Chapter 3 Mass Relationships in Chemical Reactions This chapter uses the concepts of conservation of mass to assist the student in gaining an understanding of chemical changes. Upon completion of Chapter 3, the student should be able to: 1) Convert between grams and atomic mass units (AMU s). 2) Calculate average atomic mass given the mass and natural abundance of each isotope. 3) Recall from memory Avogadro s number. 4) Determine the number of objects present in a given number of moles. 5) Convert between mass, number of moles, and number of atoms (molecules) of an element (compound). 6) Establish the molecular mass and molar mass given the molecular formula. 7) Sketch the main components of a mass spectrometer and comment on its use. 8) Compute the percent composition (mass percent) given the chemical formula for an ionic or molecular compound. 9) Describe the experimental procedure used to determine empirical formulas. 10) Establish the molecular formula given the mass of each element present (or mass percent of each element) and the compound s molar mass. 11) Balance chemical equations. 12) Interpret the meaning of chemical equations in terms of molecules, moles, and masses. 13) Distinguish between products and reactants in a chemical equation. 14) Predict the products formed by combustion reactions. 15) Use stoichiometric methods to predict the mass (number of moles) of the products formed given the mass of each reactant (number of moles of each reactant). 16) Use stoichiometric methods to deduce the limiting reagent, excess reagent, the amount of expected products produced, and the amount of excess reagent left over upon completion of the reaction given the mass (number of moles) of each reactant in the chemical equation. 17) Use stoichiometric methods to predict the theoretical yield and percent yield given the mass (number of moles) of each reactant and the actual yield of a reaction. 18) Calculate the mass (number of moles) of each reactant required given the percent yield and the mass (number of moles) of products desired. Section 3.1 Atomic Mass Sometimes atomic weight and atomic mass are used interchangeably; however, to decrease confusion and to be consistent, the term atomic weight is not used. The term atomic mass unit, or dalton, is an important concept for students who plan to enroll in materials science courses where the mass contained in a unit cell in the solid state is often determined. A helpful comparison can be made between calculating the average atomic mass and the semester grade for a course. For example, if 30% of the grade is based on the midterm, 20% on laboratory and 50% on the final and a student earns 80 on the midterm, 90 in laboratory, and a 96 on the final, the student s grade is (0.30) (80) + (0.20) (90) + (0.5) (96) = 90. A similar calculation can be done for the average atomic mass of Ne (see pages 74 and 75). Average molar mass of Ne = (0.9092) (19.9924) + (0.00257) (20.9940) + (0.0882) (21.9914) =20.2 g/mol
It is important for the student to think about the answer calculated. For example, if nearly 91% of Ne has a molar mass of 19.99 g/mol and the other two isotopes have a larger mass, then 20.2 g/mol seems reasonable. Section 3.2 Molar Mass of an Element and Avogadro s Number A fundamental concept is that 1 mole = 6.022 10 23 objects. One amu is defined as 1/12 the mass of one carbon-12 atom. We define one mole of carbon-12 (6.022 10 23 carbon-12 atoms) as 12 grams to get a molar mass of 12 g/mol. It is acceptable to say that the molar mass of carbon-12 is 12 grams; however, if 12 g/mol is used consistently, the student will find it easier to solve problems using the factor-label method introduced in Chapter 1. An error that students often make is to assume that since carbon-12 is 12 amu and also 12 grams then 1 amu = 1 gram. This error is less likely if we use 12 amu/atom and 12 g/mol since it is obvious that 1 atom is not the same as a mole. For the conversion between amu and grams, it is useful to remind the student that it takes a very large number (Avogadro s Number, 6.022 10 23 ) of amu to make one gram since the amu refers to an atom. Section 3.3 Molecular Mass As stated in Section 3.1, atomic mass and atomic weight are often used interchangeably. The same is true for molecular mass and molecular weight. Molecular mass is the correct term and should be used for consistency. In example 3.7, the factor 4 H atoms/1 molecule of (NH 2 ) 2 CO is used. Often this conversion is difficult for students to understand. If we start with a simpler example, maybe this concept will be easier. For example: How many oxygen atoms are there in one CO 2 molecule? (two oxygen atoms) How many oxygen atoms are there in one dozen CO 2 molecules? (two dozen or 24 oxygen atoms) How many oxygen atoms are there in one mole of CO 2 molecules? [two moles or 2(6.022 10 23 ) oxygen atoms] If that is clear, then how many hydrogen atoms are there in one molecule of (NH 2 ) 2 CO (four H atoms ) or how many H atoms in one mole of (NH 2 ) 2 CO [four moles or 4(6.022 10 23 ) H atoms] is easier to comprehend. Section 3.4 The Mass Spectrometer In Figure 3.3, it should be understood that the accelerating plates are negatively charged so that the positive ions will be accelerated toward them. See Section 3.1 of this manual for a discussion on calculating average molar mass. Section 3.5 Percent Composition of Compounds See Section 3.3 about a discussion on how to assist students in understanding the number of moles of an element in one mole of a compound.
In the Chemistry in Action section, Gold Fingerprinting by Mass Spectrometry, the mass spectrum shown does not include the mass spectrum of gold. Only Cd and Pb are shown to emphasize the relative abundance of these two trace elements in the gold specimen. Example 3.9 is representative of what is often done in the analysis of products formed by organic synthesis. The organic chemist makes a compound and has its empirical formula determined to give evidence that the product formed was the material desired. It should be noted that division by the smallest subscript forces at least one of the subscripts in the formula to be one. It should be recognized that 1.33 is really 4/3 thus CH 1..33 O could be written as C 3/3 H 4/3 O 3/3 and the multiplication by three (the common denominator) will result in C 3 H 4 O 3. In a similar fashion, 1.66 would represent 5/3, 1.5 would represent 3/2, etc. Section 3.6 Experimental Determination of Empirical Formulas The study of ethanol s empirical formula assumes that ethanol contains only C, H, and O. If the sample contained something other than C, H, and O (sulfur, for example), then the assumption that the difference between the amount of the starting material (11.5 g ethanol) and the calculated masses of carbon and hydrogen in the ethanol (6.00 g and 1.51 g) to give the mass of oxygen in the sample would not be correct. In example 3.11, it may be useful to describe the following relationship: (empirical mass) (integer) = molar mass where empirical mass is the mass in grams of one mole of the material written as its empirical formula. Once the integer is found, it is used to multiply the subscripts of the empirical formula to obtain the molecular formula. For example, the empirical formula for acetylene is CH (13 g/mol is its empirical mass) while the molar mass of acetylene is 26 g/mol. Therefore (13 g/mol) (integer) = 26 g/mol integer = 2 the empirical formula, CH, becomes the molecular formula C 2 H 2. A similar example is benzene with its empirical formula of CH and its molar mass of 78 g/mol: (13 g/mol) (integer) = 78 g/mol integer = 6 so the empirical formula CH becomes the molecular formula C 6 H 6 when multiplied by six. Section 3.7 Chemical Reactions and Chemical Equations When chemical equations are balanced, it is assumed that equal numbers of atoms of a given element appear as reactants and products. This is a direct result of Dalton s atomic theory (Section 2.1) which states that chemical reactions involve the combination, separation, or rearrangement of atoms, but not the creation or destruction of atoms. Students learning how to balance chemical equations have a tendency to want to change
subscripts in the molecules; therefore, the first bullet on page 85 needs to be heavily stressed. The logic used to balance the O 2 in the combustion of C 2 H 6 often escapes students. Be sure to explain that 3.5 pairs of O 2 are needed to get the desired seven oxygen atoms as products. Note that the convention in this textbook is to use the smallest possible set of whole numbers in the balanced equation. Section 3.8 Amounts of Reactants and Products It is interesting to note that CO is a flammable gas. This point can have great implications in industrial settings that use CO. Figure 3.7 shows three common types of stoichiometric calculations. A fourth calculation may include number of molecules of reactant moles of reactant moles of product number of molecules of product The method of stringing factors along as shown in example 3.14 is used by many instructors who have a great deal of experience in solving this type of problem. However, many beginning students use the logic if A then B if B then C if C then D in problem solving; therefore, stringing out factors may be very confusing to them. It is suggested that the two methods be used interchangeably so that the student can see them both.
Section 3.9 Limiting Reagents Chemists often refer to limiting reagents, which confuses students because they have been working with products and reactants, not products and reagents. Be sure to explain that this is a convention which can be understood by substituting the word reactant for reagent. For the reaction: S(l) + 3F 2 (g) SF 6 (g) it would be better to express this as sulfur reacting with fluorine instead of sulfur burning in an atmosphere of fluorine because burning is often thought of as combining with O 2. The following fun example can be used to help students understand the concept of limiting reactants. We find a recipe for party cakes that requires one pint of milk and two eggs per cake. We have plenty of flour and sugar, but our refrigerator has only 1.5 gallons of milk and 1.5 dozen eggs. Since we don t want anyone to go without cake at our party, we need to know how many cakes we can make. [This is then a limiting reactant (reagent) problem.] 1cake 1pt milk 2 pt 1qt 4 qt 1gal ( 1.5 gal) = 12 cakes if all the milk is used up 1cake 2 eggs 12 eggs 1doz ( 1.5 doz) = 9 cakes if all the eggs are used up We then can only make nine cakes because we are limited by the number of eggs we have. If we make nine cakes, then how much milk will be left? 1pt 1cake ( 9 cakes) = 9 pts of milk used up We started with: 4 qt 1gal 2 pts 1qt ( 1.5 gal) = 12 pts of milk thus 12 minus 9 results in three pints of milk left over to drink with our cakes. The logic used to solve this fun problem is identical to that used in this section to solve chemical limiting reagent problems. Note that in the fun problem we used the logic of starting with the number of cakes made to determine the amount of excess milk left over. Another way to solve for the amount of excess milk would be to do the following:
12 eggs 1doz ( 1.5 doz eggs) = 18 eggs used 1pt 2 eggs ( 18 eggs used) = 9 pts of milk used 12 pts-9pts = 3 pts of milk left over This second method is more in line with the way the author of the textbook solved the limiting reagent problem in Example 3.15. It should be emphasized that one cannot assume that the reactant with the smallest mass is limiting. example, if 12.12 grams of H 2 are reacted with 16 grams of O 2 to form water, which reactant is limiting? For 2H 2 + O 2 2H 2 O 1mol H 2 mol H O 12.12 2 = 2 2 2.02 g H 2 2 mol H 2 ( g H ) 2 2 6 moles of H O if all the H are used 1mol O 2 mol H ) 16 2 = 2 2 32 g O 2 1mol O 2 ( g O ) 2 2 1mole H O if all the O are used Thus, even though the mass of H 2 was less than the mass of O 2, the O 2 is the limiting reactant. Section 3.10 Reaction Yield As defined, it is possible that the % yield could be greater than 100%. This doesn t usually happen, but it could as a result of the following: an error in calculation measuring the product wrong contamination of the product (maybe with solvent) the reaction not following the scheme outlined (using the wrong chemical equation) Therefore, if there is no error in calculation, a percent yield greater than 100% gives valuable insight into what is actually happening in the system.