NAME PER DATE DUE ACTIVE LEARNING IN CHEMISTRY EDUCATION ALICE CHAPTER 10 THE MOLE CONCEPT (Part 3) Empirical Formulas Molecular Formulas The Ideal Gas Law 10-1 1997, A.J. Girondi
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SECTION 10.1 Laboratory Determination of Empirical Formulas An empirical formula is one which identifies which elements are present in a compound, and it gives the simplest ratio of atoms of those elements in the compound. Examples of empirical formulas include NaCl, CaCl2, and Al 2O3 in which the simplest ratios are 1:1, 1:2, and 2:3, respectively. You can write empirical formulas using oxidation numbers, but they can also be determined using experimental data from experiments. The following example will illustrate how this is done. Sample Problem: A laboratory analysis found that a compound contained 36.5 g Na, 25.4 g S, and 38.1 g of O. What is the empirical formula of the compound? Note: In these kinds of problems, you should attempt to calculate the mole quantities to at least 2 decimal places; otherwise, some problems may be solved incorrectly. Solution: Step 1: Change the mass data to moles: 36.5 g Na X 25.4 g S X 38.1 g O X 1 mole Na 22.99 g Na 1 mole S 32.06 g O 1 mole O 16.00 g O = 1.59 mole Na = 0.79 mole S = 2.38 mole O Step 2: Change the numbers of moles to a simple whole number ratio. You do this by dividing each of the mole quantities by the smallest of them. In our example, the smallest is 0.79. Thus, the calculations are: 1.59 mol Na 0.791 = 2.01 mol Na; 0.791 mol S 0.791 = 1.00 mol S; 2.38 mol O 0.791 = 3.01 mol O These results are very close to 2:1:3, so the answer is Na2SO3. In some problems of this kind, a third step is necessary. This is illustrated in the next sample problem. Sample Problem: What is the empirical formula of a compound that contains 53.73% iron, Fe, and 46.27% sulfur, S? (When the data are given in percentages, they can be changed to grams simply by assuming that you have 100 grams of the sample.) Step 1: Change the mass data to moles: 53.75 g Fe X 46.27 g S X 1 mole Fe 55.83 g Fe 1 mole S 32.07 g S = 0.9620 mol Fe = 1.443 mol S 10-3 1997, A.J. Girondi
Step 2: Change the numbers of moles to a simple whole number ratio. You do this by dividing each of the mole quantities by the smallest of them. 0.9620 mol Fe 0.9620 = 1.000 mol Fe and 1.443 mol S = 1.50 mol S 0.9260 After completing step 2, the mole ratio is 1:1.5. The 1.5 is not close enough to a whole number to be rounded. Therefore, we go to step three. Step 3: Multiply the mole ratio by the smallest integer which will give a whole number ratio. In this case, that integer is 2:2 X 1.000 = 2.000; and 2 X 1.50 = 3.00 Thus, the answer is: Fe2S3 Note: In order to be rounded to a whole number in step 2, a mole quantity should be within one-tenth of that number. For example, 2.1 can be rounded to 2, or 3.9 can be rounded to 4 and step 3 is not needed. However, 2.2 and 3.8 should not be rounded to 2 and 4. In such a case, step three is necessary. Problem 1. It is found that a sample of magnesium oxide consists of 4.04 g of magnesium (Mg) and 2.66 g of oxygen (O). Calculate the empirical formula of the magnesium oxide using this information. Problem 2. Experimental evidence reveals that a compound between iron (Fe) and oxygen (O) consists of 2.24 g Fe and 0.96 g O. Find the empirical formula. Problem 3. An investigation reveals that the percentage composition of a compound is 75.0% carbon (C) and 25.0% hydrogen (H) by mass. Find its empirical formula. 10-4 1997, A.J. Girondi
Problem 45.0% O. 4. Calculate the empirical formula of a compound which contains 32.4% Na, 22.6% S, and Problem 5. A laboratory analysis of an unknown compound revealed that 100 g of it consisted of 26.60 g K, 35.40 g Cr, and 38.08 g O. What is the empirical formula of the compound? ACTIVITY 10.2 Determination of the Formula of an Oxide of Tin In an earlier chapter you learned how to write chemical formulas for compounds by using what are known as oxidation numbers. Now that you know something about the mole concept, you are ready to try to determine the chemical formula of a compound experimentally in the lab. In this activity you are going to make a compound of tin and oxygen, and you are going to gather data which will allow you to, hopefully, determine its formula. 1. Clean and dry an evaporating dish and a watch glass cover (see figure 10.1). Determine the mass of the dish with watch glass cover to the nearest 0.01 g. 2. Place about 2 g of 30-mesh granulated tin in the dish, cover with the watch glass, and measure the mass. If the balance in your lab displays masses to three decimal places, then read the mass of the tin to the nearest 0.001 g or to the nearest 0.01 g if your balance is accurate to only two decimal places. Enter all this data in Table 10.1. 3. While your instructor is watching, under a fume hood, carefully add 5 ml of 8 M nitric acid (HNO3), and replace the watch glass. (The 8 M indicates the concentration of the acid.) Handle the acid with great care, and be sure to wear safety glasses! Do not allow the acid to drip on anything. ring stand watch glass evaporating dish wire screen iron ring Figure 10.1 Heating an Evaporating Dish 10-5 1997, A.J. Girondi
4. A reaction should begin which will result in the production of a reddish toxic gas, NO2. When this reaction has subsided, you can take the dish back to your lab station for the rest of this activity. Briefly, here is what is happening in the dish. Some of the oxygen from the nitric acid, HNO3, combines with the tin to form a compound of tin and oxygen. At the same time, the acid decomposes to form H2O and NO2. The unfinished equation is:? HNO3 +? Sn ---->? Sn?O? +? H2O +? NO2 So as the reaction goes on, you should see the formation of water, NO2 gas, and the white compound of tin and oxygen. In order to complete and balance the equation, you need to determine the correct formula for the tin oxygen compound. 5. Position the dish, watch glass and the contents on a wire gauze supported by a ring stand as shown in Figure 10.1. Begin heating the dish with a low flame. Hold the burner in your hand and rotate it slowly under the gauze. Excessive heating at this point will result in spitting and popping in the dish. Continue to heat slowly and carefully until the contents are dry. 6. When popping and spattering no longer occur, remove the watch glass with tongs and place it upside down on your lab table. Break up the solid with a stirring rod, and heat with a hot flame until the solid becomes a pale yellow. Be careful not to unnecessarily lose any material when you remove the watch glass or use the stirring rod. 7. After the dish has cooled, replace the watch glass cover and determine the mass of the dish, cover, and contents. Reheat the dish without the cover for a few minutes, allow to cool again, and determine the mass again as before. If the two masses are not very close, reheat again until there is no further loss of mass. 8. Discard the product in the waste container. Clean and return all equipment. Table 10.1 1. Mass of dish and watch glass g 2. Mass of dish, watch glass, and tin g 3. Mass of tin (subtract 1 from 2 above) g 4. Mass of dish, watch glass, and product g (after heating to a constant mass) 5. Mass of oxygen in product g (subtract 2 from 4 above) Calculations: 1. Since all of the tin you used ends up in the product, you should be able to calculate the number of moles of tin in the product. Express your answer to three decimal places. 10-6 1997, A.J. Girondi
2. From the mass of oxygen gained, calculate the number of moles of oxygen, O, in the product. (This oxygen is not in the gaseous diatomic form.) Express your answer to three decimal places. 3. Enter your results from calculations 1 and 2 above in the blanks below. Sn O This is the formula of the compound you made. However, we want to change these results to a whole number ratio. Do this by dividing each of the subscripts by the smallest of the two. For example, if you got a ratio such as 0.33 to 0.97, you would divide each value by 0.33. The result would be a ratio of 1 to 2.9. This is close enough (within ±0.1) to be rounded to 1 to 3. What if this process does not result in values which are close (within ±0.1) to whole numbers? For example, what if the division process described above gives you a ratio such as 1 to 2.5? The 2.5 cannot be rounded to a whole number. You should then multiply the values by the smallest integer which will give you a whole number ratio. In the case of 1 to 2.5, that integer would be 2. The resulting ratio would be 2 to 5. Enter the whole number ratio resulting from your lab data into the blanks below. The correct formula for the compound which you made is SnO2. The unbalanced equation representing the reaction which you used to make the SnO2 is shown below. Balance it by adding the correct coefficients. HNO 3 + Sn ----> SnO 2 + H 2 O + NO 2 In this compound the oxidation number of tin is +4 and of oxygen is -2. What are the two possible names for the tin oxygen product? {1} and {2}. SECTION 10.3 Molecular Formulas The empirical formula of a compound gives the simplest ratio of the atoms in the molecule. A molecular formula reveals the exact number of atoms of each element in the molecule. For example, the empirical formula of a certain gas is NO2, while its molecular formula is N2O4. Its empirical formula mass is 46, while its molecular formula mass is 92. Note that the molecular formula's mass is a multiple of the empirical formula's mass. That's because the molecular formula is always a multiple of the empirical formula. If you know the empirical formula mass and the molecular formula's mass, you can calculate the molecular formula as shown below: (empirical formula mass) X (M) = (molecular formula's mass) The letter M represents the multiple that relates the two formulas. For our example above: (46) X M = (92), so M = 2 Therefore, 2 X (NO2) = N2O4 With this information, you should now be able to do problem 6. 10-7 1997, A.J. Girondi
Problem 6. Your answer to problem 3 (earlier in this chapter) represents the empirical formula of the compound. If the molecular mass of that compound is 48, find its molecular formula. (Begin by calculating the formula mass of your answer to problem 3.) Problem 7. Butane is a compound of 82.7% carbon (C) and 17.3% hydrogen (H). Its molecular mass is 58 amu. What are the empirical formula and molecular formulas of butane? Problem 8. An organic (carbon) compound if found to contain 92.25 g of carbon and 7.75 g of hydrogen. If the molecular mass is 78, what is the molecular formula? SECTION 10.4 The Ideal Gas Law: PV = nrt You may remember that when you were studying the gas laws, it was mentioned that there was one other law that you would learn about later. Now's the time! It's called the ideal gas law, because it describes the behavior of an ideal gas, which is defined as a gas that always obeys the gas laws under all conditions. Of course, there is no such gas. Real gases, like H2 or O2 do not strictly obey the gas laws under extreme conditions like very high pressures or very low temperatures. Nonetheless, we don't have to be concerned about that, because our work is not usually done under such conditions. The ideal gas law is different from the others you have learned in that it deals with only one set of conditions. The problems you solved using the other gas laws included an "old" and "new" temperature, and "old" and "new" pressure, etc. Maybe you were given an "original" volume and asked to find a "new" volume. However, you will use the ideal gas law to solve problems involving only one pressure, temperature, volume, etc. The formula for the ideal gas law is: PV = nrt 10-8 1997, A.J. Girondi
where P = pressure, V = volume, n = moles, T = temperature, and R is called the ideal gas constant. R has a value of 0.0821 and its units are really crazy. They are shown below. R = 0.0821 L. atm. mole K These units are "liter-atmospheres per mole-kelvin." Scientific constants usually have strange units, but don't worry. They cancel when you use R in problems. R, like all constants, was experimentally determined, and is used to make the equation "work." We often use such constants. For example, consider the equation: P = C x A. P = number of people, and A = number of arms. What's the value of the constant C? Well, if you got O.5, that's right. In other words, if you have 16 arms and multiply by 0.5, you get 8 people. C is used to make the equation work. Its units would be people/arms, right? Anyway, PV=nRT is the only gas law that you have studied that uses such a constant. When a constant is used in an equation, the units of the other variables in the equation must be the same as those used in the constant. For that reason, when using the 0.0821 value in the ideal law you MUST express volume in liters, pressure in atmospheres, temperature in K (as always), and mass (n) in moles. Learn the relationships below which will allow you to make any needed conversions. 1 atmosphere = 760 mm Hg; 1 L = 1000 ml; oc + 273 = K To solve a variety of problems, you will have to rearrange the ideal gas equation. How s your algebra? For example, of we solve the equation for n, V, and T, respectively, we get: n = PV RT V = nrt P T = PV nr Sample Problem: What volume would 34 g of CO2 gas have at 23 o C and 800. mm pressure? (Notice that only one set of conditions is given? This is a hint that the use of the ideal gas law is appropriate.) Solution: We must change 34 g of CO2 to moles, o C to K, and mm to atm. These are the results of those conversions: 34 g CO2 = 0.77 mole; 23 o C = 296 K; 800 mm = 1.05 atm. When we do the substitutions we get: (1.05 atm)(v) = (0.77 mole)(0.0821 L.atm./mol.K.)(296 K) Solving for V we get: V = (0.77 mole)(0.0821 L.atm./Mole K)(296 K) 1.05 atm. And the answer is: 17.8, or 18 Liters Solve the problems below. Show your work, and include units with all measurements. Problem 10.0 o C? 9. What is the volume of 2.77 moles of H2 gas at a pressure of 700. mm and a temperature of 10-9 1997, A.J. Girondi
Problem 10. What is the temperature in o C of 1.06 moles of O2 gas at a volume of 670. ml and a pressure of 900. mm? Problem 11. What is the pressure (in mm of Hg) exerted by 0.80 moles of methane (CH4) gas at 67 o C and if it has a volume of 2.3 liters? In the ideal gas law, the letter "n" represents moles of gas. Moles of a substance can be calculated by dividing the number of grams (g) of the substance that you have by its molecular mass (MM). For example, 4.04 g of H2 (molecular mass = 2.02) is 2.00 moles: 4.04 g H 2 X 1 mole H 2 2.02 g H 2 = 2.00 moles H 2 In other words, moles = grams molecular mass or n = g MM Now if n = g MM, then the ideal gas law can also be written as PV = g MM RT An alternate form of the ideal gas law is: PV = g MM RT This form of the ideal gas equation can be rearranged to solve for any of the variable it contains. For example: MM = grt PV or g = (MM)PV RT Remember this form of the ideal gas law because it is particularly useful in problems which involve the mass in grams of a gas or the molecular mass of a gas. Problems which have you solved before using dimensional analysis can also be solved using the ideal gas law. Problems 11 through 14 below are identical to problems 21, 22, and 23 in chapter 9. Solve them again, but use the ideal gas law this time, and check Chapter 9 to see if you get the same answers you got before. Problem 12. A gas sample with a mass of 50.0 grams has a volume of 40.0 L at STP. What is the molecular mass of the gas? (You want to end with units of g/moles.) 10-10 1997, A.J. Girondi
Problem 13. What is the volume in liters occupied by 10.0 grams of CO2 gas at STP? (Hint: you must first find the molecular mass of CO2.) Problem 14. If 1.00 mole of a gas has a mass of 18 grams, and you have 15.0 grams of the gas, what would its volume be at STP? At this point, go back and compare the three problems above to problems 21, 22, and 23 in Chapter 9. Compare the methods of solving the problems as well as the answers. Now let's try a problem in which the gas is not at STP: Problem 15. At 600. mm Hg pressure and 35.0 o C, 2.70 L of a gas has a mass of 3.80 grams. What is its molecular mass? (Since you are using 0.0821 for the constant R, remember that you must express pressure in atm and temperature in K.) ACTIVITY 10.5 Determination of the Molar Volume of a Gas In this activity you will learn one method for determining the molar volume of a gas, and you will learn how to collect a gas in the lab by a procedure known as water displacement. You will be making measurements at room temperature and pressure, but conversion to standard conditions (STP) will be necessary to complete the calculations. Fill a large beaker (600 ml or larger) about 3 /4 full of tap water. Next, ask your instructor or others in your class whether you should follow procedure A or B below (do not follow both). You will use procedure B if you have a balance in your lab which can measure mass to three decimal places. 10-11 1997, A.J. Girondi
Procedure A: Obtain a piece of magnesium ribbon about 3.5 cm long. On an analytical balance which is accurate to three decimal places, determine the mass of your piece of Mg. It must have a mass between 0.044 and 0.046. Trim with scissors as needed. Record the mass in Table 10.3. Mg Procedure B: (Follow this only if you cannot use procedure A.) Cut a piece of magnesium ribbon about 3.5 cm long. Carefully measure the length of the piece of ribbon to the nearest millimeter, mm. Record the length in Table 10.3. 1. Obtain a piece of fine copper wire about 15 cm long and tie it to the Mg ribbon which has been rolled into a coil that will fit inside the gas-measuring tube (see Figure 10.2). Obtain a ring stand with a utility clamp to support the gas-measuring tube. 1 or 2 holed stopper Cu wire 2. Slowly pour about 10 to 15 ml of 6 M HCl (hydrochloric acid) into the gas tube. Use caution. Incline the tube slightly so air may escape and slowly fill it to the brim with tap water from a beaker. Try not to mix the acid and water any more than necessary. 3. With the tube completely full of water, insert the magnesium ribbon about 3 or 4 cm into the tube. With the wire against the side of the tube, insert a 1- hole or 2-hole stopper. The stopper should force a little water out of the tube and should hold the wire in place. 4. With your finger over the hole in the stopper, invert the tube and place the stoppered end into the beaker 3 /4 full of water. Clamp the gas-measuring tube in place so that the bottom of the rubber stopper in slightly above the bottom of the beaker. The reaction will not start immediately, since the acid has to make its way down to the metal. The acid will react with the magnesium to produce hydrogen gas: acid + water Mg(s) + 2 HCl(aq) ----> MgCl2(aq) + H2(g) 5. When the Mg has reacted completely, tap the tube with your finger to remove any bubbles you may see on the side of the tube. Adjust the height of the gas- collecting tube so that the level of the water inside the tube is equal to that in the beaker. This will make the pressure on the gas inside the tube equal to the pressure in the room. If you have trouble doing this, ask your teacher for advice. Now read the level of the water inside the tube as carefully as possible. This will be the volume of gas in the tube. Record the volume in Table 10.3. 6. Record the temperature of the water in the beaker ( o C) which we will assume to Table 10.2 Vapor Pressures of Water T ( o C) P (mm) T ( o C) P (mm) 16 13.6 22 19.8 17 14.5 23 21.2 18 15.5 24 22.4 19 16.5 25 23.8 20 17.5 26 25.2 21 18.6 27 26.7 Figure 10.2 be the temperature of the gas in the tube. In addition, measure the barometric pressure (mm Hg). Your teacher will help you to read the barometer in the lab. The temperature and pressure must be measured immediately after the gas has been collected. Record these values in the table. 10-12 1997, A.J. Girondi
7. Empty the contents of the tube and beaker and rinse both. Return all equipment to the proper place.(if you followed procedure B earlier in this experiment, ask your teacher for the precise mass of 1.000 meter of magnesium ribbon. Record this value in Table10.3. If you followed procedure A, you can disregard this value.) Table 10.3 Molar Volume of a Gas *(Skip entries 1, 2, and 3 below if you followed procedure A. Skip entry 4 if you followed procedure B.) *1. Length of Mg ribbon used mm *2. Mass of 1.000 meter of Mg ribbon g/m *3. Calculated mass of Mg ribbon used g (should be between 0.040 and 0.045 g) *4. Mass of Mg ribbon used g (from analytical balance) 5. Volume of gas in tube ml 6. Temperature o C 7. Atmospheric pressure mm Hg (Do calculation 3 before completing entries 8 and 9 below.) 8. Water vapor pressure mm Hg 9. Pressure of dry H2 mm Hg Calculations: 1. Calculate the mass of Mg ribbon used to the nearest 0.001 g by using the mass of 1.000 meter of the ribbon, which you got from the teacher. Use unit analysis. (Skip this calculation if you followed procedure A.) 2. Based on the mass of Mg used, calculate the number of moles of Mg used. Express your answer to three decimal places. That amount of precision is needed! 3. The pressure of the gases in the tube is equal to the atmospheric pressure in the room. The H2 gas in the tube was mixed with water vapor. This is a problem which results from collecting a gas in this way. Therefore, the pressure in the tube is the sum of the pressures exerted by H2 gas and H2O vapor. Dalton's law of partial pressures reveals that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures exerted by each gas. 10-13 1997, A.J. Girondi
We need to know the pressure exerted by the H2 only. We call this the pressure of the "dry" H2. To get this, we will subtract the partial pressure of water vapor from the total pressure. The formula is: Ptotal = Pwater + Phydrogen. Ptotal is the atmospheric pressure. The vapor pressure of water, Pwater, is dependent on temperature and can be found in Table 10.2. (Your ALICE reference notebook also contains a table of water vapor pressures which covers a broader temperature range.) Use the atmospheric pressure and the partial pressure of water vapor to calculate the pressure of the "dry" hydrogen: Phydrogen = Ptotal - Pwater Pressure of "dry" hydrogen = mm Hg 4. The pressure you just calculated above is the pressure to be used in your calculations. You now have the pressure, the temperature, and the volume of the H2 under these conditions. Now, you must calculate what the volume of the H2 which you collected in the tube would be (in liters) at STP. Use the combined gas law (from Chapter 4 as given below). P 1 V 1 T 1 = P 2 V 2 T 2 L H2 at STP 5. In calculation 2, you determined the number of moles of Mg used in this reaction. Enter this value in the blank below. According to the equation for the reaction, the number of moles of Mg used is the same as the number of moles of H2 produced. Fill in the blanks in the mole-to-mole ratio below, and solve for moles of H2 produced. Mg + 2 HCl ----> MgCl2 + H2 mole Mg X mole H 2 mole Mg = mole H 2 How many moles of H2 were produced in your experiment? moles H2 produced 6. With your results from steps 4 and 5 above, you have established a relationship between moles of H2 and liters of H2 at STP. Now you must use this relationship to determine what volume the hydrogen gas would have occupied if you had collected 1.00 mole of it (instead of the very small number of moles of H2 which you actually did collect). You can do this by unit analysis. The "fencepost" is set up for you below. Copy the data from calculations 4 and 5 above into the ratio below and solve the problem. 1.00 mole H 2 X L H 2 moles H 2 = L H 2 Because of the very small volume of gas collected in this activity, there is usually a fair amount of error. How does your answer for the volume of one mole of a gas at STP, compare to the generally accepted value of 22.4 L? 10-14 1997, A.J. Girondi
Problem 16. Assume that 0.0200 mole of a gas is collected in the lab by water displacement. If the temperature of the gas is 26.0 o C and the total pressure in the gas tube is 745 mm Hg (including water vapor), what is the volume in liters of the dry gas? (Use the ideal gas law.) Caution: you can t use the total pressure in your calculations. You must first find the pressure of the dry gas by subtracting the water vapor pressure at 26.0 o C from the total pressure. It is the pressure of the dry gas that is used in the calculations. Water vapor pressures can be found in Table 12 in your reference notebook. SECTION 10.6 Optional Review Problems Problem 17. Using the ideal gas law, determine the number of moles of gas which will have a volume of 2.5 liters at 1.3 atm and 303 K. Problem 18. Using the ideal gas law, determine the temperature (in o C) of 1.8 moles of a gas if it has a volume of 50.9 L at 705 mm Hg. Problem 19. Using a form of the ideal gas law, calculate the number of grams of helium gas, He, which are present in an 8.00 liter container at 345 o C and 1.33 atm. Problem 20. A 14 liter container holds 50.0 grams of a gas at 2.3 atm and 345 K. The gas is known to be one of the following: O2, Cl2, Br2, or CO2. What is the molecular mass and the identity of the gas? Problem 21. How many moles of KClO3 are required to form 37.3 g of KCl? 2 KClO3(s) ----> 2 KCl(s) + 3 O2(g) Problem 22. Analysis of a sample of propane gas (also known as LP gas) from a backyard grill is shown to contain 36.7 grams of C and 8.244 grams of H. Calculate the empirical formula of propane gas. Problem 23. Calculate the empirical formula of vitamin C if it is composed of 40.9% Carbon, 4.58% hydrogen, and 54.5% oxygen. Problem 24. A sample of a hypothetical compound contains 2.44 g of element X and 1.02 g of element Z. The atomic mass of hypothetical element X is 12.2 grams/mole, while that of hypothetical element Z is 2.04 grams/mole. The molecular mass of this hypothetical compound is 69.2. Calculate both the empirical formula and the molecular formula of this compound. Problem 25. 245 ml of a wet gas is collected by water displacement at 27.0 o C and a total pressure of 805 mm Hg. How many moles of the dry gas does this represent?. (Answer using 3 sig figs.) 10-15 1997, A.J. Girondi
SECTION 10.7 Learning Outcomes This is the end of Chapter 10. Check the learning outcomes below to be sure that you have mastered them. Take the Chapter 10 exam, and move on to Chapter 11. 1. Calculate the empirical formula of a compound from experimental data. 2. Calculate the molecular formula of a compound from experimental data. 3. Solve problems using both forms of the ideal gas law, PV = nrt and PV = grt/mm 4. Solve problems related to the collection of a gas by water displacement. SECTION 10.8 Answers to Questions and Problems Questions: {1} tin (IV) oxide; {2} stannic oxide Problems: 1. MgO 2. Fe2O3 3. CH4 4. Na2SO4 5. K2Cr2O7 6. C3H12 7. C2H5 and C4H10 8. C6H6 9. 69.9 L 10. -264 o C 11. 7379 mm Hg (rounds to 7400 mm Hg) 12. 28.0 g/mole 13. 5.09 L 14. 19 L 15. 45.1 g/mole 16. 0.518 L 17. 0.13 mole 18. 47 o C 19. 0.84 g 20. 44 g/mole, CO2 21. 0.50 mole 22. C3H8 23. C3H4O3 24. X2Z5, X4Z10 25. 0.0102 mole 10-16 1997, A.J. Girondi