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Measuring the Molar Masses of Gases - Lab Report Example

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The paper "Measuring the Molar Masses of Gases" tells us about the molar mass determination process. According to Avogadro, one mole of any gas contains 6.02 × 1023 atoms. Therefore, the molar mass is expressed in grams per mole; gmol-1…
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Measuring the Molar Masses of Gases
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Lab Report: Measuring the Molar Masses of Gases Introduction Theoretically, the molar mass of any gas is the mass of a gas containing 6.02 × 1023 atoms. According to Avogadro, one mole of any gas contains 6.02 × 1023 number of atoms. Therefore, molar mass is expressed in grams per mole; gmol-1. Practically, determining the molar mass of a gas is technically demanding, but possible. The molar mass determination process requires complemented efforts of definite gas laws, especially Avogadro’s gas principles and Archimedes’ buoyancy laws. Archimedes laws facilitate determination of floating and sinking of an object by relating the parameters of density with mass. Equation for Archimedes buoyancy force is; Fb = g(mobject – mair), where Fb = buoyancy force, g = gravitational acceleration, and m = mass. On the other hand, Avogadro’s laws allows for the determination of a gas’ volume at a given density. Equation for Avogadro’s law is; molar mass/densitySTP = 22.415L/mole. By determining the volume and mass of any gas, it becomes easy to deduce a gas’ molar mass (Peck 68). This experiment involved determining the molar masses of hydrogen and helium; gases that are less dense than air. Experimental Procedures Prior to commencement of the experiment, appropriate safety requirements were performed. First, use of safety goggles and lab coats was prioritized. Secondly, water spilled on the floor was wiped in order to prevent slippage of students during the experiment. Finally, any chance of hydrogen ignition was considered, especially adherence to the use of fumed hood and prevention of static electricity. After the safety considerations, procedural steps of the experiment were performed. The volume of the pitcher was measured using a graduated cylinder. Simultaneously, the balloon was prepared by inflating it with either hydrogen or helium and subsequently weighed. Either excess balloon was cut off or additional pieces of strings added to the balloon setup until the balloon could float in air. Next, the pitcher setup was performed by immersion into water and the balloon gas transferred into the immersed pitcher. Finally, the gas’ volume was measured together with other relevant parameters like air pressure and environmental temperature. As a means of facilitating consistency and precision, three trials were performed for both hydrogen and helium. Data and results from each step of the experiment were recorded in the sheet below. Data and Results Sheet Section: 01 Date: February 3, 2015 Volume of pitcher: 2.050 L Air Pressure: 1.01 atm Room temperature: 296.9 K Water vapor pressure: 0.029 atm Hydrogen Trial 1 Trial 2 Trial 3 Mass of empty balloon 1.10g 1.13g 1.18g Mass of clipped pieces 0.17g 0.16g 0.13g Mass of added strings 0.0g 0.0g 0.0g Final mass of balloon 1.27g 1.29g 1.31g Mass of gas 0.55g 0.67g 0.56g Volume of leftover water 1.390 L 1.485 L 1.360 L Volume of gas in pitcher 2.050 – 1.390 = 0.660 L 2.050 - 1.485 = 0.565 L 2.050 – 1.360 = 0.690 L Pressure of gas 1.01 – 0.029 = 0.98 atm Volume of gas at STP VSTPPSTP/TSTP = VGASPGAS/TGAS, Therefore, VSTP = (0.660 L × 0.98 atm × 273 K)/(296,9 K × 1 atm) VSTP = 0.595 L VSTPPSTP/TSTP = VGASPGAS/TGAS, Therefore, VSTP = (0.565 L × 0.98 atm × 273 K)/(296.9 K × 1atm) VSTP = 0.509 L VSTPPSTP/TSTP = VGASPGAS/TGAS, Therefore, VSTP = (0.609 L × 0.98 atm × 273 K)/(296.9 K × 1atm) VSTP = 0.622 L Molar mass of gas Molar mass = (D × R × T)/P Therefore, molar mass of hydrogen = (0.83 g/l × 0.0821 × 273.15 K/1 atm) = 1.861 g/mol 2.646 g/mol 1.816 g/mol Average molar mass of hydrogen (1.861 + 2.646 + 1.816)/3 = 2.107 gmol-1 Helium Trial 1 Trial 3 Trial 3 Mass of empty balloon 1.08 g 1.18 g 1.17 g Mass of clipped pieces 0.0 g 0.0 g 0.0 g Mass of added string 0.19 g 0.44 g 0.27 g Final mass of balloon (1.08 + 0.0 + 0.19) = 1.27 g 1.62 g 1.44 g Mass of gas 0.067 g 0.089 g 0.060 g Volume of leftover water 0.820 L 0.635 L 0.710 L Volume of gas in pitcher 2.050 – 0.820 = 1.230 L 2.050 – 0.635 = 1.415 L 2.050 – 0.710 = 1.340 L Pressure of gas 1.01 – 0.029 = 0.981 atm Volume of gas at STP VSTPPSTP/TSTP = VGASPGAS/TGAS, Therefore, VSTP = (1.230 L × 0.98 atm × 273 K)/(296,9 K × 1 atm) VSTP = 1.11 L VSTPPSTP/TSTP = VGASPGAS/TGAS, Therefore, VSTP = (1.415 L × 0.98 atm × 273 K)/(296,9 K × 1 atm) VSTP = 1.27 L VSTPPSTP/TSTP = VGASPGAS/TGAS, Therefore, VSTP = (1.340 L × 0.98 atm × 273 K)/(296,9 K × 1 atm) VSTP = 1.20 L Molar mass of helium Molar mass = (D × R × T)/P Therefore, molar mass of helium = (0.05 g/l × 0.0821 × 273.15 K/1 atm) = 1.21 g/mol 1.34 g/mol 0.89 g/mol Average molar mass of helium (1.21 + 1.34 + 0.89)/3 = 1.14 gmol-1 Post-lab Questions Percentage Error Based on the data collected during the experiment, the calculated average molar mass of hydrogen is 2.107 g/mol while that of helium is 1.14 g/mol. The theoretical molar mass of hydrogen gas is 2.0158 g/mol (Peck 61). However, the experimental value is 2.107 g/mol. Therefore, the experiment’s percentage error of hydrogen gas is given by [(2.0158 – 2.107)/ 2.0158] × 100; percent error = 4.524%. On the other hand, the literature mass for helium gas is 4.0002 g/mol. Percent error is [(4.0002 – 1.14)/4.0002] × 100 = 7.15%. Sources of Errors Undeniably, the substantial percentage errors in the experimental values signify presence of experimental errors. One probable cause of the deviations is parallax as an observational error. During instrument reading, there is a remote yet distinct possibility that the observer made subjective judgmental errors when deducing the smallest division of an instrument’s scale. Besides observational errors, it is possible that environmental errors contributed significantly to the deviations (Peck 73). Air pressure and environmental temperature may have unpredictably fluctuated during the experiment. Alternative Gases This experiment is used to determine the molar mass of any gas lighter than air. In this context, any gas less dense than air could be used to replace hydrogen and helium in the experiment. Among the suitable gases include methane, ammonia, neon and pure nitrogen. Neon is like helium, only that it is rear on earth. Pure nitrogen is approximately 3% less dense than air, while methane and ammonia are poisonous, but feature as suitable lifting gases that could be used in this experiment (Peck 90). Hydrogen Balloon Calculation In determining either sinking or floatation of a balloon in air, this Archimedes equation is used; Fb = g(mobject – mair), where Fb = buoyancy force, g = gravitational acceleration, and m = mass. A negative buoyancy force indicates floatation while a positive buoyancy force means sinking (Peck 77). Volume of hydrogen = 0.950 L, mass of balloon = 1.090 g. In this case, 1 mole of hydrogen = 22.4 L, therefore, 0.950 L has (0.950 L ×1 mole)/22.4 L = 0.04241 moles. The molar mass of hydrogen = 2.0158 g/mol. Therefore, 0.04241moles of hydrogen = (0.04241moles × 2.0158 g/mol) = 0.0855 g. Total mass of balloon + hydrogen = (1.090 + 0.0855) = 1.1755 g. On the other hand, the mass of air is given by density of air × volume of air equal to that of hydrogen. Mass of air = 1.205 g/L × 0.950 L) = 1.14475 g. Using Archimedes equation, Fb = 9.80 ms-2(1.1755 g – 1.14475 g) Fb = 0.3014. Since the buoyancy force is positive, it means the balloon will not float. In order for the balloon to float, Fb must be less than 0. This means the mass of balloon plus hydrogen must be equal or less than the mass of air. Therefore, at least (1.1755 g – 1.14475 g) = 0.03075 g must be removed from the balloon set up for floatation to be achieved. Reflection In conclusion, it is acknowledgeable that determination of a gas’ molar mass is a complex process that entails application of key chemistry concepts, specifically buoyancy, Avogadro’s principles, and ideal gas laws. Personally, I was impressed by all procedural steps of the experiment. Actually, I would be more than willing to reproduce any part of this experiment at any time. However, I would be pleased with the experiment if any suitable gas other than hydrogen is used in the procedures. Use of hydrogen involved substantial risks of accidental ignition and combustion. In addition, any unintentional act that could ignite the hydrogen gas may be easily misunderstood by the professor as a form of student’s misconduct, thus attracting unnecessary punitive actions. Otherwise, the entire experiment was awesome. Work Cited Peck, Larry. Experiments in General Chemistry: An Application of Gas Laws. New York, NY: Cengage Learning, 2013. Print. Read More
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