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Chapter 6 Chemical Quantities

6.6 Molecular Formulas

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

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A molecular formula• Is a multiple (or equal) of its empirical formula.

• Has a molar mass that is the empirical formula mass multiplied by a whole number.

molar mass = a whole number empirical mass

• Is obtained by multiplying the empirical formula by a whole number.

Relating Molecular and Empirical Formulas

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Diagram of Molecular and Empirical Formulas

A small integer links• A molecular formula and its empirical

formula.• A molar mass and its empirical formula

mass.

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

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Determine the molecular formula of compound thathas a molar mass of 78.11 g and an empiricalformula of CH.

STEP 1. Empirical formula mass of CH = 13.02 g

STEP 2. Divide the molar mass by the empirical mass.

78.11 g = 5.999 ~ 6 13.02 g

STEP 3. Multiply each subscript in C1H1 by 6.

molecular formula = C1x 6 H1 x 6 = C6H6

Finding the Molecular Formula

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Some Compounds with Empirical Formula CH2O

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

Table 6.5

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A compound has a molar mass of 176.1g and an empirical formula of C3H4O3. What is the molecular formula?

1) C3H4O3

2) C6H8O6

3) C9H12O9

Learning Check

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A compound has a formula mass of 176.1 and an empirical formula of C3H4O3. What is the molecular formula?

2) C6H8O6

C3H4O3 = 88.06 g/EF

176.1 g (molar mass) = 2.00 88.06 g (empirical mass)

Molecular formula = 2 x empirical formulaC3 x 2H4 x 2O3 x 2 = C6H8O6

Solution

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A compound contains C 24.27%, H 4.07%, and Cl 71.65%. The molar mass is about 99 g. What are the empirical and molecular formulas?

STEP 1. Calculate the empirical formula.

Write the mass percents as the grams in a 100.00-g sample of the compound.

C 24.27 g H 4.07 g Cl 71.65 g

Molecular Formula

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Finding the Molecular Formula (Continued)Calculate the number of moles of each element.

24.27 g C x 1 mol C = 2.021 mol C 12.01 g C

4.07 g H x 1 mol H = 4.04 mol H 1.008 g H

71.65 g Cl x 1 mol Cl = 2.021 mol Cl 35.45 g Cl

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Finding the Molecular Formula (Continued)Divide by the smallest number of moles:

2.021 mol C = 1 mol C

2.021

4.04 mol H = 2 mol H

2.021

2.02 mol Cl = 1 mol Cl

2.021

Empirical formula = C1H2Cl1 = CH2Cl

Calculate empirical mass (EM) CH2Cl = 49.48 g

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Finding the Molecular Formula (Continued)

STEP 2. Divide molar mass by empirical mass.Molar mass = 99 g = 2

Empirical mass 49.48 g

STEP 3. Multiply the empirical formula by the small integer to determine the molecular formula.

2 x (CH2Cl)

C1 x2 H2 x 2 Cl1x 2 = C2H4Cl2

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A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g, what is the molecular formula?

Learning Check

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In 100. g, there are 27.4 g S, 12.0 g N, and 60.6 g Cl.

27.4 g S x 1 mol S = 0.854 mol S 32.07 g S

12.0 g N x 1 mol N = 0.857 mol N 14.01 g N

60.6 g Cl x 1mol Cl = 1.71 mol Cl 35.45 g Cl

Solution

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Divide by the smallest number of moles

0.854 mol S /0.854 = 1.00 mol S

0.857 mol N/0.854 = 1.00 mol N

1.71 mol Cl/0.854 = 2.00 mol Cl

empirical formula = SNCl2 = 116.98 g

Molar Mass/ Empirical mass

351 g = 3

116.98 g molecular formula = (SNCl2)3 = S3N3Cl6

Solution (continued)