# 1 Chapter 6 Chemical Quantities 6.6 Molecular Formulas Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings.

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1 Chapter 6 Chemical Quantities 6.6 Molecular Formulas Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

2 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

3 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

4 Determine the molecular formula of compound that has a molar mass of 78.11 g and an empirical formula 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 C 1 H 1 by 6. molecular formula = C 1x 6 H 1 x 6 = C 6 H 6 Finding the Molecular Formula

5 Some Compounds with Empirical Formula CH 2 O Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Table 6.5

6 A compound has a molar mass of 176.1g and an empirical formula of C 3 H 4 O 3. What is the molecular formula? 1) C 3 H 4 O 3 2) C 6 H 8 O 6 3) C 9 H 12 O 9 Learning Check

7 A compound has a formula mass of 176.1 and an empirical formula of C 3 H 4 O 3. What is the molecular formula? 2)C 6 H 8 O 6 C 3 H 4 O 3 = 88.06 g/EF 176.1 g (molar mass) = 2.00 88.06 g (empirical mass) Molecular formula = 2 x empirical formula C 3 x 2 H 4 x 2 O 3 x 2 = C 6 H 8 O 6 Solution

8 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

9 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

10 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 = C 1 H 2 Cl 1 = CH 2 Cl Calculate empirical mass (EM) CH 2 Cl = 49.48 g

11 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 (CH 2 Cl) C 1 x2 H 2 x 2 Cl 1x 2 = C 2 H 4 Cl 2

12 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

13 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

14 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 = SNCl 2 = 116.98 g Molar Mass/ Empirical mass 351 g = 3 116.98 g molecular formula = (SNCl 2 ) 3 = S 3 N 3 Cl 6 Solution (continued)