To measure the pressure of the oxygen gas collected in the gas manometer, we can
ID: 509819 • Letter: T
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To measure the pressure of the oxygen gas collected in the gas manometer, we can use the atmospheric pressure in the lab. Because the water level eventually stabilizes after the reaction has finished and stopped producing oxygen, we know that the pressure of gas pushing down on the water in the right side of the manometer (where the oxygen is trapped) is equal to the pressure of gas pushing down on the water in left side of the manometer (where the air pressure in the lab is pushing on the water) P_left side(due to air) = P_right side (due to oxygen) Standard atmospheric pressure is 1 atm, which is the same as 760 torr (1 atm = 760 torr). P_ left side (due to air) = P_atmospheric pressure However, the pressure on the right side is actually a combination of the oxygen pressure plus the pressure from water vapor from the water in the gas manometer. In other words, the right side of the manometer traps both oxygen gas and water vapor, so the gas pressure in the right side is actually: P_right side(due to oxygen AND water vapor) = P_ oxygen pressure + P_water vapor To find the pressure of the oxygen gas, we need to subtract the pressure due to the water vapor: P_oxygen pressure + P_water vapor = P_right side = P_left side = P_atmospheric pressure P_oxygen pressure = P_ atmospheric pressure - P_water vapor The vapor pressure of water at different temperatures is given in Table 1. Complete the following, and enter these preliminary results in the Table 3. a. Calculate the mass (grams) of oxygen gas produced. b. Calculate the number of moles of oxygen gas by using the molar mass of O_. c. Calculate the volume of oxygen gas, and convert to units of liters (L). d. Calculate the pressure of oxygen gas by using the barometric pressure and pressure from the water vapor. e. Convert the room temperature to degrees Kelvin (K). f. Calculate the Idea gas constant R (in units of L-atm/mol -K) from your experimental results by using Equation 2. Complete the following, and enter these final results in the Table 4. a. Enter your experimental value for the ideal gas constant. b. Look up the theoretical value for the ideal gas constant, in units of L-atm/mol-k. c. Calculate the percent error between the experimental and theoretical values.Explanation / Answer
The balanced chemical equation is
2 KClO3 ------> 2 KCl + 3 O2
As per the balanced chemical equation,
2 moles KClO3 = 2 moleS KCl = 3 mole O2
1a) Mass of oxygen gas collected = (mass of pyres tube + KClO3 before heating) – (mass of pyrex tube + residue after heating) = (6.50 – 6.46) g = 0.04 g.
b) Molar mass of O2 = (2*15.9994) g/mol = 31.9988 g/mol.
Number of moles of oxygen gas corresponding to 0.04 g oxygen = (0.04 g)/(31.9988 g/mol) = 0.00125 mole.
c) Volume of oxygen gas = (final buret reading) – (initial buret reading) = (35 mL ) – (2 mL) = 33 mL = (33 mL)*(1 L/1000 mL) = 0.033 L.
d) We need to know the vapor pressure of water at the temperature of the experiment, which is 24.2C. Tables of vapor pressure of water at temperatures from 0-100C are available on the internet; I used the value from an internet source.
The vapor pressure of water at 24C is 22.4 torr; we shall use this value.
Barometric pressure is given as 0.97 atm = (0.97 atm)*(760 torr/1 atm) = 737.2 torr.
Vapor pressure of water = 22.4 torr.
Therefore, pressure of oxygen gas = (barometric pressure) – (vapor pressure of water) = (737.2 torr) – (22.4 torr) = 714.8 torr = (714.8 torr)*(1 atm/760 torr) = 0.94 atm.
e) Room temperature in Kelvin = (24.2 + 273) K = 297.2 K.
f) Use the ideal gas law: P*V = n*R*T where the symbols have their usual meanings. R is the gas constant.
(0.94 atm)*(0.033 L) = (0.00125 mol)*R*(297.2 K)
===> R = (0.94 atm)*(0.033 L)/(0.00125 mol)(297.2 K) = 0.083 L-atm/mol.K (ans).
2a) The experimental value of the ideal gas constant obtained is 0.083 L-atm/mol.K.
b) The theoretical (accepted) value is 0.082 L-atm/mol.K
c) Percent error = (theoretical value) – (experimental value)/(theoretical value)*100 = (0.082 L-atm/mol.K) – (0.083 L-atm/mol.K)/(0.082 L-atm/mol.K)*100 = 1.219% 1.22% (ans).
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