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  I.   Abstract This experiment focuses on the determination of molar mass of a volatile liquid by vapour density method. This particularly revolves around the estimation of the molar mass of volatile liquid from their vapour densities at a temperature above their boiling points using Dumas Method, which is a classical technique still used in the laboratories despite the new technological instrumentations. With the values of pressure, temperature, and density on hand, the data regarding the relative masses of the flask and water were utilized to get the unknown molar masses. After considering some sources of errors which are naturally occurring in the execution of the Dumas method for molar mass determination, the group was able to get the value of 90.33 g/mol for the average molar mass. Thus, with the data gathered and the results interpreted, it is therefore concluded by the end of the experiment that the closest match to it was that of Ethyl Acetate, also known as ethyl ethanoate, commonly abbreviated EtOAc or EA. II.   Introduction The chemical and physical methods for determining molecular and atomic formula weights or molar masses as a way of categorizing and analysing new materials have been observed through time. Many of these methods have been out of date because the modern laboratory is generally equipped with instrumentation. However, the principles used in the older methods are still considered significant and many form the basis for the prediction of chemical and physical behaviours and  properties of substances. The Dumas method is a classic technique for determining the formula weight of a volatile liquid. The concept of equal volumes of gases measured under identical conditions would contain equal numbers of gas particles was proposed by Avogadro. It was possible to describe a constant volume which would contain a gram-atomic weight of a gaseous element or compound under fixed conditions with an established relative atomic mass scale, known today as the molar volume. An application of this useful information to liquids and solids which are appreciably volatile is allowed by understanding the concept of the different gas laws.. As long as the temperature and  pressure are known, a measured volume of gas can be converted to moles since: n = PV/RT. Since massing a sample of gas is relatively simple, these two pieces of information are the minimum required for a molar mass determination. A volatile liquid is heated to a known temperature, usually above its boiling point, and allowed to escape from a container through a tiny orifice in the Dumas method. The container is cooled to room temperature once the liquid has vaporized. The vapour which gradually remained in the container at the higher temperature condenses to a liquid and is then massed. The room pressure can be used  because the system is open to the atmosphere through the orifice to calculate moles if the volume of the container is known along with the high temperature and from there a molar mass can be determined. This method depends on a lot of things to consider. One assumption is that the liquid is not so volatile that a significant amount will be lost to evaporation through the orifice as the container cools while it is volatile enough to vaporize at the elevated temperature. At the temperature and pressure at which it occupies the container, the vapour is also assumed to behave ideally. The amount of error implicit in this approximation varies from compound to compound and is tied to the variables which create deviations from ideal behavior: molecular volume and intermolecular forces. Generally speaking, the larger these are, the greater the error in the determination, which means both share a direct relationship. The situation is further complicated by the interaction of these two factors. There is another problem with the basic method which is more easily addressed. There is a fixed amount of air determined by the volume, temperature and pressure in the container when the container  is empty , or the instant before the liquid is added or the vapour forms. This air has mass and contributes to the overall mass of the container assembly, but when the liquid is present it vaporizes to some extent determined by its vapor pressure at room temperature. If the temperature and pressure remain constant, the presence of this vapour forces out some of the air through the orifice since the internal pressure is equalized with the room pressure and the number of gas particles remains constant. Thus there is air missing that should be counted in the final mass if the mass of the condensed liquid is to be determined by difference when the container is massed again. This condition results in an apparent mass of liquid that is too small which in turn makes the molar mass calculation too small. The moles of air forced out by the vapour are equal to the moles of vapour that forms. This is determined by the vapour pressure of the liquid at room temperature. If this value is known, the moles of vapour that are present in the flask above the residual condensed liquid can be calculated as: n vapour   = (P vapour x V container  )/RT Avogadro's Law tells us that these moles of vapour must equal the moles of air displaced since the conditions are the same for both. These missing moles of air can be converted to grams using the weighted average molar mass of air and added into the final mass of the assembly. All of this depends, of course, on knowing the vapour pressure of the liquid, P vapour  . III.   Methodology List of Chemicals and apparatus    2-propanol    Ethanol    Ethyl acetate    Analytical balance    600-ml beaker    125-ml Erlenmeyer flasks    Bunsen burner    Wire gauze    Thermometer/barometer Here is a photo of the Dumas set-up that we used upon  performing the experiment. An Erlenmeyer flask was immersed in a 600-ml beaker. A thermometer was used to measure the temperature as the unknown sample was vaporized. Several reminders had to be taken down throughout the experiment. A foil cap must be used. The hole in it must only be tiny, enlarging the hole could give an error into the determination. The water should be high enough not to enter through the hole. 1.   A clean and dry Erlenmeyer flask with a foil cap was weighed in an analytical balance; this was weighed as the empty flask.  2.   4 ml of an unknown sample was injected through the foil making a tiny hole. 3.   This flask was then immersed in a 600-ml beaker of water which was then quickly boiled. 4.   A gentle boil had to be achieved until all the liquid was evaporated. 5.   Boiling point of water at room pressure and the pressure itself was recorded. 6.   After allowing the flask to dry and to cool down, the mass was also then recorded. 7.   The same flask was used for water. It was filled with water almost completely with room temperature. This temperature was recorded. 8.   This flask filled with water was then weighed in an analytical balance and mass was then recorded. 9.   The volume of the flask was determined using this measurement, the first measurement and the density of the water at room temperature. IV.   Results and Discussion Barometric pressure, P: 752.5 mmHg or 0.990 atm Room Temperature, T R : 33 o C or 306.15K Density of water at T R   : 995.948 Kg/m Sample Name Ethyl Acetate Table 1.1: Determination of Estimated Molar Mass of the Unknown Sample Trial 1 Trial 2 Trial 3 Mass of “empty” flask, g  94.687 89.242 89.245 Mass of “empty” flask with Al foil and Cu wire, g  96.131 90.388 90.527 Mass of empty flask with Al foil, Cu wire, and sample 96.618 90.809 90.968 Mass of vapor (m), g 0.487 0.421 0.441 Temperature of vapor (T B ), K    368.15 366.16 368.15 Mass of flask filled with water, g 255.607 249.557 250.086 Mass of water, g 160.92 160.315 160.841 Volume of Flask based on water (V), L 0.162 0.161 0.161 Moles of vapor at T,V and P,n 0.0053062 0.0053022 0.0052734 Estimated molar mass of vapor, g/mol 91.7794 79.4010 83.6273  Table 1.2: Determination of the True Mass of Vapor that Occupied the Flask at Boiling Temperature Trial 1 Trial 2 Trial 3 Moles of air displaced by the vapor at T R   0.00092778 0.00110878 0.00092205 Molar mass of air, g/mol 28.97 28.97 28.97 Mass air displaced by the vaporized liquid at T R , g 0.02688 0.03212 0.02671 True mass of vapor that occupied the flask at the boiling temperature of water, g 0.5139 0.4531 0.4677 Table 1.3: Determination of Corrected Molar Mass of the Unknown Sample Trial 1 Trial 2 Trial 3 Corrected Molar mass of the vapor,g/mol 96.84 85.46 88.69 Corrected Molar mass of the vapor (Average),g/mol 90.33 Molar mass (literature Value), g/mol 88.11 % difference 9.45 3.05 0.66 average % difference 2.49 In this experiment, the group utilized the Dumas method in obtaining the identity of an unknown sample using the calculated molar mass. The first part involved injecting about 3-4 mL of sample leaving a tiny hole on top. This was then heated and brought to boil until all the liquid evaporated. The air was swept out of the container. What was important at that point is to be be able to judge the point at which the flask is just filled with sample vapor. This vapor was then cooled to room temperature, 306K and was weighed. This consitutes the mass of the vapor. Using the atmospheric pressure indicated in the barometer, 0.990 atm; the temperature at which the sample  boiled, 366.15K, and the volume of the flask, 0.161L, moles of the sample vapor,n, was calculated using the gas equation: n vapor   = PV/RT  boiling With the mass of the vapor and the number of its moles, estimated molecular weight had been computed, however this was way far from the literature value,and lower than expected. To correct this, the group had to account for the moles of the sample that remained as vapor at room temperature during the weighing. This is equal to the moles of air the sample vapor displaced. To determine the number of moles of this, a corrected value of P was used  –   131.531895 mmHg or 0.1731 atm. The temperature used was room temperature which was 306.15K, and the same equation was used: n displaced  = P corrected V/RT room. Using the molar mass of air, 28.97 g/mol, the mass of air displaced was obtained. This was added to the mass of the sample vapor obtained previously. The combined mass was then divided to the moles of sample vapor to obtain the correct molar mass.

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