Chemical Quantities

Chemical Quantities

Unit 6Chapter 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.

– where M is molar mass, m is mass, and n is number of mols.

M = nm n = M

m m = n × Mor or


• 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

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 molM = [(2 × 27.0) + (3 × 16.0)] g/molM = 102.0 g/molUnknown:m = ? g Al2O3

m = n × M = (9.45)(102.0) gm = 964 g Al2O3


• For another example:– The molar mass of Na2SO4 is 142.1 g/mol

– The mols of 10.0 g of Na2SO4 is given by

• Let’s try Sample Problem 10.6 (page 299)

= 7.04 × 10-2 mol n = Mm = 142.1 g/mol

10.0 g

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 gM = [(2 × 55.8) + (3 × 16.0)] g/molM = 159.6 g/molUnknown:n = ? mol Fe2O3

n = mM

92.9 g159.6 g/mol= = 0.578 mol


• Practice Problems:1. What is the mass of 2.50 mols of hydrogen gas, H2?

2. What is the mass of 0.100 mols of glucose, C6H12O6?

3. What is the mass of 12.0 mols of water, H2O?

4. What is the mass of 0.250 mols of methane gas, CH4?

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

m = n × M

• Practice Problems:1. How many mols are in 3.25 g of H2?

2. How many mols are in 9.00 g of C6H12O6?

3. How many mols are in 100. g of H2O?

4. How many mols are in 9.60 g of CH4?

The Mole-Mass Relationshipn = m

M

nH2 = = = 1.61 mol mH2 3.25 gMH2 2.02 g/mol

nC6H12O6 = = = 0.0500 mol mC6H12O6 9.00 gMC6H12O6 180 g/mol

nH2O = = = 5.56 mol mH2O 100.gMH2O 18.0 g/mol

nCH4 = = = 0.600 mol mCH4 9.60 gMCH4 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.


• 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.


• 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.”


• 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.”


• 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 L1 mol or n = V × 1 mol

22.4 L


• 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 L1 mol = 1.25 mol × 22.4 L

1 mol = 28.0 L

n = V × 1 mol22.4 L = 67.2 L × 1 mol

22.4 L = 3.00 mol

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 molUnknown:V = ? L SO2

V = 13 L SO2

V = n × 22.4 L1 mol = 0.60 mol × 22.4 L

1 mol

• Practice Problems:1. What is the volume of 3.25 mols of H2 at STP?

2. What is the volume of 0.250 mols of O2 at STP?

3. What is the volume of 12.5 mols of Cl2 at STP?

4. What is the volume of 0.0100 mols of CH4?

The Mole-Volume RelationshipV = n × 22.4 L

1 mol

VH2 = nH2× = 3.25 mol × = 72.8 L22.4 L 22.4 L1 mol 1 mol

VO2 = nO2× = 0.250 mol × = 5.60 L22.4 L 22.4 L1 mol 1 mol

VCl2 = nCl2× = 12.5 mol × = 280. L22.4 L 22.4 L1 mol 1 mol

VCH4 = nCH4× = 0.0100 mol × = 0.224 L22.4 L 22.4 L1 mol 1 mol

• Practice Problems:1. How many mols of H2 are in 1.00 L at STP?

2. How many mols of O2 are in 50.0 L at STP?

3. How many mols of C2H2 are in 0.250 L at STP?

4. How many mols of F2 are in 11.2 L at STP?

The Mole-Volume Relationshipn = V × 1 mol

22.4 L

nH2 = VH2× = 1.00 L × = 0.0446 mol1 mol 1 mol22.4 L 22.4 L

nO2 = VO2× = 50.0 L × = 2.23 mol1 mol 1 mol22.4 L 22.4 L

nC2H2 = VC2H2× = 0.250 L × = 5.60 mol1 mol 1 mol22.4 L 22.4 L

nF2 = VF2× = 11.2 L × = 0.500 mol1 mol 1 mol22.4 L 22.4 L


• 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 = mV

m = n × M 22.4 L1 molV = n ×

22.4 Ln × M

1 moln × = 22.4 L

1 mol

M = M × 22.4 L1 molm

V =ρ =


• 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.

1mol22.4 Lρ = M ×

22.4 L1 molM = ρ ×


• 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.

1mol22.4 Lρ = M × 1mol

22.4 L= 2.02 g/mol × = 0.0902 g/L

22.4 L1 molM = ρ × 22.4 L

1 mol= 3.17 g/L × = 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 compoundKnown:ρ = 1.964 g/LUnknown:M = ? g/mol

M = 44.0 g/mol

M = ρ × 22.4 L1 mol = 1.964 g/L × 22.4 L

1 mol

What do you think the gas is? Why?

• Practice Problems:1. What is the density of F2 at STP?

2. What is the density of CH4 at STP?

3. What is the density of Ne at STP?

4. What is the density of Rn at STP?

The Mole-Volume Relationshipρ = M × 1 mol

22.4 L

ρF2 = MF2× = 38.0 g/mol × = 1.70 g/L1 mol 1 mol22.4 L 22.4 L

ρCH4 = MCH4× = 16.0 g/mol × = 0.714 g/L1 mol 1 mol22.4 L 22.4 L

ρNe = MNe× = 20.2 g/mol × = 0.902 g/L1 mol 1 mol22.4 L 22.4 L

ρRn = MRn× = 222 g/mol × = 9.91 g/L1 mol 1 mol22.4 L 22.4 L

• Practice Problems:1. Find the molar mass of a gas with a density of 1.34

g/L at STP.2. Find the molar mass of a gas with a density of 2.59



g/L at STP.

The Mole-Volume RelationshipM = ρ × 22.4 L

1 mol

M = ρ × = 1.34 g/L × = 30.0 g/mol22.4 L 22.4 L1 mol 1 mol



M = ρ × = 6.52 g/L × = 146 g/mol22.4 L 22.4 L1 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.

– where M is molar mass, m is mass, and n is number of mols.

M = nm n = M

m m = n × Mor or

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

– where V is the volume of the gas in liters (L) and n is the amount of gas in mols

V = n × 22.4 L1 mol or n = V × 1 mol

22.4 L

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.

1mol22.4 Lρ = M ×

22.4 L1 molM = ρ ×

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