of bromine (80% of 100 grams) which represents 1 mole of bromine.
Coupled with the number of moles of calcium we calculated earlier, we can say that we have a compound made of
0.5 moles of calcium and 1 mole of bromine.
Ca0.5Br1
=
Ca1Br2
=
CaBr2
Therefore, our empirical formula for our unknown compound is CaBr2.
As a simple check of our work, we can examine our empirical formula to see if it is indeed a compound where the
charges, when added, equal zero. Recall that calcium (a calcium family member) has two lone valence electrons that it
would like to "get rid of' which would result in a net +2 charge while bromine (a halogen family member) has the
"desire" to gain one electron to acquire the electron configuration of its noble gas neighbor, krypton. bromine,
therefore, has a net charge of -I.
Ca+21Br-12 = net charge of 0
In this case, the charges do equal zero. We can assume that our work was done correctly and the unknown
compound has been identified as calcium bromide (CaBr2).
Before we move on to another example of how to find the empirical formula of an unknown compound, let's
summarize the steps we followed:
1. Examine the analysis provided to determine the elements present in the tested sample.
2. Assume you have been given 100 grams of the unknown sample.
3. Find the number of grams present of each ingredient (based upon the percentages found in the analysis).
4. Convert the number of grams present of each ingredient into moles of each ingredient.
5. Adjust the number of moles present of each ingredient to a whole number ratio by dividing by the smallest value
present.
6. Write the empirical formula and check your work by adding up the charges to see that they equal zero.
7. Remember that the empirical formula is often not the correct formula for a compound, it just gives you the proper
ratios of elements in a compound. This is an extremely important first step. (For example, glucose has the formula
C6H12O6, but its empirical formula is CH2O. Notice the ratios are correct - twice as many hydrogen atoms as carbon and
oxygen atoms. With further testing chemists can find the true formula of a compound if they have the empirical formula.
You may have noticed in the previous example that we rounded the atomic mass values found on the periodic table of
elements to whole numbers as we used them in the calculations (40.08 grams/mole of calcium rounded to 40
grams/mole of calcium and 79.909 grams/mole of bromine rounded to 80 grams/mole of bromine). This was done to
make the arithmetic easier. In this class we will use 3 significant figures for all atomic weight values from the periodic
table unless specified otherwise. As you explore the world of chemistry beyond the scope of this book (especially, the
field of quantitative analysis) you will learn when it is appropriate to round values to the nearest whole number and
when it is significant not to round values. For the sake of simplifying the concepts and skills required to find the
empirical formulas of unknown compounds, we will continue to round the atomic mass values found on the periodic
table of elements to whole numbers. In the practice problems that follow the next examples, you can assume that the
atomic mass values have been rounded to the nearest whole number.
Let's continue now with another example. Here is an analysis of an unknown compound that you have been asked to
identify.
The analysis for your compound
is: 11.2% hydrogen 88.8% oxygen
From the analysis, we can see that we have a compound which contains hydrogen and oxygen. Let's begin by assuming
we have 100 grams of this unknown compound. In order to find the number of moles present of each ingredient, we will
first need to find the number of grams of each ingredient present.
Next we will need to convert grams of each ingredient to moles of each ingredient. To do so, we will need to find the
atomic mass value for hydrogen and oxygen from the periodic table of elements. Hydrogen has an atomic mass of
approximately 1.0 gram/mole and oxygen has an atomic mass of approximately 16 grams/mole.
Hydrogen:
11.2% of our 100g sample
Which is the same as:
0.112 x 100 g sample = 11.2 grams of hydrogen
Oxygen:
88.8% of our 100g sample
Which is the same as:
0.888 x 100 g sample = 88.8 grams of oxygen
Find the number of moles for each element:
H: I have 11.2 g of hydrogen x 1 mole hydrogen
=
11.2 moles of hydrogen
1.00 g of hydrogen
O: I have 88.8 g of oxygen x
1 mole oxygen
=
5.55 moles of oxygen
16.0 g of oxygen
Notice that the analysis had 3 significant figures which was maintained throughout the calculations. In all problems the
answers should have the same significant figures as the number with the fewest s.f. in the original problem!
From the calculations above, we can see that we have a ratio of 11.2 moles of hydrogen for every 5.55 moles of
oxygen. We can write the compound formula as follows.
H11.2 O5.55
Our definition of empirical formula states that the ratio must be in whole numbers. To reduce the 11.2 moles : 5.55
moles ratio of ingredients in our compound formula to whole numbers, we divide by the smaller number (5.55 moles).
H11.2/5.55 O5.55/5.55
This results in the following compound:
H2O1 = H2O
Do you recognize this compound? Yes, the unknown compound we were analyzing was water! Remember to check
the stability of the compound (as well as our calculations) by seeing if the charges add up to zero.
H+12O-21 = 0
In this case, the charges do add to zero and we have written an accurate compound formula. Again, use the formula
H2O because it's easier to read. In this example you may not have immediately recognized our unknown formula as
being water (most of us recognize water as simply being H20). In an effort to simplify the writing of chemical formulas, if
only one mole of a cation or anion is present, the subscript is dropped and it is understood that one moIe of that cation
or anion is required to make a stable compound.
Reviewing the process of finding the empirical formula one more time, we can say that:
1. Assume you have been given 100 g of the sample.
2. Convert the percentage of each element to grams.
3. Convert the mass (g) amount of each element into moles.
4. Adjust the mole amount of each element to whole numbers by dividing the moles of each element by the lowest mole
amount.
A. Below are several analyses of unknown chemical compounds. Find the empirical formula for
each compound. Then write the name of that compound in the appropriate space.
I. Analysis: 15% fluorine, 85% silver
Empirical formula:
Name:
2. Analysis: 2.77% hydrogen and 97.3% chlorine
Empirical formula:
Name:
3. Analysis: 24.58% potassium, 34.81 % manganese, 40.5% oxygen
Empirical formula:
Name:
4. Analysis: 9.66% nitrogen, 87.6% iodine, 2.79% hydrogen
Empirical formula:
Name:
5. Analysis: 60.7% chlorine, 39.3% sodium
Empirical formula:
Name:
Analysis: 68% silver, 22% chlorine, 10% oxygen
Empirical formula:
Name:
Analysis: 55% oxygen, 27% phosphorous, 18% lithium
Empirical formula:
Name:
Analysis: 71% bromine, 29% copper
Empirical formula:
Name:
Analysis: 40% oxygen, 45% chlorine, 15% magnesium
Empirical formula:
Name: