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  Chem 142B, HW 6 answer key, Fall 2005  Note: There is  always  more then one way to solve a given problem. A different method to solve then indicated here, as well as different rounding at different points in the problem, may yield slightly different answers. Also, all calculations have an extra digit carried through until the final answer, which is expressed to the correct # of significant figures. 1. Discrepancies in the experimental values of the molar mass of nitrogen provided some of the first exidence for the existance of the noble gases. If pure nitrogen is collected from the decomposition of ammonium nitrate, its measured molar mass is 28.01.  NH 4  NO( s ) N 2 ( g ) + 2 H 2 O( g ) If O 2 , CO 2 , and H 2 O are removed from air, the remaining gas has an average molar mass of 28.15. Assuming this discrepancy is solely a result of contamination with argon (atomic mass = 39.95), calculate the ratio of moles of Ar to moles of N 2  in air. Solution: Here, we’re going to need to determine the mole fraction of nitrogen, and argon with respect to each other. If we let x=mole fraction of nitrogen in the atmosphere, then the mole fraction of argon in the atmosphere is going to be 1-x. This is because we have a binary system, and the total mole fraction must be equal to 1. With this, we can solve for the mole fraction of both, and determine the molar ratio between nitrogen and argon: ( ) 0119.001186.0 98827.001173.0:01173.098827.01 98827.095.3901.28 95.3915.28 /15.28195.3902.24. 2 222 ====−===−−==−+=+=  N  Ar ratio fraction fraction xmolg x x MM  MM massmolar ave  Ar  N  Ar  Ar  N  N   χ  χ    2. Metallic molybdenum can be produced from the mineral molybdenite, MoS 2 . The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide. Molybdenum trioxide is then reduced to metallic molybdenum using hydrogen gas. The balanced equations are given below. MoS 2 ( s ) + 7 / 2  O 2 ( g ) MoO 3 ( s ) + 2 SO 2 ( g ) MoO 3 ( s ) + 3 H 2 ( g ) Mo( s ) + 3 H 2 O( s ) Calculate the volumes of air and hydrogen gas at 30.°C and 1.00 atm that are necessary to produce 1.00 10 3  kg of pure molybdenum from MoS 2 . Assume air contains 21% oxygen by volume and assume 100% yield for each reaction. Solution: first, we’re going to need to determine the amount of Mo in moles that we are going to be dealing with. Once that is known, we can use the stoichiometry of the reaction to determine the moles of oxygen (and thus air) needed and hydrogen needed to perform the conversion: 222223232 44544446 10127.3 310042.1 1074.1 21.010648.3 10648.3 5.310042.1 100423.1 /94.95 1000.1  H  Mo H  Mo H air Oair O MoOO Mo MoO MoO Mo mol xmolmolmol xmolesmol xmol xmolesmol xmolmolmolmolmol xmolesmol xmolgg xmoles =⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ ====⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ ⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ ===     Now that we know how many moles of each are needed for the conversion, the ideal gas law, PV=nRT, can be used to determine the volume of each gas: 22 778.710775.7.303*08206.*10127.3 103.41032.4 1.)3015.273(*08206.0*1074.1 54665  H  H air air   L x xVolume  L x x xPnRT Volume =====+==  3. Urea (H 2  NCONH 2 ) is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide. 2 NH 3 ( g ) + CO 2 ( g ) H 2  NCONH 2 ( s ) + H 2 O( g ) Ammonia gas at 223°C and 90. atm flows into a reactor at a rate of 500. L/min. Carbon dioxide at 223°C and 45 atm flows into the reactor at a rate of 600. L/min. What mass of urea is produced per minute by this reaction assuming 100% yield? Solution: first, we’re going to want to calculate the moles/minute we have flowing into the reactor using the ideal gas law, then determine if there’s a limiting reagent:  /minmolmol/minminmol x RT PV  /minmol 23 CO NH  663)22315.273(*08206.0 .600*45 /1011.1 )22315.273(*08206.0 .500*.90 3 =+==+==  the flow of ammonia is going to be the limiting reagent, since the number of moles is not twice that of the carbon dioxide. Now, we can use the reaction stoichiometry between ammonia and urea to determine the rate that urea is fromed: ming x xmolg molmolmol xmass ureaurea NH urea NH urea /103.31033.3/06.60* 21011.1 443 33 ==⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ =   4. The nitrogen content of organic compounds can be determined by the Dumas method. The compound in question is first reacted by passage over hot CuO( s ). Compound N 2 ( g ) + CO 2 ( g ) + H 2 O( g ) The gaseous products are then passed through a concentrated solution of KOH to remove the CO 2 . After  passage through the KOH solution, the gas contains N 2  and is saturated with water vapor. In a given experiment a 0.261 g sample of a compound produced 31.8 mL N 2  saturated with water vapor at 25°C and 722. torr. What is the mass percent of nitrogen in the compound? (The vapor pressure of water at 25°C is 23.8 torr.) Solution: We’re going to need to determine the mass of nitrogen in the sample, as determined by the amount of nitrogen recovered as gas. First, we need to take out the partial pressure of water vapor to determine the partial pressure of nitrogen. Then we can use that in the ideal gas law to determine the number of moles of nitrogen gas. The partial pressure of nitrogen is going to be found by subtracting the vapor pressure of water vapor from the total pressure, ie 722-23.8=698.2 torr  nitrogen . Of course, in the ideal gas law, we’re going to need to convert to atmospheres, with the conversion factor of 760 torr=1 atm. ( ) %8.1281.12100* 261.0033458.0%033458.002.28*101941.1 101941.1 2515.273*08206.0*760 0318.0*2.698 233 22 ======+== −−  N g xmassmol x RT PV moles  N  N   5. An organic compound contains C, H, N, and O. Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2766 g of CO 2  and 0.0991 g of H 2 O. A sample of 0.4831 g of of the compound was  analyzed for nitrogen by the Dumas method. At STP, 27.6 mL of dry N 2  was obtained. In a third experiment the density of the compound as a gas was found to be 4.02 g/L at 127°C and 256 torr. What are the empirical formula and the molecular formula of the compound? (Type your answer using the format CO2 for CO 2 .) Solution: In order to determine the empirical formula, we need to first determine the mass percent of each element in the compound from the information given above. To determine the % C and H, we need to use the % mass of each in CO 2  and H  2 O respectively, so that we can determine the actual mass of each from the compound: %84.10 1023.0016.18016.20991.0%%79.73 1023.001.4401.122766.0% 2222 =⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ ==⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ = O H  H O H sampleCOC CO ggg H ggggC    Use the ideal gas law to determine the moles of nitrogen from the sample, convert to grams, and determine the %N in the sample. STP is 1 atm, and 273K: %14.7 4831.002.28*0123.0 %0123.0273*08206.0 0276.0* 22 =====  N mol Latm RT PV mol  N  N    Finally, the % mass of oxygen in the compound is the remainder, which ends up being 8.2%. Now, we assume we have 100g of the compound, and determine the moles of each we have to determine what the empirical formula is, after dividing everything through by the smallest number of moles: 513.000.1620.8;5096.0 01.1414.7;75.10 008.184.10;144.6 01.1279.73 ======== O N  H C  molmolmolmol  Which will give us an empirical formula of C  12  H  21  NO, when dividing through by about 0.51 (moles of O and N).  Now, we’ll use the density measurement to determine the molecular formula. We can rearrange the ideal gas law to get at the MM of a compound from it’s density: PdRT PV mRT  MM giveswhich ,  MM mRT nRT PV   ====  , where d=density in terms of grams/L. Plugging in what we know, we find the molar mass to be 201.1416008.1*2101.12*12 392/392760/25608206.0*400*02.4 ≈+++==≈== empericalmolecular  factor mult molgPdRT  MM   So that the molecular formula is C  24  H  42  N  2 O 2 . 6. Consider a 1.0 L container of neon gas at STP. Will the average kinetic energy, root mean square velocity, frequency of collisions of gas molecules with each other, frequency of collisions of gas molecules with the walls of the container, and energy of impact of gas molecules with the container increase, decrease, or remain the same under each of the following conditions? (a) The temperature is increased to 100°C. (b) The temperature is decreased to -50°C. (c) The volume is decreased to 0.5 L at constant temperature. (d) The number of moles of neon is doubled at constant temperature. Solution: think about what the changes mean. A change in temperature automatically mean, according to the kinetic theory of gases, that the kinetic energy of the gas is going to change, which will also correspond  to change in root mean square velocity (RMSV). Changing the temperature also will change the frequency with which the particle runs into things, and also the energy it hits it with, leading to the results of a and b. For C, if you keep the temp the same, the kinetic energy, RMSV, and impact energy must remain the same, since they are dependent on the temp. However, the smaller the volume, the particles will run into things more often then they were before, due to a smaller space to bounce around in. For d, adding more  particles is going to be the same as decreasing the volume. The more stuff there is in a container, the smaller the available volume is. 7. The effusion rate of an unknown gas is measured and found to be 31.50 mL/min. Under identical experimental conditions the effusion rate of O 2  is found to be 30.50 mL/min. Which of the following is the unknown gas: CO, CH 4 , CO 2 , NO, or NO 2 ? Solution: the one that would be just a little bit faster then oxygen, would be the molecule with a molecular weight just smaller then that for oxygen, in this case, it’s NO. According to the math, where the effusion rate is inversely proportional to the molar mass: molg MM  MM  MM  MM raterate unk unk Ounk unk O /00.3000.32* 50.3150.30;00.3250.31 50.30; 2 22 =⎟ ⎠ ⎞⎜⎝ ⎛ ===   And since MM   NO =30.01 g/mol, NO is the correct choice for the unknown gas. 8. (a) Calculate the pressure exerted by 0.5260 mol N 2  in a 1.0000 L container at 25.6°C using the ideal gas law. Solution: Plug and chug the values into the ideal gas law, solving for pressure: atmV nRT P ideal 89.12)6.2515.273(*08206.0*5260.0  =+==  (b) Calculate the pressure exerted by 0.5260 mol N 2  in a 1.0000 L container at 25.6°C using the van der Waals equation. Solution: the van der Waals equation takes the non-ideality of gases into account, by introducing some correction factors, which change between gases, to the ideal gas law: ( ) atmV nanbV nRT P real 78.121526.039.10391.0*5260.01 6.2515.273*08206.0*5260.0 22 =⎟ ⎠ ⎞⎜⎝ ⎛ −−+=⎟ ⎠ ⎞⎜⎝ ⎛ −−=  (c) Compare the results. (Write the percentage difference between the results, based on the higher value result.) Solution: subtract the two values, and divide by the higher pressure to find the percentage: %85.0853.0100* 89.1278.1289.12 %  ==−= diff    KE RMSV freq col. Wall freq coll others impact E a increase increase increase increase increase b decrease decrease decrease decrease decrease c same same increase increase same d same same increase increase same

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