Molecular Weight Of Oxygen

[Music] you have previously learned that the atomic weight of oxygen is 16 and that the molecule of oxygen gas is diatomic and therefore has a weight of 32 you've also learned that 32 G of oxygen the gram molecular weight or one mole occupies a volume of 22 and 4/10 L at standard conditions the purpose of the experiment today is to verify this statement that one mole of oxygen gas at standard conditions occupies a volume of 22 and 410 L now in order to verify this statement we need to develop a method for measuring the weight and volume and pressure and temperature of a sample of oxygen and if these four quantities are known using the gas laws we can calculate the weight of a sample of oxygen at any other temperature and pressure that we choose in this test tube we've placed a mixture of potassium chlorate and manganes dioxide the grayish color of the material is due to the manganese dioxide we'll determine the weight of this tube in its contents and then heat it with the bunson bur oxygen will be produced and will be driven through this tube which opens here into the flask as oxygen enters the flask the water now in the flask will be displaced over through this tube into the beaker by measuring the volume of the water eventually collected in the beaker we can determine the volume of oxygen produced by weighing the tube before and after heating we can determine the weight of the oxygen produced we can measure the pressure using a barometer and we can measure the temperature of the oxygen in this vessel by measuring the temperature of the water which is driven from The Vessel with these four quantities we can proceed to calculate the weight of 22 and 4/10 L of oxygen our first job then is accurately to determine the weight of this tube and its content note that we don't have to determine the weight of the empty2 because we really don't care how much potassium chlorate and manganese dioxide we've added we simply need to determine the weight of the tube and its contents before Heating and the weight of the tube and the residue after heating the difference in weight will be the weight of oxygen driven off we've placed the test tube and its contents on the left left hand pan of the balance and we've balanced the weight of the tube by adding weight to the right hand pan by placing a rer on the beam and by using the chain on the right hand pan of the balance we placed four rather large weights the large weight in bag is a 30 G weight then starting from the left in front the weight most on the left is a 10 G weight the middle weight in front is a 5 G weight and the small weight on the right in front is a 3 G weight the combined total of these four weights is 48 G on the beam at the top of the balance we've placed the rer at the .5 g mark this means that we've added 1/2 G uh to the weights by means of the rider the final adjustment is made by by adding uh a small portion of a chain to the weights already placed in the balance and the amount of chain added is read from this scale these figures represent thousands of a gram so that we see here we've added 56 and a half or 565,000 of a gram with the chain and by combining all of these weights the large weights in the pan the Rider and the chain we determine the weight of the tube and its contents before heating to be 4855 65 G first uh I blew gently in this tube so that some water was forced out of the flask through this tube which is now completely filled with water and I attached the test tube firmly to this stopper our next job is to equate the pressures inside and outside the flask this we do by raising the small Beaker until the level of the water in the small Beaker is the same as the level of the water in the flat then opening the pinch clamp we permit water to flow one way or the other until the pressures are equal then we replace the pinch clamp remove the small beaker replace it with the large Beaker and open the pinch clamp few drops of water run from the tube and then stop this indicates that we have no leaks in our system we will now light the bunson burner and heat the potassium chloride in the test tube we're now heating the chlorate mixture in the test tube oxygen is being formed here passing through this to water is being forced out of the flask and is dripping slowly into the beaker we will continue this operation until approximately 3/4 of the water in the flask has been driven into the beaker we've now heated the test tube until nearly all of the water has been forced from the flask and is now in the beaker at this point we must uh discontinue Heating and permit the test tube and its contents to cool to room temperature so that any expansion of the air in the tube which occurred at the beginning of the experiment will be counteracted by corresponding cooling of this air after we discontinue the heating being careful to keep the tip of this tube under the surface of the water we've now lowered the beaker until the level of the water in the beaker and in the flask are equal in the meantime the test tube is cooled room temperature we now know that the pressure inside the beaker is atmospheric pressure since the level of the water inside and outside is the same we'll now replace the pinch clamp and proceed to measure the volume of the water in the beaker first we will uh measure the temperature of the water in the beaker we find it to be 29° cenr then we'll carefully pour the water in the beaker into this large graduated cyinder we find the volume of the water to be not 980 ml after the tube has cooled to room temperature we replaced it in the balance and again determined its weight the tube and the residue after heating now weigh 47. 3848 G the data that we've accumulated during this experiment is now on the board the weight of the test tuol before heating was 48.5 565 gram and after heating 47. 3848 Gams the difference is 1. 1717 G and equals the weight of the oxygen produced this oxygen displaced a volume of water equal to 980 ml temperature of this water was 29° Centigrade or 302° absolute and this must also have been the temperature of the oxygen in the FL atmospheric pressure today is 729 mimet the vapor pressure of water at 29° is 30 mm and since the combined pressure of the oxygen and the water vapor in the flask equal atmospheric pressure or 729 and we know that the water vapor pressure was 30 the oxygen vapor pressure must be the difference parti pressure of oxygen in the flask then was 699 mm and we will use many of these figures in carrying out a calculation of the weight 22 and 4/10 lers of oxygen at standard condition first we must calculate the volume that this oxygen would have occupied at standard condition we had 980 mlit of oxygen or 980 L we found that the pressure exerted by this oxygen with 699 mm of mercury we need to correct this to 760 and we found that the temperature of this oxygen was 302° absolute and we need to correct this to 273 carrying out this computation we find that the corrected or STP volume of the oxygen is 815 L now we know that this oxygen weighed 1717 G and occupied a volume of 0815 L and we want to know how much oxygen what weight of oxygen would occupy a volume of 22 and 410 L and solving this expression for X we find that x equal 32.2 G now this Compares with our accepted value of 32 G of oxygen quite well our error then is some 2/10 of a gram percentage wise our error is2 * 100 over 32 or 0.6% 6/10 of 1% it's possible then using this method to calculate rather accurately a molecular weight of oxygen