1. CHM 201 General Chemistry
Answer:
Determination of the Ideal Gas Constant "R" using Carbon Dioxide. (Report 3 digits for this
exercise. No points will be deducted for significant digit errors.) (20 points)
Mass of beaker (optional)
Mass of plastic bag (optional) :
Temperature (in the bag) : 10 oC = 283.15 K
Volume of plastic bag : 4.000 L = 0.004 m3
Barometric pressure : 753.6 mm/Hg = 100471.75 Pa
Using the ideal gas equation : PV = nRT
Here, is the number of moles of CO2 .
Calculated value of R for each trial is shown below-
Trial
5. 22.16%
13.70%
Determining The Molecular Weight Of A Gas. (Significant Figures Matter Here !) (30 Pts)
Room temperature – TR : 28 oC or 301.15 K
Water temperature - Tw : 24 oC or 297.15 K
Barometric Pressure – Patm : 753.6 mm/Hg = 100471.75 Pa
Vapor Pressure of Water – Pwater : 2986.421 Pa
Pressure of Butane – Pbutane (Patm- P-water) : 95485.329 Pa
8. 35.10 mL
40.50 mL
45.00 mL
40.2 mL
Molar Mass
58.138 g/mol
Pressure/Volume Measurements. (Report 3 digits for this exercise. No points will be
deducted for significant figure errors.) (20 points)
V
12. P
414
788
1444
PV
8280
7880
7220
Please Incorporate Your Group’s Data Sheet Here.
How do you explain the large percent error in the gas constant determination? There are
both experimental and scientific factors involved. Include both.
While opening and closing the bag, a lot of dry ice leaves into the surrounding thereby
changing the value of the initial mass we took. Also, the temperature of the bag changes
since there is no proper insulation. Lastly, we assumed the gas to be ideal but in practice, no
gas in the universe is ideal. All these factors lead to large percentage error in the calculation
13. of Gas Constant.
If some water remained on the lighter at the time of the second weighing (after collecting
the gas), how would this affect your determination of molecular weight? Be specific. Show a
sample calculation to clarify your answer.
Mater remaining on the lighter will increase the measured final mass of butane. Let us
assume the water droplets weigh 2 gm, then,
Initial mass = 12.6293 g
Final Mass = 14.7084 g
Mass of butane gas = 2.0791 g
Volume of gas = 35.10 mL
Using ideal gas equation : PV = nRT
(100471.75)(0.0000351) = ( )(8.314)(297.15)
M = 1456.5 g/mol
As can be observed above, just because of the presence of water droplets on the bag surface,
the molecular mass of butane which is expected to be 52 g/mol is coming out as 1456.5
g/mol. Hence, before measurement of the mass, water droplets must be cleaned and dried.
Which Of The Individual Gas Laws Is Confirmed By The Results Of Part 3?
Since, PV = constant in the measurements, therefore, Boyle’s Law is confirmed by the results
of part-3.
