Download 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.
1
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
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
44.01 g CO2
2
x 100
=
27.29% C
32.00 g O
x 100
44.01 g CO2
=
72.71% O
100.00 %
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
Calculating Percent
Composition
3
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
Learning Check
What is the percent composition of lactic acid,
C3H6O3, a compound that appears in the blood
after vigorous activity?
4
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
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
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
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 Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
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
7
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
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
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
Solution (continued)
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
9
Basic Chemistry
= 84.07 g C x 100% = 64.58 %C
130.18 g
Copyright © 2011 Pearson Education, Inc.
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
C5H10O5  5 = C1H2O1 = CH2O
actual (molecular)
empirical formula
formula
10
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
Some Molecular and Empirical
Formulas
• The molecular formula is the same or a multiple of
the empirical.
11
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
Learning Check
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
12
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
Solution
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
13
Basic Chemistry
3) C3H6O3
Copyright © 2011 Pearson Education, Inc.
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) SN4
3) S4N4
14
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
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) 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.
15
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
Calculating Empirical Formulas
16
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
Learning Check
A compound contains 7.31 g Ni and 20.0 g Br.
Calculate its empirical (simplest) formula.
17
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
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
18
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
Solution (continued)
STEP 3 Use the lowest whole-number ratio of
moles as subscripts.
NiBr2
19
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
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
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
Percent Composition Using
100 g
21
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
Learning Check
Aspirin is 60.0% C, 4.5 % H, and 35.5 % O.
Calculate its empirical (simplest) formula.
22
Basic Chemistry
Copyright © 2011 Pearson Education, Inc.
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
1 mol H
1.008 g H
=
4.5 mol of H
1mol O
16.00 g
=
2.22 mol of O
x
35.5 g O x
23
Basic Chemistry
O
Copyright © 2011 Pearson Education, Inc.
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
24
Basic Chemistry
=
2.25 mol of C
=
2.0 mol of H
=
1.00 mol of O
Copyright © 2011 Pearson Education, Inc.
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 C
H: 2.0 mol of H x 4
= 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
C9H8O4
Basic Chemistry
25
Copyright © 2011 Pearson Education, Inc.