Transcript 7.4 Percent Composition
Chapter 7 Chemical Quantities
7.4 Percent Composition and Empirical Formulas
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 CO 2 .
CO 2 = 1C(12.01g) + 2O(16.00 g) = 44.01 g/mol 12.01 g C x 100 44.01 g CO 2 = 27.29% C 32.00 g O x 100 44.01 g CO 2 = 72.71% O 100.00 % 2
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Calculating Percent Composition
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Learning Check
What is the percent composition of lactic acid, C 3 H 6 O 3 , a compound that appears in the blood after vigorous activity?
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Solution
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 5
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Solution (continued)
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 6
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Learning Check
The chemical isoamyl acetate C 7 H 14 O 2 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 7
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Solution
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 8
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Solution (continued)
STEP 2 Divide the total mass of each element by the molar mass and multiply by 100%.
%C = total g C molar mass x 100% %C = 84.07 g C x 100% = 64.58 %C 130.18 g 9
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Empirical Formulas
• • 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 C 5 H 10 O 5 5 = C 1 H 2 O 1 = CH 2 O actual (molecular) empirical formula formula 10
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•
Some Molecular and Empirical Formulas
The molecular formula is the same or a multiple of the empirical. 11
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Learning Check
A. What is the empirical formula for C 4 H 8 ?
1) C 2 H 4 2) CH 2 3) CH B. What is the empirical formula for C 8 H 14 ?
1) C 4 H 7 2) C 6 H 12 3) C 8 H 14 C. Which is a possible molecular formula for CH 2 O?
1) C 4 H 4 O 4 2) C 2 H 4 O 2 3) C 3 H 6 O 3 12
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Solution
A. What is the empirical formula for C 4 H 8 ?
2) CH 2 C 4 H 8 4 B. What is the empirical formula for C 8 H 14 ?
1) C 4 H 7 C 8 H 14 2 C. Which is a possible molecular formula for CH 2 O?
2) C 2 H 4 O 2 3) C 3 H 6 O 3 13
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Learning Check
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) SN 4 3) S 4 N 4 14
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Solution
A compound has an empirical formula SN. If there are four atoms of N in one molecule, what is the molecular formula? Explain.
3) S 4 N 4 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. 15
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Calculating Empirical Formulas
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Learning Check
A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula. 17
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Solution
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
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Solution (continued)
STEP 3 Use the lowest whole-number ratio of moles as subscripts.
NiBr 2 19
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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 20
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Percent Composition Using 100 g
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Learning Check
Aspirin is 60.0% C, 4.5 % H, and 35.5 % O. Calculate its empirical (simplest) formula. 22
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Solution
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 = 12.01 g C 5.00 mol of C 4.5 g H x 1 mol H = 1.008 g H 4.5 mol of H 35.5 g O x 1mol O 16.00 g O = 2.22 mol of O 23
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Solution (continued)
STEP 2 Divide by the smallest number of moles.
5.00 mol C 2.22 4.5 mol H 2.22 2.22 mol O 2.22
= = = 2.25 mol of C 2.0 mol of H 1.00 mol of O 24
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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 H: 2.0 mol of H x 4 = =
9 mol of C 8 mol of H
O: 1.00 mol of O x 4 =
4 mol of O
Using these whole numbers as subscripts, the simplest formula is C 9 H 8 O 4
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