Transcript M - richardkesslerhfa

Chemical Quantities
Unit 6
Chapter 10, Section 10.2
Mole-Mass and Mole-Volume
Relationships
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).
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
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.
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.
or
M=m
n
m or m = n × M
n =M
– where M is molar mass, m is mass, and n is
number of mols.
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
m = n × M = (3.00 mol)(58.3 g/mol) = 176 g
• Let’s try Sample Problem 10.5 (page 298)
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 = 102.0 g/mol
Unknown:
m = ? g Al2O3
m = n × M = (9.45)(102.0) g
m = 964 g Al2O3
The Mole-Mass Relationship
• For another example:
– The molar mass of Na2SO4 is 142.1 g/mol
– The mols of 10.0 g of Na2SO4 is given by
10.0 g
m
= 7.04 × 10-2 mol
n =M =
142.1 g/mol
• Let’s try Sample Problem 10.6 (page 299)
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 = 159.6 g/mol
Unknown:
n = ? mol Fe2O3
92.9 g
= 0.578 mol
n= m =
M
159.6 g/mol
The Mole-Mass Relationship
• Practice Problems:
m=n×M
1. What is the mass of 2.50 mols of hydrogen gas, H2?
mH2 = nH2 × MH2 = (2.50 mol)(2.02 g/mol) = 5.05 g
2. What is the mass of 0.100 mols of glucose, C6H12O6?
mglucose = nglucose × Mglucose = (0.100 mol)(180 g/mol) = 18.0 g
3. What is the mass of 12.0 mols of water, H2O?
mH2O = nH2O × MH2O = (12.0 mol)(18.0 g/mol) = 216 g
4. What is the mass of 0.250 mols of methane gas,
CH4?
mCH4 = nCH4 × MCH4 = (2.50 mol)(16.0 g/mol) = 4.00 g
The Mole-Mass Relationship
• Practice Problems:
m
n=
M
1. How many mols are in 3.25 g of H2?
nH2 =
mH2
3.25 g
=
= 1.61 mol
MH2 2.02 g/mol
2. How many mols are in 9.00 g of C6H12O6?
nC6H12O6 =
mC6H12O6
9.00 g
=
= 0.0500 mol
MC6H12O6 180 g/mol
3. How many mols are in 100. g of H2O?
nH2O =
mH2O
100.g
=
= 5.56 mol
MH2O 18.0 g/mol
4. How many mols are in 9.60 g of CH4?
nCH4 =
mCH4
9.60 g
=
= 0.600 mol
MCH4 16.0 g/mol
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.
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.
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.”
The Mole-Volume Relationship
• Standard temperature and pressure:
– Standard temperature is 0°C (273.16 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.”
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
L
V = n × 22.4
1 mol
1 mol
or n = V × 22.4
L
– where V is the volume of the gas in liters (L) and n
is the amount of gas in mols
The Mole-Volume Relationship
• For example
– the volume of 1.25 mols of O2 gas at STP is
L = 1.25 mol × 22.4 L = 28.0 L
V = n × 22.4
1 mol
1 mol
– the number of mols of H2 in 67.2 L at STP is
1
mol
1
mol
n = V × 22.4 L = 67.2 L × 22.4 L = 3.00 mol
• Let’s try Sample Problem 10.7 (page 301).
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 = 0.60 mol × 22.4 L
1 mol
1 mol
V = 13 L SO2
The Mole-Volume Relationship
• Practice Problems:
22.4
L
V=n×
1 mol
1. What is the volume of 3.25 mols of H2 at STP?
VH2 = nH2×
22.4 L
22.4 L
= 3.25 mol ×
= 72.8 L
1 mol
1 mol
2. What is the volume of 0.250 mols of O2 at STP?
VO2 = nO2×
22.4 L
22.4 L
= 0.250 mol ×
= 5.60 L
1 mol
1 mol
3. What is the volume of 12.5 mols of Cl2 at STP?
VCl2 = nCl2×
22.4 L
22.4 L
= 12.5 mol ×
= 280. L
1 mol
1 mol
4. What is the volume of 0.0100 mols of CH4?
VCH4 = nCH4×
22.4 L
22.4 L
= 0.0100 mol ×
= 0.224 L
1 mol
1 mol
The Mole-Volume Relationship
• Practice Problems:
1
mol
n=V×
22.4 L
1. How many mols of H2 are in 1.00 L at STP?
nH2 = VH2×
1 mol
1 mol
= 1.00 L ×
= 0.0446 mol
22.4 L
22.4 L
2. How many mols of O2 are in 50.0 L at STP?
nO2 = VO2×
1 mol
1 mol
= 50.0 L ×
= 2.23 mol
22.4 L
22.4 L
3. How many mols of C2H2 are in 0.250 L at STP?
nC2H2 = VC2H2×
1 mol
1 mol
= 0.250 L ×
= 5.60 mol
22.4 L
22.4 L
4. How many mols of F2 are in 11.2 L at STP?
nF2 = VF2×
1 mol
1 mol
= 11.2 L ×
= 0.500 mol
22.4 L
22.4 L
The Mole-Volume Relationship
• Density and Molar Mass
– The density of matter is mass per unit volume.
ρ= m
V
• where ρ = density (g/L), m = mass, and V = volume.
– We can use our mole-mass and mole-volume
relationships to determine density.
L
V = n × 22.4
1 mol
m = n×M
M
=
ρ= V
L
22.4 L
n × 22.4
1 mol
1 mol
m=n×M
= M × 1 mol
22.4 L
The Mole-Volume Relationship
• Density and Molar Mass
– If we know the molar mass of a gas, then we can
calculate its density.
1mol
ρ = M × 22.4
L
– If we know the density of a gas, then we can
calculate its molar mass.
L
M = ρ × 22.4
1 mol
The Mole-Volume Relationship
• Density and Molar Mass
– For example, find the density of H2 at STP.
1mol = 2.02 g/mol × 1mol
ρ = M × 22.4
L
22.4 L = 0.0902 g/L
– For example, find the molar mass of a gas with a
density of 3.17 g/L at STP.
L = 3.17 g/L × 22.4 L
M = ρ × 22.4
1 mol
1 mol = 71.0 g/mol
Sample Problem 10.8
The density of a gaseous compound containing carbon and
oxygen is found to be 1.964 g/L at STP. What is the molar mass of
this compound
Known:
ρ = 1.964 g/L
Unknown:
M = ? g/mol
M = ρ × 22.4 L = 1.964 g/L × 22.4 L
1 mol
1 mol
M = 44.0 g/mol
What do you think the gas is? Why?
The Mole-Volume Relationship
• Practice Problems:
ρ = M × 1 mol
22.4 L
1. What is the density of F2 at STP?
ρF2 = MF2×
1 mol
1 mol
= 38.0 g/mol ×
= 1.70 g/L
22.4 L
22.4 L
2. What is the density of CH4 at STP?
ρCH4 = MCH4×
1 mol
1 mol
= 16.0 g/mol ×
= 0.714 g/L
22.4 L
22.4 L
3. What is the density of Ne at STP?
1 mol
1 mol
ρNe = MNe×
= 20.2 g/mol ×
= 0.902 g/L
22.4 L
22.4 L
4. What is the density of Rn at STP?
ρRn = MRn×
1 mol
1 mol
= 222 g/mol ×
= 9.91 g/L
22.4 L
22.4 L
The Mole-Volume Relationship
• Practice Problems:
M = ρ × 22.4 L
1 mol
1. Find the molar mass of a gas with a density of
L
22.4 L
= 1.34 g/L ×
= 30.0 g/mol
1.34 g/L at STP. M = ρ × 122.4
mol
1 mol
2. Find the molar mass of a gas with a density of
L
22.4 L
= 2.59 g/L ×
= 58.0 g/mol
2.59 g/L at STP. M = ρ × 122.4
mol
1 mol
3. Find the molar mass of a gas with a density of
L
22.4 L
= 1.78 g/L ×
= 39.9 g/mol
1.78 g/L at STP. M = ρ × 122.4
mol
1 mol
4. Find the molar mass of a gas with a density of
L
22.4 L
= 6.52 g/L ×
= 146 g/mol
6.52 g/L at STP. M = ρ × 122.4
mol
1 mol
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.
m or m = n × M
or
=
M=m
n
M
n
– where M is molar mass, m is mass, and n is
number of mols.
Summary
• To compare different gases, we put them at a
Standard Temperature and Pressure.
– This is called “STP.”
• Standard temperature is 0°C (273.16 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
22.4
L
V = n × 1 mol
1
mol
or n = V × 22.4 L
– where V is the volume of the gas in liters (L) and n is
the amount of gas in mols
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.
1mol
ρ = M × 22.4
L
– If we know the density of a gas, then we can
calculate its molar mass.
L
M = ρ × 22.4
1 mol