chapter #8 chemical quantities. chapter outline 8.2 mole relationships defined in balanced equations...

Chapter #8Chemical Quantities

Chapter Outline• 8.2 Mole relationships defined in balanced equations• 8.3 Mole conversions- Stoichiometry • 8.4 Mass Conversions Stoichiometry • 8.5 Excess and Limiting Reactants• 8.5 Theoretical Yield• 8.6 Theoretical Yield From Initial Reactant Masses• 8.7 Thermochemical Equations

8.3-8.4 STOICHIOMETRYStoichiometry is the use of balanced chemical

equations in the conversion process.

8.4 STOICHIOMETRYStoichiometry is the use of balanced chemical

equations in the conversion process.ExamplesCalculate the mass of water formed from 6.33 g of

hydrogen. A balanced equation is required. 2 H2 + O2 2H2O

8.4 STOICHIOMETRYStoichiometry is the use of balanced chemical

equations in the conversion process.ExamplesCalculate the mass of water formed from 6.33 g of

hydrogen. A balanced equation is required. 2 H2 + O2 2H2O

6.33 g H2

8.4 STOICHIOMETRYStoichiometry is the use of balanced chemical

equations in the conversion process.ExamplesCalculate the mass of water formed from 6.33 g of

hydrogen. A balanced equation is required. 2 H2 + O2 2H2O

6.33 g H2

2.016 g H2

mole H2

8.4 STOICHIOMETRYStoichiometry is the use of balanced chemical

equations in the conversion process.ExamplesCalculate the mass of water formed from 6.33 g of

hydrogen. A balanced equation is required. 2 H2 + O2 2H2O

6.33 g H2

2.016 g H2

Mole H2

8.4 STOICHIOMETRYStoichiometry is the use of balanced chemical

equations in the conversion process.ExamplesCalculate the mass of water formed from 6.33 g of

hydrogen. A balanced equation is required. 2 H2 + O2 2H2O

6.33 g H2

2.016 g H2OMole H2 2 mole H2O

2 Mole H2

8.4 STOICHIOMETRYStoichiometry is the use of balanced chemical

equations in the conversion process.ExamplesCalculate the mass of water formed from 6.33 g of

hydrogen. A balanced equation is required. 2 H2 + O2 2 H2O

6.33 g H2

2.016 g H2

Mole H2 2 mole H2O2 Mole H2 Mole H2O

18.02 g H2O

8.4 STOICHIOMETRYStoichiometry is the use of balanced chemical

equations in the conversion process.ExamplesCalculate the mass of water formed from 6.33 g of

hydrogen. A balanced equation is required. 2 H2 + O2 2 H2O

6.33 g H2

2.016 g H2

Mole H2 2 mole H2O2 Mole H2 Mole H2O

18.02 g H2O = 28.3 g H2O

8.5 Excess and Limiting ReactantsReactants are substances that can be changed into something else. For example, nails and boards are reactants for carpenters, while thread and fabric are reactants for the seamstress. And for a chemist hydrogen and oxygen are reactants for making water.

8.5 Building HousesOk, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?

8.5 Building HousesOk, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?Yes, only one house!

8.5 Building HousesOk, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?What reactant is in excess? And how many more houses could we use if we had enough boards?

8.5 Building HousesOk, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?What reactant is in excess? And how many more houses could we use if we have enough boards?

8.5 Building HousesOk, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?What reactant is in excess? And how many more houses could we use if we have enough boards?Yes, nails are in excess!

8.5 Building HousesOk, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?What reactant is in excess? And how many more houses could we use if we have enough boards?Yes, nails are in excess! Nine more houses if we have an adequate amount of boards.

8.5 Excess and Limiting ExampleIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?

8.5 Excess and Limiting ExampleIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?

Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.

8.5 Excess and Limiting ExampleIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?

Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.

2 H2 + O2 2 H2O

10.0 g O2

8.5 Excess and Limiting ExampleIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?

Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.

2 H2 + O2 2 H2O

10.0 g O2

32.0 g O2

mole O2

8.5 Excess and Limiting ExampleIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?

Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.

2 H2 + O2 2 H2O

10.0 g O2

32.0 g O2

mole O2

mole O2

2 mole H2

8.5 Excess and Limiting ExampleIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?

Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.

2 H2 + O2 2 H2O

10.0 g O2

32.0 g O2

mole O2

mole O2

2 mole H2

mole H2

2.02 g H2

8.5 Excess and Limiting ExampleIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?

Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.

2 H2 + O2 2 H2O

10.0 g O2

32.0 g O2

mole O2

mole O2

2 mole H2

mole H2

2.02 g H2 = 1.26 g H2

8.5 Excess and Limiting ExampleOnly 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.

8.5 Excess and Limiting ExampleOnly 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.

10.0 g O2mole O2

32.0 g O2

8.5 Excess and Limiting ExampleOnly 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.

10.0 g O2mole O2

32.0 g O2 mole O2

2 mole H2O

8.5 Excess and Limiting ExampleOnly 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.

10.0 g O2mole O2

32.0 g O2 mole O2

2 mole H2O 18.0 g H2Omole H2O

8.5 Excess and Limiting ExampleOnly 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.

10.0 g O2mole O2

32.0 g O2 mole O2

2 mole H2O 18.0 g H2Omole H2O

= 11.3 g H2O

8.5 Percentage YieldThe percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield?

8.5 Percentage YieldThe percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield?

percent yield = Yield (the lab amount)Theoretical Yield (by conversions)

X 100

percent yield = 8.6611.3

X 100 = 76.6%

8.7 Thermochemistry

Exothermic Reaction Endothermic Reaction

- ΔH + ΔH

Eact

8.7 ThermochemistryWhen a chemical or physical change takes place energy is either lost of gained. A Thermochemical equation describes this change. Equations gaining energy are called endothermic and equations losing energy are called exothermic.

8.7 Thermochemical EquationsWhen a chemical or physical change takes place energy is either lost of gained. A Thermochemical equation describes this change. Equations gaining energy are called endothermic and equations losing energy are called exothermic.

Examples:

C3H6O (l ) + 4O2 (g) 3CO2(g) + 3 H2O (g) ΔH = -1790 kjExothermic

H2O (l) H2O (g) ΔH = 44.01 kjEndothermic

8.7 Thermochemical EquationsHow many kj of heat are released when 709 g of C3H6O are burned?

8.7 Thermochemical EquationsHow many kj of heat are released when 709 g of C3H6O are burned?

709 g C3H6O

C3H6O (l ) + 4O2 (g) 3CO2(g) + 3 H2O (g) ΔH = -1790 kj

8.7 Thermochemical EquationsHow many kj of heat are released when 709 g of C3H6O are burned?

709 g C3H6O

C3H6O (l ) + 4O2 (g) 3CO2(g) + 3 H2O (g) ΔH = -1790 kj

58.1 g C3H6O

mole C3H6O

8.7 Thermochemical EquationsHow many kj of heat are released when 709 g of C3H6O are burned?

709 g C3H6O

C3H6O (l ) + 4O2 (g) 3CO2(g) + 3 H2O (g) ΔH = -1790 kj

58.1 g C3H6O

mole C3H6O

8.7 Thermochemical EquationsHow many kj of heat are released when 709 g of C3H6O are burned?

709 g C3H6O

C3H6O (l ) + 4O2 (g) 3CO2(g) + 3 H2O (g) ΔH = -1790 kj

58.1 g C3H6O

mole C3H6O

mole C3H6O

1790 kj= 21800 kj

The End