Chemistry 1010 Loader (Winter 2002)
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Chemistry 1010
Handout 7: Gases
1. Convert the following to the units specified:
(a) 1.723 atm to mmHg
1.723 atm x 760 mmHg atm–1 = 1.30 x 103 mmHg
Ans: 1.30 x 103 mmHg
(b) 23.11 mL to litres.
1
23.11 mL x 1000 mL
L −1 = 0.02311 L
Ans: 0.02311 L
(c) 1.223 x 105 Pa to atm
1.223 x 10 5 Pa
101.325 x 10 3 Pa atm −1 =
1.207 atm
Ans: 1.30 x 103 atm
(d) 650 mL to m3.
Ans: 6.50 x 10–4 m3.
2. (a) What volume was occupied by 0.2500 mol of CO2 gas at 25.0oC and 500 kPa
PV=nRT so V =
nRT
P
0.2500 mol x 8.314 L kPa mol −1 K −1 x 298.15 K
500 kPa
=
= 1.24 L
Ans: 1.24 L.
(b) On a cold day in St. John's a car tire is inflated so that the pressure inside the tire is 193 kPa at
-10.0oC. What will be the pressure in the tire after a fast drive down to Florida when the
temperature of the tire rises to 67.0oC?
P
P1T
P
2
If the volume is constant then T 11 = T 22 so P 2 =
since T2 > T1 the new pressure must be
T1
greater than the old.
x 340.15 K
P 2 = 193 Pa263.15
= 249 kPa
substituting,
K
Ans: 249 kPa
(c) What volume will 0.1254 g of chlorine gas occupy at STP?
0.1254 g
V=
nRT
P
=
70.9054 g mol −1
x 8.314 L kPa mol −1 K −1 x 273.15 K
101.325 kPa
0.1254 g
70.9054 g mol −1
x 0.0821 L atm
mol −1
= 0.0395 L
K −1 x 273.15 K
or V = nRT
=
= 0.03966 L
P
1 atm
Ans: 0.03966 L.
(d) What is the mole fraction of carbon dioxide inside a 5.00 L container that contains a mixture of
1.00 g of carbon dioxide gas, 1.00 g of oxygen gas and 0.500 g of argon?(Delete the
temperature given by mistake)
mol CO2 = 0.02272 mol, O2 = 0.03125 mol, Ar = 0.01252 mol
Total mole in container = 0.06650 mol so the mole fraction of carbon dioxide is =
Ans: 0.342
0.02272 mol
0.06650 mol
= 0.342
(e) What is the pressure inside a 5.00 L container in (d) above when adding 1.00 g of nitrogen
increases the total pressure by 19.55 kPa (constant temperature)?
Chemistry 1010 Loader (Winter 2002)
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mol N2 added = 0.03554 mol so the total moles becomes 0.06650 + 0.03554 = 0.10204 mol
0.03554 mol
0.10204 mol
mole fraction of N2 is =
19.55 kPa
0.3483
P total =
Ans: 36.6 kPa
= 0.3483, now P N2 = P total % x N2 so P total =
P N2
x N2
= 56.1 kPa so the total pressure was 56.13 - 19.55 = 36.5 kPa
3. A 1.2300 g sample of colourless liquid gave 420 mL of gas when vaporized at 180oC and 100.0
kPa. The liquid had the elemental analysis, C, 65.42; H, 9.15; N, 25.43. Use these data to
calculate the molecular formula for the liquid.
First we can use the gas laws to calculate the number of mole in the 1.2300 g of vaporized liquid. Since
100 kPa x 0.420 L
PV=nRT, n= PV
RT = 8.314 kPa L K −1 mol −1 x 453.15 K = 0.011148 mol
so the molar mass is =
Ans: 110 g mol–1
1.2300 g
0.011148 mol
= 110 g mol–1
4. A 0.640 g of a certain gaseous oxide of chlorine was collected in an evacuated 6.60 L vessel at
-3.0°C. The measured pressure in the vessel was 24.3 mmHg.
(a) Calculate the molar mass of the oxide of chlorine.
First we can use the gas laws to calculate the number of mole in the 0.546 g of the oxide. Since
24.3 mmHg
PV=nRT, n=
PV
RT
=
x 6.60 L
760 mmHg atm −1
0.0821 atm L K −1 mol −1
x 270.15 K
= 0.00951456 mol
0.546 g
so the molar mass is = 0.00951456 mol = 67.3 g mol–1
Ans: 57.3 g mol–1
(b)
What is the probable molecular formula for the compound? Justify your answer!
It can only contain one chlorine (why?) so the rest of the molar mass (31.8) must be oxygen. The molar
mass of oxygen is 15.9994 g mol–1 so there must be to atoms of oxygen per formula unit and the formula
is ClO2.
5. A sample of impure zinc weighing 2.552 g reacted with an excess of dilute sulfuric acid to give 545
mL of wet hydrogen gas collected over water at 15.0oC and 760 mmHg. Assume that only zinc
reacted to give the hydrogen (i.e. the impurities were inert):
Zn(s) + H2SO4(aq) = ZnSO4(aq) + H2(g).
(a) Calculate the mass of zinc that reacted.
If only the zinc reacts then for each mole of hydrogen produced there must have been one mole of zinc.
The hydrogen is collected over water so the partial pressure of hydrogen in the collected gas is 760
mmHg - 12.81 mmHg = 747.2 mmHg. so mol of dry hydrogen, n
747.2 mmHg
PV
RT
x 0.545 L
760 mmHg atm −1
0.0821 atm L K −1 mol −1
n=
= 0.022649 mol so the number of mole of zinc in the sample of
=
x 288.15 K
impure zinc is 0.022649 mol. The mass of zinc present = 0.022649 mol x 65.39 g mol–1 = 1.481 g.
Ans: 1.48 g
(b) Calculate the percentage of zinc in the impure sample.
Percentage of zinc in the impure sample is
1.481 g
2.552 g
x 100% = 58.0%
6. A 1.2300 g sample of colourless liquid gave 300.5 mL of gas when vaporized at 180oC and 100.0
kPa. The liquid had the elemental analysis, C, 71.98; H, 6.71; O, 21.31%. Use these data to
calculate the molecular formula for the liquid.
First calculate the molar mass of the liquid:
Mole of liquid, n=
PV
RT
=
100 kPa x 0.3005 L
8.314 kPa L K −1 mol −1 x 453.15 K
= 7.976 x 10–3 mol
Chemistry 1010 Loader (Winter 2002)
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1.2300 g
Molar mass of liquid = 0.00796 mol = 154.2 g mol–1
Now calculate the empirical formula of the liquid in the usual way C9H10O2 sot the empirical formula
mass is 150.2 g mol–1 this is almost the same as the experimental molar mass so in this case the
molecular and empirical formula are the same.
Ans: C9H10O2
7. A 10.68 mg sample of a compound containing only carbon, hydrogen and oxygen gave 28.16 mg
of carbon dioxide and 6.405 mg of water on combustion (in an excess of oxygen). The molar mass
of the compound was found to be 150.2 g.mol-1. Calculate the empirical and molecular formula for
the compound.
Mass of carbon in the carbon dioxide =
Mass of hydrogen in the water =
28.16 x 10 −3 g
44.0098 g mol −1
6.405 x 10 −3 g
18.015 g mol −1
x 12.011 g mol −1 = 7.6853 x 10 −3 g
x (2 x 1.0079) g mol −1 = 7.1669 x 10 −4 g
Mass of oxygen in sample = 0.01068 g - (0.0076853 g + 0.00071669 g) = 0.0022780 g
C
H
O
mass
0.0076853 g
0.00071669 g
0.0022780 g
mol
0.00063986 mol
0.00071107 mol
0.00014238 mol
mole ratio
4.49
4.99
1
X2
9
10
2
so the empirical formula is C9H10O2.
Ans: C9H10O2.
8. A gas supplier wishes to make a synthetic air mixture by adding oxygen gas to a cylinder of pure
nitrogen gas so that the cylinder contains 20% by weight of oxygen. If the pressure in the nitrogen
cylinder was initially 5254 mmHg, what must the final total pressure in the cylinder be after the
oxygen gas has been added.
required pressure =
pressure is (5254 + 1155) mmHg = 6409 mmHg
Ans: 6409 mmHg
= 1154.9 mm Hg so the final total
9. The volume of a dry mixture of carbon dioxide and nitrogen gas at 27.0oC was 3.667 L measured
at 150 kPa. The gas mixture was passed through 100.0 mL of 0.2500 mol.L-1 sodium hydroxide
solution (an excess) so that all of the carbon dioxide reacted to give sodium carbonate. The
excess sodium hydroxide required 45.50 mL of 0.1500 mol.L-1 hydrochloric acid for neutralization.
(a) Calculate the mass of carbon dioxide that was present in the gas mixtures
The reaction is 2 NaOH + CO2 = Na2CO3 + H2O, 2 mol of NaOH is used for each mol of CO2
present.
Mol HCl used to react with the excess NaOH = 45.50 x 10–3 L x 0.1500 mol.L-1 = 6.825 x x 10–3 mol
The reaction with the NaOH is NaOH + HCl = NaCl + H2O so this is the amount of NaOH that
remained. The amount of NaOH present at the start of the reaction was = 0.1000 L x 0.2500 mol.L-1 =
0.02500 mol
so the amount used = 0.02500 mol - 6.825 x x 10–3 mol = 0.018175 mol
so mol CO2 absorbed = ½ x 0.018175 mol and
mass CO2 absorbed = ½ x 0.018175 mol x 44.0098 g mol-1 = 0.3999 g
Ans: 0.3999 g
Chemistry 1010 Loader (Winter 2002)
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(b) What was the partial pressure of nitrogen gas in the original gas mixture?
We know the number of mole of CO2 so we can calculate the partial pressure of CO2 using the gas law
−3
L kPa K −1 mol −1 x 300.15 K
P CO 2 = 9.0875 x 10 mol x 8.3143.367
= 6.735 kPa
L
now, Ptotal = p CO 2 + p N 2 so p N 2 = Ptotal - p N 2 = 143 kPa
Ans: 143 kPa
10. The combustion of 1.76 L of CO(g) in oxygen gas gave carbon dioxide. If all the gas volumes
were measured at the same temperature and pressure, what volume of oxygen was used and
what volume of carbon dioxide was produced.
Ans: O2(g), 0.880 L; CO2(g), 1.76 L (Equal volumes of gases at the same temperature and pressure
contain equal numbers of mole)
11. (a)
4(g), at 85.44 kPa and 298 K.
Ans: 4.03 L
(b) Calculate the pressure inside a 3.00 litre vessel at 55.0°C that contains 1.00 g of oxygen gas,
2.00 g of nitrogen gas and 3.00 g of methane gas.
Ans: 2.60 atm
(c) How many grams of carbon dioxide gas must be added to the mixture in (b) in order to double
the total pressure. (Keeping the volume and temperature constant).
Ans: 12.7 g
12. The ozone unfriendly dichlorodifluoromethane (Freon), CCl2F2, is used as the refigerant gas in
cooling systems. Calculate the density of dichlorodifluoromethane gas at 10.0°C and 850 mmHg.
CCl2F2 has molar mass = 120.913 g mol–1, now PV = nRT so, n = PV/RT converting to the mass
and using one mole, n mol x 120.913 g mol–1 = 120.913 g mol–1 x PV/RT
rearranging then density (mass/volume) = (n mol x 120.913 g mol–1 )/V= 120.913 g mol–1 x P/RT
850 mmHg
–1
= 120.913 g mol x
0.0821 L
760 mmHg atm −1
atm K −1 mol −1 x
283.15 K
= 5.81 g L-1
Ans: 5.81 g L–1
13. When a sample of lithium reacted with water 310.7 mL of hydrogen gas was collected at 758
mmHg and 25.0°C by the downward displacement of water.
(a) Write a balanced equation for the reaction between lithium metal and water.
Ans: 2 Li(s) + 2 H2O(l) = 2 LiOH(aq) + H2(g)
(b) Calculate the number of moles of hydrogen gas produced in the reaction.
Ans: 0.0123 mol (Remember to correct for the vapour pressure of water at 25.0°C!)
(c) What was the mass of the sample of lithium metal that reacted in the reaction.
Ans: 0.171 g
14. Gaseous hydrogen chloride can be prepared by the reaction of sodium chloride with concentrated
sulfuric acid.
(a) Write down the equation for the reaction.
Ans: NaCl(s) + 2 H2SO4 (l) = NaHSO4 (s)+ HCl(g)
(b) What mass of NaCl is required to prepare enough hydrogen chloride gas to fill a 35.4 L
cylinder to a pressure of 2.25 atm at 25.0°C?
Mole of HCl required, n=
PV
RT
=
2.25 atm kPa x 35.4 L
0.0821 atm L K −1 mol −1 x 298.15 K
= 3.254 mol
Chemistry 1010 Loader (Winter 2002)
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1 mol NaCl give 1 mole of HCl so mass of NaCl required = 3.254 mol x 58.4425 g mol–1 = 190.2 g
Ans: 190 g
15. 10.00 g of magnesium metal was added to 164 mL of copper(II) sulfate solution contained in a
beaker. When reaction was complete the mixture of copper and magnesium obtained weighed
14.45 g. Calculate the original molarity of the copper(II) sulfate solution.
We need to find out how many mole of Cu are formed in the reaction
Mg(s) + Cu2+(aq) = Cu(s) + Mg2-(aq)
Hence, mole of Mg used =
1
10.00 g
24.3050 g mol −1
= 0.41144 mol, so the mass of the product
14.45 g = ((0.41144 - X) mol x 24.3050 g mol–1 ) + (X mol x 63.546 g mol–1 )
so 39.241X = 14.45 - 10.000 and X = 0.11340 and mole of Cu = 0.11340 mol. The concentration of the
mol
solution is = 0.11340
= 0.691 mol L–1
0.164 L −1
Ans: 0.691 mol L–1
16. An excess of zinc metal was allowed to react with 50.00 mL of a solution of hydrochloric acid.
When the reaction was complete 84.0 mL of dry hydrogen gas had been collected at STP.
Calculate the concentration of the hydrochloric acid solution in mol.L-1.
Ans: 0.150 mol L–1