Gas Laws Boyle\'s Law, Determining the Ideal Gas Constant R and the Molar Mass o
ID: 554942 • Letter: G
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Gas Laws Boyle's Law, Determining the Ideal Gas Constant R and the Molar Mass of an Unknown Post Lab (10 Pts) The Dumas method is one of the simplest procedures for determining the molar mass of an unknown volatile liquid. In the Dumas method, a volatile liquid sample is heated in a flask with a tiny opening until the entire sample vaporizes. Because the volume occupied by the vapor at atmospheric pressure is much larger than the volume occupied by the liquid, some of the vapor will escape from the flask. However, the vapor remaining in the flask will contain the number of moles of the substance that fills the volume of the flask at the experimental pressure and vapor 1. Why is the barometric (i.e., atmospheric) pressure considered to be the pressure of the vapor, i.e, how does the experimental procedure ensure this? Page reference Explain carefully (2 Pt) 2. Why isn't it necessary to weigh the amount of liquid initially put into the flask? Explain. (2 Pt) 3. A certain volatile hydrocarbon (a binary compound of carbon and hydrogen) is found to be 92.3 % carbon, by mass. In a separate experiment, utilizing the Dumas method, a 4.00 mL pure liquid sample of this hydrocarbon is vaporized in a 125 mL Florence flask when the barometric pressure is 768.0 torr. After the excess gas escapes, the temperature is measured as 98.0. The flask and contents are subsequently cooled to 25°C (density water at 25°C = 0.997044 g/mL) and the vapor condenses to a liquid. The flask is then emptied, cleaned, and filled with water. When weighed on a balance, the difference in weight between the flask filled to the brim with water and the dry empty flask at 2 5"C is 128.12 g The empty flask fitted with a foil cap pierced with a pinhole- weighs 25.3478 g The weight of the flask and contents is found to be 25.6803 g Please determine the following, show all work. (6 Pts) The empirical formula of this hydrocarbon. ·The volume that the vapor occupied in the Erlenmeyer flask (in L). The molecular formula of this hydrocarbon.Explanation / Answer
3. According to the ideal gas law PV= nRT
n= mass/ molecular weight = g / MW = PV/RT
MW= gRT/PV
The first equation above may be easily used to find the molecular weight of the hydrocarbon. Note that the ideal gas constant R = 0.0821 L atm. mole K . Also note that 760 torr = 1 atm.
The first part of this problem,is obtaining the empirical formula. A hydrocarbon contains only carbon and hydrogen. Thus, the mass % of H = 100 % - 92.3 % = 7.70 % H. In 100 gm of hydrocarbon there are 92.3 g of C and 7.70 g of H present (relatively).
Determining moles of each element: moles C = 92.3 g C * ! mole C / 12.01 g C = 7.69 moles C
moles H = 7.70/1.01 = 7.62 moles H.
THe empirical fromula = moles C / moles H = 7.69 moles / 7.62 moles = 1.01 / 1 = 1 :1 so, the Empirical Formula = CH.
the density of liquid water at 25ºC is listed as: 0.997044 g/mL.
V = Vwater filling flask = 128.12 g / 0.997044 g/mL = 128.50 mL = 0.12850 L
To determine the molecular formula, we need to find the molecular weight. The most useful form of the ideal gas law - considering the given information - is:
MW = gRT/ PV ; T = 98 + 273K = 371 K and V = 0.1285 L.
mass of condensed vapor = mass of gas in flask = 25.6803 g - 25.3478 g = 0.3325 g.
MW = gRT/ PV = (0.3325 g) * 0.0821 L atm. mole K * (371 K) / 768.0/ 760 (0.12850 L) = 78.0 g/mole.
The Empirical Formula Weight (EFW) for CH = 12.01 + 1.01 = 13.02 g/mole.
N = no. of empirical formula units = MW/ EFW = 78.0 g/mole / 13.02 g/mole = 5.99 = 6.
Thus, in this case, since N = 6, the molecular formula is 6 x Empirical Formula. Thus, the Molecular Formula = C6H6 . (The compound is benzene).
1. As the compound vaporizes the pressure is higher in the flask than the barometric pressure, and the excess air effuses out. When the pressure of the vapor is equal to the barometric pressure when the external pressure is equal to the atmospheric pressure the volume of the flask is considered to be the barometric pressure.
2. The liquid in excess will be vapourised in the flask and when the , the gas pressure will equal the barometric pressure, and the volume of the gas will equal the volume of the flask. Thus the vapourisation ensured that the weght measured is not erronious.
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