1. Mole-Mass and Mole-Volume Relationships
Chemical Quantities
Unit 6
Chapter 10, Section 10.2
Mole-Mass and Mole-Volume Relationships

2. Objectives When you complete this presentation, you will be able to …
describe how to convert the mass of a substance to the number of moles of a substance, and moles to mass.
identify the volume of a quantity of gas at STP (Standard Temperature and Pressure).

3. Introduction How many jelly beans are in the jar?
How might you approach the problem?
Count each jelly bean
Estimate the volume of each bean
Estimate the mass of each bean

4. Introduction
As chemists, we can’t count the number of individual atoms in a reaction (why?).
We rely on other relationships, such as …
moles → mass and mass → moles
moles → volume and volume → moles
These relationships allow us to accurately estimate the number of atoms or molecules involved in a chemical reaction.

5. The Mole-Mass Relationship
The relationship between the mass of a material and the number of moles of matter is based on the molar mass of the matter.
where M is molar mass, m is mass, and n is number of mols. M = n m n = M m or or
m = n × M

6. The Mole-Mass Relationship
For example:
The molar mass of NaCl is 58.5 g/mol
The mass of 3.00 mol of NaCl is given by
Let’s try Sample Problem 10.5 (page 298)
m = n × M
= (3.00 mol)(58.3 g/mol)
= 176 g

7. Sample Problem 10.5
The aluminum satellite dishes shown below are resistant because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass of 9.45 mol of aluminum oxide?
Known:
n = 9.45 mol
M = [(2 × 27.0) + (3 × 16.0)] g/mol
M = g/mol
Unknown:
m = ? g Al2O3
m = n × M
= (9.45)(102.0) g
m = 964 g Al2O3

8. The Mole-Mass Relationship
For another example:
The molar mass of Na2SO4 is g/mol
The mols of 10.0 g of Na2SO4 is given by
Let’s try Sample Problem 10.6 (page 299) n = M m =
142.1 g/mol
10.0 g
= 7.04 × 10-2 mol

9. Sample Problem 10.6
When iron is exposed to air, it corrodes to form red-brown rust. Rust is iron(II) oxide(Fe2O3). How many mols of iron(II) oxide are contained in 92.2 g of pure Fe2O3?
Known:
m = 92.9 g
M = [(2 × 55.8) + (3 × 16.0)] g/mol
M = g/mol
Unknown:
n = ? mol Fe2O3
n = m M
92.9 g
159.6 g/mol =
= mol

10. The Mole-Mass Relationship
Practice Problems:
What is the mass of 2.50 mols of hydrogen gas, H2?
What is the mass of mols of glucose, C6H12O6?
What is the mass of 12.0 mols of water, H2O?
What is the mass of mols of methane gas, CH4?
m = n × M
mH2 = nH2 × MH2 = (2.50 mol)(2.02 g/mol) = 5.05 g
mglucose = nglucose × Mglucose = (0.100 mol)(180 g/mol) = 18.0 g
mH2O = nH2O × MH2O = (12.0 mol)(18.0 g/mol) = 216 g
mCH4 = nCH4 × MCH4 = (2.50 mol)(16.0 g/mol) = 4.00 g

11. The Mole-Mass Relationship
Practice Problems:
How many mols are in 3.25 g of H2?
How many mols are in 9.00 g of C6H12O6?
How many mols are in 100. g of H2O?
How many mols are in 9.60 g of CH4?
nH2 = = = 1.61 mol
mH g
MH g/mol
nC6H12O6 = = = mol
mC6H12O g
MC6H12O g/mol
nH2O = = = 5.56 mol
mH2O g
MH2O g/mol
nCH4 = = = mol
mCH g
MCH g/mol

12. The Mole-Volume Relationship
We saw earlier that 1 mol of solids or liquids are not necessarily the same volume.
1 mol of C6H12O6 is a much larger volume than 1 mol of H2O.
The story is different with gases.
1 mol of O2 is nearly the same volume as 1 mol of H2 under the same conditions.
1 mol of O2 is nearly the same volume as 1 mol of CO2 under the same conditions.

13. The Mole-Volume Relationship
Why is this?
In 1811, Amadeo Avogadro proposed that equal volumes of gases, under similar conditions, contain the equal numbers of molecules.
This is called Avogadro’s hypothesis.
The individual molecules of each gas are different sizes and masses.
But, in gases, the distances between the molecules is much larger than the molecule size.

14. The Mole-Volume Relationship
When talking about gases we use the phrase “under similar conditions” a lot.
That is because temperature and pressure have a big effect on the volume of gases.
We will find out more about this in the next unit.
To compare different gases, we put them at a Standard Temperature and Pressure.
This is called “STP.”

15. The Mole-Volume Relationship
Standard temperature and pressure:
Standard temperature is 0°C ( K).
Standard pressure is 1 atm (101.3 kPa).
Don’t worry about the units right now.
The important thing is that if we say that two gases are at STP, then they are “under similar conditions.”

16. The Mole-Volume Relationship
At STP, 1 mol of a gas has a volume of 22.4 L.
The relationship between the volume of a gas and the number of mols of a gas is given by
where V is the volume of the gas in liters (L) and n is the amount of gas in mols
V = n ×
22.4 L
1 mol
n = V ×
1 mol
22.4 L or

17. The Mole-Volume Relationship
For example
the volume of 1.25 mols of O2 gas at STP is
the number of mols of H2 in 67.2 L at STP is
Let’s try Sample Problem 10.7 (page 301).
V = n ×
22.4 L
1 mol
1.25 mol ×
22.4 L
1 mol =
= 28.0 L
n = V ×
1 mol
22.4 L
67.2 L ×
1 mol
22.4 L =
= 3.00 mol

18. Sample Problem 10.7
Sulfur dioxide (SO2) is a gas produced by burning coal. It is an air pollutant and one of the causes of acid rain. Determine the volume, in liters, of 0.60 mol of SO2 gas at STP.
Known:
n = 0.60 mol
Unknown:
V = ? L SO2
V = n ×
22.4 L
1 mol
= 0.60 mol ×
22.4 L
1 mol
V = 13 L SO2

19. The Mole-Volume Relationship
V = n ×
22.4 L
1 mol
Practice Problems:
What is the volume of 3.25 mols of H2 at STP?
What is the volume of mols of O2 at STP?
What is the volume of 12.5 mols of Cl2 at STP?
What is the volume of mols of CH4?
VH2 = nH2× = 3.25 mol × = 72.8 L
22.4 L L
1 mol mol
VO2 = nO2× = mol × = 5.60 L
22.4 L L
1 mol mol
VCl2 = nCl2× = 12.5 mol × = 280. L
22.4 L L
1 mol mol
VCH4 = nCH4× = mol × = L
22.4 L L
1 mol mol

20. The Mole-Volume Relationship
n = V ×
1 mol
22.4 L
Practice Problems:
How many mols of H2 are in 1.00 L at STP?
How many mols of O2 are in 50.0 L at STP?
How many mols of C2H2 are in L at STP?
How many mols of F2 are in 11.2 L at STP?
nH2 = VH2× = 1.00 L × = mol
1 mol mol
22.4 L L
nO2 = VO2× = 50.0 L × = 2.23 mol
1 mol mol
22.4 L L
nC2H2 = VC2H2× = L × = 5.60 mol
1 mol mol
22.4 L L
nF2 = VF2× = 11.2 L × = mol
1 mol mol
22.4 L L

21. The Mole-Volume Relationship
Density and Molar Mass
The density of matter is mass per unit volume.
where ρ = density (g/L), m = mass, and V = volume.
We can use our mole-mass and mole-volume relationships to determine density.
ρ = m V
22.4 L
1 mol
V = n ×
m = n × M m V =
ρ =
22.4 L
n × M
1 mol
n × =
22.4 L
1 mol M
= M ×
22.4 L
1 mol

22. The Mole-Volume Relationship
Density and Molar Mass
If we know the molar mass of a gas, then we can calculate its density.
If we know the density of a gas, then we can calculate its molar mass.
1mol
22.4 L
ρ = M ×
22.4 L
1 mol
M = ρ ×

23. The Mole-Volume Relationship
Density and Molar Mass
For example, find the density of H2 at STP.
For example, find the molar mass of a gas with a density of 3.17 g/L at STP.
1mol
22.4 L
ρ = M ×
1mol
22.4 L
= 2.02 g/mol ×
= g/L
22.4 L
1 mol
M = ρ ×
22.4 L
1 mol
= 3.17 g/L ×
= 71.0 g/mol

24. Sample Problem 10.8
The density of a gaseous compound containing carbon and oxygen is found to be g/L at STP. What is the molar mass of this compound
Known:
ρ = g/L
Unknown:
M = ? g/mol
M = ρ ×
22.4 L
1 mol
= g/L ×
22.4 L
1 mol
M = 44.0 g/mol
What do you think the gas is? Why?

25. The Mole-Volume Relationship
Practice Problems:
What is the density of F2 at STP?
What is the density of CH4 at STP?
What is the density of Ne at STP?
What is the density of Rn at STP?
ρF2 = MF2× = 38.0 g/mol × = 1.70 g/L
1 mol mol
22.4 L L
ρCH4 = MCH4× = 16.0 g/mol × = g/L
1 mol mol
22.4 L L
ρNe = MNe× = 20.2 g/mol × = g/L
1 mol mol
22.4 L L
ρRn = MRn× = 222 g/mol × = 9.91 g/L
1 mol mol
22.4 L L

26. The Mole-Volume Relationship
Practice Problems:
Find the molar mass of a gas with a density of g/L at STP.
Find the molar mass of a gas with a density of g/L at STP.
Find the molar mass of a gas with a density of g/L at STP.
Find the molar mass of a gas with a density of g/L at STP.
M = ρ × = 1.34 g/L × = 30.0 g/mol
22.4 L L
1 mol mol
M = ρ × = 2.59 g/L × = 58.0 g/mol
22.4 L L
1 mol mol
M = ρ × = 1.78 g/L × = 39.9 g/mol
22.4 L L
1 mol mol
M = ρ × = 6.52 g/L × = 146 g/mol
22.4 L L
1 mol mol

27. Summary
The relationship between the mass of a material and the number of moles of matter is based on the molar mass of the matter.
where M is molar mass, m is mass, and n is number of mols. M = n m n = M m or or
m = n × M

28. Summary V = n × 22.4 L 1 mol n = V × 1 mol 22.4 L or
To compare different gases, we put them at a Standard Temperature and Pressure.
This is called “STP.”
Standard temperature is 0°C ( K).
Standard pressure is 1 atm (101.3 kPa).
The relationship between the volume of a gas and the number of mols of a gas is given by
where V is the volume of the gas in liters (L) and n is the amount of gas in mols
V = n ×
22.4 L
1 mol
n = V ×
1 mol
22.4 L or

29. Summary
We can use our mole-mass and mole-volume relationships to determine density.
If we know the molar mass of a gas, then we can calculate its density.
If we know the density of a gas, then we can calculate its molar mass.
1mol
22.4 L
ρ = M ×
22.4 L
1 mol
M = ρ ×