Name
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Skills Worksheet
Problem Solving
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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Problem Solving continued
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
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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
ANALYZE
What is given in the problem?
the percentage composition of the compound
What are you asked to find?
the empirical formula for the compound
Items
Data
The percentage composition of the unknown subtance
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
PLAN
What steps are needed to calculate the amount in moles of each element per
100.0 g of unknown?
State the percentage of the element in grams and multiply by the inverse of the
molar mass of the element.
What steps are needed to determine the whole-number mole ratio of the elements
in the unknown (the simplest formula)?
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
100.0 g unknown
Amount Na in mol per
100.0 g unknown
multiply by the inverse of
the molar mass of Na
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
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Problem Solving continued
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
3
4
Whole-number ratio
among the elements
Empirical formula
COMPUTE
36.5 g Na
1 mol Na
1.59 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
25.4 g S
1 mol S
0.792 mol S
100.0 g unknown 32.07 g S 100.0 g unknown
Divide the amount of each element by the amount of the least-abundant 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
100.0 g unknown 2.01 mol Na
100.0 g unknown
0.792 mol S
1 mol S
2.38 mol O
100.0 g unknown 3.01 mol O
100.0 g unknown
0.792 mol S
1 mol S
0.792 mol S
100.0 g unknown 1.00 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.
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.
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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.
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
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Sample Problem 2
Determine the empirical formula for an unknown compound composed of
38.4% oxygen, 23.7% carbon, and 1.66% hydrogen.
Solution
ANALYZE
What is given in the problem?
the percentage composition of the compound
What are you asked to find?
the empirical formula for the compound
PLAN
What steps are needed to calculate the amount in moles of each element per
100.0 g of unknown?
State the percentage of the element in grams and multiply by the inverse of the
molar mass of the element.
What steps are needed to determine the whole-number mole ratio of the elements
in the unknown (the simplest formula)?
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 wholenumber ratio.
1
Mass of K in g per
100.0 g unknown
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
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.
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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 H
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 H
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.
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.
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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 28.6% oxygen ans: Sr3P2O8 or
Sr3(PO4)2
d. 28.5% iron, 48.6% oxygen, and 22.9% sulfur ans: Fe2S3O12 or Fe2(SO4)3
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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
ANALYZE
What is given in the problem?
the empirical formula, and the experimental
molar mass
What are you asked to find?
the molecular formula of the compound
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
?
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
of unknown
Molecular formula
of unknown
multiply the experimental molar mass
by the inverse of the molar mass of the
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
COMPUTE
153 g
1 mol CH2O
5.09 mol CH2O
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
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Problem Solving continued
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.
Practice
1. Determine the molecular formula of each of the following unknown substances:
a. empirical formula CH2, experimental molar mass 28 g/mol ans: C2H4
b. empirical formula B2H5, experimental molar mass 54 g/mol ans: B4H10
c. empirical formula C2HCl, experimental molar mass 179 g/mol ans: C6H3Cl3
d. empirical formula C6H8O, experimental molar mass 290 g/mol ans:
C18H24O3
e. empirical formula C3H2O, experimental molar mass 216 g/mol ans: C12H8O4
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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
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
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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.
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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