1 chapter 7 chemical quantities 7.4 percent composition and empirical formulas basic chemistry...
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Chapter 7 Chemical Quantities
7.4 Percent Composition and
Empirical Formulas
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
The label on a bag of fertilizer states the percentages of N, P, and K.
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Percent Composition
Percent composition
• is the percent by mass of each element in a formula.
Example: Calculate the percent composition of CO2.
CO2 = 1C(12.01g) + 2O(16.00 g) = 44.01 g/mol
12.01 g C x 100 = 27.29% C 44.01 g CO2
32.00 g O x 100 = 72.71% O 44.01 g CO2 100.00 %
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Calculating Percent Composition
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What is the percent composition of lactic acid, C3H6O3, a compound that appears in the blood after vigorous activity?
Learning Check
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STEP 1 Determine the total mass of each element in the molar mass of a formula.
3C(12.01) = 36.03 g of C
+ 6H(1.008) = 6.048 g of H
+ 3O(16.00) = 48.00 g of O
= 90.08 g/mol
Solution
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STEP 2 Divide the total mass of each element by the molar mass and multiply by 100%.
%C = 36.03 g C x 100 = 40.00% C 90.08 g
%H = 6.048 g H x 100 = 6.714% H 90.08 g
%O = 48.00 g O x 100 = 53.29% O 90.08 g
Solution (continued)
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Learning Check
The chemical isoamyl acetate C7H14O2 gives the odor of pears. What is the percent of carbon in isoamyl acetate?
1) 7.102 %C
2) 35.51 %C
3) 64.58 %C
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STEP 1 Determine the total mass of each element in the molar mass of a formula.
7C(12.01) = 84.07 g of C
+ 14H(1.008) = 14.11 g of H
+ 2O(16.00) = 32.00 g of O
Molar mass = 130.18 g/mol
Solution
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STEP 2 Divide the total mass of each element by the molar mass and multiply by 100%.
%C = total g C x 100% molar mass
%C = 84.07 g C x 100% = 64.58 %C 130.18 g
Solution (continued)
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The empirical formula
• is the simplest whole number ratio of the atoms
• is calculated by dividing the subscripts in the actual (molecular) formula by a whole number to give the lowest ratio
C5H10O5 5 = C1H2O1 = CH2Oactual (molecular) empirical formula
formula
Empirical Formulas
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Some Molecular and Empirical Formulas
• The molecular formula is the same or a multiple of the empirical.
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A. What is the empirical formula for C4H8?
1) C2H4 2) CH2 3) CH
B. What is the empirical formula for C8H14?
1) C4H7 2) C6H12 3) C8H14
C. Which is a possible molecular formula for CH2O?
1) C4H4O4 2) C2H4O2 3) C3H6O3
Learning Check
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A. What is the empirical formula for C4H8?
2) CH2 C4H8 4
B. What is the empirical formula for C8H14?
1) C4H7 C8H14 2
C. Which is a possible molecular formula for CH2O?
2) C2H4O2 3) C3H6O3
Solution
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A compound has an empirical formula SN. If there are four atoms of N in one molecule, what is the molecular formula? Explain.
1) SN
2) SN4
3) S4N4
Learning Check
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A compound has an empirical formula SN. If there are four atoms of N in one molecule, what is the molecular formula? Explain.
3) S4N4
In this molecular formula four atoms of N and four atoms of S and N are related 1:1. Thus, it has an empirical formula of SN.
Solution
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Calculating Empirical Formulas
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A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula.
Learning Check
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STEP 1 Determine the moles of each element.
7.31 g Ni x 1 mol Ni = 0.125 mol of Ni
58.69 g Ni
20.0 g Br x 1 mol Br = 0.250 mol of Br
79.90 g Br
STEP 2 Divide by the smallest number of moles.
0.125 mol Ni = 1 mol of Ni
0.125
0.250 mol Br = 2 mol of Br
0.125
Solution
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STEP 3 Use the lowest whole-number ratio of moles as subscripts.
NiBr2
Solution (continued)
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Converting Decimals to Whole Numbers
When the number of moles for an element is a
decimal • greater than 0.1, but less than 0.9• multiply the moles by a small integer to obtain
whole numbers
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Percent Composition Using 100 g
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Aspirin is 60.0% C, 4.5 % H, and 35.5 % O. Calculate its empirical (simplest) formula.
Learning Check
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STEP 1 Calculate the moles of each element.
100.0 g aspirin contains 60.0% C or 60.0 g of C, 4.5% H or 4.5 g of H, and 35.5% O or 35.5 g of O.
60.0 g C x 1 mol C = 5.00 mol of C 12.01 g C
4.5 g H x 1 mol H = 4.5 mol of H 1.008 g H
35.5 g O x 1mol O = 2.22 mol of O
16.00 g O
Solution
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Solution (continued)
STEP 2 Divide by the smallest number of moles.
5.00 mol C = 2.25 mol of C 2.22
4.5 mol H = 2.0 mol of H2.22 2.22 mol O = 1.00 mol of O2.22
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Solution (continued)
STEP 3 Use the lowest whole-number ratio of moles as subscripts.
To obtain whole numbers of moles, multiply by a factor, in this case x 4.
C: 2.25 mol of C x 4 = 9 mol of CH: 2.0 mol of H x 4 = 8 mol of HO: 1.00 mol of O x 4 = 4 mol of O
Using these whole numbers as subscripts, the simplest formula is
C9H8O4
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