Stoichiometric Calculations

I. Stoichiometric

Calculations

I. Stoichiometric

Calculations

Stoichiometry – Ch. Stoichiometry – Ch. 1111

Background on things you Background on things you NEED to know how to do:NEED to know how to do:Background on things you Background on things you NEED to know how to do:NEED to know how to do:

1. Name/write correct chemical formula

2. Write chemical equations

3. Balance chemical equations

4. Predict Products

5. Mole/mass conversions

StoichiometryStoichiometryStoichiometryStoichiometryo Stoichiometry uses ratios to determine relative

amounts of reactants or products.o For example If you were to make a bicycle,

you would need one frame and two tires.o 1 frame + 2 tires 1 bicycleo If I had 74 tires, what is the most # of bicycles

I could make?

74 tires

2 tires

1 bicycle = 37 bicycles

Proportional Proportional RelationshipsRelationshipsProportional Proportional RelationshipsRelationships

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

Proportional Proportional RelationshipsRelationshipsProportional Proportional RelationshipsRelationships

StoichiometryStoichiometry• mass relationships between substances in a

chemical reaction• based on the mole ratio

Mole RatioMole Ratio• indicated by coefficients in a balanced equation• can be used to determine expected amounts of

products given amounts of reactants.

2 Mg + O2 Mg + O22 2 MgO 2 MgO

Stoichiometry StepsStoichiometry StepsStoichiometry StepsStoichiometry 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

Core step in all stoichiometry problems!!

• Mole ratio - moles moles

4. Check answer.

Molar Volume at STPMolar Volume at STPMolar Volume at STPMolar Volume at STP

Molar Mass(g/mol)

6.02 1023

particles/mol

MASSIN

GRAMSMOLES

NUMBEROF

PARTICLES

LITERSOF

SOLUTION

Molar Volume (22.4 L/mol)

LITERSOF GASAT STP

Molarity (mol/L)

Mole – Mole ConversionsMole – Mole ConversionsMole – Mole ConversionsMole – Mole Conversions The first type of problems we encounter will

go between moles and moles. For this we need to use mole ratios.

Ex: Write and balance the reaction between lead (II) nitrate and potasium iodide.

Pb(NO3)2 + 2KI 2 KNO3 + PbI2

Mole ratio of potasium iodide to lead (II) iodide:

2 moles KI

1 mole PbI2

Mole to Mole ProblemsMole to Mole ProblemsMole to Mole ProblemsMole to Mole Problems

How 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

Mole to MassMole to MassMole to MassMole to Mass

We can also convert from moles to mass, and mass to moles

For Example:

• 4 Al + 3 O2 2Al2O3

If you know how many grams of Al you start with, we can write a flow chart to show how to calculate the # of moles of oxygen need to fully react with the Al.

g Al moles Al moles of oxygen

Mass to Moles: Mass to Moles: 4 Al + 3 O4 Al + 3 O22 2Al2Al22OO33

Mass to Moles: Mass to Moles: 4 Al + 3 O4 Al + 3 O22 2Al2Al22OO33

If the reaction starts with .84 moles of aluminum, how many grams of aluminum oxide are produced?

.84 mol Al

4 mol Al

2 mol Al2O3

1 mol Al2O3

101.9 grams Al2O3

= 42.8 grams Al2O3 0.92 g of Aluminum oxide are produced from the

reaction. How much aluminum was used up?

.92 g Al2O3

101.9 g Al2O3

1 mol Al2O3

2 mol Al2O3

4 mol Al

1 mol Al

26.9 g Al

= .49 grams Al

Mass to MassMass to MassMass to MassMass to Mass How many grams of silver will be

formed from 12.0 g copper reacting with silver nitrate?

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

Stoichiometry Problems – Mole/MassStoichiometry Problems – Mole/MassStoichiometry Problems – Mole/MassStoichiometry Problems – Mole/Mass

In photosynthesis, carbon dioxide and water react to form glucose, C6H12O6 and oxygen gas.

___CO2 + ___H2O ___C6H12O6 + ___O2

If 15.6 grams of carbon dioxide react, how many moles of glucose will be produced?

How many grams of carbon dioxide must react to produce 0.25 moles of glucose?

6 6 6

15.6 g CO2

44.01 g CO2

1 mol CO2

6 mol CO2

1 mol C2 H12O6 = 0.0591 mol C2H12O6

44.01 g CO26 mol CO2

1 mol CO2

0.25 mol C2 H12O6= 66 g

CO21 mol C2 H12O6

Stoichiometry with GasesStoichiometry with GasesStoichiometry with GasesStoichiometry with Gases

If the pressure and temperature are constant, the ratio of moles in the balanced equation is the ratio of liters in an all gas reaction.

At STP 1 mol of a gas=22.4 L

Molar Volume at STPMolar Volume at STPMolar Volume at STPMolar Volume at STP

Standard Temperature & Pressure0°C and 1 atm

Molar VolumeMolar VolumeMolar VolumeMolar Volume

Hydrogen and chlorine gas react to produce hydrochloric acid. If 7.00 L of hydrogen gas react, how many liters of HCl gas are formed?

H2 (g) + Cl2 (g) 2 HCl (g)

7.00 L H2

1.0 L H2

2.0 L HCl = 14.0 L HCl

*only in all gas reactions!

Molar VolumeMolar VolumeMolar VolumeMolar Volume

In the following reaction, if 17 g of Mg react, how many L of H2 forms?

Mg (s) + 2HCl (aq) MgCl2 (aq) + H2 (g)

17.0 gMg

1 molMg

24.31 gMg

1 molH2

1 molMg

22.4 L H2

1 molH2

= 15.7 L H2

How many grams of KClO3 are req’d to

produce 9.00 L of O2 at STP?

9.00 LO2

1 molO2

22.4 L O2

= 32.8 g KClO3

2 molKClO3

3 molO2

122.55g KClO3

1 molKClO3

? g 9.00 L

Molar Volume ProblemsMolar Volume ProblemsMolar Volume ProblemsMolar Volume Problems

2KClO3 2KCl + 3O2

63.55g Cu

1 molCu

Molar Volume ProblemsMolar Volume ProblemsMolar Volume ProblemsMolar Volume Problems

How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3?

1.5L

.10 molAgNO3

1 L= 4.8 g

Cu

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

1 molCu

2 molAgNO3

? g 1.5L0.10M

MolarityMolarityMolarityMolarity

Molarity is the number of moles of solute dissolved in one liter of solution.

Units are moles per liter or moles of solute per liter of solution.

Molarity abbreviated by a capital M Molarity = moles of solute

liter of solution

Molarity ProblemsMolarity ProblemsMolarity ProblemsMolarity Problems

As an example, suppose we dissolve 23 g of ammonium chloride (NH4Cl) in enough water to make 145 mL of solution. What is the molarity of ammonium chloride in this solution?

23 g NH4Cl

53.5 g NH4Cl

1 mole NH4Cl = .43 mol NH4Cl

145 mL

1000 mL

1 L = .145 L

.43 mol NH4Cl

.145 L= 2.97 M NH4Cl

Molarity ProblemsMolarity ProblemsMolarity ProblemsMolarity Problems Now, suppose we have a beaker with 175 mL

of a 5.5 M HCl solution. How many moles of HCl is in this beaker?

1000 mL

1 L = .96 mol HCl175 mL

1 L

5.5 mol HCl

Suppose you had 70 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What would be the molarity of the solution?

70 g NaCl

58.44 g NaCl

1 mol NaCl = 1.2 mol

NaCl

1.2 mol NaCl 2.0 L

= 0.6 M NaCl

Volume of SolutionsVolume of SolutionsVolume of SolutionsVolume of Solutions

A 10.% HCl solution (soln) means:

10 g HCl (pure) 100 g HCl soln

A solution with a density of 1.5 g/mL means: 1.5 g soln 1 mL soln

Impure SubstancesImpure SubstancesImpure SubstancesImpure Substances To say a substance is 75% NaCl by mass means:

75 g NaCl (pure) 100 g of NaCl solution

or

100 g of NaCl solution

75 g of NaCL (pure) Or, Iron ore that is 15% iron by mass means:

15 g Fe 100 g ore

or

100 g of ore100 g of ore

15 g of Fe15 g of Fe

Energy & Stoichiometry

Energy & Stoichiometry

Exothermic and EndothermicExothermic and EndothermicExothermic and EndothermicExothermic and Endothermic

Exothermic process – heat is released into the surroundings

• Exo = Exit

Endothermic Process – heat is absorbed from the surroundings

• Endo = Into

HEAT

HEAT

Thermochemical EquationsThermochemical EquationsThermochemical EquationsThermochemical Equations

In a thermochemical equation, the energy of change for the reaction can be written as either a reactant or a product

Enthalpy: the heat content of a system at constant pressure (ΔH)

Endothermic (positive ΔH)

2NaHCO3 + 129kJ Na2CO3 + H2O + CO2

Exothermic (negative ΔH)

CaO + H2O Ca(OH)2 + 65.2kJ

Write the thermochemical equation for the oxidation of Iron (III) if its ΔH= -1652 kJ

Fe(s) + O2(g)→ Fe2O3(s) + 1652 kJ

How much heat is evolved when 10.00g of Iron is reacted with excess oxygen?

4 3 2

10.00g Fe

55.85g Fe

1 mol Fe

4 mol Fe

1652 kJ=73.97 kJ of heat

Exo

Write the thermochemical equation for the decomposition of sodium bicarbonate, with a ΔH = + 129 kJ:

2 NaHCO3 + 129kJ → Na2CO3(s) + H2O + CO2

How much heat is required to break down 50.0g of sodium bicarbonate?

50.0 g NaHCO3

83.9 g NaHCO3

1 mol NaHCO3

2 mol NaHCO3

129 kJ

=38.4 kJ of heat

Endo

Write the thermochemical equation for the synthesis of calcium oxide and water with a ΔH= - 65.2 kJ:

CaO + H2O → Ca(OH)2 + 65.2kJ

How much energy is released when 100 g of calcium oxide reacts?

100 g CaO

56.07 g CaO

1 mol CaO

1 mol CaO

65.2 kJ

=116 kJ of heat

Exo

Write the thermochemical equation for the decomposition of magnesium oxide with a ΔH= + 61.5 kJ:

2 MgO + 61.5 → 2 Mg + O2

How many grams of oxygen are produced when magnesium oxide is decomposed by adding 420 kJ of Energy?

420 kJ

61.5 kJ

1 mol O2

1 mol O2

31.98 g O2 =218 g of O2

Endo

Stoichiometry in the Real World

Stoichiometry in the Real World

Stoichiometry – Ch. Stoichiometry – Ch. 1111

Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants

Available 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


Available IngredientsAvailable Ingredients• 24 graham cracker squares• 1 bag of marshmallows• 12 pieces of chocolate

Limiting ReactantLimiting Reactant• chocolate

Excess ReactantsExcess Reactants• Marshmallows and

graham crackers


Limiting ReactantLimiting Reactant• one that is used up in a reaction• determines the amount of product that

can be produced

Excess ReactantExcess Reactant• added to ensure that the other reactant

is completely used up• cheaper & easier to recycle

Limiting Reactant StepsLimiting Reactant StepsLimiting Reactant StepsLimiting Reactant Steps

1. Write the balanced equation.

2. For each reactant, calculate the

amount of product formed.

3. Smaller answer indicates:

• limiting reactant

• amount of product actually possible


79.1 g of zinc react with 68.1 g HCl. Identify the limiting and excess reactants. How many grams of hydrogen can be formed?

Zn + 2HCl ZnCl2 + H2 79.1 g ? g68.1 g


2.02 g H2

1 molH2

68.1g HCl

1 molHCl

36.46 g HCl

= 1.89 g H2

1 molH2

2 molHCl

Zn + 2HCl ZnCl2 + H2 79.1 g ? g68.1 g


79.1g Zn

1 molZn

65.39g Zn

= 2.44 g H2

1 molH2

1 molZn

2.02 gH2

1 molH2

Zn + 2HCl ZnCl2 + H2 79.1 g ? g68.1 g


Zn: 2.44 g H2 HCl: 1.89 g H2

Limiting reactant: HCl

Excess reactant: Zn

Product Formed: 1.89 g H2

left over zinc

Limiting Reactants #2Limiting Reactants #2 Limiting Reactants #2Limiting Reactants #2

5.42 g of magnesium ribbon react with 4.00 g of oxygen gas. Identify the limiting and excess reactants. How many grams of magnesium oxide are formed?

2Mg + O2 2MgO 5.42 g ? g4.00 g

Limiting Reactants #2Limiting Reactants #2Limiting Reactants #2Limiting Reactants #2

5.42g Mg

1 molMg

24.31g Mg

= 8.99 g MgO

2 molMgO

2 molMg

40.31 gMgO

1 molMgO

2Mg + O2 2MgO

5.42 g ? g4.00 g

Limiting Reactants #2Limiting Reactants #2Limiting Reactants #2Limiting Reactants #2

40.31g MgO

1 molMgO

4.00g O2

1 molO2

32.00 g O2

= 10.1 g MgO

2 molMgO

1 molO2

2Mg + O2 2MgO

5.42 g ? g4.00 g

A. Limiting Reactants #2A. Limiting Reactants #2A. Limiting Reactants #2A. Limiting Reactants #2

Mg: 8.99 g MgO O2: 10.1 g MgO

Excess oxygen

Limiting reactant: Mg

Excess reactant: O2

Product Formed: 8.99 g MgO


What other information could you find in these problems?

• How much of each reactant is used – in grams, liters, moles

• How much of excess reactant is left over – in grams, liters, moles

Percent YieldPercent YieldPercent YieldPercent Yield

100yield ltheoretica

yield actualyield %

calculated on paper

measured in lab

Percent YieldPercent YieldPercent YieldPercent Yield When 45.8 g of K2CO3 react with excess HCl,

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

1 molK2CO3

138.2 g K2CO3

2 molKCl

1 molK2CO3

74.55 gKCl

1 molKCl

45.8 gK2CO3

K2CO3 + 2 HCl 2 KCl + H2CO3

= 49.4 grams KCl

Percent YieldPercent YieldPercent YieldPercent Yield

When 45.8 g of K2CO3 react with excess HCl,

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

46.3 grams KCl

49.4 grams KClx 100

% yield = 93.7 %

Stoichiometric Calculations

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