Chemistry. Zumdahl, 7th edition
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1 Chemistry Zumdahl, 7th edition
2 CH3 Stoichiometry The violent chemical reaction of bromine and phosphorus. P.76
3 Contents 3.1 Counting by Weighing 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass 3.5 Percent Composition of Compounds 3.6 Determining the Formula of a Compound 3.7 Chemical Equations 3.8 Balancing Chemical 3.9 Stoichiometric Calculations: Amounts of Reactants and Products 3.10 Calculations Involving a Limiting Reactant P.76
4 Introduction Chemical reactions have a profound effect on our lives. There are many examples: Food is converted to energy in the human body; fuels and plastics are produced from petroleum; human insulin is produced in laboratories by bacteria; The central activity of chemistry is to understand chemical changes such as these, and the study of reactions occupies a central place in this book. P.77
5 chemical stoichiometry : the quantities of materials consumed and produced in chemical reactions. P.77
6 3.1 Counting by Weighing In reality, jelly beans are not identical. For example, let s assume that you weigh 10 beans individually and get the following results: Bean Mass g g g g g g g g g g P.77
7 Jelly beans can be counted by weighing. P.77
8 The average mass of the jelly beans. Average mass total mass of beans number of beans 5.0 g For purposes of counting, the objects behave as though they were all identical, as though they each actually had the average mass. P.78
9 3.2 Atomic Masses The first quantitative information about atomic masses came from the work of Dalton, Gay-Lussac, Lavoisier, Avogadro, and Berzelius. The modern system of atomic masses, instituted in 1961, is based on 12 C ( carbon twelve ) as the standard. In this system, 12 C is assigned a mass of exactly 12 atomic mass units (amu), and the masses of all other atoms are given relative to this standard. P.78
10 The most accurate method currently available for comparing the masses of atoms involves the use of the mass spectrometer. In this instrument, diagramed in Fig. 3.1, atoms or molecules are passed into a beam of high-speed electrons. P.78
11 Figure 3.1 (left) A scientist injecting a sample into a mass spectrometer. (above) Schematic diagram of a mass spectrometer. P.78
12 when 12 C and 13 C are analyzed in a mass spectrometer, the ratio of their masses is found to be Mass 13 C Mass 12 C Since the atomic mass unit is defined such that the mass of 12 C is exactly 12 atomic mass units, Mass of 13 C ( )(12 amu) amu Exact number by definition P.79
13 The average atomic mass for carbon is computed as follows: 98.89% of 12 amu 1.11% of amu (0.9889)(12 amu) (0.0111)( amu) In this text we will call the average mass for an element the average atomic mass or, simply, the atomic mass for that element. P.79
14 Even though natural carbon does not contain a single atom with mass 12.01, for stoichiometric purposes, we can consider carbon to be composed of only one type of atom with a mass of P.79
15 It is much easier to weigh out 600 hex nuts than count them one by one. P.79
16 When a sample of natural neon is injected into a mass spectrometer, the mass spectrum shown in Fig. 3.2 is obtained. The areas of the peaks or the heights of the bars indicate the relative abundances of Ne, Ne, and Ne atoms P.80
17 Figure 3.2 P.80
18 Figure 3.2 (a) Neon gas glowing in a discharge tube. The relative intensities of the signals recorded when natural neon is injected into a mass spectrometer, represented in terms of (b) peaks and (c) a bar graph. The relative areas of the peaks are ( 20 Ne), ( 21 Ne), and ( 22 Ne); natural neon is therefore 90.92% 20 Ne, 0.257% 21 Ne, and 8.82% 22 Ne. P.80
19 Sample Exercise 3.1 The Average Mass of an Element When a sample of natural copper is vaporized and injected into a mass spectrometer, the results shown in Fig. 3.3 are obtained. Use these data to compute the average mass of natural copper. (The mass values for 63 Cu and 65 Cu are amu and amu, respectively.) P.81
20 Sample Exercise 3.1 Solution The average mass of a copper atom is Mass spectrum of natural copper. See Exercises 3.27 and 3.28 P.81
21 3.3 The Mole mole: the number equal to the number of carbon atoms in exactly 12 grams of pure 12 C. Techniques such as mass spectrometry, which count atoms very precisely, have been used to determine this number as This number is called Avogadro s number. One mole of something consists of units of that substance. P.82
22 Since atoms are so tiny, a mole of atoms or molecules is a perfectly manageable quantity to use in a reaction (see Fig. 3.4). A sample of grams of natural carbon contains the same number of atoms as grams of natural helium. Both samples contain 1 mole of atoms ( ). Table 3.1 gives more examples that illustrate this basic idea. P.82
23 Figure 3.4 Proceeding clockwise from the top, samples containing one mole each of copper, aluminum, iron, sulfur, iodine, and (in the center) mercury. P.82
24 TABLE 3.1 Comparison of 1 Mole Samples of Various Elements P.83
25 Thus the mole is defined such that a sample of a natural element with a mass equal to the element s atomic mass expressed in grams contains 1 mole of atoms amu 1 g Exact number 1 amu = x g This relationship can be used to derive the unit factor needed to convert between atomic mass units and grams. P.82
26 Sample Exercise 3.2 Determining the Mass of a Sample of Atoms Americium is an element that does not occur naturally. It can be made in very small amounts in a device known as a particle accelerator. Compute the mass in grams of a sample of americium containing six atoms. (Am = 243) Solution The mass of six atoms is amu. (243 x 6 =1458) The mass of six americium atoms in grams is g. See Exercises 3.33 P.83
27 Sample Exercise 3.3 Determining Moles of Atoms (left) Pure aluminum. (right) Aluminum alloys are used for many high-quality bicycle components, such as this chain wheel. P.84
28 Sample Exercise 3.3 Compute both the number of moles of atoms and the number of atoms in a 10.0-g sample of aluminum. Solution The number of moles of aluminum atoms in 10.0 g is mol Al atoms. (10.0/26.98 = mol) The number of atoms in 10.0 g (0.371 mol) of aluminum is atoms. (0.371 x x = 2.23 x atoms) See Exercises 3.34 P.84
29 Sample Exercise 3.5 Calculating the Number of Moles and Mass Cobalt (Co) is a metal that is added to steel to improve its resistance to corrosion. Calculate both the number of moles in a sample of cobalt containing atoms and the mass of the sample. Fragments of cobalt metal. See Exercises 3.36 P.85
30 3.4 Molar Mass A chemical compound is a collection of atoms. The mass of 1 mole of methane can be found by summing the masses of carbon and hydrogen present: Mass of 1 mole C g Mass of 4 mole H g Mass of 1 mole CH g The molar mass of a substance is the mass in grams of one mole of the compound. P.86
31 Sample Exercise 3.6 Calculating Molar Mass Juglone (, a dye known for centuries, is produced from the husks of black walnuts. It is also a natural herbicide (weed killer) that kills off competitive plants around the black walnut tree but does not affect grass and other noncompetitive plants. The formula for juglone is C 10 H 6 O 3. a. Calculate the molar mass of juglone. b. A sample of g of pure juglone was extracted from black walnut husks. How many moles of juglone does this sample represent? P.86
32 Sample Exercise 3.6 Solution a. b g of pure juglone mol juglone P.86
33 Sample Exercise 3.6 See Exercises 3.39 through 3.42 P.86
34 Sample Exercise 3.7 Calculating Molar Mass Calcium carbonate (CaCO 3 ), also called calcite, is the principal mineral found in limestone, marble, chalk, pearls, and shells of marine animals such as clams. a. Calculate the molar mass of calcium carbonate. b. A certain sample of calcium carbonate contains 4.86 moles. What is the mass in grams of this sample? What is the mass of the CO 2 3 ions present? P.86
35 Sample Exercise 3.7 Solution a. b. The exact amount is 486 g CaCO 3 Calcite crystals. See Exercises 3.43 through 3.46 P.87
36 Sample Exercise 3.8 Molar Mass and Numbers of Molecules Isopentyl acetate (C 7 H 14 O 2 ) is the compound responsible for the scent of bananas. A molecular model of isopentyl acetate is shown in the margin below. Interestingly, bees release about 1 g ( g) of this compound when they sting. The resulting scent attracts other bees to join the attack. How many molecules of isopentyl acetate are released in a typical bee sting? How many atoms of carbon are present? P.88
37 Sample Exercise 3.8 Isopentyl acetate is released when a bee stings. Solution To find the number of molecules released in a sting, we must first determine the number of moles of isopentyl acetate in g: P.88
38 Sample Exercise 3.8 See Exercises 3.47 through 3.52 P.88
39 3.5 Percent Composition of Compounds There are two common ways of describing the composition of a compound: in terms of the numbers of its constituent atoms. The percentages (by mass) of its elements. For ethanol, which has the formula C 2 H 5 OH, the mass of each element present and the molar mass are obtained as follows: Mass of C g Mass of H g Mass of O g Mass of 1 mol C 2 H 5 OH g P.89
40 The mass percent (often called the weight percent) of carbon in ethanol can be computed by comparing the mass of carbon in 1 mole of ethanol to the total mass of 1 mole of ethanol and multiplying the result by 100: mass of C in 1 mol C Mass percent of C 2 H 5 OH 100% mass of 1 mol C 2 H 5 OH Mass percent of H 13.13% Mass percent of O 34.73% P.89
41 Sample Exercise 3.10 Calculating Mass Percent Penicillin, the first of a now large number of antibiotics (antibacterial agents), was discovered accidentally by the Scottish bacteriologist Alexander Fleming in 1928, but he was never able to isolate it as a pure compound. This and similar antibiotics have saved millions of lives that might have been lost to infections. Penicillin F has the formula C 14 H 20 N 2 SO 4. Compute the mass percent of each element. P.90
42 Sample Exercise 3.10 Solution The molar mass of penicillin F is computed as follows: C: g H: g N: g S: g O: g Mass of 1 mol C 14 H 20 N 2 SO g P.90
43 Sample Exercise 3.10 Mass percent of C 53.81% Mass percent of H 6.453% Mass percent of N 8.969% Mass percent of S 10.27% Mass percent of O 20.49% Reality Check: The percentages add up to 99.99%. P.90
44 Sample Exercise 3.10 Penicillin is isolated from a mold that can be grown in large quantities in fermentation tanks. See Exercises 3.61 and 3.62 P.90
45 3.6 Determining the Formula of a Compound A device is shown in Fig Suppose a substance has been prepared that is composed of carbon, hydrogen, and nitrogen. When gram of this compound is reacted with oxygen, gram of carbon dioxide (CO 2 ) and gram of water (H 2 O) are collected. P.91
46 Figure 3.5 A schematic diagram of the combustion device used to analyze substances for carbon and hydrogen. The sample is burned in the presence of excess oxygen, which converts all its carbon to carbon dioxide and all its hydrogen to water. These products are collected by absorption using appropriate materials, and their amounts are determined by measuring the increase in masses of the absorbents. P.91
47 The molar mass of CO 2 is C: g O: g Molar mass of CO g/mol The mass percent of carbon in this compound is 38.67% C. The molar mass of H 2 O is grams. P.91
48 The remainder must be nitrogen: % (38.67% 16.22%) 45.11% N %C %H In the present case, 38.67% carbon by mass means grams of carbon per grams of compound; 16.22% hydrogen means grams of hydrogen per grams of compound; and so on. P.92
49 To determine the formula, we must calculate the number of carbon atoms in grams of carbon, the number of hydrogen atoms in grams of hydrogen, and number of nitrogen atoms in grams of nitrogen. We can find the smallest whole-number ratio of atoms in this compound : C: 1 H: 5 N: 1 P.92
50 Any molecule that can be represented as (CH 5 N) n, where is an integer, has the empirical formula CH 5 N. To be able to specify the exact formula of the molecule involved, the molecular formula, we must know the molar mass. Suppose we know that this compound with empirical formula CH 5 N has a molar mass of g/mol. P.92
51 1C: g g 5H: g g 1N: g 14.01g Formula mass of CH 5 N g/mol Some examples where this is the case are shown in Fig P.93
52 Figure 3.6 Examples of substances whose empirical and molecular formulas differ. Notice that molecular formula = (empirical formula) n, where n is an integer. P.93
53 Sample Exercise 3.11 Determining Empirical and molecule Formulas Determine the empirical and molecular formulas for a compound that gives the following percentages upon analysis (in mass percent): 71.65% Cl 24.27% C 4.07% H The molar mass is known to be g/mol. Solution First, we convert the mass percents to masses in grams. The molar mass is known to be g/mol. P.93
54 Sample Exercise 3.11 The compound C l2 C 2 H 4 is called dichloroethane. There are two forms of this compound, shown in Fig The form on the right was formerly used as an additive in leaded gasoline See Exercises 3.57 and 3.58 P.94
55 Figure 3.7 The two forms of dichloroethane. P.94
56 Sample Exercise 3.12 Determining Empirical and molecule Formulas A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. The compound has a molar mass of g/mol. What are the compound s empirical and molecular formulas? Solution In g of this compound there are g of phosphorus and g of oxygen. In terms of moles, in g of the compound we have mol P and mol O P.94
57 Sample Exercise 3.12 This yields the formula PO 2.5. The empirical formula mass for P2O5 is The molecular formula is (P 2 O 5 ) 2, or P 4 O 10. The structural formula of this interesting compound is given in Fig See Exercises 3.59 P.94
58 Figure 3.8 The structure of P 4 O 10. Note that some of the oxygen atoms act as bridges between the phosphorus atoms. This compound has a great affinity for water and is often used as a desiccant, or drying agent. P.94
59 Empirical Formula Determination Since mass percentage gives the number of grams of a particular element per 100 grams of compound, base the calculation on 100 grams of compound. Each percent will then represent the mass in grams of that element. Determine the number of moles of each element present in 100 grams of compound using the atomic masses of the elements present. Divide each value of the number of moles by the smallest of the values. If each resulting number is a whole number (after appropriate rounding), these numbers represent the subscripts of the elements in the empirical formula. If the number obtained in the previous step are not whole numbers, multiply each number by an integer so that the results are all whole numbers. P.96
60 Molecular Formula Determination Method One Obtain the empirical formula. Compute the mass corresponding to the empirical formula. Calculate the ratio The integer from the previous step represents the numbers of empirical formula units in one molecule. When the empirical formula subscripts are multiplied by this integer, the molecular formula results. This procedure is summarized by the equation: P.96
61 Method Two Using the mass percentages and the molar mass, determine the mass of each element present in one mole of compound. Determine the number of moles of each element present in one mole of compound. The integers from the previous step represent the subscripts in the molecular formula. P.96
CH3 Stoichiometry. The violent chemical reaction of bromine and phosphorus. P.76
CH3 Stoichiometry The violent chemical reaction of bromine and phosphorus. P.76 Contents 3.1 Counting by Weighing 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass 3.5 Percent Composition of Compounds 3.6
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