Gases
AP Review 2011-12
2002 Question 3
3. (d) Under identical conditions, a sample of an unknown gas effuses into a vacuum at twice the rate that a
sample of pentane gas effuses. Calculate the molar mass of the unknown gas.
2005 Form B Question 6
6. Consider two containers of volume 1.0 L at 298 K, as shown above. One container holds 0.10 mol N2(g) and
the other holds 0.10 mol H2(g). The average kinetic energy of the N2(g) molecules is 6.2  10-21 J. Assume that
the N2(g) and the H2(g) exhibit ideal behavior.
(a) Is the pressure in the container holding the H2(g) less than, greater than, or equal to the pressure in the
container holding the N2(g)? Justify your answer.
(b) What is the average kinetic energy of the H2(g) molecules?
(c) The molecules of which gas, N2 or H2, have the greater average speed? Justify your answer.
(d) What change could be made that would decrease the average kinetic energy of the N2(g) molecules in the
container?
(e) If the volume of the container holding the H2(g) was decreased to 0.50 L at 298 K, what would be the
change in each of the following variables? In each case, justify your answer.
(i) The pressure within the container
(ii) The average speed of the H2(g) molecules
2005
2. Answer the following questions about a pure compound that contains only carbon, hydrogen, and oxygen.
(a) A 0.7549 g sample of the compound burns in O2(g) to produce 1.9061 g of CO2(g) and 0.3370 g of H2O(g).
(i) Calculate the individual masses of C, H, and O in the 0.7549 g sample.
(ii) Determine the empirical formula for the compound.
(b) A 0.5246 g sample of the compound was dissolved in 10.0012 g of lauric acid, and it was determined that
the freezing point of the lauric acid was lowered by 1.68˚C. The value of Kf of lauric acid is 3.90˚C m–1.
Assume that the compound does not dissociate in lauric acid.
(i) Calculate the molality of the compound dissolved in the lauric acid.
(ii) Calculate the molar mass of the compound from the information provided.
(c) Without doing any calculations, explain how to determine the molecular formula of the compound based
on the answers to parts (a)(ii) and (b)(ii).
(d) Further tests indicate that a 0.10 M aqueous solution of the compound has a pH of 2.6. Identify the organic
functional group that accounts for this pH.
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2002 Question 3 Answer
rateunknown
1
= 2 = 72.15 g mol
rateC5H12
MM unknown
4
1
= 72.15 g mol  MMunknown = 18.04 g mol-1
MM unknown
2005 Form B Question 6 Answer
(a) The pressure in the container holding the H2(g) is equal to the pressure in the container holding the N2(g)
because there is an equal number of moles of both gases at the same temperature and volume (P = nK,
where the constant K = RT/V).
(b) The average kinetic energy of the H2(g) molecules is 6.2  10-21 J because both gases are at the same
temperature.
(c) H2(g) molecules will have the greater average speed. Both gases have the same average kinetic energy, but
H2(g) has the smaller molar mass. Therefore, the H2(g) molecules will have a greater average speed
because, at a given temperature, the average (root-mean-square) speed of gas molecules is inversely
proportional to the square root of the molar mass of the gas: urms =  3RT   1 
 M
2005 Question 2 Answer
(a)(i) massC = 1.9061 g CO2   1 mol CO 2    1 molC    12.01g C  = 0.5202 g C

 

 
 44.01g CO 2   molCO 2   molC 
massH = 0.3370gH 2 O   1 molH 2 O    2 molH    1.008g H  = 0.03771 g H
 18.016g H O   molH O   molH 

2
2

 
 
massO = 0.7549g – 0.5202 g – 0.03771 g = 0.1970 g O
(ii) nC = 0.5202 g C   1 mol C  = 0.04331 mol C
 12.01gC 


nH = 0.03771 g H   1 mol H  = 0.03741 g H
 1.008 g H 


nO = 0.1970 g O   1 mol O  = 0.02131 mol O
 16.00 g O 


 0.04331molC   0.03741mol H   0.01231molO 

:
:

 0.01231   0.01231   0.01231 
3.518 l C : 3.039 mol H : 1.000 mol O; the empirical formula is C7H6O2
(b)(i) Tf = i  Kf  molality (i = 1 since compound does not dissociate)
o
molality = T f = 1.68 C = 0.431 molal
Kf
3.90 o Cm 1
(ii) ncpd = 10.0012 g lauric acid   1 kg    0.431 mol cpd  = 0.00431 mol
 1,000 g   1 kg lauric acid 

 

-1
molar mass = 0.5246 g cpd/0.00431 mol = 122 g mol
(c) The molar mass should be divided by (or compared to) the empirical mass to obtain a whole number. Each
subscript in the empirical formula is multiplied by this whole number.
(d) Since an aqueous solution of the compound is acidic, the compound must be an organic acid. The
functional group in an organic acid is the carboxyl group –COOH.
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2009
2. A student was assigned the task of determining the molar mass of an unknown gas. The student measured the
mass of a sealed 843 mL rigid flask that contained dry air. The student then flushed the flask with the unknown
gas, resealed it, and measured the mass again. Both the air and the unknown gas were at 23.0°C and 750. torr.
The data for the experiment are shown in the table below.
Volume of sealed flask 843 mL
Mass of sealed flask and dry air 157.70 g
Mass of sealed flask and unknown gas 158.08 g
(a) Calculate the mass, in grams, of the dry air that was in the sealed flask. (The density of dry air is 1.18 g L-1
at 23.0°C and 750 torr.)
(b) Calculate the mass, in grams, of the sealed flask itself (i.e., if it had no air in it).
(c) Calculate the mass, in grams, of the unknown gas that was added to the sealed flask.
(d) Using the information above, calculate the value of the molar mass of the unknown gas.
After the experiment was completed, the instructor informed the student that the unknown gas was carbon
dioxide (44.0 g mol-1).
(e) Calculate the percent error in the value of the molar mass calculated in part (d).
(f) For each of the following two possible occurrences, indicate whether it by itself could have been responsible
for the error in the student’s experimental result. You need not include any calculations with your answer.
For each of the possible occurrences, justify your answer.
Occurrence 1: The flask was incompletely flushed with CO2(g), resulting in some dry air remaining in the
flask.
Occurrence 2: The temperature of the air was 23.0°C, but the temperature of the CO2(g) was lower than the
reported 23.0°C.
2011 Form B
2. An 8.55 mol sample of methanol, CH3OH , is placed in a 15.0 L evacuated rigid tank and heated to 327˚C. At
that temperature, all of the methanol is vaporized and some of the methanol decomposes to form carbon
monoxide gas and hydrogen gas, as represented in the equation below.
CH3OH(g)  CO(g) + 2 H2(g)
(a) The reaction mixture contains 6.30 mol of CO(g) at equilibrium at 327˚C.
(i) Calculate the number of moles of H2(g) in the tank.
(ii) Calculate the number of grams of CH3OH(g) remaining in the tank.
(iii) Calculate the mole fraction of H2(g) in the tank
(iv) Calculate the total pressure, in atm, in the tank at 327˚C
(b) Consider the three gases in the tank at 327˚C: CH3OH(g), CO(g), and H2(g)
(i) How do the average kinetic energies of the molecules of the gases compare? Explain.
(ii) Which gas has the highest average molecular speed? Explain.
(c) The tank is cooled to 25˚C, which is well below the boiling point of methanol. It is found that small amounts
of H2(g) and CO(g) have dissolved in the liquid CH3OH. Which of the two gases would you expect to be
more soluble in methanol at 25˚C? Justify your answer.
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2009 Question 2 Answer
(a) m=DV = (1.18 g L-1)(0.843 L) = 0.995 g
(b) 157.70 g − 0.995 g = 156.71 g
(c) 158.08 g − 156.71 g = 1.37 g
 750

atm0.843 L 

PV
 760

(d) n =

 0.0342 mol
RT 0.0821 L atmmol-1 K -1 296 K 

molar mass =

1.37 g
= 40.1 g mol-1
0.0342 mol
1
1
(e) percent error = | 44.0 g mol  40.1 g mol |  100 = 8.9%
44.0 g mol 1
(f) Occurrence 1: This occurrence could have been responsible. The dry air left in the flask is less dense (or
has a lower molar mass) than CO2 gas at the given T and P. This would result in a lower mass of gas in the
flask and a lower result for the molar mass of the unknown gas.
Occurrence 2: This occurrence could not have been responsible. The density of CO2 is greater at the lower
temperature. A larger mass of CO2 would be in the flask than if the CO2 had been at 23.0°C, resulting in a
higher calculated molar mass for the unknown gas.
(g) Valid methods include the following:
(i) Find the mass of the empty flask. Fill the flask with a liquid of known density (e.g., water at 23°C),
and measure the mass of the liquid-filled flask. Subtract to find the mass of the liquid. Using the
known density and mass, calculate the volume.
(ii) Measure 843 mL of a liquid (e.g., water) in a 1000 mL graduated cylinder and transfer the liquid
quantitatively into the flask to see if the water fills the flask completely.
2011 Form B Question 2 Answer
(a) (i) 6.30mol CO x = 12.6 mol H2
(a) (ii) 72.1 g
(a) (iii)0.596
(a) (iv) 69.5 atm
(b) (i) The average kinetic energies are the same because all three gases are at the same temperature.
(b) (ii) KE = mv2, so at a given temperature the molecules with the lowest mass have the highest average speed.
Therefore the molecules in H2 gas have the highest average molecular speed.
(c) The only attractive forces between molecules of H2 and CH3OH would be due to weak London dispersion
forces (LDFs). In contrast, the LDFs are stronger between CO molecules and CH3OH molecules because CO
has more electrons than H2. In addition CO is slightly polar; thus intermolecular dipole-dipole attractions can
form between CO molecules and CH3OH molecules. With stronger intermolecular interactions between
molecules of CO and CH3OH, CO would be expected to be more soluble in CH3OH than H2
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