1. Chapter 5
Gases

2. Calculate the Volume Occupied by 637 g of SO2 (MM 64. 07) at 6
Calculate the Volume Occupied by 637 g of SO2 (MM 64.07) at 6.08 x 103 mmHg and –23 °C.
Given:
Find:
mSO2 = 637 g, P = 6.08 x 103 mmHg, t = −23 °C,
V, L
Solution Map:
Relationships:
1 atm = 760 mmHg
T(K) = t(°C) + 273, 1 mol SO2 = g g n
P, n, T, R V
Solution:

3. Calculate the Density of a Gas at 775 torr and 27 °C if 0
Calculate the Density of a Gas at 775 torr and 27 °C if moles Weighs g
Given:
Find:
m = 9.988g, n = mol, P = atm, T = 300. K
density, g/L
m=9.988g, n=0.250 mol, P=775 mmHg, t=27°C,
density, g/L
P, n, T, R V
V, m d
Solution Map:
Relationships:
1 atm = 760 mmHg,
T(K) = t(°C) + 273
Solution:

4. Molar Mass of a Gas
One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law.

5. Calculate the Molar Mass of a Gas with Mass 0
Calculate the Molar Mass of a Gas with Mass g that Has a Volume of L at 55 °C and 886 mmHg.
Given:
Find:
m = 0.311g, V = L, P = atm, T = 328 K
Molar Mass, g/mol
m=0.311g, V=0.225 L, P=886 mmHg, t=55°C,
Molar Mass, g/mol
P, V, T, R n
n, m MM
Solution Map:
Relationships:
1 atm = 760 mmHg,
T(K) = t(°C) + 273
Solution:

6. Example: A sample of a gas has a mass of 0. 311 g. Its volume is 0
Example: A sample of a gas has a mass of g. Its volume is L at a temperature of 55 °C and a pressure of 886 mmHg. Find its molar mass.
Information:
Given: V = L, P = 886 mmHg,
t = 55 °C, m = g
Find: molar mass, (g/mol)
Equation:
PV = nRT; MM = mass/moles
Write a solution map:
mass
P,V,T,R n
Molar Mass
PV = nRT
When using the ideal gas equation, the units of V must be L;
and the units of P must be atm, or you will have to convert.
The units of T must be kelvin, K.

7. What Is the Molar Mass of a Gas if 12
What Is the Molar Mass of a Gas if 12.0 g Occupies 197 L at 380 torr and 127 °C?
Given:
Find:
m = 12.0g, V = 197 L, P = 0.50 atm, T =400 K,
molar mass, g/mol
m=12.0 g, V= 197 L, P=380 torr, t=127°C,
molar mass, g/mol
P, V, T, R n
n, m MM
Solution Map:
Relationships:
1 atm = 760 torr,
T(K) = t(°C) + 273
Solution:

8. Partial Pressure
Each gas in the mixture exerts a pressure independent of the other gases in the mixture.
The pressure of a component gas in a mixture is called a partial pressure.
The sum of the partial pressures of all the gases in a mixture equals the total pressure.
Dalton’s law of partial pressures.
Ptotal = Pgas A + Pgas B + Pgas C +...

9. A Mixture of He, Ne, and Ar Has a Total Pressure of 558 MmHg
A Mixture of He, Ne, and Ar Has a Total Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.
Given:
Find:
PHe= 341 mmHg, PNe= 112 mmHg, Ptot = 558 mmHg
PAr, mmHg
Ptot, PHe, PNe
PAr
Solution Map:
Relationships:
PAr = Ptot – (PHe + PNe)
Ptot = Pa + Pb + etc.
Solution:

10. Finding Partial Pressure
To find the partial pressure of a gas, multiply the total pressure of the mixture by the fractional composition of the gas.
For example, in a gas mixture that is 80.0% He and 20.0% Ne that has a total pressure of 1.0 atm, the partial pressure of He would be:
PHe = (0.800)(1.0 atm) = 0.80 atm
Fractional composition = percentage divided by 100.

11. The Partial Pressure of Each Gas in a Mixture, or the Total Pressure of a Mixture, Can Be Calculated Using the Ideal Gas Law 11

12. Find the Partial Pressure of Neon in a Mixture of Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe.
Given:
Find:
Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol
PNe, atm
nXe, V, T, R
PXe
Ptot, PXe
PNe
Solution Map:
Relationships:
Solution:

13. Deep Sea Divers and Partial Pressure
It is also possible to have too much O2, a condition called oxygen toxicity.
PO2 > 1.4 atm.
Oxygen toxicity can lead to muscle spasms, tunnel vision, and convulsions.
It is also possible to have too much N2, a condition called nitrogen narcosis.
Also known as rapture of the deep.
When diving deep, the pressure of the air that divers breathe increases, so the partial pressure of the oxygen increases.
At a depth of 55 m, the partial pressure of O2 is 1.4 atm.
Divers that go below 50 m use a mixture of He and O2 called heliox that contains a lower percentage of O2 than air.

14. Partial Pressure vs. Total Pressure
At a depth of 30 m, the total pressure of air in the divers
lungs, and the partial pressure of all the gases in the air,
are quadrupled!

15. Gas Stoichiometry

16. Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
Information:
Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 305 K
Find: VO2, L
Equation: PV = nRT
Conversion Factors:
1 mole KClO3 = g
2 mole KClO3  3 moles O2
Write a solution map:
P,T,R
g KClO3
nKClO3
nO2
VO2
PV=nRT
When using the ideal gas equation, the units of V must be L;
and the units of P must be atm, or you will have to convert.
The units of T must be kelvin, K.

17. Apply the solution map:
Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
Information:
Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 305 K
Find: VO2, L
Equation: PV = nRT
Conversion Factors:
1 mole KClO3 = g
2 mole KClO3  3 moles O2
Solution Map:
g → mol KClO3 → mol O2 → L O2
Apply the solution map:
Find moles of O2 made.

18. Apply the solution map:
Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
Information:
Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 305 K,
nO2 = 3.60 moles
Find: VO2, L
Equation: PV = nRT
Conversion Factors:
1 mole KClO3 = g
2 mole KClO3  3 moles O2
Solution Map:
g → mol KClO3 → mol O2 → L O2
Apply the solution map:
Convert the units.

19. Apply the solution map:
Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
Information:
Given: 294 g KClO3
PO2 = mmHg, TO2 = 305 K,
nO2 = 3.60 moles
Find: VO2, L
Equation: PV = nRT
Conversion Factors:
1 mole KClO3 = g
2 mole KClO3  3 moles O2
Solution Map:
g → mol KClO3 → mol O2 → L O2
Apply the solution map:

20. Practice—What Volume of O2 at 0
Practice—What Volume of O2 at atm and 313 K is Generated by the Thermolysis of 10.0 g of HgO? 2 HgO(s)  2 Hg(l) + O2(g), Continued
Given:
Find:
nO2 = mol, P = atm, T = 313 K
VO2, L
mHgO = 10.0g, P=0.750 atm, T=313 K
VO2, L
g HgO
mol HgO
mol O2
P, n, T, R V
Solution Map:
Relationships:
1 atm = 760 mmHg, HgO = g/mol
2 mol HgO : 1 mol O2
Solution:

21. Calculate the Volume Occupied by 1.00 Mole of an Ideal Gas at STP.
P x V = n x R x T
(1.00 atm) x V = (1.00 moles)( )(273 K)
L∙atm
mol∙K
V = 22.4 L
1 mole of any gas at STP will occupy 22.4 L.
This volume is called the molar volume and can be used as a conversion factor.
As long as you work at STP.
1 mol  22.4 L

22. Molar Volume There is so much empty space between molecules
in the gas state that
the volume of the
gas is not effected
by the size of the
molecules (under
ideal conditions).

23. Example —How Many Grams of H2O Form When 1
Example —How Many Grams of H2O Form When 1.24 L H2 Reacts Completely with O2 at STP? O2(g) + 2 H2(g) → 2 H2O(g)
Given:
Find:
VH2 = 1.24 L, P = 1.00 atm, T = 273 K
massH2O, g
g H2O
L H2
mol H2
mol H2O
Solution Map:
Relationships:
H2O = g/mol, 1 mol = 22.4 STP
2 mol H2O : 2 mol H2
Solution: 23

24. Example: How many grams of water will form when 1
Example: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
Information:
Given: 1.24 L H2
Find: g H2O
Conversion Factors:
1 mol H2O = g
2 mol H2O  2 mol H2
1 mol H2  22.4 L
Write a solution map:
L H2
mol H2
mol H2O
g H2O

25. Apply the solution map:
Example: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
Information:
Given: 1.24 L H2
Find: g H2O
Conversion Factors:
1 mol H2O = g
2 mol H2O  2 mol H2
1 mol H2  22.4 L
Solution Map:
L → mol H2 → mol H2O → g H2O
Apply the solution map:

26. Practice—What Volume of O2 at STP is Generated by the Thermolysis of 10.0 g of HgO? 2 HgO(s)  2 Hg(l) + O2(g), Continued
Given:
Find:
mHgO = 10.0 g, P = 1.00 atm, T = 273 K
VO2, L
L O2
g HgO
mol HgO
mol O2
Solution Map:
Relationships:
HgO = g/mol, 1 mol = 22.4 L at STP
2 mol HgO : 1 mol O2
Solution: 26