Section 6.7: Empirical Formulas
Tutorial 1 Practice, page 292
1. (a) Given: % Al = 20.2 %; % Cl = 79.8%
Required: empirical formula of compound
Solution: A 100.0 g sample of this compound contains 20.2 g of aluminum and 79.8 g of
chlorine.
Step 1. Calculate the amount of each element in the 100.0 g sample.
! 1 mol $
nAl = (20.2 g) #
&
" 26.98 g %
nAl = 0.748 70 mol [2 extra digits carried]
! 1 mol $
nCl = (79.8 g) #
&
" 35.45 g %
nCl = 2.2511 mol [2 extra digits carried]
Step 2. Divide the amount of each element by the smallest amount.
nAl 0.748 70 mol
=
= 1.00
nAl 0.748 70 mol
nCl
=
2.2511 mol
= 3.01
nAl 0.748 70 mol
The ratio of aluminum to chlorine is 1:3. This ratio suggests an empirical formula of AlCl3.
Statement: The empirical formula of the compound is AlCl3.
(b) Given: % C = 18.4 %; % N = 21.5%; remainder is K
Required: empirical formula of compound
Solution: A 100.0 g sample of this compound contains 18.4 g of carbon, 21.5 g of nitrogen, and
60.01 g of potassium.
Step 1. Calculate the percentage of potassium in the compound.
% K = 100 % ! (% C + % N)
= 100 % ! (18.4 % + 21.5 %)
= 100 % ! 39.9 %
% K = 60.1 %
A 100.0 g sample of this compound contains 18.4 g of carbon, 21.5 g of nitrogen, and 60.01 g of
potassium.
Step 2. Calculate the amount of each element in the 100.0 g sample.
! 1 mol $
nC = (18.4 g) #
&
" 12.01 g %
nC = 1.5321 mol [2 extra digits carried]
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6.7-1
! 1 mol $
nN = (21.5 g) #
&
" 14.01 g %
nN = 1.5346 mol [2 extra digits carried]
! 1 mol $
nK = (60.1 g) #
&
" 39.10 g %
nK = 1.5371 mol [2 extra digits carried]
Since the amount of each element, to two decimal places, is the same, the three elements are in
the ratio 1:1:1. This ratio suggests an empirical formula of KCN.
Statement: The empirical formula of the compound is KCN.
(c) Given: % Al = 52.9 %; remainder is O
Required: empirical formula of compound
Solution: A 100.0 g sample of this compound contains 52.9 g of aluminum and 47.1 g oxygen.
Step 1. Calculate the amount of each element in the 100.0 g sample.
! 1 mol $
nAl = (52.9 g) #
&
" 26.98 g %
nAl = 1.9607 mol [2 extra digits carried]
! 1 mol $
nO = (47.1 g) #
&
" 16.00 g %
nO = 2.9438 mol [2 extra digits carried]
Step 2. Divide the amount of each element by the smallest amount.
nAl 1.9607 mol
=
= 1.00
nAl 1.9607 mol
nO
=
2.9438 mol
= 1.50
nAl 1.9607 mol
The ratio of aluminum to oxygen is 1:1.50. This ratio suggests an empirical formula of Al1O1.5.
Multiplying the subscripts of this formula by 2 gives Al2O3.
Statement: The empirical formula of the compound is Al2O3.
(d) Given: % C = 50.85 %; % H = 8.47 %; % O = 40.68 %
Required: empirical formula of compound
Solution: A 100.0 g sample of this compound contains 50.85 g of carbon, 8.47 g of hydrogen,
and 40.68 g of oxygen.
Step 1. Calculate the amount of each element in the 100.0 g sample.
! 1 mol $
nC = (50.85 g) #
&
" 12.01 g %
nC = 4.233 97 mol [2 extra digits carried]
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Chapter 6: Quantities in Chemical Formulas
6.7-2
! 1 mol $
nH = (8.47 g) #
&
" 1.01 g %
nH = 8.3861 mol [2 extra digits carried]
! 1 mol $
nO = (40.68 g) #
&
" 16.00 g %
nO = 2.542 50 mol [2 extra digits carried]
Step 2. Divide the amount of each element by the smallest amount.
nC 4.233 97 mol
=
= 1.67
nO 2.542 50 mol
nH
8.3861 mol
=
= 3.30
nO 2.542 50 mol
The ratio of carbon to hydrogen to oxygen is 1.67 : 3.30 : 1. This ratio suggests an empirical
formula of C1.67H3.3O1. Multiplying the subscripts of this formula by 3 gives C5H10O3.
Statement: The empirical formula of the compound is C5H10O3.
Section 6.7 Questions, page 293
1. (a) I would need the percentage composition of the compound.
(b) I would need the molar masses of the elements in the compound.
2. The compounds H2CO3, K2Cr2O7, C3H6O3N are empirical formulas because the elements in
each compound are in the simplest whole-number ratio. C2H4O2 is not an empirical formula
because its elements are not in the simplest whole-number ratio. The empirical formula of this
compound is CH2O.
3. (a) Both compounds have the same empirical formula NO2.
(b) The two compounds do not have the same empirical formula. The empirical formula of C3H6
is CH2 and C4H7 is already an empirical formula.
(c) Both compounds have the same empirical formula CH.
(d) Both compounds have the same empirical formula C6H5O.
4. (a) The multiplier is 2. The empirical formula is C2H7.
(b) The multiplier is 3. The empirical formula is C2H2O3.
(c) The multiplier is 2. The empirical formula is Na2S2O3.
(d) The multiplier is 3. The empirical formula is C4H11O3.
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Chapter 6: Quantities in Chemical Formulas
6.7-3
5. Given: % C = 40.9 %; % H = 4.55 %; remainder is O
Required: empirical formula of ascorbic acid
Solution: A 100.0 g sample of this compound contains 40.9 g of carbon, 4.55 g of hydrogen, and
54.55 g of oxygen.
Step 1. Calculate the amount of each element in the 100.0 g sample.
! 1 mol $
nC = (40.9 g) #
&
" 12.01 g %
nC = 3.4055 mol [2 extra digits carried]
! 1 mol $
nH = (4.55 g) #
&
" 1.01 g %
nH = 4.5050 mol [2 extra digits carried]
! 1 mol $
nO = (54.55 g) #
&
" 16.00 g %
nO = 3.409 38 mol [2 extra digits carried]
Step 2. Divide the amount of each element by the smallest amount.
nO 3.409 38 mol
=
= 1.00
nC
3.4055 mol
nH 4.5050 mol
=
= 1.32
nC 3.4055 mol
The ratio of carbon to hydrogen to oxygen is 1:1.32:1. This ratio suggests an empirical formula
of C1H1.32O1. Multiplying the subscripts of this formula by 3 gives C3H4O3.
Statement: The empirical formula of ascorbic acid is C3H4O3.
6. Given: Compound 1: % N = 46.7 %; % O = 53.3 %
Compound 2: % N = 30.4 %; % O = 69.6 %
Required: empirical formulas of Compound 1 and Compound 2
Solution: A 100.0 g sample of Compound 1 contains 46.7 g of nitrogen and 53.3 g oxygen.
A 100.0 g sample of Compound 2 contains 30.4 g of nitrogen and 69.6 g oxygen.
Step 1. Calculate the amount of each element in the 100.0 g sample.
Compound 1:
! 1 mol $
nN = (46.7 g) #
&
" 14.01 g %
nN = 3.3333 mol [2 extra digits carried]
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Chapter 6: Quantities in Chemical Formulas
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! 1 mol $
nO = (53.3 g) #
&
" 16.00 g %
nO = 3.3313 mol [2 extra digits carried]
Since the amount of each element, to two decimal places, is the same, nitrogen and oxygen are in
the ratio 1:1. This ratio suggests an empirical formula of NO.
Compound 2:
! 1 mol $
nN = (30.4 g) #
&
" 14.01 g %
nN = 2.1699 mol [2 extra digits carried]
! 1 mol $
nO = (69.6 g) #
&
" 16.00 g %
nO = 4.3500 mol [2 extra digits carried]
Step 2. Divide the amount of each element in Compound 2 by the smallest amount.
nO 4.3500 mol
=
= 2.00
nN 2.1699 mol
The ratio of nitrogen to oxygen is 1:2. This ratio suggests an empirical formula of NO2.
Statement: The empirical formula of Compound 1 is NO and the empirical formula of
Compound 2 is NO2.
7. Given: % C = 63.68 %; % H = 9.80 %; % N = 12.38 %; remainder is O
Required: empirical formula of nylon-6
Solution: A 100.0 g sample of this compound contains 63.68 g of carbon, 9.80 g of hydrogen,
12.38 g of nitrogen, and 14.14 g of oxygen.
Step 1. Calculate the amount of each element in the 100.0 g sample.
Compound 1:
! 1 mol $
nC = (63.68 g) #
&
" 12.01 g %
nC = 5.302 25 mol [2 extra digits carried]
! 1 mol $
nH = (9.80 g) #
&
" 1.01 g %
nH = 9.7030 mol [2 extra digits carried]
! 1 mol $
nN = (12.38 g) #
&
" 14.01 g %
nN = 0.883 655 mol [2 extra digits carried]
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Chapter 6: Quantities in Chemical Formulas
6.7-5
! 1 mol $
nO = (14.14 g) #
&
" 16.00 g %
nO = 0.883 750 mol [2 extra digits carried]
Step 2. Divide the amount of each element by the smallest amount.
nC
5.302 25 mol
=
= 6.00
nN 0.883 655 mol
nH
9.7030 mol
=
= 11.0
nN 0.883 655 mol
nO
=
0.883 750 mol
= 1.00
nN 0.883 655 mol
The ratio of carbon to hydrogen to oxygen to nitrogen is 6 : 11 : 1 : 1. This ratio suggests an
empirical formula of C6H11ON.
Statement: The empirical formula of Nylon-6 is C6H11ON.
8. Given: % Al = 75 %; remainder is C
Required: empirical formula of aluminum carbide
Solution: A 100.0 g sample of this compound contains 75 g of aluminum and 25 g of carbon.
Step 1. Calculate the amount of each element in the 100.0 g sample.
! 1 mol $
nAl = (75 g) #
&
" 26.98 g %
nAl = 2.780 mol [2 extra digits carried]
! 1 mol $
nC = (25 g) #
&
" 12.01 g %
nC = 2.082 mol [2 extra digits carried]
Step 2. Divide the amount of each element by the smallest amount.
nAl 2.780 mol
=
= 1.34
nC 2.082 mol
The ratio of aluminum to carbon is 1:1.34. This ratio suggests an empirical formula of Al1.34C.
Multiplying the subscripts of this formula by 3 gives Al4C3.
Statement: The empirical formula of aluminum carbide is Al4C3.
9. Given: mcompound = 21.7 g ; mHg = 20.1 g
Required: empirical formula of compound
Solution:
Step 1. Calculate the mass of oxygen in the compound.
mO = 21.7 g ! 20.1 g = 1.6 g
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Chapter 6: Quantities in Chemical Formulas
6.7-6
Step 2. Calculate the amount of each element in the 21.7 g sample.
! 1 mol $
nHg = (20.1 g) #
&
" 200.59 g %
nHg = 0.100 20 mol [2 extra digits carried]
! 1 mol $
nO = (1.6 g) #
&
" 16.00 g %
nO = 0.1000 mol [2 extra digits carried]
Since the amount of each element is the same to three decimal places, mercury and oxygen are in
the ratio 1 : 1. This ratio suggests an empirical formula of HgO.
Statement: The empirical formula of the compound is HgO.
10. (a) Given: % O = 25.8 %; remainder is X
Required: percentage of element X
Solution:
% X = 100 % – 25.8 % = 74.2 %
Statement: The percentage of element X in the compound is 74.2 %.
(b) Given: % X = 74.2 %; % O = 25.8 %; formula of compound = X2O
Required: identity of element X
Solution: A 100.0 g sample of X2O contains 74.2 g X and 25.8 g of oxygen.
Step 1. Calculate the amount of oxygen in the 100.0 g sample.
! 1 mol $
nO = (25.8 g) #
&
" 16.00 g %
nO = 1.6125 mol [2 extra digits carried]
Since the formula of the compound is X2O, the amount of X in 100.0 g sample must be twice the
amount of oxygen, which is 3.2250 mol.
m
Step 2. Use the formula n =
to calculate the molar mass of element X.
M
m
nX = X
MX
MX =
mX
nX
74.2 g
3.2250 mol
M X = 23.0 g/mol
The element with molar mass closest to this value is sodium.
Statement: The unknown element is sodium.
=
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Chapter 6: Quantities in Chemical Formulas
6.7-7
11. Answers may vary. Sample answers:
(a)
Glucose
empirical formula
CH2O
molecular formula
C6H12O6
structural formula
Fructose
CH2O
C6H12O6
(b) There is considerable evidence to suggest a link or correlation between the consumption of
fructose and obesity. In one study, laboratory rats were given water sweetened with high-fructose
corn syrup. The concentration of the fructose was half the concentration of fructose in soft
drinks. Another group of rats was given water sweetened with sucrose (table sugar) at twice the
concentration used for the fructose group. Researchers found that the fructose group showed
greater signs of obesity than the sucrose group. Sucrose consists of one molecule of glucose and
one molecule of fructose bonded together. Glucose in the bloodstream stimulates the production
of the hormone leptin, which helps to suppress the appetite and insulin, which helps the body
process glucose. Fructose, however, has no effect on leptin or insulin. Some human studies have
shown that drinking beverages sweetened with fructose led to greater buildup of fat around the
belly than the consumption of glucose-sweetened beverages.
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Chapter 6: Quantities in Chemical Formulas
6.7-8