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CHEMFILE MINI-GUIDE TO PROBLEM SOLVING
CHAPTER 7
Empirical Formulas
Suppose you analyze an unknown compound that is a white powder and
find that it is composed of 36.5% sodium, 38.1% oxygen, and 25.4% sulfur. You can use those percentages to determine the mole ratios among
sodium, sulfur, and oxygen and write a formula for the compound.
To begin, the mass percentages of each element can be interpreted as
“grams of element per 100 grams of compound.” To make things simpler,
you can assume you have a 100 g sample of the unknown compound.
The unknown compound contains 36.5% sodium by mass. Therefore
100.0 g of the compound would contain 36.5 g of sodium. You already
know how to convert mass of a substance into number of moles, so you
can calculate the number of moles of sodium in 36.5 g. After you find the
number of moles of each element, you can look for a simple ratio among
the elements and use this ratio of elements to write a formula for the
compound.
The chemical formula obtained from the mass percentages is in the
simplest form for that compound. The mole ratios for each element,
which you determined from the analytical data given, are reduced to the
smallest whole numbers. This simplest formula is also called the empirical formula. The actual formula for the compound could be a multiple of
the empirical formula. For instance, suppose you analyze a compound
and find that it is composed of 40.0% carbon, 6.7% hydrogen, and 53.3%
oxygen. If you determine the formula for this compound based only on
the analytical data, you will determine the formula to be CH2O. There
are, however, other possibilities for the formula. It could be C2H4O2 and
still have the same percentage composition. In fact, it could be any multiple of CH2O.
It is possible to convert from the empirical formula to the actual chemical formula for the compound as long as the molar mass of the compound is known. Look again at the CH2O example. If the true compound
were CH2O, it would have a molar mass of 30.03 g/mol. If you do more
tests on the unknown compound and find that its molar mass is 60.06,
you know that CH2O cannot be its true identity. The molar mass 60.06 is
twice the molar mass of CH2O. Therefore, you know that the true chemical formula must be twice the empirical formula, (CH2O) ⫻ 2, or
C2H4O2 . Any correct molecular formula can be determined from an empirical formula and a molar mass in this same way.
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General Plan for Determining Empirical Formulas
and Molecular Formulas
1
Percentage of
element expressed
as grams of
element per 100 g
unknown
2
Convert using the
molar mass of
each element.
Amount of each
element per 100 g
of unknown
Use the amount of
the least-abundant
element to calculate
the simplest wholenumber ratio among
the elements.
3
4
Empirical formula
of the compound
The calculated ratio
is the simplest
Convert using
formula.
the experimental
molar mass of the
unknown and the
molar mass of the
simplest formula.
5
Calculated
whole-number
ratio among the
elements
Molecular formula
of the compound
SAMPLE PROBLEM 1
Determine the empirical formula for an unknown compound
composed of 36.5% sodium, 38.1% oxygen, and 25.4% sulfur
by mass.
SOLUTION
1. ANALYZE
• What is given in the
problem?
• What are you asked to find?
the percentage composition of the
compound
the empirical formula for the compound
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Items
Data
The percentage composition of the
unknown substance
36.5% sodium
38.1% oxygen
25.4% sulfur
The molar mass of each element*
22.99 g Na/mol Na
16.00 g O/mol O
32.07 g S/mol S
Amount of each element per 100.0 g
of the unknown
? mol
Simplest mole ratio of elements in
the unknown
?
* determined from the periodic table
2. PLAN
• What steps are needed to
calculate the amount in
moles of each element per
100.0 g of unknown?
• What steps are needed to
determine the wholenumber mole ratio of the
elements in the unknown
(the simplest formula)?
State the percentage of the element
in grams and multiply by the inverse
of the molar mass of the element.
Divide the amount of each element
by the amount of the least-abundant
element. If necessary, multiply the
ratio by a small integer that will
produce a whole-number ratio.
1
2
Mass of Na per
multiply by the inverse of
100.0 g unknown the molar mass of Na
Amount Na in mol per
100.0 g unknown
1
percent of Na stated as grams
Na per 100 g unknown
molar mass Na
36.5 g Na
1 mol Na
mol Na
⫻
⫽
100.0 g unknown 22.99 g Na 100.0 g unknown
Repeat this step for the remaining elements.
2
Amount of Na in mol per
100.0 g unknown
divide by the
amount of the
least-abundant
element
4
Empirical formula
3
Whole-number ratio
among the elements
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3. COMPUTE
36.5 g Na
1.59 mol Na
1 mol Na
⫽
⫻
100.0 g unknown
22.99 g Na
100.0 g unknown
38.1 g O
1 mol O
2.38 mol O
⫻
⫽
100.0 g unknown
16.00 g O
100.0 g unknown
0.792 mol S
1 mol S
25.4 g S
⫽
⫻
100.0 g unknown
32.07 g S
100.0 g unknown
Divide the amount of each element by the amount of the leastabundant element, which in this example is S. This can be accomplished by multiplying the amount of each element by the inverse of
the amount of the least abundant element.
1.59 mol Na
2.01 mol Na
100.0 g unknown
⫽
⫻
100.0 g unknown
0.792 mol S
1 mol S
3.01 mol O
100.0 g unknown
2.38 mol O
⫽
⫻
100.0 g unknown
0.792 mol S
1 mol S
100.0 g unknown
1.00 mol S
0.792 mol S
⫻
⫽
100.0 g unknown
0.792 mol S
1 mol S
From the calculations, the simplest mole ratio is 2 mol Na : 3 mol
O : 1 mol S.
The simplest formula is therefore Na2O3S. Seeing the ratio 3 mol
O : 1 mol S, you can use your knowledge of chemistry to suggest that
this possibly represents a sulfite group, ⫺SO3 and propose the formula Na2SO3 .
4. EVALUATE
• Are the units correct?
Yes; units canceled throughout the
calculation, so it is reasonable to assume that the resulting ratio is accurate.
• Is the number of significant
figures correct?
Yes; ratios were calculated to three
significant figures because percentages were given to three significant
figures.
• Is the answer reasonable?
Yes; the formula, Na2SO3 is
plausible, given the mole ratios and
considering that the sulfite ion has a
2⫺ charge and the sodium ion has a
1⫹ charge.
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PRACTICE
1. Determine the empirical formula for compounds that have the following analyses:
a. 28.4% copper, 71.6% bromine
ans: CuBr2
b. 39.0% potassium, 12.0% carbon,
1.01% hydrogen, and 47.9% oxygen
ans: KHCO3
c. 77.3% silver, 7.4% phosphorus, 15.3%
oxygen
ans: Ag3PO4
d. 0.57% hydrogen, 72.1% iodine, 27.3%
oxygen
ans: HIO3
SAMPLE PROBLEM 2
Determine the empirical formula for an unknown compound
composed of 38.4% potassium, 36.3% oxygen, 23.7% carbon,
and 1.66% hydrogen.
2. PLAN
• What steps are needed to
calculate the amount in
moles of each element per
100.0 g of unknown?
• What steps are needed to
determine the whole-number
mole ratio of the elements in
the unknown (the simplest
formula)?
1
Mass of K in g per
100.0 g unknown
State the percentage of the element
in grams and multiply by the inverse
of the molar mass of the element.
Divide the amount of each element
by the amount of the least-abundant
element. If necessary, multiply the
ratio by a small integer to produce
a whole-number ratio.
multiply by the inverse of
the molar mass of K
2
Amount of K in mol per
100.0 g unknown
divide by the amount of the
least-abundant element,
and multiply by an integer
that will produce a wholenumber ratio
3
4
Whole-number ratio
among the elements
Empirical formula
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3. COMPUTE
38.4 g K
1 mol K
0.982 mol K
⫻
⫽
100.0 g unknown
39.10 g K
100.0 g unknown
Proceed to find the amount in moles per 100.0 g of unknown for
the elements carbon, oxygen, and hydrogen, as in Sample Problem 1.
When determining the formula of a compound having more than
two elements, it is usually advisable to put the data and results in a
table.
Element
Mass per
100.0 g
of unknown
Molar mass
Amount in
mol per
100.0 g
of unknown
Potassium
38.4 g K
39.10 g/mol
0.982 mol K
Carbon
23.7 g C
12.01 g/mol
1.97 mol C
Oxygen
36.3 g O
16.00 g/mol
2.27 mol O
Hydrogen
1.66 g H
1.01 g/mol
1.64 mol
Again, as in Sample Problem 1, divide each result by the amount in
moles of the least-abundant element, which in this example is K.
You should get the following results:
Element
Amount in mol
of element
per 100.0 g
of unknown
Amount in mol
of element
per mol of potassium
Potassium
0.982 mol K
1.00 mol K
Carbon
1.97 mol C
2.01 mol C
Oxygen
2.27 mol O
2.31 mol O
Hydrogen
1.64 mol
1.67 mol H
In contrast to Sample Problem 1, this calculation does not give a
simple whole-number ratio among the elements. To solve this problem, multiply by a small integer that will result in a whole-number
ratio. You can pick an integer that you think might work, or you can
convert the number of moles to an equivalent fractional number. At
this point, you should keep in mind that analytical data is never perfect, so change the number of moles to the fraction that is closest to
the decimal number. Then, choose the appropriate integer factor to
use. In this case, the fractions are in thirds so a factor of 3 will change
the fractions into whole numbers.
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Amount in
mol of
element
per mole
of potassium
Fraction
nearest the
decimal
value
Integer
factor
Whole-number
mole ratio
1.00 mol K
1 mol K
⫻3
3 mol K
2.01 mol C
2 mol C
⫻3
6 mol C
2.31 mol O
2 1/3 mol O
⫻3
7 mol O
1.67 mol H
1 2/3 mol H
⫻3
5 mol H
Thus, the simplest formula for the compound is K3C6H5O7 , which
happens to be the formula for potassium citrate.
4. EVALUATE
• Is the answer reasonable?
Yes; the formula, K3C6H5O7 is plausible, considering that the potassium
ion has a 1⫹ charge and the citrate
polyatomic ion has a 3⫺ charge.
PRACTICE
1. Determine the simplest formula for compounds that have the following analyses. The data may not be exact.
a. 36.2% aluminum and 63.8% sulfur
ans: Al2S3
b. 93.5% niobium and 6.50% oxygen
ans: Nb5O2
c. 57.6% strontium, 13.8% phosphorus, and
ans: Sr3P2O8 or
28.6% oxygen
Sr3(PO4)2
d. 28.5% iron, 48.6% oxygen, and 22.9%
ans: Fe2S3O12 or
sulfur
Fe2(SO4)3
SAMPLE PROBLEM 3
A compound is analyzed and found to have the empirical formula CH2O. The molar mass of the compound is found to be
153 g/mol. What is the compound’s molecular formula?
SOLUTION
1. ANALYZE
• What is given in the
problem?
• What are you asked to find?
the empirical formula, and the
experimental molar mass
the molecular formula of the compound
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Items
Data
Empirical formula of unknown
CH2O
Experimental molar mass of unknown
153 g/ mol
Molar mass of empirical formula
30.03 g/mol
Molecular formula of the compound
?
2. PLAN
• What steps are needed to
determine the molecular
formula of the unknown
compound?
Multiply the experimental molar
mass by the inverse of the molar
mass of the empirical formula. The
subscripts of the empirical formula
are multiplied by the whole-number
factor obtained.
4
5
Empirical formula
multiply the experimental molar mass
of unknown
by the inverse of the molar mass of the
Molecular formula
of unknown
empirical formula, and multiply each
subscript in the empirical formula by
the resulting factor
given
factor that shows the number
of times the empirical formula
1
molar mass of
must be multiplied to get the
empirical formula
molecular formula
153 g
1 mol CH2O
mol CH2O
⫻
⫽
1 mol unknown
30.03 g
1 mol unknown
3. COMPUTE
5.09 mol CH2O
1 mol CH2O
153 g
⫽
⫻
1 mol unknown
30.03 g
1 mol unknown
Allowing for a little experimental error, the molecular formula
must be five times the empirical formula.
Molecular formula ⫽ (CH2O) ⫻ 5 ⫽ C5H10O5
4. EVALUATE
• Is the answer reasonable?
Yes; the calculated molar mass of
C5H10O5 is 150.15, which is close to
the experimental molar mass of the
unknown. Reference books show
that there are several different compounds with the formula C5H10O5 .
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PRACTICE
1. Determine the molecular formula of each of the following unknown
substances:
ans: C2H4
a. empirical formula CH2
experimental molar mass 28 g/mol
b. empirical formula B2H5
ans: B4H10
experimental molar mass 54 g/mol
c. empirical formula C2HCl
ans: C6H3Cl3
experimental molar mass 179 g/mol
d. empirical formula C6H8O
ans: C18H24O3
experimental molar mass 290 g/mol
e. empirical formula C3H2O
ans: C12H8O4
experimental molar mass 216 g/mol
ADDITIONAL PROBLEMS
1. Determine the empirical formula for compounds that have the following analyses:
a. 66.0% barium and 34.0% chlorine
b. 80.38% bismuth, 18.46% oxygen, and 1.16% hydrogen
c. 12.67% aluminum, 19.73% nitrogen, and 67.60% oxygen
d. 35.64% zinc, 26.18% carbon, 34.88% oxygen, and 3.30% hydrogen
e. 2.8% hydrogen, 9.8% nitrogen, 20.5% nickel, 44.5% oxygen,
and 22.4% sulfur
f. 8.09% carbon, 0.34% hydrogen, 10.78% oxygen, and 80.78%
bromine
2. Sometimes, instead of percentage composition, you will have the
composition of a sample by mass. Use the same method shown in
Sample Problem 1, but use the actual mass of the sample instead of
assuming a 100 g sample. Determine the empirical formula for compounds that have the following analyses:
a. a 0.858 g sample of an unknown substance is composed of
0.537 g of copper and 0.321 g of fluorine
b. a 13.07 g sample of an unknown substance is composed of
9.48 g of barium, 1.66 g of carbon, and 1.93 g of nitrogen
c. a 0.025 g sample of an unknown substance is composed of
0.0091 g manganese, 0.0106 g oxygen, and 0.0053 g sulfur
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3. Determine the empirical formula for compounds that have the following analyses:
a. a 0.0082 g sample contains 0.0015 g of nickel and 0.0067 g of
iodine
b. a 0.470 g sample contains 0.144 g of manganese, 0.074 g of nitrogen, and 0.252 g of oxygen
c. a 3.880 g sample contains 0.691 g of magnesium, 1.824 g of sulfur, and 1.365 g of oxygen
d. a 46.25 g sample contains 14.77 g of potassium, 9.06 g of oxygen, and 22.42 g of tin
4. Determine the empirical formula for compounds that have the following analyses:
a. 60.9% As and 39.1% S
b. 76.89% Re and 23.12% O
c. 5.04% H, 35.00% N, and 59.96% O
d. 24.3% Fe, 33.9% Cr, and 41.8% O
e. 54.03% C, 37.81% N, and 8.16% H
f. 55.81% C, 3.90% H, 29.43% F, and 10.85% N
5. Determine the molecular formulas for compounds having the following empirical formulas and molar masses:
a. C2H4S; experimental molar mass 179
b. C2H4O; experimental molar mass 176
c. C2H3O2 ; experimental molar mass 119
d. C2H2O, experimental molar mass 254
6. Use the experimental molar mass to determine the molecular formula for compounds having the following analyses:
a. 41.39% carbon, 3.47% hydrogen, and 55.14% oxygen; experimental molar mass 116.07
b. 54.53% carbon, 9.15% hydrogen, and 36.32% oxygen; experimental molar mass 88
c. 64.27% carbon, 7.19% hydrogen, and 28.54% oxygen; experimental molar mass 168.19
7. A 0.400 g sample of a white powder contains 0.141 g of potassium,
0.115 g of sulfur, and 0.144 g of oxygen. What is the empirical formula for the compound?
8. A 10.64 g sample of a lead compound is analyzed and found to be
made up of 9.65 g of lead and 0.99 g of oxygen. Determine the empirical formula for this compound.
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9. A 2.65 g sample of a salmon-colored powder contains 0.70 g of
chromium, 0.65 g of sulfur, and 1.30 g of oxygen. The molar mass
is 392.2. What is the formula of the compound?
10. Ninhydrin is a compound that reacts with amino acids and proteins
to produce a dark-colored complex. It is used by forensic chemists
and detectives to see fingerprints that might otherwise be invisible.
Ninhydrin’s composition is 60.68% carbon, 3.40% hydrogen, and
35.92% oxygen. What is the empirical formula for ninhydrin?
11. Histamine is a substance that is released by cells in response to injury, infection, stings, and materials that cause allergic responses,
such as pollen. Histamine causes dilation of blood vessels and
swelling due to accumulation of fluid in the tissues. People sometimes take antihistamine drugs to counteract the effects of histamine. A sample of histamine having a mass of 385 mg is composed
of 208 mg of carbon, 31 mg of hydrogen, and 146 mg of nitrogen.
The molar mass of histamine is 111 g/mol. What is the molecular
formula for histamine?
12. You analyze two substances in the laboratory and discover that each
has the empirical formula CH2O. You can easily see that they are
different substances because one is a liquid with a sharp, biting odor
and the other is an odorless, crystalline solid. How can you account
for the fact that both have the same empirical formula?
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