Chemical Reactions: Mole and Mass Relationships Chapter 6 Lecture Fundamentals of General, Organic, and Biological Chemistry 8th Edition McMurry, Ballantine, Hoeger, Peterson Chapter Six Chemical Reactions: Mole and Mass Relationships Christina A. Johnson University of California, San Diego © 2017 Pearson Education, Inc.

Outline 6.1 The Mole and Avogadro’s Number 6.2 Gram-Mole Conversions 6.3 Mole Relationships and Chemical Equations 6.4 Mass Relationships and Chemical Equations 6.5 Limiting Reagent and Percent Yield

Concepts to Review Problem Solving: Unit Conversions and Estimating Answers Section 1.10 Molecular Formulas and Formula Units Sections 3.8 and 4.6 Balancing Chemical Equations Section 5.2

6.1 The Mole and Avogadro’s Number Atomic weight is the average mass of an element’s atoms. Learning Objective: Define the mole, and calculate the molar mass of a compound from the molecular formula.

6.1 The Mole and Avogadro’s Number Balanced chemical equations indicate what is happening at the molecular level during a reaction. To obtain the correct ratio of reactant molecules, the reactants must be weighed.

6.1 The Mole and Avogadro’s Number Molecular weight (MW) is the average mass of a substance’s molecules. A substance’s molecular weight (or formula weight for an ionic compound) is the sum of the atomic weights for all the atoms in the molecule or formula unit.

6.1 The Mole and Avogadro’s Number The molecular weight of ethylene (C2H4) is 28.0 amu:

6.1 The Mole and Avogadro’s Number Samples of different substances always contain the same number of molecules or formula units whenever their mass ratio is the same as their molecular or formula weight ratio.

6.1 The Mole and Avogadro’s Number A mole is the amount of a substance whose mass in grams is numerically equal to its molecular or formula weight. Molar mass: The mass in grams of 1 mole of a substance, numerically equal to molecular weight.

6.1 The Mole and Avogadro’s Number One mole of any substance contains 6.022 × 1023 formula units. This value is called Avogadro’s number (NA) after the Italian scientist who first recognized the importance of the mass/number relationship in molecules.

6.1 The Mole and Avogadro’s Number How big is Avogadro’s number?

6.1 The Mole and Avogadro’s Number These samples of sulfur, copper, mercury, and helium each contain 1 mol. They will not have the same mass even though the amount of moles is the same.

Worked Example 6.1 Pseudoephedrine hydrochloride (C10H16ClNO) is a nasal decongestant commonly found in cold medication. (a) What is the molar mass of pseudoephedrine hydrochloride? (b) How many molecules of pseudoephedrine hydrochloride are in a tablet that contains a dose of 30.0 mg of this decongestant?

Worked Example 6.1 Cont. ANALYSIS: We are given a mass and need to convert it to a number of molecules. This is most easily accomplished by using the molar mass of pseudoephedrine hydrochloride calculated in part (a) as the conversion factor from mass to moles and realizing that this mass (in grams) contains Avogadro’s number of molecules (6.022 × 1023).

Worked Example 6.1 Cont. Solution: (a) The molecular weight of pseudoephedrine hydrochloride is found by summing the atomic weights of all atoms in the molecule as follows:

30.0 mg pseudoephedrine hydrochloride Worked Example 6.1 Cont. Solution Continued: (b) This problem will involve unit conversions: STEP 1: Identify known information. We are given the mass of pseudoephedrine hydrochloride (in mg). 30.0 mg pseudoephedrine hydrochloride STEP 2: Identify answer and units. We are looking for the number of molecules of pseudoephedrine hydrochloride in a 30 mg tablet. ?? = molecules

Worked Example 6.1 Cont. Solution Continued: STEP 3: Identify conversion factors. Since the molecular weight of pseudoephedrine hydrochloride is 201.70 amu, 201.70 g contains 6.022 × 1023 molecules. We can use this ratio as a conversion factor to convert from mass to molecules. We will also need to convert 30 mg to grams.

Worked Example 6.1 Cont. Solution Continued: STEP 4: Solve. Set up an equation so that unwanted units cancel.

Worked Example 6.2 A tiny pencil mark just visible to the naked eye contains about 3 × 1017 atoms of carbon. What is the mass of this pencil mark in grams?

Worked Example 6.2 Cont. ANALYSIS: We are given a number of atoms and need to convert to mass. The conversion factor can be obtained by realizing that the atomic weight of carbon in grams contains Avogadro’s number of atoms (6.022 × 1023).

Worked Example 6.2 Cont. Solution STEP 1: Identify known information. We know the number of carbon atoms in the pencil mark. 3 × 1017 atoms of carbon STEP 2: Identify answer and units. Mass of carbon = ?? g

Worked Example 6.2 Cont. Solution Continued: STEP 3: Identify conversion factors. The atomic weight of carbon is 12.01 amu, so 12.01 g of carbon contains 6.022 × 1023 atoms.

Worked Example 6.2 Cont. Solution Continued: STEP 4: Solve. Set up an equation using the conversion factors so that unwanted units cancel.

6.2 Gram-Mole Conversions Molar mass = Mass of 1 mole of substance = Mass of 6.022 × 1023 molecules = Molecular weight in grams Learning Objective: Convert between mass and moles using the molar mass of a substance.

6.2 Gram-Mole Conversions Molar mass serves as a conversion factor between numbers of moles and mass.

Worked Example 6.3 The nonprescription pain relievers Advil and Nuprin contain ibuprofen (C13H18O2), whose molecular weight is 206.3 amu. If all the tablets in a bottle of pain reliever together contain 0.082 mol of ibuprofen, what is the number of grams of ibuprofen in the bottle?

Worked Example 6.3 Cont. ANALYSIS: We are given a number of moles and asked to find the mass. Molar mass is the conversion factor between the two.

Worked Example 6.3 Cont. Solution: STEP 1: Identify known information. 0.082 mol ibuprofen in bottle STEP 2: Identify answer and units. mass ibuprofen in bottle = ?? g STEP 3: Identify conversion factor. We use the molecular weight of ibuprofen to convert from moles to grams.

Worked Example 6.3 Cont. Solution Continued: STEP 4: Solve. Set up an equation using the known information and conversion factor so that unwanted units cancel.

Worked Example 6.4 The maximum dose of sodium hydrogen phosphate (NaHPO4, MW = 142.0 amu) that should be taken in one day for use as a laxative is 3.8 g. How many moles of sodium hydrogen phosphate, how many moles of Na+ ions, and how many total moles of ions are in this dose?

Worked Example 6.4 Cont. ANALYSIS: Molar mass is the conversion factor between mass and number of moles. The chemical formula Na2HPO4 shows that each formula unit contains 2 Na+ ions and 1 HPO42- ion.

Worked Example 6.4 Cont. Solution: STEP 1: Identify known information. We are given the mass and molecular weight of Na2HPO4. 3.8 g Na2HPO4; MW = 142.0 amu STEP 2: Identify answer and units. We need to find the number of moles of Na2HPO4 and the total number of moles of ions. Moles of Na2HPO = ?? mol Moles of Na+ ions = ?? mol Total moles of ions = ?? mol

Worked Example 6.4 Cont. Solution Continued: STEP 3: Identify conversion factor. We can use the molecular weight of Na2HPO4 to convert from grams to moles.

Worked Example 6.4 Cont. Solution Continued: STEP 4: Solve. We use the known information and conversion factor to obtain moles of Na2HPO4; since 1 mol of Na2HPO4 contains 2 mol of Na+ ions and 1 mol of HPO42– ions, we multiply these values by the number of moles in the sample.

6.3 Mole Relationships and Chemical Equations In chemical reactions, the unit to specify the relationship between reactants and products is the mole. Learning Objective: Determine molar ratios of reactants and products using balanced chemical equations.

6.3 Mole Relationships and Chemical Equations Coefficients in a chemical equation tell how many molecules, and thus how many moles, of each reactant are needed and how many of each product are formed.

6.3 Mole Relationships and Chemical Equations Coefficients can be put in the form of mole ratios, which act as conversion factors when setting up factor-label calculations. For the synthesis of ammonia 3 H2 + 2 N2  2 NH3 the mole ratios would be:

Worked Example 6.5 Rusting involves the reaction of iron with oxygen to form iron(III) oxide, Fe2O3: 4 Fe (s) + 3 O2 (g)  2 Fe2O3 (s) What are the mole ratios of the product to each reactant and of the reactants to each other? How many moles of iron(III) oxide are formed by the complete oxidation of 6.2 mol of iron?

Worked Example 6.5 Cont. ANALYSIS and Solution: (a) The coefficients of a balanced equation represent the mole ratios.

Worked Example 6.5 Cont. ANALYSIS and Solution Continued: (b) To find how many moles of Fe2O3 are formed, write down the known information—6.2 mol of iron—and select the mole ratio that allows the quantities to cancel, leaving the desired quantity.

6.4 Mass Relationships and Chemical Equations In chemical reactions, the unit to specify the relationship between reactants and products is the mole. Learning Objective: Using mole ratios, calculate the mass of product that can be formed from a given mass of reactant.

6.4 Mass Relationships and Chemical Equations Coefficients in a balanced chemical equation represent molecule-to-molecule or mole-to-mole relationships between reactants and products. Actual amounts of substances used in the laboratory are weighed out in grams. Three types of conversions are needed when doing chemical arithmetic.

6.4 Mass Relationships and Chemical Equations Mole-to-mole conversions are carried out using mole ratios as conversion factors.

6.4 Mass Relationships and Chemical Equations Mole-to-mass and mass-to-mole conversions are carried out using molar mass as a conversion factor.

6.4 Mass Relationships and Chemical Equations Mass-to-mass conversions cannot be carried out directly.

6.4 Mass Relationships and Chemical Equations There are four steps for determining mass relationships among reactants and products: STEP 1: Write the balanced chemical equation. STEP 2: Choose molar masses and mole ratios to convert the known information into the needed information. STEP 3: Set up the factor-label expression. STEP 4: Calculate the answer and check the answer against the ballpark estimate you made before you began your calculations.

Worked Example 6.6 In the atmosphere, nitrogen dioxide reacts with water to produce NO and nitric acid, which contributes to pollution by acid rain. 3 NO2 (g) + H2O (l)  2 HNO3 (aq) + NO (g) How many grams of HNO3 are produced for every 1.0 mole of NO2 that reacts? The molecular weight of HNO3 is 63.0 amu.

Worked Example 6.6 Cont. ANALYSIS: We are given the number of moles of a reactant and are asked to find the mass of a product. Problems of this sort always require working in moles and then converting to mass.

Worked Example 6.6 Cont. Solution: STEP 1: Write balanced equation. 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) STEP 2: Identify conversion factors. We need a mole to mole conversion to find the number of moles of product, and then a mole to mass conversion to find the mass of product. For the first conversion, we use the mole ratio of HNO3 to NO2 as a conversion factor, and for the mole to mass calculation, we use the molar mass of HNO3 (63.0 g/mol) as a conversion factor.

Worked Example 6.6 Cont. Solution Continued: STEP 3: Set up factor labels. Identify appropriate mole ratio factor labels to convert moles NO2 to moles HNO3 and moles HNO3 to grams. STEP 4: Solve. = 42 g HNO3

Worked Example 6.7 The following reaction produced 0.022 g of calcium oxalate (CaC2O4). What mass of calcium chloride was used as reactant? (The molar mass of CaC2O4 is 128.1 g/mol, and the molar mass of CaCl2 is 111.0 g/mol.) CaCl2 (aq) + Na2C2O4 (aq)  CaC2O4 (s) + 2NaCl (aq)

Worked Example 6.7 Cont. ANALYSIS: Both the known information and that to be found are masses, so this is a mass-to-mass conversion problem. The mass of CaC2O4 is first converted into moles, a mole ratio is used to find moles of CaCl2, and the number of moles of CaCl2 is converted into mass.

Worked Example 6.7 Cont. Solution: STEP 1: Write balanced equation. CaCl2(aq) + Na2C2O4(aq) CaC2O4(s) + 2 NaCl(aq) STEP 2: Identify conversion factors. Convert the mass of CaC2O4 into moles, use a mole ratio to find moles of CaCl2, and convert the number of moles of CaCl2 to mass. We will need three conversion factors.

Worked Example 6.7 Cont. Solution Continued: STEP 3: Set up factor-labels. We will need to perform gram to mole and mole to mole conversions to get from grams CaC2O4 to grams CaCl2. STEP 4: Solve. = 0.019 g CaCl2

6.5 Limiting Reagent and Percent Yield Only rarely are all reactants converted to products. Learning Objective: Using mole ratios and the mass of reactants, calculate the theoretical yield and percent yield for a reaction.

6.5 Limiting Reagent and Percent Yield When running a chemical reaction, we don’t always have the exact amounts of reagents to allow all of them to react completely. The limiting reagent is the reactant that runs out first.

6.5 Limiting Reagent and Percent Yield Theoretical yield is the amount of product formed assuming complete reaction of the limiting reagent. Chemical reactions do not always yield the exact amount predicted. Actual yield is the amount of product actually formed in a reaction.

6.5 Limiting Reagent and Percent Yield Percent yield is the percent of the theoretical yield actually obtained from a chemical reaction.

Worked Example 6.8 The combustion of acetylene gas (C2H2) produces carbon dioxide and water as indicated in the following reaction: 2 C2H2 (g) + 5 O2 (g) → 4 CO2 (g) + 2 H2O (g) When 26.0 g of acetylene is burned in sufficient oxygen for complete reaction, the theoretical yield of CO2 is 88.0 g. Calculate the percent yield for this reaction if the actual yield is only 72.4 g CO2.

Worked Example 6.8 Cont. ANALYSIS: The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100. Solution:

Worked Example 6.9 The element boron is produced commercially by the reaction of boric oxide with magnesium at high temperature. B2O3 (l) + 3 Mg (s) → 2 B (s) + 3 MgO (s) What is the theoretical yield of boron when 2350 g of boric oxide is reacted with 3580 g of magnesium? The molar masses of boric oxide and magnesium are 69.6 g/mol and 24.3 g/mol, respectively.

Worked Example 6.9 Cont. ANALYSIS: To calculate theoretical yield, we first have to identify the limiting reagent. The theoretical yield in grams is then calculated from the amount of limiting reagent used in the reaction.

Worked Example 6.9 Cont. Solution: STEP 1: Identify known information. We have the masses and molar masses of the reagents. 2350 g B2O3, molar mass 69.6 g/mol 3580 g Mg, molar mass 24.3 g/mol STEP 2: Identify answer and units. We are solving for the theoretical yield of boron. Theoretical mass of B = ?? g

Worked Example 6.9 Cont. Solution Continued: STEP 3: Identify conversion factors. We can use the molar masses to convert from masses to moles of reactants (B2O3, Mg). From moles of reactants, we can use mole ratios from the balanced chemical equation to find the number of moles of B produced, assuming complete conversion of a given reactant. B2O3 is the limiting reagent, since complete conversion of this reagent yields less product (67.6 mol B formed) than does complete conversion of Mg (98.0 mol B formed).

Worked Example 6.9 Cont. Solution Continued: STEP 4: Solve. Once the limiting reagent has been identified (B2O3), the theoretical amount of B that should be formed can be calculated using a mole to mass conversion.

Worked Example 6.10 The reaction of ethylene with water to give ethyl alcohol (CH3CH2OH) occurs in 78.5% actual yield. How many grams of ethyl alcohol are formed by reaction of 25.0 g of ethylene? (For ethylene, MW = 28.0 amu for ethyl alcohol, MW = 46.0 amu.) H2C=CH2 + H2O → CH3CH2OH

Worked Example 6.10 Cont. ANALYSIS: Treat this as a typical mass relationship problem to find the amount of ethyl alcohol that can theoretically be formed from 25.0 g of ethylene, and then multiply the answer by 78.5% to find the amount actually formed.

Worked Example 6.10 Cont. Solution: The theoretical yield of ethyl alcohol is So, the actual yield is as follows:

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