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CHAPTER H ousehold ammonia, chlorine bleach, and oxygen bleach can be found in almost everyone's cleaning closet. But ammonia, chlorine, and oxygen have many other uses. They are some of the most important reactants and products of the chemical industry. When ammonia gas is synthesized, nitrogen gas and hydrogen gas react by volume in the following way. 1 L Nig) + 3 L Hig) -+ 2 L NH 3 (g) According to the reaction above, 1 L of nitrogen and 3 L of hydrogen combine to form 2 L of ammonia. According to Avogadro's principle, the volumes of gas combine in simple whole-number ratios because 1 L of nitrogen contains as many particles as 1 L of hydrogen. But how many particles is that? The volumes of nitrogen, hydrogen, and ammonia gas are easily measured, but the particles in a gas are too small and too numerous to be counted easily. How can you determine the number of particles in a gas without counting? Concept Check Review the following concepts before studying this chapter. Chapter 4: the types of compounds Chapter 5: naming compounds; writing chemical formulas Chapter 6: writing and balancing chemical reactions Chapter 10: the factor label method; volume, pressure, and temperature relationships of gases 4 Reading Chemistry Go through the chapter, and write down the sections heads. As you read, jot down several key points or vocabulary under each section head. Also, for each section, write down any questions you form about the topics. Chemistry Around You Is it possible to count the atoms, formula units, and molecules of substances? When a liter of methanol, for example, is formed from hydrogen gas and carbon dioxide gas, how many molecules of methanol are formed? How many molecules of hydrogen gas and carbon dioxide gas reacted? To find out more about molar mass, visit the Glencoe Science Web site. science.glencoe.com 403 Counting Particles ofMatter L Objectives Compare and contrast the mole as a number and the mole as a mass. Relate counting particles to weighing samples of substances. Solve stoichiometric problems using molar mass. Key Terms stoichiometry mole Avogadro constant molar mass molecular mass formula mass Chapter 12 ouis Staffilino of Dillonvale, Ohio, saved pennies for 65 years. When he deposited them in a bank in 1994, he had 40 large drums of pennies. His wealth represented an enormous counting job for the bank teller. Just as a bank teller counts coins and bills, a chemist counts atoms, molecules, and formula units of substances. Unlike a bank teller, a chemist cannot count individual items because the particles of matter are too small and just too numerous. How do you determine the number of particles in a sample of matter without counting? Stoichiometry In Chapter 11, you learned that volumes of gases always combine in definite ratios. This observation, called the law of combining volumes, is based on measurements of the gas volumes. When Avogadro suggested that gases combine in fixed ratios because equal volumes of gases at the same temperature and pressure contain equal numbers of particles, he may have been thinking of particles rearranging themselves. Individual gas particles are so small that their rearranging cannot be observed, but the volumes of gases can be measured directly. Avogadro's principle is one of the earliest attempts to relate the number of particles in a sample of a substance to a direct measurement made on the sample. Today, by using the methods of stoichiometry> we can measure the amounts of substances involved in chemical reactions and relate them to one another. For example, a sample's mass or volume can be converted to a count of the number of its particles, such as atoms, ions, or molecules. Pennies are not small, but counting a drum full of pennies, like counting the number of particles in a gas, is a formidable task. Can we measure the pennies by grouping them in some conveniently large quantity and then counting the number of groups to find the total number of pennies in the drum? Can methods of stoichiometry help? Figure 12.1 presents two ways to count large numbers of pennies. To count the pennies directly, you would have to handle everyone-all 24 216 pennies. But you Chemical Quantities could also just make three measurements: the weight of the drum of pennies) the weight of the empty drum, and the weight of a group of 1000 pennies. Now, if you want to count atoms, what size group is most suitable? You will need a larger group than you used for the pennies-much larger than a thousand or even a million. Atoms are so tiny that an ordinary-sized sample of a substance contains so many of these submicroscopic particles that counting them by grouping them in thousands would be unmanageable. Even grouping them by millions would not help. The group or unit of measure used to count numbers of atoms, molecules) or formula units of substances is the mole (abbreviated mol). The number of things in one mole is 6.02 X 1023 , This big number has a short name: the Avogadro constant, as illustrated in Figure 12.2. The most precise value of the Avogadro constant is 6.0221367 x 1023 • For most purposes, rounding to 6.02 x 1023 is sufficient. Figure 12.1 Two Ways to Count Pennies You could count the pennies directly. ~ • You could calculate the number of pennies in a drum by using the mass of all the pennies, 70125 g, and the mass of 1000 pennies. The mass of 1000 pennies is 2890.7 g. Use the ratio 1000 pennies to find the total number of pennies. 2890.7 g 70 125 Jt X 1000 pennies 2890.7 g' = . 24259 pennies The drum contains 24259 pennies, worth $242.59. Figure 12.2 One Mole Is a BIG Number. If 6.02 X 1023 sheets of paper were placed in a stack, they would reach from Earth to the sun-more than a million times. The thickness of a sheet of paper is small, but an atom is much smaller. One mole of magnesium atoms is hardly a handful. 12.1 Counting Particles of Matter 405 All kinds of submicroscopic particles can be conveniently counted using the Avogadro constant. There are 6.02 X 10 23 carbon atoms in a mole of carbon, and there are 6.02 X 10 23 carbon dioxide molecules in a mole of carbon dioxide. There are 6.02 X 1023 sodium ions in a mole of sodium ions. For that matter) there are 6.02 X 1023 eggs in a mole of eggs, but eggs are so large compared with the particles that interact in chemistry that you would have no need to estimate how many there are in a typical sample. Can you see a relationship here? For counting many small things, use a large unit of measure; for counting fewer things or larger things, use a small unit of measure, as shown in Figure 12.3. Molar Mass stoichiometry stoichen (G K) element or part metreon (GK) measure Stoichiometry allows you to determine the number of atoms in a substance by relating other measurable quantities of that substance. , How many moles of methanol are in 500 g of methanol? Methanol is formed from CO 2 gas and hydrogen gas according to the balanced chemical equation below. CO 2 (g) ~ CH 30H(g) + H 20(g) Suppose you wanted to produce 500 g of methanol. How many grams of CO 2 gas and H 2 gas would you need? How many grams of water would be produced as a by-product? Those are questions about the masses of reactants and products. But the balanced chemical equation shows that three molecules of hydrogen gas react with one molecule of carbon dioxide gas. The equation relates molecules, not masses, of reactants and products. Like Avogadro, you need to relate the macroscopic measurements-the masses of carbon dioxide and hydrogen-to the number of molecules of methanol. To find the mass of carbon dioxide and the mass of hydrogen needed to produce 500 g of methanol, you first need to know how many molecules of methanol are in 500 g of methanol. • figure 12.3 Grouping Items for Counting The size of the group depends upon how many items we would buy or use at one time. Would you rather count the sheets of paper or the reams? 406 + 3H 2 (g) Chapter 12 Chemical Quantities Remember the drum of pennies? By knowing the mass of one group of 1000 pennies and knowing the mass of all the pennies, you can find the number of groups of 1000 pennies. Then, finding the total number of pennies is an easy matter. You have a suitable unit of measure, the mole, and you know the mass of methanol you want to produce. But, you still need to know the mass of 1 mol of methanol molecules. Six iron atoms Six carbon atoms Molar Mass of an Element Figure 12.4 You know from Chapter 2 that average atomic masses of the elements are given on the periodic table. For example, the average mass of one iron atom is 55.8 u, where u means "atomic mass units." The atomic mass unit is defined so that the atomic mass of an atom of the most common carbon isotope is exactly 12 u, and the mass of 1 mol of the most common isotope of carbon atoms is exactly 12 g. The mass of 1 mol of a pure substance is called its molar mass. For example, the molar mass of iron is 55.847 g, and the molar mass of platinum is 195.08 g. Relative masses of elements are demonstrated in Figure 12.4. The molar mass is the mass in grams of the average atomic mass. If an element exists as a molecule, remember that the particles in 1 mol of that element are themselves composed of atoms. For example, the element oxygen exists as molecules composed of two oxygen atoms, so a mole of oxygen molecules contains 2 mol of oxygen atoms. Therefore, the molar mass of oxygen molecules is twice the molar mass of oxygen atoms: 2 X 16.00 g = 32.00 g. Figure 12.5 shows the molar masses of some elements. Relative Masses of Iron and Carbon An average iron atom is 4.65 times as heavy as an average carbon atom. 5ix iron atoms are 4.65 times as heavy as six carbon atoms. One mole of iron atoms is 4.65 times as heavy as 1 mol of carbon atoms. Figure 12.5 Molar Masses of Some Elements Each sample contains 6.02 x 1023 atoms. Note that the masses of the samples are all different. Moles relate counts of atoms, molecules, or ions to mass because a molar mass always contains an Avogadro constant of particles. 12.1 Counting Particles of Matter 407 Determining Number Without Counting Chemists and chemical engineers usually need to control the number of atoms, molecules, and ions in their reactions carefully. These particles are too small to be seen and too numerous to count, but their numbers can be determined by measuring their masses. Simulate this process by determining the approximate number of small, identical items in a large bag by weighing instead of counting. Procedure ~; 1. Count out ten items and weigh them. Record the mass. 2. Weigh and record the mass of an empty, self-sealing plastic storage bag. 3. Completely fill the bag with items and seal it. 4. Weigh and record its mass. 5. Develop and carry out an experi- ment asking yourself the following question, "How can I determine the number of buttons in the bag without opening it?" Analysis 1. How many items are in the bag? 2. Explain how you determined that number. Number of Atoms in a Sample of an Element SAMPLE PROBLEM The mass of an iron bar is 16.8 g. How many Fe atoms are in the sample? Problem-Solving HI NT • Use the periodic table to find the molar mass of iron. The average mass of an Remember that the units of molar mass iron atom is 55.8 u. Then the mass of ore grams per mole, which con be used 1 mol of iron atoms is 55.8 g. as a conversion factor. Set Up • To convert the mass of the iron bar to the number of moles of iron, use the mass of 1 mol of iron atoms as a conversion factor. •• •• 16.8 g Fe 1 mol Fe • 55.8 g Fe • • • • Now, use the number of atoms in a mole to find the number of iron • • atoms in the bar. • • 23 • 16.8 g Fe 11 mol Fe 16.02 X 10 Fe atoms • 1 mol Fe 55.8 g Fe • • • Solve : • Simplify the expression above. 23 • 16.8.g-Fe'11~ 16.02 X 10 Fe atoms _ • 55.8.g-Fe' l~ ••• 23 • 16.8 X 6.02 X 10 Fe atoms 23 • • = 1.81 X 10 Fe atoms 55.8 • • • Check : • Notice how the units of measure cancel to leave only Fe atoms. Analyze t · · 4D8 Chapter 12 Chemical Quantities .-. Molar Mass of aCompound As you learned in Chapter 4, covalent compounds are composed of molecules, and ionic compounds are composed of formula units. The molecular mass of a covalent compound is the mass in atomic mass units of one molecule. Its molar mass is the mass in grams of 1 mol of its molecules. The formula mass of an ionic compound is the mass in atomic mass unlts of one formula unit. Its molar mass is the mass in grams of 1 mol of its formula units. How to calculate the molar mass for ethanol, a covalent compound, and for calcium chloride, an ionic compound, is shown below. Ethanol, C 2H 60, a covalent compound 2 C atoms 6 H atoms 10 atom molecular mass of C2H 60 2 X 12.0 u = 6 X 1.00 u = 1 X 16.0 u = 24.0 u 6.00 u + 16.0 u 46.0 u 46.0 g 46.0 glmol mass of 1 mol C 2H 6 0 molecules molar mass of ethanol Calcium chloride, CaC1 2 , an ionic compound 40.1 u 1 Ca atom 1 X 40.1 u = 2 Cl atoms 2 X 35.5 u = +71.0 u 111.1 u formula mass of CaCl 2 = mass of 1 mol CaC1 2 formula units molar mass of calcium chloride IlLIg 111.1 glmol By making calculations with molar masses, you can find the number of molecules of methanol in your 500 g; the number of molecules of carbon dioxide and hydrogen that reacted; and the masses of carbon dioxide, hydrogen, methanol, and water in grams. Figure 12.6 shows the molar masses of some compounds. Figure 12.6 Molar Masses of Some Compounds Each sample contains 6.02 X 1023 molecules of a covalent compound or 6.02 x 1023 formula units of an ionic compound. Each compound has its own molar mass. 12.1 - ..- Counting Particles of Matter 409 SAMPLE PROBLEM Number of Formula Units in a Sample of a Compound • • • • ' The mass of a quantity of iron(III) oxide is 16.8 g. How many formula units are in the sample? • •• Analyze : • Use the periodic table to calculate the mass of one formula unit of Fe 20 3. 2 X 55.8 u = 3 X 16.0 u = 2 Fe atoms : 3 0 atoms formula mass of Fe 2 0 3 111.6 u + 48.0 u 159.6 u Therefore, the molar mass of Fe203 (rounded off) is 160 g. • • • • • •• •• Solve 16.8 g FeZ03 1 mol FeZ03 160 g Fe203 • Set Up • : • •• • • • • • • • · Now, multiply the number of moles of iron oxide by the number in a mole. 23 16.8 g FeZ03 1 mol Fe203 6.02 X 10 Fe203 formula units 160 g Fe Z03 mol FeZ03 . 23 16.8 ~ 1 ~ 6.02 X 10 FeZ03 formula units 160~ l~ 23 16.8 X 6.02 X 10 FeZ03 formula units = 160 22 6.32 X 10 FeZ03 formula units • • Check •• • The units of measure show that the calculation is set up correctly. The • • • • • • • • • • • • • •• • • • • •• • •• • • •• · •• • • • multiplication also checks. 1. Without calculating, decide whether 50.0 g of sulfur or 50.0 g of tin represents the greater number of atoms. Verify your answer by calculating. 2. Determine the number of atoms in each sample below. a) b) c) d) 98.3 g mercury, Hg 45.6 g gold, Au 10.7 g lithium, Li 144'.6 g tungsten, W 3. Determine the number of moles in each sample below. a) b) c) d) 6.84 16.0 68.0 17.5 g sucrose, C 12 H 22 0 11 g sulfur dioxide, S02 g ammonia, NH 3 g copper(II) oxide, CuO In the previous problems, you used the molar mass to convert a mass measurement to a number of moles. Now, you will learn to convert a number of moles to a mass measurement. 41Z Chapter 12 Chemical Quantities Mass of a Number of Moles of a Compound SAMPLE PROBLEM . • • What mass of water must be weighed to obtain 7.50 mol of H 20? • • Analyze : • The molar mass of water is obtained from its molecular mass. • • 2 H atoms 2 X 1.00 u = 2.00 u • • 1 X 16.0 u = 16.0 u 1 0 atom • • • molecular mass of H 20 18.0 u • • The molar mass of water is 18.0 g/mo1. •• Set Up •• • Use the molar mass to convert the number of moles to a mass mea• surement. • • • 7.50 mol H20 118.0 g H 20 • • 1 mol H 20 • • : Solve : • 7.5~118.0gH20 1~ = 7.50 X 18.0gH20 = 135gH20 • • Check : • The final units are grams of H 20, as expected. The multiplication also · : • •• • • • • • • • ••• checks. 4. Determine the mass of the following molar quantities. a) 3.52 mol 5i b) 1.25 mol aspirin, C9 H s0 4 c) 0.550 mol F2 d) 2.35 mol barium iodide, Ba1 2 The concept of molar mass makes it easy to determine the number of particles in a sample of a substance by simply measuring the mass of the sample. The concept is also useful in relating masses of reactants and products in chemical reactions. rpR.lemental Problems For more practice with solving problems, see Supplemental Practice Problems, Appendix B. Understanding Concepts 1. How is counting a truckload of pennies like counting the atoms in 10.0 g of aluminum? How is it different? 2. The average atomic mass of nitrogen is 14 times greater than the average atomic mass of hydrogen. Would you expect the molar mass of nitrogen gas, N 2, to be 14 times greater than the molar mass of hydrogen gas, H/ Why? 3. Why is the following statement meaningless? An industrial process requires 2.15 mol of a sugar-salt mixture. Thinking Critically 4. Analyzing Determine the mass in grams of one average atomic mass unit. Applying Chemistry 5. Balances If the mole is truly central to quantitative work in chemistry, why do balances in chemistry labs measure mass and not moles? 12.1 Counting Particles of Matter 413 Using Moles H~·'llI;.'r" T he rose's distinctive scent is mostly the odor of one compound, geraniol. The balanced chemical equation for the formation of geraniol shows the ratios of carbon, hydrogen, and oxygen that interact. Although the particles are too small to be seen or weighed, a mole of carbon, hydrogen, or oxygen is large enough. Can you use the chemical equation to predict the masses of carbon, hydrogen, and oxygen that react and the mass of geraniol that is formed? Objectives Predict quantities of reactants and prod· I ~·;~::. ucts in chemical reac, ,.... tions. I·,,..••~,- Determine mole ratios from formulas for compounds. Identify formulas of compounds by using mass ratios. Using Molar Masses in Stoichiometric Problems Key Terms molar volume ideal gas law empirical formula In Chapter 11, you saw how balanced chemical equations indicate the volume of gas required for a reaction or the volume of gas produced. Similarly, you can use balanced chemical equations and numbers of moles of each substance to predict the masses of reactants or products. Predicting Mass of a Reactant SAMPLE PROBLEM I As you review this problem, note that nitrogen and hydrogen are related in terms of moles, not mass. In chemical reactions, when substances react, their particles react. Ammonia gas is synthesized from nitrogen gas and hydrogen gas according to the balanced chemical equation below. N 2 (g) + 3 H 2 (g) -+ 2 NH 3 (g) How many grams of hydrogen gas are required for 3.75 g of nitrogen gas to react completely? • The amount of hydrogen needed depends upon the number of Analyze nitrogen molecules present in 3.75 g and the mole ratio of hydrogen gas to nitrogen gas in the balanced chemical equation. • • Find the number of moles of N 2 molecules by using the molar Set Up : mass of nitrogen. • •• 3.75 g N211 mol N 2 • 28.0 g N2 · · Chapter 1Z Chemical Quantities To find the mass of hydrogen needed, first find the number of moles of H 2 molecules needed to react with all the moles of N 2 molecules. The balanced chemical equation shows that 3 mol of H 2 molecules react with 1 mol of N 2 molecules. Multiply the number of moles of N 2 molecules, as shown in Set Up, by this ratio. •• •• • • • • • • • • • • The units in the expression above simplify to moles of H 2 molecules. To find the mass of hydrogen, multiply the number of moles of hydrogen molecules by the mass of 1 mol of H 2 molecules, which is 2.00 g. • •• ·••• • Solve • • • •• •• • 3.75.g-N211~13~12.00gH2 28.0.g.Ni l~ l~ = 3.75 X 1 X 3 X 2.00 g H 2 28.0 = 0.804 g H 2 • Check •• • Check the multiplication to verify that the result is correct. Predicting Mass of a Product • : : I What mass of ammonia is formed when 3.75 g of nitrogen gas react with hydrogen gas according to the balanced chemical equation below? • : N 2 (g) + 3 H 2 (g) ~ 2 NH 3(g) • Analyze : • The amount of ammonia formed '" Problem-Solving : depends upon the number of nitrogen HI NT • : molecules present and the mole ratio of : nitrogen and ammonia in the balanced The balanced chemical equation is the : chemical equation. source of conversion factors relating • Set Up • • As in Sample Problem 4, the number of moles of one substance to moles of another substance. moles of nitrogen molecules is given by the expression below. 3.75 g N 2 1 mol N 2 28.0 g N2 •• •• To find the mass of ammonia produced, first find the number of moles of ammonia molecules that form from 3.75 g of nitrogen. Use the mole ratio of ammonia molecules to nitrogen molecules to find the number of moles of ammonia formed. • •• • ·••• • • • • • • • • Use the molar mass of ammonia, 17.0 g, to find the mass of ammonia formed. 17.0 g NH 3 I mol NH 3 12.2 Using Moles 415 , -• • Solve • • .. · • 3.75..g.Nzll~12~117.0 g NH 3 28.0~ • • • 1~ = 1~ 3.75 X 1 X 2 X 17.0gNH3 • 28.0 = 4.55 g NH3 • • Check : • Convince yourself that the factors are set up correctly and that the • multiplication is correct. • • Supplemental problems •• •• • • 5. The combustion of propane, C 3H s ) a fuel used in backyard grills and camp stoves, produces carbon dioxide and water vapor. •• C3H s(g) + + 4H 2 0(g) 50 2 (g) -+ 3COig) W'hat mass of carbon dioxide forms when 95.6 g of propane burns? For more practice with solving problems, see Supplemental Practice Problems, Appendix B. 6. Solid xenon hexafluoride is prepared by allowing xenon gas and fluorine gas to react. Xe(g) •• • • • + 3F2 (g) -+ XeF 6 (s) How many grams of fluorine are required to produce 10.0 g of XeF 6? 7. Using the reaction in Practice Problem 6, how many grams of xenon are required to produce 10.0 g of XeF 6 ? Using Molar Volumes in Stoichiometric Problems r Figure 12.7 Molar Volumes of Gases One mole of any gas at STP occupies 22.4 L. How large is that? It is the volume of a cube that is 28.2 cm on each edge. Each such volume contains 6.02 X 1023 atoms of a gaseous element or 6.02 X 10 23 molecules of a molecular element or compound, but the mass of 1 mol is different for each element. One mole of helium floats because its mass is less than the mass of 22.4 L of air. 416 Chapter 12 In Chapter 11, you used the law of combining volumes. Avogadro inferred from that law that equal volumes of gases contain equal numbers of particles. In terms of moles, Avogadro's principle states that equal volumes of gases at the same temperature and pressure contain equal numbers of moles of gases. The molar volume of a gas is the volume that a mole of a gas occupies at a pressure of one atmosphere (equal to 101 kPa) and a temperature of O.Oo°e. Under these conditions of STP, the volume of 1 mol of any gas is 22.4 L, as shown in Figure 12.7. Like the molar mass, the molar volume is used in stoichiometric calculations. He 22.4 L Chemical Quantities 6.02 x 10 23 Particles One E u N 00 22.4 Liters N 28.2 cm mole Air Bags In 1990, on a Virginia hilltop, two cars collided in a head-on crash. Both drivers walked away with only minor injuries. A chemical reaction, along with seat belts, saved their lives. This was the first recorded head-on collision between two cars with air bags. How air bags function Although air bags seem to inflate instantaneously, the process occurs in steps. 1. When a car collides with a rigid barrier at a speed of 12 mph or greater, two or three sensors on the front of the car send an electric current to fire the control unit 0.01 s after impact. Reliability During a ten-year period, one car in a million may have an air-bag defect. Why are they so successful? There are no moving parts to wear out, all exposed components are tightly sealed, and the gold-plated electrical connectors corrode slowly. Gas-volume relationships The driver's air bag requires 0.0650 m 3 of nitrogen to inflate-no more, no less. The passenger's air bag needs 0.1340 m 3 • The pellet must have the exact amount of sodium azide needed to produce the correct amount of nitrogen. As with all expanding gases, pressure and temperature affect the amount of sodium azide needed. Because the nitrogen gas is formed in an explosion, it has to be cooled before it goes into the air bag. 2. After 0.05 s, a chemical reaction in the stored air bag creates a gaseous product that inflates it and pushes open the cover on the steering wheel or on the passenger-side dash. 3. The driver or passenger strikes the inflated bag. 4. The bag deflates in 0.045 s as the gas escapes through holes at the base of the bag. The chemical reactions Sodium azide (NaN 3) is the chemical that produces nitrogen gas to inflate the air bag. Sodium azide pellets, an igniter, inflator, and a tightly folded nylon air bag are stored under a breakaway cover in the steering wheel or dash. The igniter provides a current to decompose the sodium azide into nitrogen gas and sodium. 2NaN 3 (s) electricity ---+ 3N 2(g) + 2Na(s) The sodium immediately reacts with iron(llI) oxide in the pellet to form sodium oxide and iron. 6Na(s) + Fe 20 3 (s) -+ 3Na20(S) + 2Fe(s) The sodium oxide reacts with carbon dioxide and water vapor in the air to form sodium hydrogen carbonate. Na 20(s) + 2C0 2(g) + H 20(g) -+ 2NaHC0 3 (s) 1. Applying If 130 g of sodium azide are needed for the driver's air bag, how much is needed for the passenger's bag? Explain. 2. Inferring What effect does the heat from the sodium azide reaction have on the pressure and volume of the nitrogen gas formed? To find out more about ·~~lIJmtflJ"'~~~air bags, visit the , Glencoe Science Web site. science.glencoe.com 12.2 Using Moles 417 Using Molar Volume SAMPLE PROBLEM c In the space shuttle, exhaled carbon dioxide gas is removed from the air by passing it through canisters of lithium hydroxide. The following reaction takes place. CO 2(g) + 2LiOH(s) ~ Li 2C0 3(S) + H 20(g) How many grams of lithium hydroxide are required to remove 500.0 L • of carbon dioxide gas at 101 kPa pressure and 25.0°C? Analyze : • In this problem, you use the molar volume to find the number of : moles~ • Set Up : • The volume of gas at 25°C must be converted to a volume at STP. • :• V = 500.0 L CO 2 (273K) 298K = 458 L CO 2 • •• Now, find the number of moles of CO 2 gas as below. CD-ROM • • Use the Interac•• 458 L CO 2 1 mol CO 2 tive Virtual Exploration •• 22.4 L CO2 Predicting Mass ofProduct • • The chemical equation shows that the ratio of moles of LiOH to • found on Disc 2 of • • CO the CD-ROM. 2 is 2 to 1. Therefore, the number of moles of lithium hydroxide is • given by the expression below. •• • 458 L CO 2 1 mol CO 2 2 mol LiOH •• • 22.4 L CO 2 1 mol CO 2 • • • To convert the number of moles of LiOH to mass, use its molar mass, • • 23.9 g/mo!. • • • 458 L CO 2 1 mol CO 2 2 mol LiOR 23.9 g LiOH • • 22.4 L CO 2 1 mol CO 2 1 mol LiOH • • Solve : • 458J...GE1211~ 12~123.9gLiOH _ • 22.4.J-GB2 1 jP.Ol-€e)i 1 ~ • • • 458 x 2 X 23.9 g LiOH . • • = 977 g LtOH 22.4 • • • Check : • A mass of 1000 g of lithium hydroxide is about 40 mol of the com:• pound. According to the chemical equation, about half as much, or 20 mol of CO 2, will be removed from the air. At STP, 20 mol of any gas : : occupy about 450 L, which will expand to about 500 L at 25°C. · · · 8. What mass of sulfur must burn to produce 3.42 L of S02 at 273°C and 101 kPa? The reaction is S(s) + 02(g) ~ S02(g). 9. What volume of hydrogen gas can be produced by reacting 4.20 g of sodium in excess water at 50.0°C and 106 kPa? The reaction is 2 Na + 2 H 20 ~ 2 NaOH + H 2• 418 Chapter 12 Chemical Quantities Gases are much less dense than solids, as illustrated by Figure 12.8. Ideal Gas Law Exactly how the pressure p, volume V, temperature T, and number of particles n of gas are related is given by the ideal gas law shown here. PV= nRT The value of the constant R can be determined using the definition of molar volume. At STP, 1 mol of gas occupies 22.4 L. Therefore, when P = 101.3 kPa, V = 22.4 L, n = 1 mol, and T = 273.15 K, the equation for the ideal gas law can be shown as follows. Figure 12.8 Explosives Gases produced in powerful chemical reactions of explosives expand with great energy. One mole of a gas occupies much more space than 1 mol of any solid. 101.3 kPa X 22.4 L = 1 mol X R X 273.15 K Now, we can solve for R. R = 101.3 kPa 1 22 .4 L 11 1 mol -2-7-3.-1-5=K 8.31 kPa·L mol·K Now, you can find the volume, pressure, temperature, and number of moles of a gas. Using the Ideal Gas Law SAMPLE PROBLEM • •• • •• How many moles are contained in a 2.44-L sample of gas at 25.0°C and 202 kPa? • Analyze : • Solve the ideal gas law for n, the number of moles. • • PV • n=• RT • • • 202 kPa X 2.44 L • Set Up •• • n = 8.31 kPa· L) X 298 K • ( mol·K • · ·•• • Solve •• • n • •• • ·••• • •• 202 kPa 2.44 L = 1 8.31 L· kPa mol·K 202 J>Pa X 2.44X X 1 mol· jt' 8.31X· ~ X 298 jt' 1 298 K = 202 mol X 2.44 8.31 X 298 = 0.199 mol Check • • First, find the volume that 2.44 L of a gas would occupy at STP. • • • • • • • • •• • •• • • ••• • Then, find the number of moles in this volume. 4.47 ~ 11 mol 22.4)':' = 0.200 mol 0.200 mol is close to the calculated value. 1Z.Z Using Moles 419 Bagging the Gas Chemists and chemical engineers often need to determine amounts of reactants and products that will react efficiently and cost effectively. Use the molar volume to determine the amount of baking soda required to react with vinegar to yield just enough carbon dioxide to fill a one-quart, self-sealing plastic bag. n Procedure ~; Squeeze the air from the bag and seal the zipper top. 1. Write the balanced equation for the reaction of baking soda 7. Place the bag in a trash can (sodium hydrogen carbonate) or behind an explosion shield. and vinegar (acetic acid) that 8. Release the twist tie, quickly produces sodium acetate, water, mix the reactants, and allow the and carbon dioxide. reaction to proceed. 2. Find the volume of a I-quart, Analysis self-sealing plastic bag by filling 1. Show and explain the calculait with water and then pouring tions that you used to deterthe water into a graduated mine the required mass of sodicylinder or measuring cup. um hydrogen carbonate. 3. Calculate the mass of sodium 2. What mass of sodium hydrogen hydrogen carbonate that will carbonate would be required to fill the bag with CO 2 gas when react with excess acetic acid to the compound reacts with produce 20 000 L of carbon excess acetic acid. dioxide gas at STP for a water4. Put on an apron and goggles. treatment plant? 5. Weigh the calculated amount of 3. What would have happened in sodium hydrogen carbonate step 8 if the amount of acetic and place it in a bottom corner acid was insufficient to react of the bag. Use a plastic-coated with all of the baking soda? twist tie to seal off this corner. 4. Suppose the pressure of the gas 6. Pour about 60 mL of 1M acetic in the bag was measured to be acid into the other bottom cor101.5 kPa. Is this pressure conner of the bag. Be careful not to sistent with the ideal gas law? allow the reactants to mix. Assume T = 20°C and P = 101 kPa. •• Supplemental prDblems For more practice with solving problems, see Supplemental -Practice Problems, Appendix B. 420 Chapter 12 ·•• ·••• • • • • •• • 10. How many moles of helium are contained in a 5.00-L canister at 101 kPa and 30.0°C? 11. What is the volume of 0.020 mol Ne at 0.505 kPa and 27.0°C? 12. How much zinc must react in order to form 15.5 L of hydrogen, H 2 (g), at 32.0°C and 115 kPa? Chemical Quantities Zn(s) + H 2S0 4 -+ ZnS0 4 (aq) + H 2 (g) Theoretical Yield and Actual Yield The amount of product of a chemical reaction predicted by stoichiometry is called the theoretical yield. As shown earlier, if 3.75 g of nitrogen completely react, a theoretical yield of 4.55 g of ammonia would be produced. The actual yield of a chemical reaction is usually less than predicted. The collection techniques and apparatus used, time, and the skills of the chemist may affect the actual yield. When actual yield is less than theoretical yield, you express the efficiency of the reaction as percent yield. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield expressed as a percent. Suppose the actual yield was 3.86 g of ammonia. Then, the percent yield could be calculated by using the following equation. . ld actual yield percen t Yle - theoretical yield X 100% 3.86~ 4.55..g..Ntr; X 100% = 84.8% This means that 84.8 percent of the possible yield of ammonia was obtained from the reaction. Calculating percentage yield is similar to calculating a baseball player's batting average, as shown in Figure 12.9. A manufacturer is interested in producing chemicals as efficiently and inexpensively as possible. High yields make commercial manufacturing of substances possible. For example, in human experiments, taxol, a naturally occurring complex compound, is a strong agent against cancer. For ten years, chemists tried to synthesize this compound in the lab. In 1994, two independent academic research groups succeeded. However, the process is so complicated and time-consuming that the percent yield is probably not even one percent. CD·ROM Use the Interactive Virtual Experiment How much oxygen is available? found on Disc 2 of the CD-ROM. Figure 12.9 Batting Average and Percent Yield In 1995, Alfredo Spinelli batted .352, which means: hits 352 ---=-attempts 1000 . Just as batting averages measure a hitter's efficiency, percent yield measures a reaction's efficiency. 12.2 Using Moles 421 Improving Percent Yield in Chemical Synthesis The single most important industrial chemical in the world is probably sulfuric acid. In the United States, its production exceeds 40 million tons per year. The strength of the U.S. economy can be gauged by the amount ofsulfuric acid produced annually. Over the years, the manufacturing process ofsulfuric acid has been improved to provide a higher and more economical yield. Step 1 Sulfur is burned in air to produce sulfur dioxide, a stable compound. This reaction takes place quickly and readily. 5(s) + Oig) --+ 50 2(g) Impurities produced in combustion are removed from the Sulfur sulfur dioxide so they melter will not react with and impede the catalyst in the next step. Melted sulfur Uses of Sulfuric Acid Sixty percent of all manufactured sulfuric acid is used to make fertilizers. It is also used in manufacturing detergents, photographic film, synthetic fibers, pigments, paints, drugs, and other acids. It is the electrolyte in some batteries, acts as a catalyst and a dehydrating agent, and is a component in refining petroleum and metals. The Lead·Chamber Process-A Low·Yield Process The first industrial method, the lead-chamber process, is not commonly used now because its purity is low and its percent yield is only 60 to 80 percent. But it is much cheaper than the later and more productive contact process. The lead-chamber process is used for manufacturing sulfuric acid for applications that do not demand high purity. The Contact Process-The High-Yield Process The contact process is the most widely used commercial method. It is more expensive than the lead-chamber process, but it is simple and it produces high-purity sulfuric acid at a high percent yield-about 98 percent. In addition, it creates no by-products that pollute the atmosphere. The contact process has four steps. l:=====Ai=r-:,-:,-:,: [ Water --=::..:....--=----+ ~-- ~::±======- ~=~=====:::J Maximizing Percent Yield Reaction yields in the contact process were increased in several ways-by selecting an optimum temperature, using an efficient catalyst, removing a product from a reaction that does not go to completion, and by controlling the rate of reaction of S03 with water. Keeping operating pressures at the correct values also increases yield. Using Scientific Methods to Attain the Best Yields To develop higher yields that are closer to the theoretical values, chemists adjust temperatures, pressures, or other conditions in their industrial processes. They look for better catalysts or new ways to deal with undesirable side reactions. Unwanted by-products are serious concerns to the chemist. Because they may cause environmental damage or be expensive to dispose of, byproducts may raise production costs. Step 3 Step 4 Because sulfur trioxide reacts violently with water, it is bubbled through 98 percent concentrated sulfuric acid and forms pyrosulfuric acid, H2 S20 7 • When water is added to the pyrosulfuric acid, high-quality, 98 percent concentrated sulfuric acid is formed. 50 3(g) + H2S04 (1) ---* H2S20 7(1) H2 S2 0 7 (1) + H20(I) ---* 2 H1SOil) Step 1 Sulfur burner ) S02 S + O2 t ) Step 2 Step 2 Because sulfur dioxide reacts slowly with excess oxygen, a catalyst, either vanadium pentoxide (V 20 S) or finely divided platinum at a temperature of 400°C, is used. The reaction produces sulfur trioxide. 2S0 ig) + Gig) c~t 250 3(g) Sulfur trioxide is quickly removed from the contact chamber because it tends to form sulfur dioxide and oxygen again. Removing sulfur trioxide promotes the production of more sulfur trioxide. 1. Hypothesizing Roasting iron pyrite, FeS 2> can replace burning sulfur in Step 1 to produce sulfur dioxide. Write an equation to show what you think the reaction with pyrite is. Green Chemistry The contact process is economically sound because the reactants are abundant and inexpensive and because it produces no unwanted byproducts, which may be expensive to store or dispose of and which may pollute the environment. Ecologically sound chemical manufacturing is called green chemistry or green technology. 2. Acquiring Information Find out how sulfuric acid is used to manufacture hydrochloric acid and write the equation(s) of the reactions. Determining Mass Percents As you know, the chemical formula of a compound tells you the elements that comprise it. For example, the formula for geraniol (the main compound that gives a rose its scent) is CIOH1SO. The formula shows that geraniol is comprised of carbon, hydrogen, and oxygen. Because all these elements are nonmetals, geraniol is probably covalent and comprised of molecules. In addition, the formula CIQH1SO tells you that each molecule of geraniol contains ten carbon atoms, 18 hydrogen atoms, and one oxygen atom. In terms of numbers of atoms, hydrogen is the major element in geraniol. How can you tell whether it is the major element by mass? You can answer this question by determining the mass percents of each element in geraniol, which are shown in Figure 12.10. o Figure 12.10 10.4% Mass Percents of Elements In Geraniol This pie graph shows the composition of geraniol in terms of mass percents of the elements. C 77.9% Suppose you have a mole of geraniol. Its molar mass is 154 glmol. Of this mass, how many grams do the carbon atoms contribute? The formula shows that one molecule of geraniol includes ten atoms of carbon. Therefore, 1 mol of geraniol contains 10 mol of carbon. Multiply the mass of 1 mol of carbon by 10 to get the mass of carbon in 1 mol of geraniol. 10~112.0gC j:OOt = 120 g C Now, use this mass of carbon to find the mass percent of carbon in geraniol. %C = 120 g C 154 g CIQH 18 0 X 100% = 77.9% The mass percents of the other elements are calculated below in a similar fashion. Mass of hydrogen in 1 mol geraniol: 18~X 1.~H Mass percent of H = = 18.0 gH mass ofH f . 1 X 100% mass a geramo 18.0 g H 426 Chapter 12 Chemical Quantities X 100% = 11.7% H Mass of oxygen in 1 mol geraniol: 1~ 116.0g0 ~ Mass percent of 0 = = 16.0g0 mass of 0 f . 1 X 100% mass 0 geramo = 16.0 g 0 154 g C LO HUIO X 100% = 10.4% 0 In the example above, you learned how to use the chemical formula and the molar masses to find the mass percent of a compound. You can also solve the reverse problem. You can use the mass percent to find the chemical formula of an unknown compound. Determining Chemical Formulas Suppose you analyzed an unknown compound and found that, by mass, it was 18.8 percent sodium, 29.0 percent chlorine, and 52.2 percent oxygen. Because this compound contains some metal and nonmetal elements, it may be an ionic compound. To determine its chemical formula, find the relative numbers of sodium, chlorine, and oxygen atoms in the formula unit of the compound. Suppose you have 100.0 g of the unknown compound. Because you know the sample includes 18.8 g of sodium, 29.0 g of chlorine, and 52.2 g of oxygen, use the molar mass to find the number of moles of each element. 18.8~ 11 mol Na . = 0.817 mol Na 23.0~ 29.0~ll mol CI - 52.2~11 malO - 35.5~ - 0.817 rna 16.0~ - 3.26 mo I Cl 10 You know that atoms combine in ratios of small whole numbers to form compounds. To find the whole-number ratios, divide the mole numbers by the smallest one. 0.817 mol Na = 1.00 mol Na 0.817 0.817 mol Cl = 1.00 mol CI 0.817 3.26 malO = 3.99 malO 0.817 These values are whole numbers or very close to whole numbers. Therefore, the mole ratio for this compound is 1 mol Na :1 mol Cl :4 mol O. Similarly, the ratio of atoms in this compound is 1 Na :1 Cl :4 O. 12.2 ... - Using Moles 421 The formula of a compound having the smallest whole-number ratio of atoms in the compound is called the empirical formula. The empirical formula of this unknown compound is NaCI0 4 • What is the chemical formula for this compound? You have learned that the formula for an ionic compound represents the simplest possible ratio of the ions present and is called a formula unit. Chemical formulas for most ionic compounds are the same as their empirical formulas. Because the unknown compound is ionic, the chemical formula for a formula unit of the compound' is the same as its empirical formula, NaCI0 4 . The compound is called sodium perchlorate. As another example, suppose the mass percents of a compound are 40.0 percent carbon, 6.70 percent hydrogen, and 53.3 percent oxygen. Because all its elements are nonmetals, the compound is covalent. Imagine you have 100 g of the compound. Then you have 40.0 g of carbon, 6.70 g of hydrogen, and 53.3 g of oxygen. Use the molar masses of these elements to find the number of moles of each element. empirical empeirikos (GK) working from experience From experimental work, the empirical formula of an unknown compound can be derived. 4o.0~11 mol C _ 12.0,..g.e- - 3.33 mol C I' 6.70,.g.-H' 11 mol H 1.00~ = 6.70 mol H J ° 53.3,g.e--\1 mol 16.0,g.e-- = 3.33 malO ~~~~,~ Now, divide all the mole numbers by the smallest one. 3.33 mol C 3.33 - - - - = 1.00 mol C 6.70 mol H = 2.01 mol H 3.33 3.33 mol 0 3.33 = 1.00 mol 0 Because the ratio of moles is the same as the ratio of atoms, CH 20 is the empirical formula for this compound. But, the empirical formula is not always the chemical formula. Many different covalent compounds have the same empirical formula, as demonstrated in Figure 12.11, because atoms can share electrons in different ways. Figure 12.11 Compounds Having the Empirical Formula CH20 For each of these compounds, the ratio of atoms is 1C:2H:l0. Because one molecule of each compound has a different number of each atom, it is a different compound with its own molecular formula. 428 Chapter 12 Chemical Quantities To decide which multiple of the empirical formula is the correct molecular formula, you need the molar mass of the compound. Suppose a separate analysis shows that the molar mass of the compound is 90.0 glmol. Therefore, the molecular mass of the compound is 90.0 u. The molecular mass of a CH 20 molecule is 30.0 u. By dividing the molecular mass of the compound by 30.0 u, you find the multiple. 90.0 u --=3 30.0 u The molecular formula of the compound contains three empirical formula units. The molecular formula is C3 H 6 0 3 • The compound is lactic acid, the sour-tasting substance in spoiled milk. Lactic acid is also produced by active muscles, causing them to feel sore after strenuous activity. Connecting Ideas In this chapter, you learned to solve several kinds of stoichiometric problems. For each kind, you used the mole concept because when substances react, their particles interact. The number of particles at this submicroscopic level controls what happens macroscopically. In the next chapter, you will use the mole concept and the particle nature of matter to study the mixtures of substances called solutions. Understanding Concepts 1. Octane, C sH 1s ' is one of the many components of gasoline. Write the balanced chemical equation for the combustion of octane to form carbon dioxide gas and water vapor. Identify as many mole ratios among the substances in the combustion as you can. If 20.0 g of each reactant are used, which is the limiting reactant? 2. If 25.0 g of octane burn as in Problem 1, how many grams of water will be produced? How many grams of carbon dioxide? 3. A manufacturer advertises a new synthesis reaction for methane with a percent yield of 110 percent. Comment on this claim. Thinking Critically 4. Predicting The reaction of iron(III) oxide and aluminum is called the thermite reaction because of its intense heat. The iron produced is molten and was formerly used to weld railroad tracks in remote areas. Its balanced chemical equation is shown below. Applying Chemistry 5. Blue Jeans Indigo, the dye used to color blue jeans, is prepared using sodium amide. Sodium amide contains the following mass percents of elements: hydrogen, 5.17 percent; nitrogen, 35.9 percent; and sodium, 58.9 percent. Find an empirical formula for sodium amide. 12.2 Using Moles 429 12.1 Counting Particles of Matter The number of particles (atoms, molecules, or ions) in macroscopic matter controls the consumption and formation of substances in chemical reactions. • One mole equals 6.02 X 10 23 • • Use the molar mass to convert mass to moles or moles to mass. 12.2 Using Moles • A balanced chemical equation provides mole ratios of the substances in the reaction. • The mole is a central concept in making chemical calculations. • The ideal gas law is expressed in the following equation. PV= nRT • Percent yield measures the efficiency of a chemical reaction. actual yield X 100% theoretical yield • Percent composition can be determined from the chemical formula of a compound. Percent yield = • The empirical formula of a compound can be determined from its percent composition. • The chemical formula of a compound can be determined if the molar mass and the empirical formula are known. Key Terms For each of the following terms, write a sentence that shows your understanding of its meaning. Avogadro constant empirical formula formula mass ideal gas law molar mass molar volume mole molecular mass stoichiometry UNDERSTANDING CONCEPTS 3. Explain how you would use weighing to count 40 000 washers. 1. Your uncle leaves you a barrel of nickels. You determine that the nickels weigh 1345 pounds and that 50 nickels weigh 232 g. How many dollars is the barrel of nickels worth? 4. What is the molar mass of the following sub- 2. A manufacturer must supply 20 000 connector units, each consisting of a bolt, two washers, and three nuts. How many of each part are needed? stances? a) bromine b) argon c) lead(II) nitrate d) dinitrogen tetroxide 5. A reaction requires 0.498 mol of CU2S04' How many grams must you weigh to obtain this number of formula units? 6. Which has the largest mass? a) three atoms of magnesium b) one molecule of sucrose, C 12 H n O Il c) ten atoms of helium 7. Which has the largest mass? a) 3 mol of magnesium b) 1 mol of sucrose, C 12 H n O\l c) 10 mol of helium 8. What is the molecular mass of UF 6? What is the molar mass of UF 6 ? 430 Chapter 12 Chemical Quantities I - , • ,~CHAPTER 12 ASSESSMENT ~.""'.I.:' __ J~~'_-----=-=---. 9. What mass of copper contains the same number of atoms as 68.7 g of iron? 10. Explain why the mass of copper is not equal to the mass of iron in question 9. 11. Determine the molar mass of each of the following compounds. a) C 6H sBr b) K 2Cr 20 7 c) (NH4)3P04 d) Fe(N0 3)3 12. Which molar mass in question 11 contains the largest number of atoms? Largest number of ions? Largest number of formula units? 13. Calculate the mass of 0.345 mol of sodium nitrite, NaN0 2 • 14. Calculate each of the following: a) the mass in grams of 0.254 mol of calcium sulfate b) the number of atoms in 2.0 g of helium c) the number of moles in 198 g of glucose, C6H 12 0 6 d) the total number of ions in 10.8 g of magnesium bromide 15. What are the molar masses of the following elements? a) nitrogen b) iodine c) oxygen d) nickel 16. Sodium nitrate decomposes upon heating to form sodium nitrite and oxygen gas. This reaction is sometimes used to produce small quantities of oxygen in the lab. How many grams of sodium nitrate must be heated to produce 128 g of oxygen? 17. Calculate the mass of silver chloride, Agel, and dihydrogen monosulfide, H 2S, formed when 85.6 g of silver sulfide, Ag 2S, reacts with excess hydrochloric acid, HCl. 18. Aluminum nitrite and ammonium chloride react to form aluminum chloride, nitrogen, and water. What mass of aluminum chloride is present after 43.0 g of aluminum nitrite and 43.0 g of ammonium chloride have reacted completely? - I'i' -----~ 19. A 1O.O-g sample of magnesium reacted with excess hydrochloric acid to form magnesium chloride by the reaction below. When the reaction was complete, 30.8 g of magnesium chloride were recovered. What was the percent yield? Mg(s) + 2HCI(aq) -+ MgCI 2(aq) + H 2(g) 20. Calcium carbide, CaC2> reacts with water to form calcium oxide, CaO, and acetylene, C 2H 2. Acetylene is a fuel that combines with oxygen in an exothermic reaction to produce water and carbon dioxide. This combination of reactions was used to produce light in lanterns known as "carbide lamps." A certain lamp produces 10.0 L of CO 2 (g) at STP. How many grams of calcium carbide must have reacted in the lamp? 21. What is the molecular formula for each of the following compounds? a) empirical formula: CH 2i molar mass: 42 glmol b) empirical formula: CHi molar mass: 78 g/mol c) empirical formula: N0 2i molar mass: 92 glmol 22. An oxide of nitrogen is 26 percent nitrogen by mass. The molar mass of the oxide is approximately 105 g/mol. What is the formula of the compound? 23. How many moles are in each of the following samples? a) 43.6 g NH 3 b) 5.0 g of aspirin, C 9H 80 4 c) 15.0 g CuO APPLYING CONCEPTS 24. Hydrogen fuels are rated with respect to their hydrogen content. Determine the percent hydrogen for the followir...g fuels. a) ethane, C 2H 6 b) methane, CH4 c) whale oil, C32H6402 Chapter 12 Assessment 431 25. Oxidizers are generally compounds that contain relatively large mass percentages of oxygen. Hydrogen peroxide and sodium nitrate are good oxidizing agents. Compare the mass percent of oxygen in each of these compounds. THINKING CRITICALLY Applying Concepts 33. What is the mass in grams of a single molecule of water? 26. Oleic acid is a component in olive oil. It is 76.5 percent C, 12.1 percent H, and 11.3 percent O. The molar mass of the compound is approximately 282 g/mol. What is the molecular formula of oleic acid? 34. Explain how a mole is used in chemistry as both a number and a mass. 27. Serotonin is a compound that conducts nerve impulses in the brain. Serotonin is 68.2 percent C, 6.86 percent H, 15.9 percent N, and 9.08 percent 0. The molar mass is 176 g/mol. What is the molecular formula of serotonin? 36. All of the group 2 metals form metal oxides with the general formula MO, where M is a group 2 metal. How do the mass percentages of oxygen in these compounds compare? 28. Gas masks used by firefighters often contain potassium superoxide, K0 2• K0 2 reacts with carbon dioxide and water in exhaled breath to produce oxygen by the following reaction: 4K0 2(s) + 2H 20(g) + 4C0 2 (g) 4KHC0 3 (s) + 30 2 (g) ~ Assume that a person exhales 14.0 g of CO 2 (g) over 20 minutes. How many grams of potassium superoxide must the mask contain in order to react with the carbon dioxide? 29. Suppose that the percentage yield of the reaction in the last problem is 80.0 percent. How many grams of K0 2 would need to be in the mask to insure enough oxygen? Art Connection 30. How does the United States standardize measurements of mass and weight? Everyday Chemistry 31. How many grams of sodium azide, NaN 31 are needed to fill a 0.0650-m 3 driver's air bag with nitrogen if the pressure required is 2.00 atmospheres and the gas temperature is 40.0°C? (Hint: 1 m 3 = 1000 L) Chemistry and Technology 32. What are some uses of sulfuric acid? 432 Chapter 12 Comparing and Contrasting 35. Distinguish between molar mass, formula mass, and molecular mass. 37. ChemLab Suggest a procedure to determine the mass percent of silver nitrate, AgN0 3, in a mixture of silver nitrate and sodium nitrate, NaN0 3 · Comparing and Contrasting 38. Mini Lab 1 Explain how your procedure in this activity is like using the molar mass to determine the number of particles in a sample of a known substance. Relating Cause and Effect 39. MiniLab 2 If you used calcium carbonate instead of baking soda, how would the amount of reactants needed change? CUMULATIVE REVIEW 40. Distinguish between atomic number and mass number. How do these two numbers compare for isotopes of an element? (Chapter 2) 41. Answer the following questions about sugar and table salt. (Chapter 4) a) What type of compound is each substance? b) Which compound is an electrolyte? c) Which compound is made up of molecules? 42. Why does the second period of the periodic table contain eight elements? (Chapter 7) 43. Explain why the metals zinc, aluminum, and magnesium are resistant to corrosion, while iron is not. (Chapter 8) Chemical Quantities - .... . I~ • ~CHAPTER 12 ASSESSMENT '.1 .i ~ .; l' '_ , ;". . 44. The volume of a gas at STP is 8.50 L. What is its volume if the pressure is 1250 mm Hg and the temperature is 75.0°C? (Chapter 11) SKILL REVIEW 45. Using a Table Complete the table below. In the last column, rank the number of particles from smallest to largest. Did you use the number of moles or the number of individual particles for your ranking? Explain. WRITING IN CHEMISTRY 46. A mole is such a large number that it is hard to envision just how big 6.02 X 1023 is. Write three of your own examples of the number of things in a mole. One example must use time, one must use distance, and the third is up to you. PROBLEM SOLVING 47. A student carries out the following sequence of chemical reactions on a 0.635-g sample of pure copper. Cu(s) + 2HN0 3 (aq) ---7 CU(N0 3 )2(aq) + H 2(g) Cu(N0 3 h(aq) + 2NaOH(aq) ---7 Cu(OHMs) + 2NaN0 3 (aq) CU(OH)2(S) ---7 CuO(s) + H 20(l) CuO(s) + H 2S0 4(aq) ---7 CuS04(aq) + H 20(l) CuS04(aq) + Mg(s) ---7 Cu(s) + MgS0 4(aq) The student isolates and weighs the product at the end of each step before proceeding to the next. What is the theoretical yield of the copper product in each step? The formula units in 10.0 9 of calcium fluoride The sodium ions in 10.0 9 of sodium chloride The water molecules in 10.0 9 of water The hydrogen atoms in 10.0 9 of water The carbon dioxide molecules in 10.0 9 of carbon dioxide The carbon monoxide molecules in 10.0 9 of carbon monoxide The aspirin molecules, C9 Hs04,o in 10.0 g of aspirin The carbon atoms contained within 10.0 g of aspirin The valence electrons in 10.0 9 of aspirin Chapter 12 .... - Assessment 433