1 Chapter 6 Chemical Quantities 6.5 Percent Composition and Empirical Formulas Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings.

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Chapter 6 Chemical Quantities

6.5 Percent

Composition and

Empirical Formulas

Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings

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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 = 1 C(12.01g) + 2 O(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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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

3C(12.01) + 6H(1.008) + 3O(16.00) = 90.08 g/mol

36.03 g C + 6.048 g H + 48.00 g O

STEP 2

%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

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Learning Check

The chemical isoamyl acetate C7H14O2 gives the odor of pears. What is the percent carbon in isoamyl acetate?

1) 7.102 %C

2) 35.51 %C

3) 64.58 %C Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings

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3) 64.58 %C

Molar mass C7H14O2 = 7C(12.01) + 14H(1.008)

+ 2O(16.00) = 130.18 g/mol

Total C = 7C(12.01) = g

% C = total g C x 100 total g

% C = 84.07 g C x 100 = 64.58 % C 130.18 g

Solution

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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.

Table 6.3

Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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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 4 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 4 atoms of N in one molecule, what is the molecular formula? Explain.

3) S4N4

In this molecular formula 4 atoms of N and 4 atoms of S and N are related 1:1. Thus, it has an empirical formula of SN.

Solution

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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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Convert 7.31 g Ni and 20.0 g Br to moles.

7.31 g Ni x 1 mol Ni = 0.125 mol Ni

58.69 g Ni

20.0 g Br x 1 mol Br = 0.250 mol Br

79.90 g Br

Divide by smallest:

0.125 mol Ni = 1 Ni 0.250 mol Br = 2 Br

0.125 0.125

Write ratio as subscripts: NiBr2

Solution

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Converting Decimals to Whole Numbers

When the number of moles for an element is adecimal, all the moles are multiplied by a small

integer to obtain whole number.

Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Table 6.4

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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 in 100 g.

100 g aspirin contains 60.0% C or 60.0 g C, 4.5% H or 4.5 g H, and 35.5% O or 35.5 g O.

60.0 g C x 1 mol C = 5.00 mol C 12.01 g C

4.5 g H x 1 mol H = 4.5 mol H 1.008 g H

35.5 g O x 1mol O = 2.22 mol O

16.00 g O

Solution

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Solution (continued)

STEP 2. Divide by the smallest number of mol.

5.00 mol C = 2.25 mol C (decimal)

2.22 4.5 mol H = 2.0 mol H2.222.22 mol O = 1.00 mole O2.22

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Solution (continued)

3. Use the lowest whole number ratio as subscripts

When the moles are not whole numbers, multiply by a factor to give whole numbers, in this case x 4.

C: 2.25 mol C x 4 = 9 mol CH: 2.0 mol H x 4 = 8 mol HO: 1.00 mol O x 4 = 4 mol O

Using these whole numbers as subscripts the simplest formula is

C9H8O4