Molecular Weight of a Volatile Liquid

Page 1 Physical Basis of Chemistry Third Experiment Molecular Weight of a Volatile Liquid Purpose: The Purpose of this experiment is to determine the molecular weight of a compound that is a volatile liquid. By vaporizing a measured amount of the sample, and determining the pressure, volume, and temperature of the sample, the approximate molecular weight may be determined by presuming ideal gas behavior. Background: • # 2 - Pressure, Gas Laws, Ideal Gas Law, Partial Pressures - Some Examples • Gases - Experiments and Relationships - PowerPoint Summary (PowerPoint summary of our discussion of gases.) These handouts are on Bb. Introduction: In Experiment # 2, you had the opportunity to determine the empirical formula of a binary compound. In order to determine the molecular formula of a compound - given the empirical formula - you need the molecular weight of the compound (in addition to the empirical formula weight). In this experiment, you will utilize one of many experimental procedures to determine the molecular weight of a compound. This method, first developed by Dumas, exploits the ideal gas law in order to determine the molecular weight of a compound that can be vaporized (without significant decomposition). Specifically - via the ideal gas law - you will determine the molecular weight of a substance by determining the pressure exerted, volume occupied, temperature, and mass of a sample of its vapor. The example below will hopefully enable you to understand how this is accomplished. Page 2 Materials (per pair): Various samples of volatile liquids (with boiling points < 100ºC) 125 mL Erlenmeyer flask 400 mL beaker 10 mL & 50 mL graduated cylinders 2” aluminum squares Rubber bands Pin Buret clamp Thermometer Bunsen burner Ring stand & ring Balances - Triple-beam & Analytical Lab barometer for reading of the atmospheric pressure (A crude sketch of the set-up is displayed on the following page.) Page 3 Set-up – an “adapted” Dumas method: Se t - up f or t o de t e rmine MW of a vaporiz e d volit ile liquid - Dumas me t hod - adapt e d Ring st and The rmome t e r se cure d wit h bure t clamp. Al f oil cap wit h small pinhole se cure d wit h a rubbe r band. 12 5 mL Erle nme ye r f lask Volat ile liquid - e nough t o cove r bot t om of Erle nme ye r f lask. 4 0 0 mL be ake r 3 " diame t e r ring Page 4 Experimental Procedure (work in pairs): 1. Place a small pinhole in the Al foil square. 2. Assemble the pictured apparatus – a clean, dry 125 mL Erlenmeyer flask, fitted with the Al foil square containing the pinhole. Carefully secure it over the mouth of the Erlenmeyer flask with the rubber band. 3. On an analytical balance, weigh (_.xxxx g) the apparatus assembled in step 2. 4. Place a 3” diameter ring on a ring stand at a height that will be comfortably above the tip of the Bunsen burner when it is lit. You will probably have to adjust the level of the ring once you position the lit Bunsen burner. 5. Partially fill (about ¾) a clean 400 mL beaker with water (tap water is fine) and place it on a wire gauze on the 3” diameter ring. Bring the water just to boiling. 6. Meanwhile, remove the rubber band and foil cap from the Erlenmeyer flask, and place 4-5 mL of the volatile liquid in the Erlenmeyer flask - the exact amount is not important – as long it completely covers the bottom of the flask. Then replace the foil cap and rubber band. 7. Once the water in the beaker is brought to boiling, remove the Bunsen burner and clamp the assembled apparatus to the ring stand by its neck with the buret clamp. [The amount of water in the beaker can be adjusted such that when the final set-up is fully assembled, the Erlenmeyer flask is submerged up to its neck – just an inch or so below the top - in the water bath in the beaker.] 8. Using a second buret clamp, secure a thermometer and position it in the water bath in the beaker such that it can monitor the water bath temperature without touching either the sides of the beaker nor the Erlenmeyer flask. 9. Return the Bunsen burner beneath the set-up and gently maintain the flame such that the thermometer registers a steady temperature somewhere between 85ºC and 95ºC. It should not be boiling. Be sure that the Erlenmeyer flask is submerged up to its neck. 10. The volatile liquid should steadily begin to vaporize. You can monitor the escaping excess vapor by looking at its vapor trail as it exits the pinhole. [Another way to monitor the escaping vapor is to place the blade of the spatula over the pinhole and notice the condensation – like your breath on a mirror.] 11. When you notice the vapor trail has ceased to escape from the pinhole, remove the Bunsen burner and immediately record the temperature on the thermometer to the tenth’s place (_.xºC). 12. Allow the system to cool to room temperature. Then carefully unclamp the the Erlenmeyer flask – still covered with the foil cap and rubber band – from the water bath. If it is still warm to the touch, you may set it on the bench top on the wire gauze. 13. Once cooled, completely dry the outside of the covered Erlenmeyer flask and weigh it on the same analytical balance (_.xxxx g). [By the way, when the system is cooled, you should notice that there is some liquid on the bottom of the Erlenmeyer flask – but less than the initial amount. Why is this?] 14. If you have time and are so inclined, please carry out another trial. That is, remove the rubber band and foil cap and add some more volatile liquid the bottom of the Erlenmeyer flask is covered once again. Repeat steps 8-13 and obtain a second weighing. Page 5 15. Once you have completed step 13 (or 14, if you choose to perform more than one trial), remove the foil cap and rubber band from the Erlenmeyer flask and properly dispose of any remaining volatile liquid. 16. Rinse out the Erlenmeyer flask be sure that it is dry. 17. Weigh the clean, dry Erlenmeyer flask on a triple-beam balance and record its mass to the second decimal place (_.xx). 18. Now fill the Erlenmeyer flask with DI water to the brim. Dry the outside if it is wet. Weigh the water-filled Erlenmeyer flask on the same triple-beam balance and record its mass to the second decimal place (_.xx). Also, record the temperature of the water in the filled Erlenmeyer flask on the thermometer to the tenth’s place (_.xºC). 19. Clean Up – Discard the liquid in the appropriate container, clean and return glassware to appropriate location. 20. Be sure to obtain and record the barometric pressure before you leave the lab. Lab Report – Submit electronically (digital drop box) by ___________________ Note: When submitting to digital Drop Box label you files with your name first and then a brief description of what the document is. Presuming that the vaporized volatile liquid follows the ideal gas law (PV = nRT), we may determine the moles of the contained vapor via: PV , i.e., its measured pressure (P), volume (V), n = moles of contained vapor = RT and temperature (T). Of course, the gas constant (R) is tabulated and equals 0.0821 L• atm•mol-1•K-1. Then, the molecular weight of the gas (MW in g/mol) can be determined from the measured mass of the contained vapor (g in grams) divided by the calculated number of moles. MW = g g gRT n = PV = PV . RT   Before you carry out the determination of the molecular weight (MW) of the volatile liquid, a worked example may help. Please refer to the Appendix at the end of this experiment. Page 6 DATA/CALCULATIONS Note: Data and Calculations may be completed on an Excel spreadsheet. Molecular Weight of a Volatile Liquid (If you performed more than one trial, adjust your data table accordingly.) Weight of flask + foil cap + rubber band g 93.2511g 93.2586 Temperature of water bath when (excess) vapor ceased to escape 93.8 Weight of flask + foil cap + rubber band + condensed vapor 93.6622g Weight of condensed vapor 0.5879 93.8465 g 96.4714g g 244. 8 Weight of clean, dry, and empty Erlenmeyer flask 92.8803 Weight of water that completely filled Erlenmeyer flask g g 151.9 Temperature of water in the filled Erlenmeyer flask 21 ºC g Weight of water in the filled Erlenmeyer flask 770.05 87.2 g .4036 Weight of Erlenmeyer flask filled with water Barometric pressure 86.5 ºC mm Hg gT & compare both runs Before you carry out the determination of the molecular weight (MW) of the volatile liquid, a worked example may help. Please refer to the Appendix at the end of this experiment. Page 7 Table 1: Density of pure liquid water at various temperatures: Temperature (ºC) 15ºC 16ºC 17ºC 18ºC 19ºC 20ºC 21ºC density (d) in g/mL 0.999099 0.998943 0.998774 0.998595 0.998405 0.998203 0.997992 Temperature (ºC) 22ºC 23ºC 24ºC 25ºC 26ºC 27ºC 28ºC density (d) in g/mL 0.997770 0.997538 0.997296 0.997044 0.996783 0.996512 0.996232 Determination of the volume occupied by the vapor in the Erlenmeyer flask: (Using the mass of water that completely filled the Erlenmeyer flask and the above table of water densities as needed – determine the volume (in L) occupied by the vapor in the Erlenmeyer flask. Also, explain your reasoning. In order to calculate the volume that the gas occupied, the water must be filled to the top to take up the total volume. The flask then needs to be massed. The mass of the flask needs to be subtracted from the total to calculate the mass of the water occupying the volume of the flask. The density of the water equals the mass of the water divided by the volume the water occupies. When we take the mass of the water divided by the volume of the water equals the density of the water. When we set the density of water at 294 Kelvin (0.9980), divide the mass of the water by it, it gives us the volume that the gas occupied. Density _ of _ water _ at _ 294 K = 0.9980 g mL Mass _ of _ water = 151.9 g 151.9 g 151.9 g 0.9980 g mL g = 0.9980 mL → ∗V = ∗ V → 152.2ml V V ∗ 0.9980 g mL 0.9980 g mL Volume of vapor .1522 L Determination of molecular weight of the volatile liquid: (Assuming ideal gas behavior, use the pressure of the vapor (P), its volume (V), and its temperature (T) to determine the moles of vapor in the Erlenmeyer flask. Then, using the mass of vapor which occupied the Erlenmeyer flask, determine the molecular weight (MW in g/mole) of the volatile liquid. If you performed more than one trial, show all work for each trial and then compute the average molecular weight.) Page 8 MW = gRT PV MW = gRT PV L ∗ atm ∗ 294 K ∗ mole K → = 91.94 g mol 1.014atm ∗ .1522 L L ∗ atm ∗ 294 K 0.4036 g ∗ 0.0821 mole K ∗ → = 63.12 g mol 1.014atm ∗ .1522 L 0.5879 g ∗ 0.0821 (Average) Molecular Weight 77.53 g/mole CONCLUSION QUESTIONS General Questions: 1. Why is the barometric (i.e., atmospheric) pressure considered to be the pressure of the vapor, i.e., how does the experimental procedure ensure this? Explain carefully. As it vaporizes the pressure is higher in the flask than the barometric pressure, and pushes out the air as it effuses. It stops pushing out the air when the pressure of the vapor is equal to the barometric pressure. 2. Why isn't it necessary to weigh the amount of liquid initially put into the flask? Explain. Any excess liquid will escape when it is vaporized. When the pressure is equal, the gas pressure will equal the barometric pressure, and the volume of the gas will equal the volume of the flask. 3. An allotrope is one of several forms of the same element with different structure. Consider the element carbon (C). Several years ago an allotrope of carbon - neither graphite nor diamond - was discovered and named buckminsterfullerine or “buckyball”, CX , where “X” is a whole number. A sample of CX(g) at 400.0oC and 680.0 mm Hg is found to have a density of 11.67 g/L. Determine “X” and hence the formula CX for one of these buckyball molecules. Extra – It is sometimes stated that the above method (Dumas method) relies on the presumption that the investigated gas follows the ideal gas law (PV = nRT). This allows one to determine the moles of the contained vapor via: PV , i.e., its measured pressure (P), volume (V), n = moles of contained vapor = RT and temperature (T). Of course, the gas constant (R) is tabulated and equals 0.0821 L• atm•mol-1•K-1. Then, the molecular weight of the gas (MW in g/mol) Page 9 can be determined from the measured mass of the contained vapor (g in grams) divided by the calculated number of moles. g g gRT = = . In other words, if the gas did not behave ideally, then the n PV PV RT   calculated molecular weight would be a crude approximation at best. Please explain why this is commonly not a big problem, i.e., why an approximate value of the MW = molecular weight is often sufficient when determining the actual chemical formula of a volatile compound. What other information is commonly obtained for a compound in the process of determining its chemical formula such that an approximate molecular weight is often “good enough”? It is often not a problem because the formula is always in whole numbers, and the approximation is rounded off to a whole number for the formula. Appendix - A worked Example: A certain volatile hydrocarbon (a binary compound of carbon and hydrogen) is found to be 92.3 % carbon, by mass. In a separate experiment, utilizing the Dumas method, a 4.00 mL pure liquid sample of this hydrocarbon is vaporized in a 125 mL Erlenmeyer flask when the barometric pressure is 768.0 torr. The empty flask - fitted with a foil cap pierced with a pinhole - weighs 25.3478 g. After the excess gas escapes, the temperature is measured as 98.0oC. The flask and contents are subsequently cooled to 25oC and the vapor condenses to a liquid. The weight of the flask and contents is found to be 25.6803 g. The flask is then emptied, cleaned, and filled with water. When weighed on a triple-beam balance, the difference in weight between the flask filled to the brim with water and the dry empty flask – at 25ºC - is 128.12 g. Please determine the following: • The empirical formula of this hydrocarbon. • The volume that the vapor occupied in the Erlenmeyer flask (in L). • The molecular formula of this hydrocarbon. Solution to Example: The ideal gas law in terms of mass (g = grams), molecular weight (MW), temperature (T), volume (V), and pressure (P). mass g = . PV = nRT and n = moles = molecular weight MW Page 10 n = g PV MW = RT Solving for MW: MW = gRT . PV The first equation above may be easily used to find the molecular weight of the L atm. . hydrocarbon. Note that the ideal gas constant R = 0.0821 mole K Also note that 760 torr = 1 atm. The kelvin (K) temperature is obtained by adding 273 to the temperature in oC. The first part of this problem, obtaining the empirical formula. A hydrocarbon contains only carbon and hydrogen. Thus, the mass % of H = 100 % - 92.3 % = 7.70 % H. We assume 100 g of the hydrocarbon. Then, there are 92.3 g of C and 7.70 g of H present (relatively). Determining moles of each element:  1 mole C  = 7.69 moles C. moles C = (92.3 g C)•  12.01 g C moles H = 1 mole H (7.70 g H)• 1.01 g H    = 7.62 moles H. Divide to obtain the mole ratio: 7.69 moles 1.01 moles C 1 = = = so, the Empirical Formula = CH. 7.62 moles 1 moles H 1 As was presumed when we carried out this experiment, the vapor completely fills the flask - once the excess escapes (also flushing out the air). The volume of water that fills the flask equals the full volume of the flask and the volume occupied by the gas in the ideal gas law (V). V = volume of vapor = full volume of the flask = volume of water that fills it to the brim. V = Vwater filling flask = mass of water filling flask . dwater Recall that density is temperature -dependent. So, we must use the density of water at its temperature when it filled the flask. By the info in the problem, this temperature is 25ºC. Referring to the table of densities (Table 1), the density of liquid water at 25ºC is listed as: 0.997044 g/mL. V = Vwater filling flask = 128.12 g = 128.50 mL = 0.12850 L 0.997044 g/mL To determine the molecular formula, we need to find the molecular weight. The most useful form of the ideal gas law - considering the given information - is: Page 11 MW = gRT ; T = 98 + 273 = 371 K and V = 0.1285 L. PV mass of condensed vapor = mass of gas in flask = 25.6803 g - 25.3478 g = 0.3325 g. MW = L atm.   (371 K) (0.3325 g) 0.0821 mole K gRT  = = 78.0 g/mole. PV 768.0  760  (0.12850 L)   The EFW = Empirical Formula Weight for CH = 12.01 + 1.01 = 13.02 g/mole. N = # of empirical formula units = 78.0 g/mole MW = = 5.99 = 6. 13.02 g/mole EFW Thus, in this case, since N = 6, the molecular formula is 6 x Empirical Formula. Thus, the Molecular Formula = C6H6 . (The compound is benzene).