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Problem Solving
Percentage Composition
Suppose you are working in an industrial laboratory. Your supervisor gives you a
bottle containing a white crystalline compound and asks you to determine its
identity. Several unlabeled drums of this substance have been discovered in a
warehouse, and no one knows what it is. You take it into the laboratory and carry
out an analysis, which shows that the compound is composed of the elements
sodium, carbon, and oxygen. Immediately, you think of the compound sodium
carbonate, Na2CO3 , a very common substance found in most laboratories and
used in many industrial processes.
Before you report your conclusion to your boss, you decide to check a reference book to see if there are any other compounds that contain only the elements
sodium, carbon, and oxygen. You discover that there is another compound,
sodium oxalate, which has the formula Na2C2O4 . When you read about this compound, you find that it is highly poisonous and can cause serious illness and even
death. Mistaking sodium carbonate for sodium oxalate could have very serious
consequences. What can you do to determine the identity of your sample? Is it
the common industrial substance or the dangerous poison?
Fortunately, you can determine not only which elements are in the compound,
but also how much of each element is present. As you have learned, every compound has a definite composition. Every water molecule is made up of two
hydrogen atoms and one oxygen atom, no matter where the water came from. A
formula unit of sodium chloride is composed of one sodium atom and one chlorine atom, no matter whether the salt came from a mine or was obtained by evaporating sea water.
Likewise, sodium carbonate always has two sodium atoms, one carbon atom,
and three oxygen atoms per formula unit, giving it the formula Na2CO3 ; and a formula unit of sodium oxalate always contains two sodium atoms, two carbon
atoms, and four oxygen atoms, giving it the formula Na2C2O4 . Because each atom
has a definite mass, each compound will have a distinct composition by mass.
This composition is usually expressed as the percentage composition of the compound — the percentage by mass of each element in a compound. To identify a
compound, you can compare the percentage composition obtained by laboratory
analysis with a calculated percentage composition of each possible compound.
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General Plan for Determining Percentage
Composition of a Compound
1
Molar mass
of
element
2
Mass of
element per
mole of
compound
Convert using the
formula of the compound.
Convert by multiplying
by the inverse of the
molar mass of the
compound. Then convert
to a percentage by
multiplying by 100.
5
Mass of
element in a
sample of
compound
3
Convert by expressing
percentage as a fraction
and then multiplying by
the mass of the sample.
Percentage
element in the
compound
Repeat 1, 2,
and 3 for each
remaining
element in the
compound.
4
Percentage
composition of
the compound
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Sample Problem 1
Determine the percentage composition of sodium carbonate, Na2CO3 .
Solution
ANALYZE
What is given in the problem?
the formula of sodium carbonate
What are you asked to find?
the percentage of each element in sodium carbonate (the percentage composition)
Items
Data
Formula of sodium carbonate
Na2CO3
Molar mass of each element*
Na 22.99 g/mol
C 12.01 g/mol
O 16.00 g/mol
Molar mass of sodium carbonate
105.99 g/mol
Percentage composition of sodium carbonate
?%
* determined from the periodic table
PLAN
What step is needed to determine the mass of each element per mole of compound?
Multiply the molar mass of each element by the ratio of the number of moles of
that element in a mole of the compound (the subscript of that element in the
compound’s formula).
What steps are needed to determine the portion of each element as a percentage
of the mass of the compound?
Multiply the mass of each element by the inverse of the molar mass of the
compound, and then multiply by 100 to convert to a percentage.
Step 1
1
2
Molar mass of Na
Mass Na per mole
Na2CO3
multiply by the subscript of Na
in Na2CO3
molar mass Na
ratio of mol Na per mol
Na2CO3 from formula
22.99 g Na
2 mol Na
g Na
1 mol Na
1 mol Na2CO3 1 mol Na2CO3
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Step 2
2
3
Mass Na per
mole Na2CO3
from Step 1
Percentage Na in
Na2CO3
multiply by the inverse of the
molar mass of Na2CO3 and
multiply by 100
1
molar mass Na2CO3
1 mol Na2CO3
g Na
100 percentage Na in Na2CO3
1 mol Na2CO3 105.99 g Na2CO3
Now you can combine Step 1 and Step 2 into one calculation.
combining Steps 1 and 2
2 mol Na
1 mol Na2CO3
22.99 g Na
100 percentage Na in Na2CO3
1 mol Na2CO3 105.99 g Na2CO3
1 mol Na
Finally, determine the percentage of carbon and oxygen in Na2CO3 by repeating
the calculation above with each of those elements.
3
4
Percentage of each
element in Na2CO3
Percentage composition
repeat Steps 1 and 2 for
each remaining element
COMPUTE
percentage sodium
22.99 g Na
2 mol Na
1 mol Na2CO3
100 43.38% Na
1 mol Na
1 mol Na2CO3 105.99 g Na2CO3
percentage carbon
12.01 g C
1 mol C
1 mol Na2CO3
100 11.33% C
1 mol C
1 mol Na2CO3 105.99 g Na2CO3
percentage oxygen
16.00 g O
3 mol O
1 mol Na2CO3
100 45.29% O
1 mol O
1 mol Na2CO3 105.99 g Na2CO3
Element
Percentage
sodium
43.38% Na
carbon
11.33% C
oxygen
45.29% O
EVALUATE
Are the units correct?
Yes; the composition is given in percentages.
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Is the number of significant figures correct?
Yes; four significant figures is correct because the molar masses have four
significant figures.
Yes; the percentages add up to 100 percent.
Practice
1. Determine the percentage composition of each of the following compounds:
a. sodium oxalate, Na2C2O4 ans: 34.31% Na, 17.93% C, 47.76% O
b. ethanol, C2H5OH ans: 52.13% C, 13.15% H, 34.72% O
c. aluminum oxide, Al2O3 ans: 52.92% Al, 47.08% O
d. potassium sulfate, K2SO4 ans: 44.87% K, 18.40% S, 36.72% O
2. Suppose that your laboratory analysis of the white powder discussed at the
beginning of this chapter showed 42.59% Na, 12.02% C, and 44.99% oxygen.
Would you report that the compound is sodium oxalate or sodium carbonate
(use the results of Practice Problem 1 and Sample Problem 1)? ans: sodium
carbonate
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Sample Problem 2
Calculate the mass of zinc in a 30.00 g sample of zinc nitrate, Zn(NO3)2.
Solution
ANALYZE
What is given in the problem?
the mass in grams of zinc nitrate
What are you asked to find?
the mass in grams of zinc in the sample
Items
Data
Mass of zinc nitrate
30.00 g
Formula of zinc nitrate
Zn(NO3)2
Molar mass of zinc nitrate
189.41 g/mol
Mass of zinc in the sample
?g
PLAN
What steps are needed to determine the mass of Zn in a given mass of Zn(NO3)2?
The percentage of Zn in Zn(NO3)2 can be calculated and used to find the mass of
Zn in the sample.
1
5
Molar mass of Zn
Mass Zn in g in sample
multiply
by the mole
ratio of Zn to
Zn(NO3)2
express percentage
as a fraction and
multiply by the
mass of the sample
2
3
Mass Zn per
mole Zn(NO3 )2
Percentage Zn
in Zn(NO3 )2
molar mass Zn
multiply by the inverse of
the molar mass of
Zn(NO3)2 , then multiply
by 100
ratio of mol Zn per mol
Zn(NO3)2 from formula
1
molar mass Zn(NO3)2
65.39 g Zn
1 mol Zn
1 mol
100 percentage Zn
1 mol Zn
1 mol Zn(NO3)2 189.41 g Zn(NO3)2
percentage Zn
expressed as a fraction
given
g Zn
g Zn(NO3)2 g Zn in sample
100 g Zn(NO3)2
COMPUTE
65.39 g Zn
1 mol Zn
1 mol Zn(NO3)2
100 34.52% Zn
1 mol Zn
1 mol Zn(NO3)2 189.41 g Zn(NO3)2
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Note that mass percentage is the same as grams per 100 g, so 34.52% Zn in
Zn(NO3)2 is the same as 34.52 g Zn in 100 g Zn(NO3)2 .
34.52 g Zn
30.00 g Zn(NO3)2 10.36 g Zn
100 g Zn(NO3)2
EVALUATE
Are the units correct?
Yes; units cancel to give the correct units, grams of zinc.
Is the number of significant figures correct?
Yes; four significant figures is correct because the data given have four significant figures.
Yes; the molar mass of zinc is about one third of the molar mass of Zn(NO3)2,
and 10.36 g Zn is about one third of 30.00 g of Zn(NO3)2.
Practice
1. Calculate the mass of the given element in each of the following compounds:
a. bromine in 50.0 g potassium bromide, KBr ans: 33.6 g Br
b. chromium in 1.00 kg sodium dichromate, Na2Cr2O7 ans: 397 g Cr
c. nitrogen in 85.0 mg of the amino acid lysine, C6H14N2O2 ans: 16.3 mg N
d. cobalt in 2.84 g cobalt(II) acetate, Co(C2H3O2)2 ans: 0.945 g Co
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HYDRATES
Many compounds, especially ionic compounds, are produced and purified by
crystallizing them from water solutions. When this happens, some compounds
incorporate water molecules into their crystal structure. These crystalline compounds are called hydrates because they include water molecules. The number of
water molecules per formula unit is specific for each type of crystal. When you
have to measure a certain quantity of the compound, it is important to know how
much the water molecules contribute to the mass.
You may have seen blue crystals of copper(II) sulfate in the laboratory. When
this compound is crystallized from water solution, the crystals include five water
molecules for each formula unit of CuSO4 . The true name of the substance is
copper(II) sulfate pentahydrate, and its formula is written correctly as
CuSO4 5H2O. Notice that the five water molecules are written separately. They
are preceded by a dot, which means they are attached to the copper sulfate molecule. On a molar basis, a mole of CuSO4 5H2O contains 5 mol of water per mole
of CuSO4 5H2O. The water molecules contribute to the total mass of
CuSO4 5H2O. When you determine the percentage water in a hydrate, the water
molecules are treated separately, as if they were another element.
Sample Problem 3
Determine the percentage water in copper(II) sulfate penta hydrate,
CuSO4 5H2O.
Solution
ANALYZE
What is given in the problem?
the formula of copper(II) sulfate pentahydrate
What are you asked to find?
the percentage water in the hydrate
Items
Data
Formula of copper(II) sulfate pentahydrate
CuSO4 5H2O
Molar mass of H2O
18.02 g/mol
Molar mass of copper(II) sulfate pentahydrate*
249.72 g/mol
Percentage H2O in CuSO4 5H2O
?%
* molar mass of CuSO4 mass of 5 mol H2O
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PLAN
What steps are needed to determine the percentage of water in CuSO4 5H2O?
Find the mass of water per mole of hydrate, multiply by the inverse molar mass
of the hydrate, and multiply that by 100 to convert to a percentage.
1
3
Molar mass of H2O
Percentage H2O in CuSO4 5H2O
multiply by the
mole ratio of
H2O to
CuSO4 5H2O
multiply by the inverse of
the molar mass of
CuSO4 5H2O; then
multiply by 100
2
Mass H2O per mole
CuSO4 5H2O
molar mass H2O
ratio of moles H2O per mole
CuSO4 5H2O from formula
1
molar mass CuSO4 5H2O
18.01 g H2O
5 mol H2O
1 mol CuSO4 5H2O
1 mol H2O
1 mol CuSO4 5H2O 249.72 g CuSO4 5H2O
100 percentage H2O
COMPUTE
18.01 g H2O
5 mol H2O
1 mol CuSO4 5H2O
100 36.08% H2O
1 mol H2O
1 mol CuSO4 5H2O 249.72 g CuSO4 5H2O
EVALUATE
Are the units correct?
Yes; the percentage of water in copper(II) sulfate pentahydrate was needed.
Is the number of significant figures correct?
Yes; four significant figures is correct because molar masses were given to at
least four significant figures.
Yes; five water molecules have a mass of about 90 g, and 90 g is a little more
than 1/3 of 250 g; the calculated percentage is a little more than 1/3.
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Practice
1. Calculate the percentage of water in each of the following hydrates:
a. sodium carbonate decahydrate, Na2CO3 10H2O ans: 62.97% H2O in
Na2CO310H2O
b. nickel(II) iodide hexahydrate, NiI26H2O ans: 25.71% H2O in NiI26H2O
c. ammonium hexacyanoferrate(III) trihydrate (commonly called ammonium
ferricyanide), (NH4)2Fe(CN)6 3H2O ans: 17.89 % H2O in
(NH4)2Fe(CN)63H2O
d. aluminum bromide hexahydrate ans: 28.85% H2O in AlBr36H2O
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1. Write formulas for the following compounds and determine the percentage
composition of each:
a. nitric acid
b. ammonia
c. mercury(II) sulfate
d. antimony(V) fluoride
2. Calculate the percentage composition of the following compounds:
a. lithium bromide, LiBr
b. anthracene, C14H10
c. ammonium nitrate, NH4NO3
d. nitrous acid, HNO2
e. silver sulfide, Ag2S
f. iron(II) thiocyanate, Fe(SCN)2
g. lithium acetate
h. nickel(II) formate
3. Calculate the percentage of the given element in each of the following compounds:
a. nitrogen in urea, NH2CONH2
b. sulfur in sulfuryl chloride, SO2Cl2
c. thallium in thallium(III) oxide, Tl2O3
d. oxygen in potassium chlorate, KClO3
e. bromine in calcium bromide, CaBr2
f. tin in tin(IV) oxide, SnO2
4. Calculate the mass of the given element in each of the following quantities:
a. oxygen in 4.00 g of manganese dioxide, MnO2
b. aluminum in 50.0 metric tons of aluminum oxide, Al2O3
c. silver in 325 g silver cyanide, AgCN
d. gold in 0.780 g of gold(III) selenide, Au2Se3
e. selenium in 683 g sodium selenite, Na2SeO3
f. chlorine in 5.0 104 g of 1,1-dichloropropane, CHCl2CH2CH3
5. Calculate the percentage of water in each of the following hydrates:
a. strontium chloride hexahydrate, SrCl2 6H2O
b. zinc sulfate heptahydrate, ZnSO4 7H2O
c. calcium fluorophosphate dihydrate, CaFPO3 2H2O
d. beryllium nitrate trihydrate, Be(NO3)2 3H2O
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6. Calculate the percentage of the given element in each of the following
hydrates. You must first determine the formulas of the hydrates.
a. nickel in nickel(II) acetate tetrahydrate
b. chromium in sodium chromate tetrahydrate
c. cerium in cerium(IV) sulfate tetrahydrate
7. Cinnabar is a mineral that is mined in order to produce mercury. Cinnabar is
mercury(II) sulfide, HgS. What mass of mercury can be obtained from 50.0 kg
of cinnabar?
8. The minerals malachite, Cu2(OH)2CO3 , and chalcopyrite, CuFeS2 , can be
mined to obtain copper metal. How much copper could be obtained from
1.00 103 kg of each? Which of the two has the greater copper content?
9. Calculate the percentage of the given element in each of the following
hydrates:
b. tin in potassium stannate trihydrate, K2SnO3 3H2O
c. chlorine in calcium chlorate dihydrate, CaClO3 2H2O
10. Heating copper sulfate pentahydrate will evaporate the water from the crystals, leaving anhydrous copper sulfate, a white powder. Anhydrous means
“without water.” What mass of anhydrous CuSO4 would be produced by
heating 500.0 g of CuSO4 5H2O?
11. Silver metal may be precipitated from a solution of silver nitrate by placing a
copper strip into the solution. What mass of AgNO3 would you dissolve in
water in order to get 1.00 g of silver?
12. A sample of Ag2S has a mass of 62.4 g. What mass of each element could be
obtained by decomposing this sample?
13. A quantity of epsom salts, magnesium sulfate heptahydrate, MgSO4 7H2O, is
heated until all the water is driven off. The sample loses 11.8 g in the process.
What was the mass of the original sample?
14. The process of manufacturing sulfuric acid begins with the burning of sulfur.
What mass of sulfur would have to be burned in order to produce 1.00 kg of
H2SO4 ? Assume that all of the sulfur ends up in the sulfuric acid.
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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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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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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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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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Problem Solving continued
EVALUATE
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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Problem Solving continued
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
given
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
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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EVALUATE
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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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?
Holt Chemistry
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TEACHER RESOURCE PAGE
Concept Review:
Molar Conversions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
FOUR STEPS FOR SOLVING
QUANTITATIVE PROBLEMS
1.
2.
3.
4.
5.
6.
2.41 1024 atoms Al
9.33 1023 atoms Na
44.0 mol F
70.7 mol H2
291 mol K
877 g Sb
119 g U
167 g Fr
0.7239 mol Pb
2.54 x 105 mol Au
Concept Review: Relative
Atomic Mass and Chemical
Formulas
1.
2.
3.
4.
5.
6.
7.
8.
85.47 amu
35.45 amu
310.18 g/mol
79.88 g/mol
84.01 g/mol
94.12 g/mol
26.04 g/mol
18.02 g/mol
7.
8.
9.
10.
11.
12.
13.
14.
15.
0.026 mm
3.21 L
0.80 g/cm3
21.4 g/cm3
30 boxes
a. 1.73 L
0.120 m 0.120 m 0.120 m
b. 9.2 g; 5.0 cm3
c. 60.4 kg; 1.88 104 dm3
d. 0.94 g/cm3; 5.3 104 m3
e. 2.5 103 kg; 2.7 106 cm3
2.8 g/cm3
a. 0.72 m
b. 2.5 103 atoms
1300 L/min
1.3 106 cal/h
5.44 g/ cm3
2.24 104 cm3
32 000 uses
2500 L
9.5 L/min
MOLE CONCEPT
1. a.
b.
c.
d.
e.
f.
2. a.
b.
c.
d.
e.
f.
3. a.
b.
c.
d.
e.
f.
Concept Review: Formulas
and Percentage Composition
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
CdS
AlF3
K2Cr2O7
CaSO3
CaSO4
C6H6
C2H4O2
P4O10
B2H6
C2H2
Si 46.75%, O 53.25%
19.99% C, 26.64% O, 46.65% N, 6.73% H
39.99% C, 6.73% H, 53.28%
(NH4)3PO4 has the greater percentage
of nitrogen.
15. Sphalerite, ZnS, has the greater percentage of zinc.
3.7 104 mol Pd
150 mol Fe
0.040 mol Ta
5.38 105 mol Sb
41.1 mol Ba
3.51 108 mol Mo
52.10 g Cr
1.5 104 g or 15 kg Al
8.23 107 g Ne
3 102 g or 0.3 kg Ti
1.1 g Xe
2.28 105 g or 228 kg Li
1.02 1025 atoms Ge
3.700 1023 atoms Cu
1.82 1024 atoms Sn
1.2 1030 atoms C
1.1 1021 atoms Zr
1.943 1014 atoms K
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4. a.
b.
c.
d.
e.
f.
5. a.
b.
c.
d.
e.
f.
6. a.
b.
c.
d.
e.
f.
7. a.
b.
c.
d.
e.
f.
8. a.
b.
c.
d.
e.
f.
9. a.
b.
c.
d.
e.
f.
10. a.
b.
c.
d.
e.
11. a.
b.
c.
d.
e.
f.
10.00 mol Co
0.176 mol W
4.995 105 mol Ag
1.6 1015 mol Pu
7.66 107 mol Rn
1 1011 mol Ce
2.5 1019 atoms Au
5.10 1024 atoms Mo
4.96 1020 atoms Am
3.011 1026 atoms Ne
2.03 1018 atoms Bi
9.4 1016 atoms U
117 g Rb
223 g Mn
2.11 105 g Te
2.6 103 g Rh
3.31 108 g Ra
8.71 105 g Hf
0.749 mol CH3COOH
0.0213 mol Pb(NO3)2
3 104 mol Fe2O3
2.66 104 mol C2H5NH2
1.13 105 mol C17H35COOH
378 mol (NH4)2SO4
764 g SeOBr2
4.88 104 g CaCO3
2.7 g C20H28O2
9.74 106 g C10H14N2
529 g Sr(NO3)2
1.23 103 g UF6
2.57 1024 formula units WO3
1.81 1021 formula units Sr(NO3)2
4.37 1025 molecules C6H5CH3
3.08 1017 molecules C29H50O2
9.0 1026 molecules N2H4
5.96 1023 molecules C6H5NO2
1.14 1024 formula units FePO4
6.4 1019 molecules C5H5N
6.9 1020 molecules
(CH3)2CHCH2OH
8.7 1017 formula units
Hg(C2H3O2)2
5.5 1019 formula units Li2CO3
52.9 g F2
1.19 103 g or 1.19 kg BeSO4
1.388 105 g or 138.8 kg CHCl3
9.6 1012 g Cr(CHO2)3
6.6 104 g HNO3
2.38 104 g or 23.8 kg C2Cl2F4
12. 0.158 mol Au
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
0.159 mol Pt
0.288 mol Ag
0.234 mol C6H5OH
3.8 g I2
1.00 1022 atoms C
a. 0.0721 mol CaCl2
55.49 mol H2O
b. 0.0721 mol Ca2
0.144 mol Cl
a. 1.325 mol C12H22O11
b. 7.762 mol NaCl
0.400 mol ions
4.75 mol atoms
a. 249 g H2O
b. 13.8 mol H2O
c. 36.1 mL H2O
d. 36.0 g H2O
The mass of a sugar molecule is much
greater than the mass of a water molecule. Therefore, the mass of 1 mol of
sugar molecules is much greater than
the mass of 1 mol of water molecules.
1.52 g Al
0.14 mol O2
a. 0.500 mol Ag
0.250 mol S
b. 0.157 mol Ag2S
0.313 mol Ag
0.157 mol S
c. 33.8 g Ag
5.03 g S
PERCENTAGE COMPOSITION
1. a. HNO3
1.60% H
22.23% N
76.17% O
b. NH3
82.22% N
17.78% H
c. HgSO4
67.616% Hg
10.81% S
21.57% O
d. SbF5
56.173% Sb
43.83% F
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2. a. 7.99% Li
92.01% Br
b. 94.33% C
5.67% H
c. 35.00% N
3.
4.
5.
6.
7.
8.
5.05% H
59.96% O
d. 2.15% H
29.80% N
68.06% O
e. 87.059% Ag
12.94% S
f. 32.47% Fe
13.96% C
16.29% N
37.28% S
g. LiC2H3O2
10.52% Li
36.40% C
4.59% H
48.49% O
h. Ni(CHO2)2
39.46% Ni
16.15% C
1.36% H
43.03% O
a. 46.65% N
b. 23.76% S
c. 89.491% Tl
d. 39.17% O
e. 79.95% Br in CaBr2
f. 78.767% Sn in SnO2
a. 1.47 g O
b. 26.5 metric tons Al
c. 262 g Ag
d. 0.487 g Au
e. 312 g Se
f. 3.1 104 g Cl
a. 40.55% H2O
b. 43.86% H2O
c. 20.70% H2O
d. 28.90% H2O
a. Ni(C2H3O2)24H2O
23.58% Ni
b. Na2CrO44H2O
22.22% Cr
c. Ce(SO4)24H2O
34.65% Ce
43.1 kg Hg
malachite: 5.75 102 kg Cu
chalcopyrite: 3.46 102 kg Cu
malachite has a greater Cu content
9. a. 25.59% V
b. 39.71% Sn
c. 22.22% Cl
10. 319.6 g anhydrous CuSO4
11. 1.57 g AgNO3
12. 54.3 g Ag
8.08 g S
13. 23.1 g MgSO47H2O
14. 3.27 102 g S
EMPIRICAL FORMULAS
1. a. BaCl2
b. BiO3H3 or Bi(OH)3
c. AlN3O9 or Al(NO3)3
d. ZnC4H6O4 or Zn(CH3COO)2
e. NiN2S2H8O8 or Ni(NH4)2SO4
f. C2HBr3O2 or CBr3COOH
2. a. CuF2
b. Ba(CN)2
c. MnSO4
3. a. NiI2
b. MgN2O6 or Mg(NO3)2
c. MgS2O3, magnesium thiosulfate
d. K2SnO3, potassium stannate
4. a. As2S3
b. Re2O7
c. N2H4O3 or NH4NO3
d. Fe2Cr3O12 or Fe2(CrO4)3
e. C5H9N3
f. C6H5F2N or C6H3F2NH2
5. a. C6H12S3
b. C8H16O4
c. C4H6O4
d. C12H12O6
6. a. C4H4O4
b. C4H8O2
c. C9H12O3
7. K2S2O5, potassium metabisulfite
8. Pb3O4
9. Cr2S3O12 or Cr2(SO4)3, chromium(III)
sulfate
10. C9H6O4
11. C5H9N3, the empirical formula and the
molecular formula are the same
12. The molecular formulas of the
compounds are different multiples of
the same empirical formula. (FYI: The
first could be acetic acid, C2H4O2, and
the second could be glucose, C6H12O6,
or some other simple sugar.)