Chapter 8Chapter 8

Quantities in Chemical Quantities in Chemical ReactionsReactions

Day 1Day 1Dec. 12Dec. 12thth

Section: 8.1-8.4Section: 8.1-8.4

Warm-upWarm-up

Write the balanced equation for Write the balanced equation for when Sodium hydroxide reacts with when Sodium hydroxide reacts with copper(II) sulfate.copper(II) sulfate.

Write the balanced equation for Write the balanced equation for when calcium nitrate and when calcium nitrate and magnesium chloride react.magnesium chloride react.

8.1- Global warming: Too much 8.1- Global warming: Too much Carbon DioxideCarbon Dioxide

The amount of every substance used and The amount of every substance used and made in a chemical reaction is related to made in a chemical reaction is related to the amounts of all the other substances in the amounts of all the other substances in the reaction.the reaction. Law of Conservation of Mass.Law of Conservation of Mass. Balancing equations by balancing atoms.Balancing equations by balancing atoms.

The study of the numerical relationship The study of the numerical relationship between chemical quantities in a chemical between chemical quantities in a chemical reaction is called reaction is called stoichiometrystoichiometry..

8.2- Making Pancakes8.2- Making Pancakes The number of pancakes you can make depends The number of pancakes you can make depends

on the amount of the ingredients you use.on the amount of the ingredients you use.

• This relationship can be expressed mathematically.1 cup flour 2 eggs ½ tsp baking powder 5 pancakes

1 cup flour + 2 eggs + ½ tsp baking powder 5 pancakes

Making Pancakes, Making Pancakes, ContinuedContinued

If you want to make more or less than 5 If you want to make more or less than 5 pancakes, you can use the number of eggs pancakes, you can use the number of eggs you have to determine the number of you have to determine the number of pancakes you can make.pancakes you can make. Assuming you have enough flour and baking Assuming you have enough flour and baking

powder.powder.

pancakes 20 eggs 2

pancakes 5eggs 8

Relating Pancakes to Making Relating Pancakes to Making MoleculesMolecules

Mole-to-Mole ConversionsMole-to-Mole Conversions The balanced equation is the “recipe” for a The balanced equation is the “recipe” for a

chemical reaction.chemical reaction. The equation 3 HThe equation 3 H22((gg) + N) + N22((gg) ) 2 NH 2 NH33((gg) tells ) tells

us that 3 molecules of Hus that 3 molecules of H22 react with exactly 1 react with exactly 1 molecule of Nmolecule of N22 and make exactly 2 molecules and make exactly 2 molecules of NHof NH3 3 or:or:

3 molecules H3 molecules H22 1 molecule N 1 molecule N2 2 2 molecules 2 molecules NHNH33

Since we count molecules by moles:Since we count molecules by moles:

3 moles H3 moles H22 1 mole N 1 mole N2 2 2 moles NH 2 moles NH33

Example Two:Example Two: According to the following equation, According to the following equation,

how many moles of water are made how many moles of water are made in the combustion of 0.10 moles of in the combustion of 0.10 moles of glucose?glucose?

CC66HH1212OO66 + 6 O + 6 O22 6 CO 6 CO22 + 6 H + 6 H22OO Solution = Solution =

Day #2Day #2Dec 13Dec 13thth

Section 8.3Section 8.3

Warm up 12/13Warm up 12/13

Worksheet 8.1 # 1 and 3Worksheet 8.1 # 1 and 3

1212

8.3- Making Molecules8.3- Making MoleculesMass-to-Mass ConversionsMass-to-Mass Conversions

We know there is a relationship between the We know there is a relationship between the mass and number of moles of a chemical.mass and number of moles of a chemical.

1 mole = Molar Mass in grams.1 mole = Molar Mass in grams. The molar mass of the chemicals in the reaction The molar mass of the chemicals in the reaction

and the balanced chemical equation allow us to and the balanced chemical equation allow us to convert from the amount of any chemical in the convert from the amount of any chemical in the reaction to the amount of any other.reaction to the amount of any other.

Example 2 - How Many Grams of OExample 2 - How Many Grams of O22 Can Be Can Be Made from the Decomposition of 100.0 g of Made from the Decomposition of 100.0 g of

PbOPbO22??2 PbO2 PbO22((ss)) →→ 2 PbO(2 PbO(ss) + O) + O22((gg))

(Given these formula masses:PbO(Given these formula masses:PbO2 2 == 239.2, O239.2, O22 = 32.00) = 32.00)

Day 3Day 3

Check and Review WS 8.1-8.2Check and Review WS 8.1-8.2

Warm up 12/14Warm up 12/14

WS 8.1 #5,7, & 9WS 8.1 #5,7, & 9

Day 4Day 4

Check WS 8.3Check WS 8.3

Quiz Section 8.1-8.4Quiz Section 8.1-8.4

Warm-up 12/15Warm-up 12/15

Have worksheet 8.3 outHave worksheet 8.3 out Get a periodic table, calculator, Get a periodic table, calculator,

pencil outpencil out

Day 5Day 5

Section 8.5 and 8.6Section 8.5 and 8.6

Limiting reactantsLimiting reactants

Warm-up 12/16Warm-up 12/16

Given the following reaction, how Given the following reaction, how many grams of silver will be made if many grams of silver will be made if 4.3g of silver nitrate were used?4.3g of silver nitrate were used?

Silver nitrate and copper produce, Silver nitrate and copper produce, silver and copper nitratesilver and copper nitrate

More Making PancakesMore Making Pancakes We know that:We know that:

• But what would happen if we had 3 cups of flour,

10 eggs, and 4 tsp of baking powder?

1 cup flour + 2 eggs + ½ tsp baking powder 5 pancakes

More Making Pancakes, More Making Pancakes, ContinuedContinued

pancakes 25 eggs 2

pancakes 5eggs 10

pancakes 15 flour cups 1

pancakes 5flour cups 3

pancakes 40 powder baking tsp2

1pancakes 5

powder baking tsp4

2323

More Making Pancakes, More Making Pancakes, ContinuedContinued Each ingredient could potentially make a Each ingredient could potentially make a

different number of pancakes.different number of pancakes. But all the ingredients have to work together!But all the ingredients have to work together! We only have enough flour to make 15 We only have enough flour to make 15

pancakes, so once we make 15 pancakes, pancakes, so once we make 15 pancakes, the flour runs out no matter how much of the the flour runs out no matter how much of the other ingredients we have.other ingredients we have.

More Making Pancakes, More Making Pancakes, ContinuedContinued

The flour The flour limits the amount of limits the amount of pancakespancakes we can make. In chemical we can make. In chemical reactions we call this the reactions we call this the limiting limiting reactantreactant.. Also known as limiting reagent.Also known as limiting reagent.

The maximum number of pancakes we can The maximum number of pancakes we can make depends on this ingredient. In make depends on this ingredient. In chemical reactions, we call this the chemical reactions, we call this the theoretical yieldtheoretical yield.. It also determines the amounts of the other It also determines the amounts of the other

ingredients we will use!ingredients we will use!

Example 2 - How Many Moles of SiExample 2 - How Many Moles of Si33NN44 Can Be Made from 1.20 Moles of Si and Can Be Made from 1.20 Moles of Si and

1.00 Moles of N1.00 Moles of N22 in the Reaction in the Reaction 3 Si + 2 N 3 Si + 2 N22 Si Si33NN44??

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More Making PancakesMore Making Pancakes Let’s now assume that as we are making pancakes, Let’s now assume that as we are making pancakes,

we spill some of the batter, burn a pancake, drop we spill some of the batter, burn a pancake, drop one on the floor, or other uncontrollable events one on the floor, or other uncontrollable events happen so that we only make 11 pancakes. The happen so that we only make 11 pancakes. The actual amount of product made in a chemical actual amount of product made in a chemical reaction is called the reaction is called the actual yieldactual yield..

We can determine the efficiency of making We can determine the efficiency of making pancakes by calculating the percentage of the pancakes by calculating the percentage of the maximum number of pancakes we actually make. maximum number of pancakes we actually make. In chemical reactions, we call this the In chemical reactions, we call this the percent percent yieldyield..

YieldPercent 100Yield lTheoretica

Yield Actual % %% 37100

pancakes 15

pancakes 11

Theoretical and Actual Theoretical and Actual YieldYield As we did with the pancakes, in order to As we did with the pancakes, in order to

determine the theoretical yield, we should determine the theoretical yield, we should use reaction stoichiometry to determine use reaction stoichiometry to determine the amount of product each of our the amount of product each of our reactants could make. reactants could make.

The theoretical yield will always be the The theoretical yield will always be the least possible amount of product.least possible amount of product. The theoretical yield will always come from the The theoretical yield will always come from the

limiting reactant.limiting reactant. Because of both controllable and Because of both controllable and

uncontrollable factors, uncontrollable factors, the actual yield of the actual yield of product will always be less than the product will always be less than the theoretical yield.theoretical yield.

Measuring Amounts in the Measuring Amounts in the LabLab

In the lab, our balances do not measure In the lab, our balances do not measure amounts in moles, unfortunately, they amounts in moles, unfortunately, they measure amounts in grams.measure amounts in grams.

This means we must add two steps to This means we must add two steps to each of our calculations: first convert each of our calculations: first convert the amount of each reactant to moles, the amount of each reactant to moles, then convert the amount of product then convert the amount of product into grams.into grams.

Since the percentage yield is < 100, Since the percentage yield is < 100, the answer makes sense.the answer makes sense.

Check:Check:

Solution:Solution:

Cu g 221 Cu mol 1

Cu g 3.546

C mol 1

Cu mol 2

C g 12.01

C mol 1C g 1.51

Cu g .7011 Cu mol 1

Cu g 3.546

OCu mol 1

Cu mol 2

OCu g 143.02

OCu mol 1OCu g 5.114

22

22

Example 8.6—When 11.5 g of C Are Allowed to React with 114.5 g of Example 8.6—When 11.5 g of C Are Allowed to React with 114.5 g of CuCu22O in the Reaction Below, 87.4 g of Cu Are Obtained.O in the Reaction Below, 87.4 g of Cu Are Obtained.

Cu Cu22O(O(ss) + C() + C(ss) ) 2 Cu( 2 Cu(ss) + CO() + CO(gg))

• The smallest amount is 101.7 g Cu, therefore that is the theoretical yield.

•The reactant that produces 101.7 g Cu is the Cu2O, Therefore, Cu2O is the limiting reactant.

YieldPercent 100Yield lTheoretica

Yield Actual % Yield 85.9% %100

Cu g 101.7

Cu g 87.4

Example 2- —How Many Grams of NExample 2- —How Many Grams of N22((gg) Can Be ) Can Be Made from 9.05 g of NHMade from 9.05 g of NH33 Reacting with 45.2 g Reacting with 45.2 g

of CuO?of CuO?2 NH2 NH33((gg) + 3 CuO() + 3 CuO(ss) ) → N→ N22((gg) + 3 Cu() + 3 Cu(ss) + 3 ) + 3

HH22O(O(ll))If 4.61 g of NIf 4.61 g of N22 Are Made, What Is the Percent Are Made, What Is the Percent

Yield?Yield?

Day 7Day 7Dec 19Dec 19thth

Copper/ Silver LabCopper/ Silver Lab

Warm –up 12/19Warm –up 12/19

Have your prelab out and complete.Have your prelab out and complete.

Day 8Day 8

Lab WU and Quiz 8.5-8.6Lab WU and Quiz 8.5-8.6

Day 9Day 9Jan 4Jan 4

Section 8.7Section 8.7

Enthalpy ChangeEnthalpy Change

We previously described processes as We previously described processes as exothermic if they released heat, or exothermic if they released heat, or endothermic if they absorbed heat.endothermic if they absorbed heat.

The The enthalpy of reactionenthalpy of reaction is the is the amount of thermal energy that flows amount of thermal energy that flows through a process. through a process. At constant pressure.At constant pressure. HHrxnrxn

3737

Sign of Enthalpy ChangeSign of Enthalpy Change For exothermic reactions, the sign of the For exothermic reactions, the sign of the

enthalpy change is negative when:enthalpy change is negative when: Thermal energy is produced by the reaction.Thermal energy is produced by the reaction. The surroundings get hotter.The surroundings get hotter. H = H = ── For the reaction CHFor the reaction CH44((ss) + 2 O) + 2 O22((gg) ) CO CO22((gg) + 2 ) + 2

HH22O(O(ll), the ), the HHrxnrxn = = −−802.3 kJ per mol of 802.3 kJ per mol of CHCH44..

For endothermic reactions, the sign of the For endothermic reactions, the sign of the enthalpy change is positive when:enthalpy change is positive when: Thermal energy is absorbed by the reaction.Thermal energy is absorbed by the reaction. The surroundings get colder.The surroundings get colder. H = H = ++ For the reaction NFor the reaction N22((ss) + O) + O22((gg) ) 2 NO( 2 NO(gg), the ), the

HHrxnrxn = = ++182.6 kJ per mol of N182.6 kJ per mol of N22..

Enthalpy and Enthalpy and StoichiometryStoichiometry The amount of energy change in a reaction The amount of energy change in a reaction

depends on the amount of reactants.depends on the amount of reactants. You get twice as much heat out when you burn You get twice as much heat out when you burn

twice as much CHtwice as much CH44..

Writing a reaction implies that amount of Writing a reaction implies that amount of energy changes for the stoichiometric energy changes for the stoichiometric amount given in the equation.amount given in the equation.

For the reaction CFor the reaction C33HH88((ll) + 5 O) + 5 O22((gg) ) 3 CO 3 CO22((gg) + 4 ) + 4 HH22O(O(gg) ) HHrxnrxn = = −2044 kJ−2044 kJ

So 1 mol CSo 1 mol C33HH88 5 mol O 5 mol O22 3 mol CO 3 mol CO22 4 mol H 4 mol H22O O −−2044 kJ.2044 kJ.

Example 8.7—How Much Heat Is Associated Example 8.7—How Much Heat Is Associated with the Complete Combustion of 11.8 x with the Complete Combustion of 11.8 x

101033 g of C g of C33HH88((gg)?)?

1 mol C1 mol C33HH88 = -2044 kJ, Molar mass = 44.11 = -2044 kJ, Molar mass = 44.11 g/molg/mol

The sign is correct and the value is The sign is correct and the value is reasonable.reasonable.

11.8 x 1011.8 x 1033 g C g C33HH88, ,

heat, heat, kJkJ

Check:Check:

Solution:Solution:

Solution Map:Solution Map:

RelationshipsRelationships::

Given:Given:Find:Find:

g 09.44

HC mol 1 83

kJ 1047.5HC mol 1

kJ 2044-

HC g 44.11

HC mol 1 HC g 1011.8 5

8383

8383

3

83HC mol 1

kJ 2044-mol C3H8 kJg C3H8

Example 2- How Much Heat Is Evolved When Example 2- How Much Heat Is Evolved When a 0.483 g Diamond Is Burned?a 0.483 g Diamond Is Burned?((HHcombustioncombustion = = −395.4 kJ/mol C)−395.4 kJ/mol C)

Day 10 & 11Day 10 & 11

Review and TestReview and Test