Stoichiometric Calculations Stoichiometry – Ch. 9.

Stoichiometric Stoichiometric

CalculationsCalculations

Stoichiometric Stoichiometric

CalculationsCalculations

Stoichiometry – Ch. 9

StoichiometryStoichiometryGreek for “measuring elements”The calculations of quantities in

chemical reactions based on a balanced equation.

We can interpret balanced chemical equations several ways.

Stoichiometry DefinitionStoichiometry DefinitionComposition stoichiometry

deals with the mass relationships of elements in compounds.

Reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction.

Click below to watch the Visual Concept.

Visual Concept

StoichiometryStoichiometry

A. Proportional RelationshipsA. Proportional Relationships

I have 5 eggs. How many cookies can I make?

3/4 c. brown sugar1 tsp vanilla extract2 eggs2 c. chocolate chipsMakes 5 dozen cookies.

2 1/4 c. flour1 tsp. baking soda1 tsp. salt1 c. butter3/4 c. sugar

5 eggs 5 doz.

2 eggs= 12.5 dozen cookies

Ratio of eggs to cookies

A. Proportional A. Proportional RelationshipsRelationshipsMole RatioMole Ratio is a conversion factor

that relates the amounts in moles of any two substances involved in a chemical reaction

Example: 2Al2O3(l) → 4Al(s) + 3O2(g)

Mole Ratios: 2 mol Al2O3 2 mol Al2O3 4 mol Al

4 mol Al 3 mol O2 3 mol O2

B. Stoichiometry StepsB. Stoichiometry Steps

1. Write a balanced equation.2. Identify known & unknown.3. Line up conversion factors.

◦Mole ratio - moles moles◦Molar mass - moles grams◦Molarity - moles liters soln◦Molar volume - moles liters gas

4. Check answer.

Converting Between Amounts in Converting Between Amounts in MolesMoles

Conversions of Quantities in Conversions of Quantities in MolesMoles

D. Stoichiometry ProblemsD. Stoichiometry ProblemsHow many moles of KClO3 must

decompose in order to produce 9 moles of oxygen gas?

9 mol O2 2 mol KClO3

3 mol O2

= 6 mol KClO3

2KClO3 2KCl + 3O2 ? mol 9 mol

Periodic Table

MolesA MolesB Massg B

Periodic Table

Balanced Equation

Massg A

• Decide where to start based on the units you are given• Stop based on what unit you are asked for

D. Stoichiometry ProblemsD. Stoichiometry ProblemsHow many grams of silver will

be formed from 12.0 g copper?

12.0g Cu

1 molCu

63.55g Cu

= 40.7 g Ag

Cu + 2AgNO3 2Ag + Cu(NO3)2

2 molAg

1 molCu

107.87g Ag

1 molAg

12.0 g ? g

For example...For example...If 10.1 g of Fe are added to a

solution of Copper (II) Sulfate, how much solid copper would form?

2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu

10.1 g Fe

55.85 g Fe

1 mol Fe

2 mol Fe

3 mol Cu

1 mol Cu

63.55 g Cu

= 17.2 g Cu

How do you get good at How do you get good at this?this?

A. Limiting ReactantsA. Limiting ReactantsAvailable IngredientsAvailable Ingredients

◦4 slices of bread◦1 jar of peanut butter◦1/2 jar of jelly

Limiting ReactantLimiting Reactant• bread

Excess ReactantsExcess Reactants• peanut butter and jelly

Limiting ReactantsLimiting ReactantsThe limiting reactant is the reactant

that limits the amount of the other reactant that can combine and the amount of product that can form in a chemical reaction.

The excess reactant is the substance that is not used up completely in a reaction.

How do you find out?How do you find out?Do two stoichiometry problems.

The one that makes the least product is the limiting reagent.

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Visual Concept

Limiting Reactants and Limiting Reactants and Excess ReactantsExcess Reactants

If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?

2Cu + S Cu2S

10.6 g Cu 63.55g Cu

1 mol Cu

2 mol Cu

1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 13.3 g Cu2S

3.83 g S 32.06g S

1 mol S

1 mol S

1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 19.0 g Cu2S

= 13.3 g Cu2S

Cu is Limiting Reagent

Percentage Yield Percentage Yield The theoretical yield is the

maximum amount of product that can be produced from a given amount of reactant.

The actual yield of a product is the measured amount of that product obtained from a reaction.

.

Section 3 Limiting Reactants and Percentage Yield

Percent yieldPercent yield = Actual x 100 % Theoretical

Percent YieldPercent YieldWhen 45.8 g of K2CO3 react with excess

HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.

K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g

actual: 46.3 g

Percent YieldPercent Yield

45.8 gK2CO3

1 molK2CO3

138.21 gK2CO3

= 49.4g KCl

2 molKCl

1 molK2CO3

74.55g KCl

1 molKCl

K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g

actual: 46.3 gTheoretical Yield:

Percent YieldPercent Yield

Theoretical Yield = 49.4 g KCl

% Yield =46.3 g

49.4 g 100 =93.7%

K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g 49.4 g

actual: 46.3 g