Mass of pyrex tube + KClO before heating (g) Mass of pyrex tube + residue after
ID: 509299 • Letter: M
Question
Mass of pyrex tube + KClO before heating (g)
Mass of pyrex tube + residue after heating (g)
Initial buret reading (mL)
Final buret reading (mL)
Barometric pressure (torr)
Temperature ()
6.3494 g
6.3018 g
0.4 mL
35.1 mL
762.5 torr
23.6
NOTE:
Equation 2:
R = PV / nT --> units in atm-L / mol-K --> L-atm / mol-K
Equation 4:
P (oxygen pressure) = P (atmospheric pressure) - P (water vapor)
The vapor pressure of water at different temperatures is given in Table 1.
Table 1: Vapor pressure of water at different temperatures
22.4
*Explanations would be extremely appreciated and helpful as lab professor does not teach. Thanks.
Mass of pyrex tube + KClO before heating (g)
Mass of pyrex tube + residue after heating (g)
Initial buret reading (mL)
Final buret reading (mL)
Barometric pressure (torr)
Temperature ()
6.3494 g
6.3018 g
0.4 mL
35.1 mL
762.5 torr
23.6
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 o2. 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. Note: you will need to use Equation 4 and convert to units of atm. 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. Determination of the Ideal Gas ConstantExplanation / Answer
2 KClO3 ------------> 2 KCl (s) + 3 O2
a) Mass of oxygen = (6.3494 - 6.3018 )g = 0.0476 g [difference in weight will give the mass of oxygen gas because after decomposition of potassium cholrate, it will form KCl (solid) and O2 (gas ). O2 gas will escape so differnce in weight will give the mass of oxygen]
b) Molar mass of O2 = 16 x 2 = 32 g/mol
Number of moles = given mass/molar mass = 0.0476/32 = 0.0015 mol
c) Volume of the gas = 35.1 - 0.4 mL = 34.7 mL = 34.7 x 10-3 L [ 1mL = 10-3 L]
d) Poxygen = Patm - P water vapor = 762.5 - 22.4 = 740.1 torr = 740.1 / 760 atm = 0.974 atm [1 atm = 760 torr]
e) T = 23.6 C = (23.6 + 273.15 ) K = 296.8 K [ 0C = 273.15 K]
f) R = PV/nT = (0.974 x 34.7 x 10-3)/(0.0015 x 296.8 ) = (33.8 x 10-3)/ 0.4452 = 0.0759 L atm K-1mol-1
Theoretical value of R = 0.0821 L atm K-1mol-1
error = (0.0821 - 0.0759) = 0.0062
% error = 0.0062 *100% / 0.0821 = 7.6%