Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 1 Chapter 7 Chemical Quantities 7.1 The Mole 7.2 Molar Mass 7.3 Calculations Using Molar Mass 7.4 Percent Composition and Empirical Formulas

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 2 A mole contains 6.02 x particles (atoms, ions, molecules, formula unit) The number 6.02 x is known as Avogadro’s number. One mole of any element contains Avogadro’s number of atoms. 1 mole Na = 6.02 x Na atoms 1 mole Au= 6.02 x Au atoms 7.1 A Mole

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 3 One mole of a covalent compound contains Avogadro’s number of molecules. 1 mole CO 2 = 6.02 x CO 2 molecules 1 mole H 2 O = 6.02 x H 2 O molecules One mole of an ionic compound contains Avogadro’s number of formula units. 1 mole NaCl = 6.02 x NaCl formula units A Mole of Molecules

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 4 Samples of One Mole Quantities

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 5 Avogadro’s number is written as conversion factors x particles and 1 mole 1 mole 6.02 x particles The number of molecules in 0.50 mole of CO 2 molecules is calculated as 0.50 mole CO 2 molecules x 6.02 x CO 2 molecules 1 mole CO 2 molecules = 3.0 x CO 2 molecules Avogadro’s Number

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 6 A. Calculate the number of atoms in 2.0 moles of Al. B. Calculate the number of moles of S in 1.8 x S. Learning Check

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 7 A. Calculate the number of atoms in 2.0 moles of Al. 2.0 moles Al x 6.02 x Al atoms 1 mole Al =1.2 x Al atoms B. Calculate the number of moles of S in 1.8 x S. 1.8 x S atoms x 1 mole S 6.02 x S atoms = 3.0 mole S atoms Solution

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 8 The mass of one mole is called molar mass (g/mole). The molar mass of an element is the atomic mass expressed in grams. 7.2 Molar Mass

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 9 Give the molar mass to the nearest 0.1 g. A. 1 mole of K atoms =________ B. 1 mole of Sn atoms =________ Learning Check

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 10 Give the molar mass to the nearest 0.1 g. A. 1 mole of K atoms =39.1 g B. 1 mole of Sn atoms =118.7 g Solution

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 11 Molar Mass of CaCl 2 For a compound, the molar mass is the sum of the molar masses of the elements in the formula. We calculate the molar mass of CaCl 2 to the nearest 0.1 g as follows. Formula mass of CaCl 2 = [ (35.45)] = 111.1amu Formula mass = molar mass, so 111.1amu = 111.1g/mol

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 12 Molar Mass of K 3 PO 4 Determine the molar mass of K 3 PO 4 to 0.1 g. Molar Mass = [3(39.1) (16)] = 212.3g/mol

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 13 One-Mole Quantities 32.1 g 55.9 g 58.5 g g g

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 14 A. 1 mole of K 2 O = ______g B. 1 mole of antacid Al(OH) 3 = ______g Learning Check

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 15 A. 1 mole of K 2 O 2 (39.1) + 1 (16.0) = 94.2 g/mol 1mole K 2 O x 94.2g/mol = 94.2g B. 1 mole of antacid Al(OH) 3 1 (27.0) + 3 (16.0) + 3 (1.0) = 78.0 g/mol 1mole Al(OH) 3 x 78.0 g/mol = 78.0g Solution

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 16 Prozac, C 17 H 18 F 3 NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac? Learning Check

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 17 Prozac, C 17 H 18 F 3 NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac? 17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0) = = 309 g/mole Solution

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 18 Methane CH 4 known as natural gas is used in gas cook tops and gas heaters. 1 mole CH 4 = 16.0 g The molar mass of methane can be written as conversion factors g CH 4 and 1 mole CH 4 1 mole CH g CH 4 Molar Mass Factors

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 19 Acetic acid C 2 H 4 O 2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid. Learning Check

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 20 Acetic acid C 2 H 4 O 2 gives the sour taste to vinegar. Write two molar mass factors for acetic acid. 2(12.0) + 4(1.0) + 2(16) = 60.0g/mol 1 mole of acetic acid = 60.0 g acetic acid 1 mole acetic acid and 60.0 g acetic acid 60.0 g acetic acid 1 mole acetic acid Solution

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 21  Mole factors are used to convert between the grams of a substance and the number of moles. 7.3 Calculations with Molar Mass Grams Mole factor Moles

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 22 Aluminum is often used for the structure of lightweight bicycle frames. How many grams of Al are in 3.00 moles of Al? 3.00 moles Al x 27.0 g Al = 81.0 g Al 1 mole Al mole factor for Al Calculating Grams from Moles

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 23 The artificial sweetener aspartame (Nutri-Sweet) C 14 H 18 N 2 O 5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame? Learning Check

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 24 Calculate the molar mass of C 14 H 18 N 2 O (12.0) + 18 (1.0) + 2 (14.0) + 5(16.0) = 294 g/mole Set up the calculation using a mole factor. 225 g aspartame x 1 mole aspartame 294 g aspartame mole factor(inverted) = mole aspartame Solution

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings Percent Composition In a compound, the percent composition is the percent by mass of each element in the formula. In one mole of CO 2 there are 12.0 g of C and 32.0 g of O (molar mass 44.0 g/mol), 12.0 g C x 100 = 27.3 % C 44.0 g CO g O x 100 = 72.7 % O 44.0 g CO %

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 26 What is the percent carbon in C 5 H 8 NNaO 4 (MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? 1) 7.10 %C 2) 35.5 %C 3) 60.0 %C Learning Check

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 27 2) 35.5 %C Molar mass = g % = total g C x 100 total g MSG = 60.0 g C x 100 = 35.5 % C g MSG Solution

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 28 The molecular formula is the true or actual number of the atoms in a molecule. The empirical formula is the simplest whole number ratio of the atoms. The empirical formula is calculated by dividing the subscripts in the 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 molecular empirical formula formula Types of Formulas

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 29 Some Molecular and Empirical Formulas

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 30 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 Learning Check

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 31 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 Solution

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 32 If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 1) SN 2) SN 4 3) S 4 N 4 Learning Check

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 33 If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 3) S 4 N 4 If the molecular formula has 4 atoms of N, and S and N are related 1:1, then there must also be 4 atoms of S. Solution

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 34 A molecular formula is equal to or a multiple of the empirical formula. Thus, the molar mass is equal to or a multiple of the empirical mass. molar mass = a whole number empirical mass Multiply the empirical formula by the whole number to determine the molecular formula. Relating Empirical and Molecular Formulas

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 35 Determine the molecular formula of a compound that has a molar mass of 78.0 and an empirical formula of CH. 1. Empirical mass of CH = 13.0 g/mol 2. Divide the molar mass by the empirical mass g/mol = g/mol 4. Multiply the subscripts in CH by Molecular formula = (CH) 6 = C 6 H 6 Finding the Molecular Formula

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 36 A compound has a formula mass of and an empirical formula of C 3 H 4 O 3. What is the molecular formula? Learning Check

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 37 A compound has a formula mass of and an empirical formula of C 3 H 4 O 3. What is the molecular formula? C 3 H 4 O 3 = 3(12.0) + 4(1.0) + 3(16.0) = 88.0 g/mol g/mol (molar mass) = g/mol(empirical mass) Molecular formula = 2 (empirical formula) (C 3 H 4 O 3 ) 2 = C 6 H 8 O 6 Solution

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 38 A compound is C 24.27%, H 4.07%, and Cl 71.65%. The molar mass is known to be 99.0 g. What are the empirical and molecular formulas? 1. Write the mass percents as the grams in a g sample of the compound. C g H 4.07 g Cl g Finding the Molecular Formula

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 39 Finding the Molecular Formula Continued 2. Calculate the number of moles of each element g C x 1 mole C = 2.02 moles C 12.0 g C 4.07 g H x 1 mole H = 4.03 moles H 1.01 g H g Cl x 1 mole Cl = 2.02 moles Cl 35.5 g Cl

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 40 Finding the Molecular Formula (continued) 3. Divide each by the smallest 2.02 moles C =1 mole C moles H =2 moles H moles Cl =1 mole Cl 2.02 Empirical formula = C 1 H 2 Cl 1 = CH 2 Cl

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 41 Finding the Molecular Formula (continued) 4. Calculate empirical mass (EM) empirical mass CH 2 Cl = 49.5 g/mol 5. Divide molar mass by empirical mass Molar mass = 99.0 g/mol = 2 Empirical mass 49.5 g/mol 6. Determine Molecular formula (CH 2 Cl) 2 = C 2 H 4 Cl 2

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 42 Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical (simplest) formula. Learning Check

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 43 In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O g C x 1 mole C = 5.00 moles C 12.0 g C 4.5 g H x 1 mole H = 4.5 moles H 1.01 g H 35.5 g O x 1mole O = 2.22 moles O 16.0 g O Solution (continued)

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 44 Solution (continued) Divide by the smallest number of moles moles C= 2.25 moles C moles H = 2.0 moles H moles O = 1.00 mole O 2.22 Note that the results are not all whole numbers. To obtain whole numbers, multiply by a factor to give whole numbers.

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 45 Solution (continued) Multiply each number of moles by 4 C: 2.25 moles C x 4 = 9 moles C H: 2.0 moles Hx 4 = 8 moles H O: 1.00 mole O x 4 = 4 moles O Use the whole numbers as subscripts to obtain the simplest formula C 9 H 8 O 4

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 46 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

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 47 In g, there are 27.4 g S, 12.0 g N and 60.6 g Cl g S x 1 mole S = mole S 32.1 g S 12.0 g N x 1 mole N = moles N 14.0 g N 60.6 g Cl x 1mole Cl = 1.71 moles Cl 35.5 g Cl Solution

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 48 Dividing by the smallest number of moles mole S /0.854 = 1.00 mole S mole N/0.854 = 1.00 mole N 1.71 moles Cl/0.854 = 2.00 moles Cl Empirical formula = SNCl 2 = g/mol Molar Mass/ Empirical mass 351 = Molecular formula = (SNCl 2 ) 3 = S 3 N 3 Cl 6 Solution (continued)