1. Gases and the Mole

2. Gay-Lussac formulated the law of combining volumes.

3. Law of Combining Volumes
Gases at the same temperature and pressure react with one another in volume ratios of small whole numbers.

4. Law of Combining Volumes
Example 1:
H2 + Cl2
2HCl
1 L + 1 L
2 L

5. Law of Combining Volumes
Example 2:
2H2 + O2
2H2O
2 L + 1 L
2 L

6. Avogadro’s law
The volume of a gas, maintained at a constant temperature and pressure, is directly proportional to the number of moles of the gas.

7. Molar Volume
the volume that a mole of gas occupies at standard temperature and pressure

8. 1 mole of ANY gas at STP occupies 22.4 L.
Molar Volume
1 mole of ANY gas at STP occupies 22.4 L.

9. Sample Problem 1
What volume would 7 moles of carbon dioxide occupy at STP?
7 mol CO2
22.4 L
= 157 L CO2
1 mol

10. Question Change 228 L of O2 at STP to moles. 470 moles 426 moles

11. 228 L
1 mol
= 10.2 mol
22.4 L

12. Sample Problem 2 What volume will 4.00 mol of ammonia occupy at STP?
4.00 mol NH3
22.4 L
= 89.6 L NH3 at STP
1 mol

13. Sample Problem 3
What volume will 2.50 mol of hydrogen gas occupy at K and 400. torr?
2.50 mol H2
22.4 L
= 56.0 L H2 at STP
1 mol

14. Sample Problem 3
The combined gas law is used to adjust this volume to nonstandard conditions.
P1V1 T1
P2V2 T2 =

15. Sample Problem 3 117 L = V2 = = V2 (760. torr) (56.0 L) (400. torr)
273 K
300. K
(760. torr)
(56.0 L)
(300. K)
= V2
(273 K)
(400. torr)
117 L = V2

16. Sample Problem 4
A sample of oxygen gas occupies 1.00 L when its temperature is 190. K and its pressure is 129 kPa.
How many moles of oxygen are present?

17. The given solution must first be adjusted to standard conditions.
Sample Problem 4
The given solution must first be adjusted to standard conditions.
P1V1 T1
P2V2 T2 =

18. Sample Problem 4 V2 = 1.83 L at STP = = V2 (129 kPa) (1.00 L)

19. Sample Problem 4 1.83 L 1 mol O2 = 0.0817 mol O2 22.4 L
Once the gas’s volume at STP has been calculated, the number of moles can be found.
1.83 L
1 mol O2
= mol O2
22.4 L

20. 1 mol
1 mol
Image adapted from Chemistry textbook, p. 261
Every gas has the same number of particles in one mole; but, the mass of each gas is different.

21. Gas Density Recall that density is mass/volume.
For a gas, its density would have units of grams/liter.

22. We can expand this definition to include moles:
Gas Density
We can expand this definition to include moles:
grams/mole
L/mole
D =

23. Gas Density
However, what is the number of liters/mole for all gasses at STP?
22.4
Therefore, the denominator is always 22.4 (at STP).

24. Sample Problem 5 grams/mole L/mole D = 70.9 grams/mole 22.4 L/mole D =
What is the density of the gas Cl2 at STP?
D =
grams/mole
L/mole
70.9 grams/mole
22.4 L/mole
D =
= 3.17 g/L

25. Sample Problem 6 What is the density of hydrogen gas? molar mass
D of H2 =
molar mass
molar volume
2.016 g/mol
22.4 L/mol
= g/L

26. Sample Problem 7 32.00 g/mol 1.429 g/L = 22.39 L/mol
If the density of oxygen gas at STP is g/L, find the molar volume of the gas.
molar volume of O2 =
molar mass
density
32.00 g/mol
1.429 g/L
= L/mol

27. Sample Problem 8 = 2.144 g/L × 22.4 L/mol = 48.0 g/mol molar mass =
One mole of an unknown gas has a density of g/L at STP. What is its molar mass?
molar mass =
density ×
molar volume
= g/L × 22.4 L/mol
= 48.0 g/mol

28. Sample Problem 8
One mole of an unknown gas has a density of g/L at STP. What is its molar mass?
Further tests could prove this gas to be ozone, which has a molar mass of g/L.

29. The “Math” of Chemistry
Stoichiometry
The “Math” of Chemistry

30. Start with “X” grams of A.
“Skeleton Equation”
A + B
C + D
Start with “X” grams of A.
How many grams of B?
X g A
1 mol A
# mol B
# g B
# g A
# mol A
1 mol B

31. Question Periodic table Coefficient ratio Combined gas law
How do you change from moles of one substance to moles of another substance?
Periodic table
Coefficient ratio
Combined gas law
Number of particles

32. Question Periodic table Coefficient ratio Combined gas law
How do you change from grams to moles?
Periodic table
Coefficient ratio
Combined gas law
Number of particles

33. 1 2 3 4 5 6 7 8 9 10
Chemistry textbook, p. 263

34. Question Combined gas law Coefficient ratio Number of particles
How do you change from one set of conditions to another?
Combined gas law
Coefficient ratio
Number of particles
Use 22.4

35. Question Combined gas law Coefficient ratio Number of particles
How do you change from volume of gas to moles of gas?
Combined gas law
Coefficient ratio
Number of particles
Use 22.4

36. Sample Problem 9
When 2.00 mol of calcium react with water to form calcium hydroxide and hydrogen gas, what volume of hydrogen gas will be produced at STP?

37. Sample Problem 9
Solution: First, write out the reaction and balance it.
Ca + 2 H2O
Ca(OH)2 + H2
Moles of calcium can be used to find moles of hydrogen, which can then be converted to the volume of hydrogen.

38. Sample Problem 9 = 44.8 L H2 at STP 2.00 mol Ca 1 mol H2 22.4 L

39. Sample Problem 10
How many grams of water will be produced if L of oxygen gas at STP is burned with hydrogen?
Solution: The balanced equation for the reaction is
2H2 + O2
2H2O.

40. Sample Problem 10
The volume of oxygen must be converted to the moles of oxygen. The moles of oxygen can be used to find moles of water, which can then be converted to the mass of water.

41. Sample Problem 10 0.500 L O2 1 mol O2 2 mol H2O 22.4 L O2 1 mol O2
g H2O
= g H2O
1 mol H2O

42. Sample Problem 11 Xe + 3 F2 XeF6
If 15.9 g of Xenon gas reacts with excess fluorine, how many liters of xenon hexafluoride will be produced at 500°C?
Xe + 3 F2
XeF6

43. Plan:
gXe mXe mXeF6 lXeF6 (at STP)
lXeF6 (at STP) lXeF6 (at 500°C)

44. 15.9 g Xe
1 mol Xe
1 mol XeF6
131.3 g Xe
1 mol Xe
22.4 L XeF6
= 2.71 L XeF6
1 mol XeF6

45. 2.71 L XeF6 V2 =
273 K
773 K
V2 = 7.67 L XeF6 at 500 °C

46. Sample Problem 12
For 2 H2 + O2 2 H2O, if you start with 45 L of H2 at 300 K and 700 mm Hg, how many g of O2 will be needed?
Plan: Change H2 to STP (combined gas law)
Change H2 to g O2 (unit analysis)

47. (700)(45 L)(273) = (300)(760)V2 V2 = 37.7 L H2 (700 mmHg)(45 L)
(300 K)
(760 mmHg)V2
(273 K) =
(700)(45 L)(273) =
(300)(760)V2
V2 = 37.7 L H2

48. 37.7 L H2
1 mol
22.4 L
1 mol O2
2 mol H2
32 g O2
1 mol O2
= g O2

49. When 17 grams of calcium react with water,
Sample Problem 13
When 17 grams of calcium react with water,
Ca + 2 H2O(l) Ca(OH)2 + H2.
What volume of H2 will be produced at 300 K and a pressure of 560 mm Hg?

50. Plan:
Use “skeleton equation” to change to L of H2 at STP.
Use combined gas law to change to L of H2 not at STP.

51. 17 g Ca
1 mol Ca
40 g Ca
1 mol H2
1 mol Ca
22.4 L H2
1 mol H2
= 9.52 L H2

52. (760)(9.52 L)
(273)
(560)(V2)
(300) =
V2 = 14.2 L H2 not at STP

53. Ideal Gas
A hypothetical gas that behaves exactly according to the kinetic theory

54. Ideal Gas Law
Relates pressure, volume, amount, and temperature for any gas at moderate conditions.

55. Ideal Gas Law PV = nRT n = number of moles R = Universal gas constant
(1 atm)
(22.4 L) = = n T
(1 mol)
(273 K)
= L·atm / mole·K

56. Sample Problem 14
How many moles of a gas are present in a 7.92 L sample at 2.34 atm pressure and 58 K?
PV = nRT
(2.34 atm)
(7.92 L) = n
( L·atm/mole·K)
(58 K)

57. Sample Problem 14
How many moles of a gas are present in a 7.92 L sample at 2.34 atm pressure and 58 K?
PV = nRT
(2.34)
(7.92) = n
= mol
( /mole)
(58)

58. Sample Problem 15 PV RT = n PV = nRT
How many moles of a gas are present in a 2.4 L sample at 1.25 atm of pressure and 27°C?
Variables: PV RT
= n
PV = nRT
V = 2.4 L
P = 1.25 atm
T = 27 °C
R = L·atm/mole·K

59. Sample Problem 15 = n n = 0.12 mol
How many moles of a gas are present in a 2.4 L sample at 1.25 atm of pressure and 27°C?
(1.25 atm)
(2.4 L) = n
[( L·atm/(mol·K)]
(300 K)
n = 0.12 mol

60. Sample Problem 16
A 5.04 × 10−4 kg sample of gas occupies 4.57 x 10−4 m3 at 9.63 × 104 Pa and 293 K. What is the molar mass of this unknown gas?

61. V = 4.57 x 10−4 m3 P = 9.63 × 104 Pa T = 293 K R = 8.31 m3·Pa/(mole·K)
Variables:
V = 4.57 x 10−4 m3
P = 9.63 × 104 Pa
T = 293 K
R = 8.31 m3·Pa/(mole·K) PV RT
= n
PV = nRT

62. = n = (9.63 × 104 Pa ) (4.57 x 10−4 m3 ) [8.31 m3·Pa/(mole·K)] (293 K)
n = mol
Convert 5.04 × 10−4 kg to grams.
0.504 g
27.8 g =
= 27.8 g/mol
mol
mol