INTRODUCTORY CHEMISTRY Concepts and Critical Thinking


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1 INTRODUCTORY CHEMISTRY Concepts and Critical Thinking Sixth Edition by Charles H. Corwin Chapter 9 The Mole Concept by Christopher Hamaker 2011 Pearson Education, Inc. Chapter 9 1
2 Avogadro s Number Avogadro s number (symbol N) is the number of atoms in grams of carbon. Its numerical value is 6.02 x Therefore, a g sample of carbon contains 6.02 x carbon atoms Pearson Education, Inc. Chapter 9 2
3 The Mole The mole (mol) is a unit of measure for an amount of a chemical substance. A mole is Avogadro s number of particles, which is 6.02 x particles. 1 mol = Avogadro s number = 6.02 x units We can use the mole relationship to convert between the number of particles and the mass of a substance Pearson Education, Inc. Chapter 9 3
4 Analogies for Avogadro s Number The volume occupied by one mole of softballs would be about the size of Earth. One mole of Olympic shot put balls has about the same mass as that of Earth. One mole of hydrogen atoms laid side by side would circle Earth about 1 million times Pearson Education, Inc. Chapter 9 4
5 One Mole of Several Substances C 12 H 22 O 11 H 2 O lead mercury K 2 Cr 2 O 7 copper sulfur NaCl 2011 Pearson Education, Inc. Chapter 9 5
6 Mole Calculations We will be using the unit analysis method again. Recall the following steps: 1. Write down the unit requested. 2. Write down the given value. 3. Apply unit factor(s) to convert the given units to the desired units Pearson Education, Inc. Chapter 9 6
7 Mole Calculations I How many sodium atoms are in mol Na? 1. We want atoms of Na. 2. We have mol Na mole Na = 6.02 x atoms Na mol Na x 6.02 x atoms Na 1 mol Na = 7.22 x atoms Na 2011 Pearson Education, Inc. Chapter 9 7
8 Mole Calculations I, Continued How many moles of potassium are in 1.25 x atoms K? 1. We want moles K. 2. We have 1.25 x atoms K mole K = 6.02 x atoms K x atoms K x 1 mol K 6.02 x atoms K = 2.08 x 103 mol K 2011 Pearson Education, Inc. Chapter 9 8
9 Molar Mass The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance. The atomic mass of iron is amu. Therefore, the molar mass of iron is g/mol. Since oxygen occurs naturally as a diatomic, O 2, the molar mass of oxygen gas is two times g or g/mol Pearson Education, Inc. Chapter 9 9
10 Calculating Molar Mass The molar mass of a substance is the sum of the molar masses of each element. What is the molar mass of magnesium nitrate, Mg(NO 3 ) 2? The sum of the atomic masses is as follows: ( ) = (62.01) = amu The molar mass for Mg(NO 3 ) 2 is g/mol Pearson Education, Inc. Chapter 9 10
11 Mole Calculations II Now we will use the molar mass of a compound to convert between grams of a substance and moles or particles of a substance x particles = 1 mol = molar mass If we want to convert particles to mass, we must first convert particles to moles, and then we can convert moles to mass Pearson Education, Inc. Chapter 9 11
12 Mole Mole Calculations What is the mass of 1.33 moles of titanium, Ti? 1. We want grams. 2. We have 1.33 moles of titanium. 3. Use the molar mass of Ti: 1 mol Ti = g Ti mole Ti x g Ti 1 mole Ti = 63.7 g Ti 2011 Pearson Education, Inc. Chapter 9 12
13 Mole Calculations III What is the mass of 2.55 x atoms of lead? 1. We want grams. 2. We have atoms of lead. 3. Use Avogadro s number and the molar mass of Pb atoms Pb x 1 mol Pb atoms Pb x g Pb 1 mole Pb = 87.9 g Pb 2011 Pearson Education, Inc. Chapter 9 13
14 Mole Calculations III, Continued How many O 2 molecules are present in g of oxygen gas? 1. We want molecules O We have grams O Use Avogadro s number and the molar mass of O g O 2 x 1 mol O x10 23 molecules O x g O 2 1 mole O x molecules O Pearson Education, Inc. Chapter 9 14
15 Mass of a Single Molecule What is the mass of a single molecule of sulfur dioxide? The molar mass of SO 2 is g/mol. We want mass/molecule SO 2, we have the molar mass of sulfur dioxide. Use Avogadro s number and the molar mass of SO 2 as follows : g SO 2 1 mole SO x 2 1 mol SO x molecules SO x g/molecule 2011 Pearson Education, Inc. Chapter 9 15
16 Molar Volume At standard temperature and pressure, 1 mole of any gas occupies 22.4 L. The volume occupied by 1 mole of gas (22.4 L) is called the molar volume. Standard temperature and pressure are 0 C and 1 atm Pearson Education, Inc. Chapter 9 16
17 Molar Volume of Gases We now have a new unit factor equation: 1 mole gas = 6.02 x molecules gas = 22.4 L gas 2011 Pearson Education, Inc. Chapter 9 17
18 One Mole of a Gas at STP The box below has a volume of 22.4 L, which is the volume occupied by 1 mole of a gas at STP Pearson Education, Inc. Chapter 9 18
19 Gas Density The density of gases is much less than that of liquids. We can easily calculate the density of any gas at STP. The formula for gas density at STP is as follows: molar mass in grams molar volume in liters = density, g/l 2011 Pearson Education, Inc. Chapter 9 19
20 Calculating Gas Density What is the density of ammonia gas, NH 3, at STP? First we need the molar mass for ammonia (1.01) = g/mol The molar volume NH 3 at STP is 22.4 L/mol. Density is mass/volume g/mol 22.4 L/mol = g/l 2011 Pearson Education, Inc. Chapter 9 20
21 Molar Mass of a Gas We can also use molar volume to calculate the molar mass of an unknown gas g of an unknown gas occupies 1.00 L at STP. What is the molar mass? We want g/mol; we have g/l g 1.00 L x 22.4 L 1 mole = 43.9 g/mol 2011 Pearson Education, Inc. Chapter 9 21
22 Mole Unit Factors We now have three interpretations for the mole: 1. 1 mol = 6.02 x particles mol = molar mass mol = 22.4 L at STP for a gas. This gives us three unit factors to use to convert among moles, particles, mass, and volume Pearson Education, Inc. Chapter 9 22
23 Calculating Molar Volume A sample of methane, CH 4, occupies 4.50 L at STP. How many moles of methane are present? We want moles; we have volume. Use molar volume of a gas: 1 mol = 22.4 L L CH 4 x = mol CH 4 1 mol CH L CH Pearson Education, Inc. Chapter 9 23
24 Calculating Mass Volume What is the mass of 3.36 L of ozone gas, O 3, at STP? We want mass O 3 ; we have 3.36 L O 3. Convert volume to moles, then moles to mass. 1 mol O 3.36 L O 3 x g O x L O 3 1 mol O 3 = 7.20 g O Pearson Education, Inc. Chapter 9 24
25 Calculating Molecule Volume How many molecules of hydrogen gas, H 2, occupy L at STP? We want molecules H 2 ; we have L H 2. Convert volume to moles, and then moles to molecules L H 2 x 1 mol H x10 23 molecules H x L H 2 1 mole H 2 = 1.34 x molecules H Pearson Education, Inc. Chapter 9 25
26 Percent Composition The percent composition of a compound lists the mass percent of each element. For example, the percent composition of water, H 2 O is 11% hydrogen and 89% oxygen. All water contains 11% hydrogen and 89% oxygen by mass Pearson Education, Inc. Chapter 9 26
27 Calculating Percent Composition There are a few steps to calculating the percent composition of a compound. Let s practice using H 2 O. 1. Assume you have 1 mole of the compound. 2. One mole of H 2 O contains 2 mol of hydrogen and 1 mol of oxygen. Therefore, 2(1.01 g H) + 1(16.00 g O) = molar mass H 2 O 2.02 g H g O = g H 2 O 2011 Pearson Education, Inc. Chapter 9 27
28 Calculating Percent Composition, Continued Next, find the percent composition of water by comparing the masses of hydrogen and oxygen in water to the molar mass of water g H g H 2 O g O g H 2 O x 100% = 11.2% H x 100% = 88.79% O 2011 Pearson Education, Inc. Chapter 9 28
29 Percent Composition Problem TNT (trinitrotoluene) is a white crystalline substance that explodes at 240 C. Calculate the percent composition of TNT, C 7 H 5 (NO 2 ) 3. 7(12.01 g C) + 5(1.01 g H) + 3 (14.01 g N g O) = g C 7 H 5 (NO 2 ) g C g H g N g O = g C 7 H 5 (NO 2 ) Pearson Education, Inc. Chapter 9 29
30 Percent Composition of TNT g C g TNT 1.01 g H g TNT g N g TNT g O g TNT x 100% = 37.01% C x 100% = 2.22% H x 100% = 18.50% N x 100% = 42.26% O 2011 Pearson Education, Inc. Chapter 9 30
31 Empirical Formulas The empirical formula of a compound is the simplest whole number ratio of ions in a formula unit or atoms of each element in a molecule. The molecular formula of benzene is C 6 H 6. The empirical formula of benzene is CH. The molecular formula of octane is C 8 H 18. The empirical formula of octane is C 4 H Pearson Education, Inc. Chapter 9 31
32 Calculating Empirical Formulas We can calculate the empirical formula of a compound from its composition data. We can determine the mole ratio of each element from the mass to determine the empirical formula of radium oxide, Ra? O?. A g sample of radium metal was heated to produce g of radium oxide. What is the empirical formula? We have g Ra and = g O Pearson Education, Inc. Chapter 9 32
33 Calculating Empirical Formulas, Continued The molar mass of radium is g/mol, and the molar mass of oxygen is g/mol. 1 mol Ra g Ra x = mol Ra g Ra 1 mol O g O x = mol O g O We get Ra O Simplify the mole ratio by dividing by the smallest number. We get Ra 1.01 O 1.00 = RaO is the empirical formula Pearson Education, Inc. Chapter 9 33
34 Empirical Formulas from Percent Composition We can also use percent composition data to calculate empirical formulas. Assume that you have 100 grams of sample. Acetylene is 92.2% carbon and 7.83% hydrogen. What is the empirical formula? If we assume 100 grams of sample, we have 92.2 g carbon and 7.83 g hydrogen Pearson Education, Inc. Chapter 9 34
35 Empirical Formula for Acetylene Calculate the moles of each element. 1 mol C 92.2 g C x = 7.68 mol C g C 1 mol H 7.83 g H x = 7.75 mol H 1.01 g H The ratio of elements in acetylene is C 7.68 H Divide by the smallest number to get the following formula: C H 7.68 = C 1.00 H 1.01 = CH 2011 Pearson Education, Inc. Chapter 9 35
36 Molecular Formulas The empirical formula for acetylene is CH. This represents the ratio of C to H atoms on acetylene. The actual molecular formula is some multiple of the empirical formula, (CH) n. Acetylene has a molar mass of 26 g/mol. Find n to find the molecular formula: (CH) n CH = 26 g/mol 13 g/mol n = 2 and the molecular formula is C 2 H Pearson Education, Inc. Chapter 9 36
37 Chapter Summary Avogadro s number is 6.02 x 10 23, and is 1 mole of any substance. The molar mass of a substance is the sum of the atomic masses of each element in the formula. At STP, 1 mole of any gas occupies 22.4 L Pearson Education, Inc. Chapter 9 37
38 Chapter Summary, Continued We can use the following flow chart for mole calculations: 2011 Pearson Education, Inc. Chapter 9 38
39 Chapter Summary, Continued The percent composition of a substance is the mass percent of each element in that substance. The empirical formula of a substance is the simplest whole number ratio of the elements in the formula. The molecular formula is a multiple of the empirical formula Pearson Education, Inc. Chapter 9 39
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