Partner: Jack 17 November Determination of the Molar Mass of Volatile Liquids

Transcription

1 Partner: Jack 17 November 2011 Determination of the Molar Mass of Volatile Liquids Purpose: The purpose of this experiment is to determine the molar mass of three volatile liquids. The liquid is vaporized in a Beral-type pipet with capillary head then rapidly cool to condense the vapor. The mass of the condensed vapor is determined then used in a form of ideal gas equation to find the molar mass of each volatile liquid. The significance of this lab is that forensic scientist could use the molar mass of the volatile liquid together with the empirical formula of a compound to determine the molecular formula of that compound. Hypothesis: The hypothesis is that the molar mass of the volatile liquid can be found by vaporizing the volatile liquid in a Beral-type pipet. The mass of the condensed vapor can be used in a form of the ideal gas equation to find the molar mass of the volatile liquid. The temperature of the pipet was heated so it was high, so the molecules behave like an ideal gas because they have a lot of energy to overcome intermolecular forces. Materials: Materials Trial 1 Trial 2 Ethyl Alcohol (Ethanol) (C 2 H 5 OH) 9 ml 9 ml Acetone (CH 3 COCH 3 ) 9 ml 9 ml Isopropyl Alcohol ((CH 3 ) 2 CHOH) 9 ml 9 ml Distilled water 2 L Tap water 1.2 L 15-mL Beral-type pipets 18 pipets 4-decimals electric balance 1 balance Hot plate with magnetic stirrer 2 plates Magnetic bar 3 bars 250-mL beaker 2 beakers 150-mL beaker 3 beakers 600-mL beaker 5 beakers Ring stand 3 ring stands Thermometer 3 thermometers Scissors 1 pair Paper towel 3 blocks Permanent marker 4 markers Plastic tubing pipet holder 4 tubing pipet holders Universal clamp 3 clamps Boiling stones 90 ml 10-mL graduated cylinder 2 cylinders 25-mL graduated cylinder 2 cylinders Buret clamp 3 clamps Procedures:

2 1.) Add about 30 ml of boiling stones to the 600-mL beaker 2.) Add about 500 ml of distilled water to the beaker 3.) Put it on a hot plate with magnetic stirrer 4.) Start heating the water using heat level of 9 5.) Pull the thin part of the Beral-type pipet 6.) Keep pulling until it stretch, creating a capillary tube where it s pulled 7.) Try to leave as much thin part as possible 8.) Cut the capillary part of the pipet, leaving it about 1 cm long on the side of the bulb head 9.) Repeat step 5.) to 8.) for two more pipets 10.) Label the pipets as 1, 2, and 3 11.) Mass each pipet 12.) Using the graduated cylinder, measure 3 ml of ethyl alcohol 13.) Transfer the alcohol to pipet number 1 14.) Repeat step 12.) and 13.) for pipet number 2 and 3 15.) Insert the tube part of all three pipets through a plastic tubing pipet holder 16.) Clamp the plastic tubing 17.) Put the bulb part of the pipet in the heated water 18.) Put the thermometer in the beaker in a way that the thermometer does not touch the side of the beaker 19.) Add about 500 ml of tap water to another 600-mL beaker 20.) Wait until all the volatile liquid (ethyl alcohol) had vaporized 21.) As soon as all the liquid had vaporized, check the temperature 22.) Transfer the pipets to the tap water beaker 23.) Check the temperature of the tap water 24.) After a minute, take the pipets out of the water 25.) Gently dry each pipet with paper towel 26.) Mass each pipet 27.) Rinse the inside of each pipet with distilled water 28.) Fill the pipets completely with water 29.) Mass each water-filled pipet 30.) Squeeze all the water from inside the pipet to a 25-mL graduated cylinder 31.) Record the volume of water 32.) Repeat step 30.) and 31.) for the other two pipets 33.) Repeat step 5.) to 32.) using acetone 34.) Repeat step 5.) to 32.) using isopropyl alcohol 35.) Repeat step 1.) to 34.) for one more trial Results: The longer the thin tube part of the Beral-type pipet was left, the easier it is to get them through the plastic tubing pipet holder because the bulb part of the pipet would not be pushed against each other. The boiling stones caused the water to blur from the stone dust. After heating the pipet for a while, the top of the hot plate turn yellow. The brown color of the permanent marker turned to blue after it was heated along with the pipets in the water bath. Some of the permanent ink was washed

3 away with heated water. The ink is easy to be washed with the volatile liquid. All the volatile liquid feels cold when touched and exposed to air. Some liquid take longer to vaporize than the other. Determination of Molar Mass of Ethanol (Ethyl Alcohol) Trial 1 Temperature of boiling water bath (K) Mass of empty pipet (g) Mass of pipet and water (g) Mass of water in filled pipet (g) Volume of pipet (ml) Ethyl Alcohol Trial 1 Trial 2 Trial 3 ethyl alcohol (g) Mass of condensed ethyl alcohol (g) Molar mass of ethyl alcohol Average molar mass of ethyl alcohol Determination of Molar Mass of Ethanol (Ethyl Alcohol) Trial 2 Temperature of boiling water bath (K) Mass of empty pipet (g) Mass of water in filled pipet (g) Volume of pipet (ml) Ethyl Alcohol Trial 1 Trial 2 Trial 3 ethyl alcohol (g) Mass of condensed ethyl alcohol (g) Molar mass of ethyl alcohol Average molar mass of ethyl alcohol Determination of Molar Mass of Acetone Trial 1

4 Temperature of boiling water bath (K) Mass of empty pipet (g) Mass of pipet and water (g) Mass of water in filled pipet (g) Volume of pipet (ml) Acetone Trial 1 Trial 2 Trial acetone(g) Mass of condensed acetone(g) Molar mass of acetone Average molar mass of acetone Determination of Molar Mass of Acetone Trial 2 Temperature of boiling water bath (K) Mass of empty pipet (g) Mass of water in filled pipet (g) Volume of pipet (ml) Acetone Trial 1 Trial 2 Trial acetone(g) Mass of condensed acetone(g) Molar mass of acetone Average molar mass of acetone Determination of Molar Mass of Isopropyl Alcohol Trial 1 Temperature of boiling water bath (K) Mass of empty pipet (g) Mass of pipet and water (g)

5 Mass of water in filled pipet (g) Volume of pipet (ml) Isopropyl Alcohol Trial 1 Trial 2 Trial 3 isopropyl alcohol(g) Mass of condensed isopropyl alcohol(g) Molar mass of isopropyl alcohol Molar mass of isopropyl alcohol Determination of Molar Mass of Isopropyl Alcohol Trial 2 Temperature of boiling water bath (K) Mass of empty pipet (g) Mass of water in filled pipet (g) Volume of pipet (ml) Isopropyl Alcohol Trial 1 Trial 2 Trial 3 isopropyl alcohol(g) Mass of condensed isopropyl alcohol(g) Molar mass of isopropyl alcohol Molar mass of isopropyl alcohol Volatile Liquid Average Molar Mass of Trial 1 Molar Mass Table Average Molar Mass of Trial 2 Average Molar Mass of Both Trials Literature Molar Mass of the Volatile Liquids (Wikipedia) Percent Errors (%) Ethyl Alcohol Acetone Isopropyl Alcohol Analysis:

6 Mass of water in filled pipet = (Mass of water and pipet) (Mass of pipet) Mass of water in filled pipet equals mass of water and pipet minus the mass of pipet Trial 1 (Ethanol Pipet 1): = g Trial 1 (Ethanol Pipet 2): = g Trial 1 (Ethanol Pipet 3): = g Trial 1 (Acetone Pipet 1): = g Trial 1 (Acetone Pipet 2): = g Trial 1 (Acetone Pipet 3): = g Trial 1 (Isopropyl alcohol Pipet 1): = g Trial 1 (Isopropyl alcohol Pipet 2): = g Trial 1 (Isopropyl alcohol Pipet 3): = g Volume of the pipet = (Mass of water in filled pipet) (Density of water at room temperature) Volume of the pipet equals mass of the water in filled pipet divide by density of water at room temperature Trial 1 (Ethanol Pipet 1): = ml ethanol Trial 1 (Ethanol Pipet 2): = ml ethanol Trial 1 (Ethanol Pipet 3): = ml ethanol Trial 1 (Acetone Pipet 1): = ml acetone Trial 1 (Acetone Pipet 2): = ml acetone Trial 1 (Acetone Pipet 3): = ml acetone Trial 1 (Isopropyl alcohol Pipet 1): = ml isopropyl alcohol Trial 1 (Isopropyl alcohol Pipet 2): = ml isopropyl alcohol Trial 1 (Isopropyl alcohol Pipet 3): = ml isopropyl alcohol Mass of condensed ethyl alcohol = ( ethyl alcohol) (Mass of pipet) Mass of condensed ethanol equals mass of pipet and condensed ethanol minus mass of the pipet Trial 1 (Pipet 1): = g

7 Trial 1 (Pipet 2): = g Trial 1 (Pipet 3): = g Trial 2 (Pipet 1): = g Trial 2 (Pipet 2): = g Trial 2 (Pipet 3): = g Mass of condensed acetone = ( acetone) (Mass of pipet) Mass of condensed acetone equals mass of pipet and condensed acetone minus the mass of the pipet Trial 1 (Pipet 1): = g Trial 1 (Pipet 2): = g Trial 1 (Pipet 3): = g Trial 2 (Pipet 1): = g Trial 2 (Pipet 2): = g Trial 2 (Pipet 3): = g Mass of condensed isopropyl alcohol = ( isopropyl alcohol) (Mass of pipet) Mass of condensed isopropyl alcohol equals mass of pipet and condensed isopropyl alcohol minus mass of pipet Trial 1 (Pipet 1): = g Trial 1 (Pipet 2): = g Trial 1 (Pipet 3): = g Trial 2 (Pipet 1): = g Trial 2 (Pipet 2): = g Trial 2 (Pipet 3): = g Molar mass of ethanol = [(Mass of condensed ethanol) (Gas constant) (Temperature of the boiling water bath)] [(Barometric pressure) (Volume of the pipet)] Molar mass of ethanol equals product of mass of condensed ethanol, gas constant, and temperature of boiling water bath divide by the product of barometric pressure and volume of the pipet

8 Trail 1 (Pipet 1): ( ) [( ) ] = g/mol ethanol Trial 1 (Pipet 2): ( ) [( ) ] = g/mol ethanol Trial 1 (Pipet 3): ( ) [( ) ] = g/mol ethanol Trial 2 (Pipet 1): ( ) ( ) = g/mol ethanol Trial 2 (Pipet 2): ( ) ( ) = g/mol ethanol Trial 2 (Pipet 3): ( ) ( ) = g/mol ethanol Molar mass of acetone = [(Mass of condensed acetone) (Gas constant) (Temperature of the boiling water bath)] [(Barometric pressure) (Volume of the pipet)] Molar mass of acetone equals product of mass of condensed acetone, gas constant, and temperature of boiling water bath divide by the product of barometric pressure and volume of the pipet Trail 1 (Pipet 1): ( ) [( ) ] = g/mol acetone Trial 1 (Pipet 2): ( ) [( ) ] = g/mol acetone Trial 1 (Pipet 3): ( ) [( ) ] = g/mol acetone Trial 2 (Pipet 1): ( ) ( ) = g/mol acetone Trial 2 (Pipet 2): ( ) ( ) = g/mol acetone Trial 2 (Pipet 3): ( ) ( ) = g/mol acetone Molar mass of isopropyl alcohol = [(Mass of condensed isopropyl alcohol) (Gas constant) (Temperature of the boiling water bath)] [(Barometric pressure) (Volume of the pipet)] Molar mass of isopropyl alcohol equals product of mass of condensed isopropyl alcohol, gas constant, and temperature of boiling water bath divide by the product of barometric pressure and volume of the pipet Trial 1 (Pipet 1): ( ) [( ) ] = g/mol isopropyl alcohol Trial 1 (Pipet 2): ( ) [( ) ] = g/mol isopropyl alcohol Trial 1 (Pipet 3): ( ) [( ) ] = g/mol isopropyl alcohol Trial 2 (Pipet 1): ( ) ( ) = g/mol isopropyl alcohol Trial 2 (Pipet 2): ( ) ( ) = g/mol isopropyl alcohol

9 Trial 2 (Pipet 3): ( ) ( ) = g/mol isopropyl alcohol Average molar mass of ethyl alcohol = Σ(Molar mass of ethyl alcohol) 3 Average molar mass of ethyl alcohol equals the sum of molar mass of ethyl alcohol divide by 3 Trial 1: ( ) 3 = g/mol ethanol Trial 2: ( ) 3 = g/mol ethanol Average molar mass of acetone = Σ(Molar mass of acetone) 3 Average molar mass of acetone equals the sum of molar mass of acetone divide by 3 Trial 1: ( ) 3 = g/mol acetone Trial 2: ( ) 3 = g/mol acetone Average molar mass of isopropyl alcohol = Σ(Molar mass of isopropyl alcohol) 3 Average molar mass of isopropyl alcohol equals the sum of molar mass of isopropyl alcohol divide by 3 Trial 1: ( ) 3 = g/mol isopropyl alcohol Trial 2: ( ) 3 = g/mol isopropyl alcohol Average molar mass of both trial = [(Average molar mass of trial 1) + (Average molar mass of trial 2)] 2 Average molar mass of both trial equals the sum of average molar mass of trial 1 and trial 2 then divide by 2 Ethanol: ( ) 2 = g/mol Acetone: ( ) 2 = g/mol Isopropyl Alcohol: ( ) 2 = g/mol Percent Error = (Average molar mass of both trial) (Literature molar mass of the volatile liquids) 1 Percent error equals the absolute value of the average molar mass of both trial divide by the literature molar mass of the volatile liquids minus 1 Ethanol: ( ) 1 = % Acetone: ( ) 1 = % Isopropyl Alcohol: ( ) 1 = % The hypothesis was not confirmed true. This is because the molar mass of each volatile liquid calculated from the experiment and the literature molar mass is significantly different from each other. Volatile

10 liquids with lower boiling point often give better results than those with higher boiling point. This is because the volatile liquid would get more energy than it needed to vaporize. Thus, it is quite certain that all liquid was vaporized. So when the vapor condenses, the mass inside the pipet would be accurate since there s no liquid left in the pipet that would be added to the mass of the condensed vapor. If vapor condenses in the neck of the 15-mL Beral-type pipet, the mass of the condensed vapor would be greater. Thus, the mass that should have been of the vapor that was in the ratio of its molar volume would be increase because there was some mass that did not evaporate. This could cause a considerably big error because the mass of the gas for a certain volume is much lower than the mass of a liquid for a certain volume because the density of a liquid is greater than gas. Thus, the mass of little amount of liquid could affect the mass of volume significantly. If the liquids have enough attraction to form dimers, the molar mass would increase. This is because the molecules combined from two to one so the volume to that was previously taken by each molecule will be decrease in half. Thus, the molar mass would be doubled since each molecule now has twice the mass of before. Conclusion: The hypothesis cannot be verified from the results acquired from the experiment. The hypothesis did not work possibly because not all liquid was vaporized because the water bath absorbed a lot of heat so the heat might have been too low to vaporize all the liquid. If not all the liquid was vaporized, the molar mass would be greater than it should have been. Also, some errors could have happened during the experiment. The pipets might have been moved to the room temperature water bath too slowly because there were some hindrance such as the Buret clamp above the universal clamp which was clamping on to the pipets. This might have caused the air that was pushed out to get back inside and increase the mass of the vapor. If the mass increased, the molar mass that would also increased because the molar mass uses the mass in the calculation. Another error that could have happened was that the vapor might have condensed at the neck of the pipet. The neck of the pipet was cut so that it was long in order to put it through the plastic tube pipet holder. However, the longer the neck also meant that it was further away from the heat of the boiling water bath. Thus, there is a higher chance that the vapor condensed at the neck of the pipet. If the vapor condensed at the neck of the pipet, the mass would be greater because instead of getting the mass of the volatile vapor, the mass found would have been of the volatile vapor and the volatile liquid. Since the total volume was the same whether it is purely vapor or vapor plus liquid, the mass of the vapor plus liquid would be greater because the liquid is denser than gas. The increase in mass would also results in the increase in molar mass. Some ways to prevent the errors in the future experiment would be to make sure there would be nothing to obstruct the transfer of pipet from the boiling water bath to the room temperature water bath. For example, the Buret clamp holding the thermometer could be put on other ring stand so that the universal clamp holding the pipets could slide up and down the stand with no hindrance. Another way to prevent the future errors would be to cut the neck of the pipet just the right amount so that it would not be too long that it was too far from the water to allow the volatile vapor to condense. The neck should be measured and cut so that all three pipets could easily slip through the plastic tube and ended right above the tube

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