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Chapter 6B

Chemical

Quantities

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CHAPTER OUTLINE���

§  The Mole Concept §  Molar Mass §  Calculations Using the Mole §  Stoichiometry & Molar Ratios §  Mole-Mole & Mass-Mole Calculations §  Mass-Mass Calculations §  Limiting Reactant §  Percent Yield §  Energy Changes in Chemical Reactions

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THE MOLE���CONCEPT���

q  Chemists find it more convenient to use mass relationships in the laboratory, while chemical reactions depend on the number of atoms present.

q  In order to relate the mass and number of atoms, chemists use the SI unit mole (abbreviated mol).

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THE MOLE���CONCEPT���

q  The number of particles in a mole is called Avogadro’s number and is 6.02x1023.

1 mole equals to 6.02 x 1023

Avogadro’s number (NA)

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THE MOLE���CONCEPT���

A mole is a very large quantity 6.02x1023

If 10,000 people started to count Avogadro’s number and counted at the rate of 100 numbers per minute each

minute of the day, it would take over 1 trillion years to count the total number.

MOLE

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THE MOLE���CONCEPT���

1 mole H2 molecules = 6.02x1023 H2 molecules = 2 x (6.02x1023) H atoms

1 mole Na+ ions

1 mole H2O molecules = 6.02x1023 H2O molecules = 2 x (6.02x1023) H atoms = 6.02x1023 O atoms

= 6.02x1023 Na+ ions

1 mole H atoms = 6.02x1023 H atoms

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THE MOLE���CONCEPT���

q  The atomic mass of one atom expressed in amu is numerically the same as the mass of 1 mole of atoms of the element expressed in grams.

Mass of 1 H atom = 1.008 amu

Mass of 1 mol H atoms = 1.008 grams

Mass of 1 Mg atom = 24.31 amu

Mass of 1 mol Mg atoms = 24.31 g

Mass of 1 Cl atom = 35.45 amu

Mass of 1 mol Cl atoms = 35.45 g

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MOLAR MASS���

1 mol O atom =

2 mol H atoms =

Mass of one mole of H2O

2 (1.01 g) = 2.02 g

1 (16.00 g) = 16.00 g

18.02 g Molar mass

q  The mass of one mole of a substance is called molar mass, and is measured in grams.

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MOLAR MASS���

2 mol O atoms = 2 mol H atoms =

Mass of one mole of Ca(OH)2

2 (1.01 g) = 2.02 g 2 (16.00 g) = 32.00 g

74.10 g Molar mass

1 mol Ca atom = 1 (40.08 g) = 40.08 g

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Calculate the molar mass of each compound shown below:

Example 1:���

Lithium carbonate Li2CO3

Li 2 x 6.94 = 13.88 g C 1 x 12.01 = 12.01 g O 3 x 16.00 = 48.00 g

Molar mass = 73.89 g/mol

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Calculate the molar mass of each compound shown below:

Example 2:���

Salicylic acid C7 H6O3

C 7 x 12.01 = 84.07 g H 6 x 1.01 = 6.06 g O 3 x 16.00 = 48.00 g

Molar mass = 138.13 g/mol

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CALCULATIONS���USING THE MOLE���

q  Conversions between mass, mole and particles can be done using molar mass and Avogadro’s number.

Mass of a substance

Moles of a substance MM

Particles of a substance NA

Molar mass

Avogadro’s number

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How many moles of iron are present in 25.0 g of iron?

Example 1:���

3 significant figures

mole25.0 g Fe x mol Fe g

=55.85 1

Molar mass

0.448

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What is the mass of 5.00 mol of water?

Example 2:���

3 significant figures

2 2 g5.00 mol H O x g H O mol

=1 18.02 90.1

Molar mass

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How many Mg atoms are present in 5.00 g of Mg?

Example 3:���

mol5.00 g Mg x g1

24.30

mass → mol → atoms

atomsx = mol

Avogadro’s number

1 6.02 x 1023

1.24x1023 atoms Mg Molar mass 3 significant figures

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How many molecules of HCl are present in 25.0 g of HCl?

Example 4:���

mol25.0 g HCl x g

1

36.46

mass → mol → molecules

moleculesx = mol1

6.02 x 1023

4.13 x 1023 molecules HCl

3 significant figures

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MOLES OF ELEMENTS���IN A FORMULA���

q  The subscripts in a chemical formula of a compound indicate the number of atoms of each type of element.

q  For example, 1 molecule of aspirin contains:

C9H8O4

9 C atoms 8 H atoms

4 O atoms

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MOLES OF ELEMENTS���IN A FORMULA���

q  The subscripts also indicate the number of moles of each element in one mole of the compound.

q  For example, 1 mole of aspirin contains:

C9H8O4

9 mole C atoms

8 mole H atoms

4 mole O atoms

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MOLES OF ELEMENTS���IN A FORMULA���

q  Using the subscripts from the aspirin formula, one can write the following conversion factors for each of the elements in 1 mole of aspirin.

C9H8O4

9 moles C 1 mole C9H8O4

8 moles H 1 mole C9H8O4

4 moles O 1 mole C9H8O4

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Determine the number of moles of C atoms in 1 mole of each of the following substances:

Example 1:���

a) Acetoaminophen used in Tylenol C8 H9NO2

8 moles C 1 mole C8H9 NO2

b) Zinc dietary supplement Zn(C2 H3O2)2

4 moles C 1 mole Zn(C2 H3O2)2

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How many carbon atoms are present in 1.50 moles of aspirin, C9H8O4?

Example 2:���

9 moles C 1 mole C9H8 O4

1.50 mol C9H8O4 x x 6.02x1023 atoms 1 mole C

= 8.13x1024 atoms

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SUMMARY OF ���MASS-MOLE CALCULATIONS���

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STOICHIOMETRY���

q  Stoichiometry is the quantitative relationship between the reactants and products in a balanced chemical equation.

q  A balanced chemical equation provides several important information about the reactants and products in a chemical reaction.

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MOLAR���RATIOS���

For example:

1 N2 (g) + 3 H2 (g) → 2 NH3 (g)

1 molecule 3 molecules 2 molecules

100 molecules 300 molecules 200 molecules

106 molecules 3x106 molecules 2x106 molecules

1 mole 3 moles 2 moles

This is the molar ratios between

the reactants and products

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Example 1:���

2

2

mol O =mol CO

13 8

4 10

2

mol C H =mol H O

Determine each mole ratio below based on the reaction shown:

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

2 10

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STOICHIOMETRIC���CALCULATIONS���

q  Stoichiometric calculations can be classified as one of the following:

MOLES of compound A

MOLES of compound B

molar ratio

Mole-mole calculations

MASS of compound A

MM

Mass-mole calculations

MASS of compound B

MM

Mass-mass calculations

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MOLE-MOLE���CALCULATIONS���

q  Relates moles of reactants and products in a balanced chemical equation

MOLES of compound A

MOLES of compound B

molar ratio

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How many moles of nitrogen will react with 2.4 moles of hydrogen to produce ammonia as shown in the reaction below?

Example 1:���

1 N2 (g) + 3 H2 (g) → 2 NH3 (g)

2.4 mol H2 2

2

mol Nx mol H

1

3 = 0.80 mol N2

Mole ratio

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How many moles of ammonia can be produced from 32 moles of hydrogen? (Assume excess N2 present)

Example 2:���

1 N2 (g) + 3 H2 (g) → 2 NH3 (g)

32 mol H2 3

2

mol NHx mol H2 3 = 21 mol NH3

Mole ratio

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In one experiment, 6.80 mol of ammonia are prepared. How many moles of hydrogen were used up in this experiment?

Example 3:���

1 N2 (g) + 3 H2 (g) → 2 NH3 (g)

6.80 mol NH3 2

3

mol Hx mol NH

3 2 = 10.2 mol H2

Mole ratio

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MASS-MOLE���CALCULATIONS���

q  Relates moles and mass of reactants or products in a balanced chemical equation

MOLES of compound A

MOLES of compound B

molar ratio

MASS of compound A

MM

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How many grams of ammonia can be produced from the reaction of 1.8 moles of nitrogen with excess hydrogen as shown below?

Example 1:���

3

2

mol NHx mol N

1 N2 (g) + 3 H2 (g) → 2 NH3 (g)

1.8 mol N2 2 1 = 61 g NH3

Mole ratio

3

3

g NHx mol NH1 17.04

Molar mass

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How many moles of hydrogen gas are required to produce 75.0 g of ammonia?

Example 2:���

3

3

mol NHx g NH

1 N2 (g) + 3 H2 (g) → 2 NH3 (g)

75.0 g NH3 1

17.04 = 6.60 mol H2 2

3

mol Hx mol NH2 3

Molar mass Mole ratio

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How many moles of ammonia can be produced from the reaction of 125 g of nitrogen?

Example 3:���

2

2

mol Nx g N

1 N2 (g) + 3 H2 (g) → 2 NH3 (g)

125 g N2 1

28.02 = 8.92 mol NH3 3

2

mol NHx mol N1 2

Molar mass Mole ratio

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MASS -MASS���CALCULATIONS���

q  Relates mass of reactants and products in a balanced chemical equation

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What mass of oxygen will be required to react completely with 96.1 g of propane, C3H8, according to the equation below?

Example 1:���

1 C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)

Mass of propane

Moles of propane

Moles of oxygen

Mass of oxygen

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Example 1:���

3 8

3 8

mol C Hx g C H

2

3 8

mol Ox mol C H

1 C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)

96.1 g C3H8 44.11 1

= 349 g O2

1 5

2

2

g Ox mol O

Molar mass Mole ratio

32.00 1

Molar mass

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What mass of carbon dioxide will be produced from the reaction of 175 g of propane, as shown?

Example 2:���

1 C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)

Mass of propane

Moles of propane

Moles of carbon dioxide

Mass of carbon dioxide

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Example 2:���

3 8

3 8

mol C Hx g C H

2

3 8

mol COx mol C H

1 C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)

175 g C3H8 44.11 1

= 524 g CO2

1 3

2

2

g COx mol CO

Molar mass Mole ratio

44.01 1

Molar mass

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LIMITING���REACTANT���

q  When 2 or more reactants are combined in non-stoichiometric ratios, the amount of product produced is limited by the reactant that is not in excess.

q  This reactant is referred to as limiting reactant. q  When doing stoichiometric problems of this

type, the limiting reactant must be determined first before proceeding with the calculations.

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LIMITING REACTANT ANALOGY���

Consider the following recipe for a sundae:

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LIMITING REACTANT ANALOGY���

How many sundaes can be prepared from the following ingredients: The number of sundaes possible is limited by the amount of syrup, the limiting reactant.

Limiting reactant

Excess reactants

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GUIDE TO���LIMITING REACTANT CALC.���

Assume reactant 1 is limiting

Assume reactant 2 is limiting

Compare

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How many moles of H2O can be produced by reacting 4.0 mol of hydrogen and 3.0 mol of oxygen gases as shown below:

Example 1:���

2 H2 (g) + 1 O2 (g) → 2 H2O (g)

Limiting reactant

Mole-mole calculations

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Example 1:���

2

2

mol H Ox mol H

4.0 mol H2 2 2

= 4.0 mol H2O

2 H2 (g) + 1 O2 (g) → 2 H2O (g)

Assume H2 is LR

Assume O2 is LR

3.0 mol O2 2

2

mol H Ox mol O2 1

= 6.0 mol H2O

Correct answer

Correct assumption

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A fuel mixture used in the early days of rocketry was a mixture of N2H4 and N2O4, as shown below. How many grams of N2 gas is produced when 100. g of N2H4 and 200. g of N2O4 are mixed?

Example 2:���

2 N2H4 (l) + 1 N2O4 (l) → 3 N2 (g) + 4 H2O (g)

Limiting reactant

Mass-mass calculations

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Example 2:���

Assumes N2H4 is LR

Assumes N2O4 is LR

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Example 2:���

2 4

2 4

mol N Hx g N H

2

2 4

mol Nx = mol N H100. g N2H4 32.06

4.68 mol N2

2 3

2 N2H4 (l) + 1 N2O4 (l) → 3 N2 (g) + 4 H2O (g)

Assume N2H4 is LR

1

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Example 2:���

2 4

2 4

mol N Ox g N O

2

2 4

mol Nx = mol N O200. g N2O4 92.02

6.52 mol N2

1 3

2 N2H4 (l) + 1 N2O4 (l) → 3 N2 (g) + 4 H2O (g)

Assume N2O4 is LR

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Example 2:���

6.52 mol N2

2 N2H4 (l) + 1 N2O4 (l) → 3 N2 (g) + 4 H2O (g)

Assume N2O4 is LR

Assume N2H4 is LR 4.68 mol N2

Correct amount N2H4 is

LR

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Example 2:���

2

2

g Nx mol N4.68 mol N2

28.02 = 131 g N2

2 N2H4 (l) + 1 N2O4 (l) → 3 N2 (g) + 4 H2O (g)

Calculate mass of N2

1

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How many moles of Fe3O4 can be produced by reacting 16.8 g of Fe with 10.0 g of H2O as shown below:

Example 3:���

3 Fe (s) + 4 H2O (l) → Fe3O4 (s) + 4 H2 (g)

Limiting reactant

Mass-mol calculations

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Example 3:���

mol Fex g Fe

3 4 mol Fe Ox = mol Fe

16.8 g Fe

0.100 mol Fe3O4

3 1

Assume Fe is LR

55.85 1

3 Fe (s) + 4 H2O (l) → Fe3O4 (s) + 4 H2 (g)

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Example 3:���

2

2

mol H Ox g H O

3 4

2

mol Fe Ox = mol H O10.0 g H2O

0.139 mol Fe3O4

4 1

Assume H2O is LR

18.02 1

3 Fe (s) + 4 H2O (l) → Fe3O4 (s) + 4 H2 (g)

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Example 3:���

0.139 mol Fe3O4 Assume H2O is LR

Assume Fe is LR 0.100 mol Fe3O4

Correct amount Fe is

LR

3 Fe (s) + 4 H2O (l) → Fe3O4 (s) + 4 H2 (g)

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How many grams of AgBr can be produced when 50.0 g of MgBr2 is mixed with 100.0 g of AgNO3, as shown below:

Example 4:���

MgBr2 + 2 AgNO3 → 2 AgBr + Mg(NO3)2

Limiting Reactant

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Example 4:���

2

2

mol MgBrx g MgBr 2

mol AgBrx mol MgBr50.0 g MgBr2 184.10 1 2

Assume MgBr2 is LR

1

MgBr2 + 2 AgNO3 → 2 AgBr + Mg(NO3)2

g AgBrx = mol AgBr

187.77 1 102 g AgBr

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Example 4:���

3

3

mol AgNOx g AgNO 3

mol AgBrx mol AgNO100.0 g AgNO3 169.88 2 2

Assume AgNO3 is LR

1

MgBr2 + 2 AgNO3 → 2 AgBr + Mg(NO3)2

g AgBrx = mol AgBr

187.77 1 111 g AgBr

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Example 4:���

111 g AgBr Assume AgNO3 is LR

Assume MgBr2 is LR 102 g AgBr

Correct amount MgBr2

is LR

MgBr2 + 2 AgNO3 → 2 AgBr + Mg(NO3)2

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PERCENT YIELD���

q  The amount of product calculated through stoichiometric ratios are the maximum amount product that can be produced during the reaction, and is thus called theoretical yield.

q  The actual yield of a product in a chemical reaction is the actual amount obtained from the reaction.

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PERCENT YIELD���

q  The percent yield of a reaction is obtained as follows:

Actual yield x100 = Percent yieldTheoretical yield

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In an experiment forming ethanol, the theoretical yield is 50.5 g and the actual yield is 46.8 g. What is the percent yield for this reaction?

Example 1:���

Actual yield% yield = x100Theoretical yield

46.8 g= x100 =50.5 g 92.7 %

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Silicon carbide can be formed from the reaction of sand (SiO2) with carbon as shown below:

Example 2:���

1 SiO2 (s) + 3 C (s) → 1 SiC (s) + 2 CO (g)

When 125 g of sand are processed, 68.4 of SiC is produced. What is the percent yield of SiC in this reaction?

Actual yield

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Example 2:���

2

2

mol SiOx g SiO 2

mol SiCx mol SiO125 g SiO2 60.09

83.4 g SiC

1 1

Calculate theoretical yield

1

1 SiO2 (s) + 3 C (s) → 1 SiC (s) + 2 CO (g)

g SiCx = mol SiC1

40.10

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Example 2:���

Actual yield% yield = x100Theoretical yield

68.4 g= x100 =83.4 g 82.0 %

Calculate percent yield

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In an experiment to prepare aspirin, the theoretical yield is 153.7 g and the actual yield is 124.0 g. What is the percent yield of this reaction?

Example 3:���

Actual yield% yield = x100Theoretical yield

124.0 g= x100 =153.7 g 80.68 %

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Carbon tetrachloride (CCl4) was prepared by reacting 100.0 g of Cl2 with excess carbon disulfide (CS2), as shown below. If 65.0 g was prepared in this reaction, calculate the percent yield.

Example 4:���

CS2 + 3 Cl2 → CCl4 + S2Cl2

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Example 4:���

2

2

mol Clx g Cl

4

2

mol CClx mol Cl100.0 g Cl2 70.90 3 1

Calculate theoretical yield

1

4

4

g CClx = mol CCl1

153.81

CS2 + 3 Cl2 → CCl4 + S2Cl2

72.31 g

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Example 4:���

Actual yield% yield = x100Theoretical yield

65.0 g= x100 =72.31 g

89.9 %

Calculate percent yield

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ENERGY CHANGES���IN CHEMICAL REACTIONS���

q  Heat is energy change that is lost or gained when a chemical reaction takes place.

q  The system is the reactants and products that we are observing. The surroundings are all the things that contain and interact with the system, such as the reaction flask, the laboratory room and the air in the room.

q  The direction of heat flow depends whether the products in a reaction have more or less energy than the reactants.

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ENERGY CHANGES���IN CHEMICAL REACTIONS���

q  For a chemical reaction to occur, the molecules of the reactants must collide with each other with the proper energy and orientation.

q  The minimum amount of energy required for a chemical reaction to occur is called the activation energy.

q  The heat of reaction is the amount of heat absorbed or released during a reaction and is designated by the symbol ΔH.

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ENERGY CHANGES���IN CHEMICAL REACTIONS���

q  When heat is released during a chemical reaction, it is said to be exothermic.

q  For exothermic reactions, ΔH is negative and is included on the right side of the equation.

3 H2 (g) + N2 (g) 2 NH3 (g) + 22.0 kcal ΔH= -22.0 kcal

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ENERGY CHANGES���IN CHEMICAL REACTIONS���

q  When heat is gained during a chemical reaction, it is said to be endothermic.

q  For endothermic reactions, ΔH is positive and is included on the left side of the equation.

2 H2O (l) + 137 kcal 2 H2 (g) + O2 (g) ΔH= +137 kcal

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CALCULATING HEAT���IN A REACTION���

q  The value of ΔH in a chemical reaction refers to the heat lost or gained for the number of moles of reactants and products in a balanced chemical equation.

q  For example, based on the chemical equation shown below:

2 H2O (l) 2 H2 (g) + O2 (g) ΔH= +137 kcal

137 kcal 2 mol H2

or 137 kcal 2 mol H2O

or 137 kcal 1 mol O2

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Given the reaction shown below, how much heat, in kJ, is released when 50.0 g of NH3 form?

Example 1:���

3 H2 (g) + N2 (g) 2 NH3 (g) ΔH= -92.2 kJ

50.0 g NH3 x 1 mol NH3

17.04 g NH3 92.2 kJ

2 mol NH3 x = 135 kJ

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How many kJ of heat are needed to react 25.0 g of HgO according to the reaction shown below:

Example 2:���

2 HgO (s) 2 Hg (l) + O2 (g) ΔH= 182 kJ

25.0 g HgO x 1 mol HgO

216.59 g HgO 182 kJ

2 mol HgO x = 10.5 kJ

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THE END