Name: Ngo Dang Quang Hieu Trieu Vinh Khang Dinh Xuan Phuoc Duong Khac Trieu 22125084 22125112 22125226 22125333 Experiment 12 Report Sheet Molar Mass of a Volatile Liquid A Preparing the Sample Unknown number: A (Ethanol) Trial 1 Mass of dry flask, foil, and rubber band (g) Trial Trial 76.36(g) B Vaporize the Sample Temperature of boiling water (oC, K) 98.5 oC 93 oC 95 oC Mass of dry ask, foil, rubber band and vapor (g) 76.63(g) 76.53(g) 76.72(g) C Determine the Volume and Pressure of the Vapor Volume of 125 mL flask (L) 168.00(mL) 100+20+ 48=Total volume Atmospheric Pressure (torr, atm) 1.00(atm) D Calculations Moles of vapor, nvapor (mol) 6.54 x 10−3 (mol) Mass of vapor, mvapor (g) Molar mass of compound (g/mol) 0.27(g) 0.17(g) 0.36(g) 41.28(g/mol) 25.99(g/mol) 55.04(g/mol) Average molar mass (g/mol) 40.77(g/mol) Standard deviation of molar mass 14.53 Relative standard deviation of molecular mass (%RSD) Calculations for RSD: S t andard Deviation= %RSD= √ ∑ ( x− x´ ) n−1 = √ 35.63% ( 41.28−40.77 )2 + ( 25.99−40.07 )2 + ( 55.04−40.77 )2 =14.53 3−1 Standard Deviation 14.53 x 100 %= x 100=35.63 % Arverage molar mass 40.77 Class data/group Molar mass Sample unknown no 41.28 25.99 55.04 81.04 48.92 67.27 A Preparing the Sample Unknown number: B (Acetone) Trial Mass of dry flask, foil, and rubber band (g) Trial Trial 76.49(g) B Vaporize the Sample Temperature of boiling water (oC, K) Mass of dry ask, foil, rubber band and vapor (g) 96 oC 93 oC 98.5 oC 77.02(g) 76.81(g) 76.93(g) C Determine the Volume and Pressure of the Vapor Volume of 125 mL flask(L) 168.00(mL) 100+20+ 48=Total volume Atmospheric Pressure (torr, atm) 1.00(atm) D Calculations Moles of vapor, nvapor (mol) 6.54 x 10−3 (mol) Mass of vapor, mvapor (g) Molar mass of compound (g/mol) 0.53(g) 0.32(g) 0.44(g) 81.04(g/mol) 48.92(g/mol) 67.27(g/mol) 10 Average molar mass (g/mol) 65.74(g/mol) 11 Standard deviation of molar mass 16.11 12 Relative standard deviation of molecular mass (%RSD) Calculations for RSD: Standard Deviation= %RSD= √ ∑ ( x−´x ) n−1 24.50% ( 81.04−65.74 )2+ ( 48.92−65.74 )2+ ( 67.27−65.74 )2 = =16.1 3−1 √ Standard Deviation 6.11 x 100 %= x 100=24.50 % Arveragemolar mass 65.74 Laboratory Answers Part A.1 The mass of the flask (before the sample is placed into the flask) is measured when the outside of the flask is wet However, in Part B.3, the outside of the flask is dried before its mass is measured a Will the mass of vapor in the flask be reported as too high or too low, or will it be unaffected? The mass of vapor in the flask be reported will too high by when subtracting the mass of the dry flask, foil, rubber band from the mass of dry flask, foil, rubber band and vapor, the first part will be much higher because of the wetness, so it'll seem as though there was a greater mass of vapor than there was b Will the molar mass of the vapor in the flask be reported as too high or too low or will it be unaffected? The molar mass of the vapor in the flask be reported will too high; when calculating molar mass there'll be a much higher grams than there was supposed to be From the time the mass of the flask is first measured in Part A.1 until the time it is finally measured in Part B.3, it is handled a number of times with oily fingers Does this lack of proper technique result in the molar mass of the vapor in the flask being reported as too high or too low or as unaffected? The molar mass of the vapor in the flask being reported as too high since the oil from the fingers will increase the amount of mass overall on the flask, and during calculations it'll seem like there a greater mass of vapor than there actually is Part B.2 The flask is completely filled with vapor only when it is removed from the hot water bath in Part B.3 However, when the flask cools, some of the vapor condenses in the flask As a result of this observation, will the reported molar mass of the liquid be too high, too low, or unaffected? The reported molar mass of the liquid will be unaffected because the mass of the vapor should be equivalent to the mass of the vapor that condenses Part B.2 Suppose the thermometer is miscalibrated to read 0.3°C higher than actual Does this error in calibration result in the molar mass of the vapor in the flask being reported as too high, too low, or as unaffected? Explain The molar mass of the vapor in the flask being reported as too high by when calculating moles of vapor using PV =nRT if the temperature is too high then the calculated mole will be too low Then when you divide the mass by these too low moles then the molar mass will be too high Part C.1 If the volume of the flask is assumed to be 125ml instead of the measured volume, would the calculated molar mass of the unknown liquid be too high, too low, or unaffected by this experimental error? Explain The calculated molar mass of the unknown liquid will be too high as when calculating moles of vapor using PV =nRT if the volume is too low then the calculated moles will be too low Then when you divide the mass by these too low moles then the molar mass will be too high The pressure reading from the barometer is recorded higher than it actually is How does this affect the reported molar mass of the liquid: too high, too low, or unaffected? Explain The reported molar mass of the liquid will too low because if calculate moles of vapor using PV =nRT if the pressure if too high then the calculated moles will be too high Then when you divide the mass by these too high moles then the molar mass will be too low Experiment Prelaboratory Assignment Molar Mass of a Volatile Liquid The following data were recorded in determining the molar mass of a volatile liquid following the Experimental Procedure for this experiment Mass of dry ask, oil, and rubber band (g) 74.722 Temperature of boiling water (℃, K), K) 98.7 Mass of dry ask, foil, rubber band and vapor (g) 74.921 Volume of 125 mL ask (L) 0.152 Atmospheric pressure (torr, atm) 752 a How many moles of vapor are present? We have: P= 752 =0.989 atm 760 V = 0.152 L T=371.85 ° K R = 0.08206 n= PV = 4.92x 10-3 mol RT b What is the molar mass of the vapor? Mass of vapor: 0.199g Molar mass = 0.199 = 40.36 g/mol 4.92 x 10−3 a If the atmospheric pressure of the flask is assumed to be 760 torr in question 1, what is the reported molar mass of the vapor? n= PV x 0.152 = = 4.98 x 10-3 RT 0.08206 x 371.85 Molar mass of vapor = 0.199 =39.94 g/mol 4.98 x 10−3 b What is the percent error caused by the error in the recording of the pressure of the vapor? %erro r = M differnce 40.36−39.94 x 100= x 100=1.04 % Mactual 40.36 The ideal gas law equation (equation 12.1) is an equation used for analyzing ideal gases According to the kinetic molecular theory that defines an ideal gas, no ideal gases exist in nature, only real gases Van der Waals’ equation is an attempt to make corrections to real gases that not exhibit ideal behavior Describe the type of gaseous molecules that are most susceptible to nonideal behavior According to the ideal gas law, no intermolecular attraction exists between the gas molecules and the specific size of the gas molecules is negligible compared to the overall volume occupied by the gases For example, polar gases that can have stronger dipole-dipole interactions often behave less ideally than nonpolar gases Larger gases are also susceptible to the non-ideal state because the specific volume of the gas becomes more significant a How is the pressure of the vaporized liquid determined in this experiment? Pressure of liquid vaporized = Atmospheric pressure – water vapor pressure at that temperature b How is the volume of the vaporized liquid determined in this experiment? Filling the Erlenmeyer flask with water, then the volume of the flask is measured by transferring the water into the graduated tube c How is the temperature of the vaporized liquid determined in this experiment? After the water boils slightly, the temperature is recorded with a thermometer d How is the mass of the vaporized liquid determined in this experiment? Let cool until room temperature, then dry the outside with a scale e The molar mass of a compound is measured to be 30.7, 29.6, 31.1 and 32.0 g/mol in four trials a What is the average molar mass of the compound? 30.85 g/mol b Calculate the standard deviation and the relative standard deviation (as %RSD) for the determination of the malar mass The Standard Deviation = 0.86 %RSD = standard deviation 0.86 = = 27.8% the average molar mass 30.85