Chapter 11 – Stoichiometry 11.1What is Stoichiometry? 11.2 Stoichiometric Calculations...

Chapter 11 – Stoichiometry

11.1 What is Stoichiometry?

11.2 Stoichiometric Calculations

11.3 Limiting Reactants

11.4 Percent Yield

Section 11.1 Defining Stoichiometry

• Describe the types of relationships indicated by a balanced chemical equation.

• Explain the relationship between stoichiometry and the principle of the conservation of mass.

• State the mole ratios from a balanced chemical equation.

The amount of each reactant present at the start of a chemical reaction determines how much product can form.

Key Concepts

• Balanced chemical equations can be interpreted in terms of moles, mass, and representative particles (atoms, molecules, formula units).

• The law of conservation of mass applies to all chemical reactions.

• Mole ratios are derived from the coefficients of a balanced chemical equation. Each mole ratio relates the number of moles of one reactant or product to the number of moles of another reactant or product in the chemical reaction.

Section 11.1 Defining Stoichiometry

Stoichiometry

Study of quantitative relationships between amounts of reactants used and amounts of products formed by a chemical reactionCombo of 2 Greek words – “to measure the elements”Really a tool (system) for accounting for the conservation of matter

• What goes in must come out

Rests on principle of conservation of matter (mass)

2 Al(s) + 3 Br2(l) Al2Br6(s)

Stoichiometry

Stoichiometry - Questions

If I have this much reactant, how much product can I form?

If I have this much of one reactant, how much of a second reactant do I need so that after the reaction is completed I won’t have any of either reactant left over?

Mass reactant

StoichiometricfactorMoles

reactantMoles

product

Mass product

Requires balanced equation!

Conversion Map for Stoichiometry Calculations

Stoichiometry & Balanced Equation

4Fe(s) + 3O2(g) 2Fe2O3(s)

Can interpret as• 4 atoms Fe + 3 molecules O2 2

formula units Fe2O3

• 4 mol Fe + 3 mol O2 2 mol Fe2O3

No direct information on masses• Can use molar mass to get mass info

Stoichiometry & Balanced Equation

4Fe(s) + 3O2(g) 2Fe2O3(s)

4 mol Fe 55.85 g Fe/mol Fe = 223.4 g Fe

3 mol O2 32.00 g O2/mol O2 = 96.00 g O2

2 mol Fe2O3 159.7 g Fe2O3/mol Fe2O3 = 319.4 g Fe2O3

Total mass reactants = 319.4 g

Total mass products = 319.4 g

Stoichiometry & Balanced EquationExample problem 11.1, p. 370

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

Interpret as molecules, moles & mass1 molecule C3H8 + 5 molecules O2 3 molecules CO2 + 4 molecules H2O

1 mol C3H8 + 5 mol O2 3 mol CO2 + 4 mol H2O

1 mol C3H8 44.09 g C3H8 /mol C3H8 = 44.09 g C3H8

5 mol O2 32.00 g O2/mol O2 = 160.0 g O2

3 mol CO2 44.01 g CO2/mol CO2 = 132.0 g CO2

4 mol H2O 18.02 g H2O /mol H2O = 72.08 g H2O

Mass reactants = 44.09 + 160.0 = 204.1 gMass products = 132.0 + 72.08 = 204.1 g

Practice

Problems 1 (a-c), 2(a-b) page 371

Problems 44 - 45 page 392

Problems 1- 3, page 982

Mole Ratio

Ratio between the numbers of moles of any two substances in a balanced chemical equation

2Al(s) + 3Br2(l) 2AlBr3(s)

Possible to write 3 unique mole ratios:Al:Br2 (2:3) Al: AlBr3 (2:2) Br2:AlBr3 (3:2)

Get 3 more ratios from inverses of above for a total of 6 for this reaction

Mole Ratio

In general, for n species (reactant, product) in chemical equation, have n(n-1) mole ratios

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

4 species, 4 3 = 12 possible mole ratios (6 + inverse of each)

Fe:H2O Fe:Fe3O4 Fe:H2

H2O:Fe3O4 H2O:H2 Fe3O4: H2

Mole Ratio

How many mole ratios can be written for the following reaction?

4H2(g) + O2(g) → 2H2O(l)

A. 6 B. 4 C. 3 D. 2

?

Practice

Problems 3 (a-c), 4 (a-b) page 372

Problems 47 - 53, page 392

Problems 4 - 5, page 982-3





11.4 Percent Yield

Section 11.2 Stoichiometric Calculations

• List the sequence of steps used in solving stoichiometric problems.

• Solve stoichiometric problems (mole-to-mole, mole-to-mass, mass-to-mass and variations including number of particles) using the factor-label method.

The solution to every stoichiometric problem requires a balanced chemical equation.

Key Concepts

• Chemists use stoichiometric calculations to predict the amounts of reactants used and products formed in specific reactions.

• The first step in solving stoichiometric problems is writing the balanced chemical equation.

• Mole ratios derived from the balanced chemical equation are used in stoichiometric calculations.

• Stoichiometric problems make use of mole ratios to convert between mass and moles.

Section 11.2 Stoichiometric Calculations

Stoichiometric CalculationsStart with balanced equation

Determine required mole ratios

Use mole-to-mass and mass-to-mole relations for 3 basic conversions:

1) Mole to mole

2) Mole to mass

3) Mass to mass

Once have moles, can get # particles

Mass reactant


reactantMoles

product

Mass product



nA + mB xC + yDnA + mB xC + yD

Mass reactant

Molesreactant

# part. reactant

Mass product

Molesproduct

# part. product

1 4

3 6

52

Mole to Mole Conversions

Given number of moles of a reactant (product), how many moles of a product (reactant) must be formed?

Use initial moles and mole ratio

mol known x [mol unknown/mol known] = mol unknown

mole ratio from equation


2K(s) + 2H2O(l) 2KOH(aq) + H2(g)

Mol H2 if start with 4.00x10-2 mol K?

4.00x10-2 mol K [1 mol H2/2 mol K]= 2.00x10-2 mol H2



How many moles of CO2 will be

produced in following reaction if initial amount of reactants = 0.50 moles?

2NaHCO3(s) →

Na2CO3(s) + CO2(g) + H2O(g)

A.0.25 B. 0.3 C. 0.5 D. 1.0

?

Practice

Problems 11,12 page 375

Problems 61- 62, page 393


Mole to Mass Conversions

Step 1 – Do mole to mole conversion using mole ratio

Step 2 – Do mole to mass conversion using molar mass

Mass reactant


reactantMoles

product

Mass product



Mole to Mass Conversions2Na(s) + Cl2(g) 2NaCl(s)

Mass NaCl produced from 1.25 mol Cl2?

1.25 mol Cl2 [2 mol NaCl/1 mol Cl2]


= 2.50 mol NaCl

2.50 mol NaCl [58.44 g NaCl/mol NaCl]

= 146 g NaCl

molar mass of NaCl

Practice




Mass to Mass Conversions

Step 1 – Do mass to mole conversion using molar mass

Step 2 – Do mole to mole conversion using mole ratio

Step 3 – Do mole to mass conversion using molar mass

Mass reactant


reactantMoles

product

Mass product



Mass to Mass ConversionsNH4NO3(s) N2O(g) + 2H2O(g)

Mass of H2O from 25.0 g NH4NO3(s)?

AN= NH4NO3(s)

25.0 g AN [1 mol AN/80.04 g AN]

1/molar mass of AN

= 0.312 mol AN

0.312 mol AN [2 mol H2O/1 mol AN]

mole ratio from equation= 0.624 mol H2O

Mass to Mass ConversionsNH4NO3(s) N2O(g) + 2H2O(g)

Mass of H2O from 25.0 g NH4NO3(s)?0.624 mol H2O [18.02 g H2O/mol H2O]

molar mass of H2O

= 11.2 g H2O

Practice


Problems 60, 67- 72, pages 393-4






11.4 Percent Yield

Section 11.3 Limiting Reactants

• Identify the limiting reactant in a chemical equation.

• Identify the excess reactant, and calculate the amount remaining after the reaction is complete.

• Calculate the mass of a product when the amounts of more than one reactant are given.

A chemical reaction stops when one of the reactants is used up.

Key Concepts

• The limiting reactant is the reactant that is completely consumed during a chemical reaction. Reactants that remain after the reaction stops are called excess reactants.

• To determine the limiting reactant, the actual mole ratio of the available reactants must be compared with the ratio of the reactants obtained from the coefficients in the balanced chemical equation.

• Stoichiometric calculations must be based on the limiting reactant.

Section 11.3 Limiting Reactants

Limiting Reactants

Typical case: reactants not present in exact stoichiometric ratios

• Mole ratio given by coefficients of reactants in balanced chemical equation

N2 + 3H2 2NH3

Stoichiometric ratio of reactants for ammonia synthesis is 1 N2: 3 H2

If ratio 1:3 then one reactant is limiting

Limiting and Excess ReactantsAmmonia Synthesis

N2(g) + 3H2(g) 2NH3(g)

Case of 1:1 N2:H2 mole ratio

Before Reaction After Reaction

3N2 3H2 2NH3 2N2

H2 is limiting reactantN2 is excess reactant

Limiting and Excess Reactants

Limiting reactant• Limits the extent of reaction• Determines the amount of product

Excess reactant• Always have “left over”• Often a reactant is deliberately used in

excess to: Drive reaction to completion (equilibrium) Speed up reaction (kinetics)

Limiting and Excess ReactantsLimiting and excess air (oxygen) in

combustion reaction

Air limiting – soot (carbon)

Air excess – CO2 & H2O

Limiting Reactant - Identifying

Step 1 – If masses given, convert to moles

Step 2 – Compare actual mole ratio of reactants to mole ratio determined from balanced equation

• Ratios same - no limiting reactant• Ratios different - compare to identify

limiting reactant

Limiting Reactant

S8(l) + 4Cl2(g) 4S2Cl2(l)

200.0 g sulfur reacts with 100.0 g Cl2Part a: g of product produced?

Step 1 – Convert masses to moles

200.0 g S8 [1 mol S8 /256.5 g S8]

= 0.7797 mol S8

100.0 g Cl2 [1 mol Cl2 /70.91 g Cl2]

= 1.410 mol Cl21/molar mass of Cl2

1/molar mass of S8

Limiting ReactantS8(l) + 4Cl2(g) 4S2Cl2(l)

200.0 g sulfur reacts with 100.0 g Cl2Part a: g product ?

Step 2 – Compare actual mole ratio to mole ratio from equation

= 1.808:1

Equation: Cl2:S8 = 4:1

Actual: 1.410 mol Cl2:0.7797 mol S8

Compare ratios Cl2 is limiting


200.0 g sulfur reacts with 100.0 g Cl2Part a: g product ?

Cl2 limiting reactant, have 1.410 mol

Calculate mol product using mole ratio1.410 mol Cl2 [4 mol S2Cl2 /4 mol Cl2]

= 1.410 mol S2Cl2

1.410 mol S2Cl2 [135.0 g S2Cl2/mol S2Cl2] Mole ratio molar mass of S2Cl2

= 190.4 g S2Cl2


200.0 g sulfur reacts with 100.0 g Cl ?

Part b: g of sulfur actually reacted?

Cl2 limiting reactant, have 1.410 mol

Calculate mol sulfur using mole ratio

1.410 mol Cl2 [1 mol S8 /4 mol Cl2]

= 0.3525 mol S8

0.3525 mol S8 [256.5 g S8 /mol S8] Mole ratio

molar mass of S8

= 90.42 g S8 reacts 109.6 g excess


200.0 g sulfur reacts with 100.0 g Cl ?

Part b: g of sulfur actually reacted?

Know that all Cl2(g) reacted (100.0 g)

Previously calculated 190.4 g product S2Cl2(l) produced

Cons. mass: 300.0 g = starting mass = final mass = m[S8(l)] + m[4S2Cl2(l)]

m[S8(l)]= 300.0 g -m[4S2Cl2(l)] =109.6 g

Limiting Reactant

Excess reactant in following reaction if start with 50.0g of each reactant?

P4(s) + 5 O2(g) → P4O10(s)

A. O2 B. P4

C. None – quantities are stoichiometric

?50.0 g P4 [1 mol P4 /123.90 g P4]= 0.404 mol P4

50.0 g O2 [1 mol O2 /32.00 g O2] = 1.56 mol O2

Need 5 0.404 mol = 2.02 mol O2 O2 limiting

Practice

Problems 23, 24 page 383







11.4 Percent Yield

Section 11.4 Percent Yield

• Calculate the theoretical yield of a chemical reaction from data.

• Explain the difference between a theoretical yield of a chemical reaction and an actual yield.

• Provide reasons why the theoretical yield and actual yield can be different from each other.

• Determine the percent yield for a chemical reaction.

• Explain why the percent yield can exeed 100%

Percent yield is a measure of the efficiency of a chemical reaction.

Key Concepts• The theoretical yield of a chemical reaction is the

maximum amount of product that can be produced from a given amount of reactant. Theoretical yield is calculated from the balanced chemical equation.

• The actual yield is the amount of product produced. Actual yield must be obtained through experimentation.

• Percent yield is the ratio of actual yield to theoretical yield expressed as a percent. High percent yield is important in reducing the cost of every product produced through chemical processes.

Section 11.4 Percent Yield

Percent Yield

Actual (real world) reactions don’t produce amount of product expected from masses of reactant and the balanced chemical equation (stoichiometry)

Types of Yields

Theoretical Yield• Maximum amount of product that can be

produced from given amounts of reactants

• Calculated using stoichiometry

Actual Yield• Amount of product actually obtained

when the reaction is carried out

Types of Yields

Theoretical Yield and Actual Yield can differ because

• Reaction stops before limiting reactant all used up

• Product is not all collected – sticks to reaction vessel, filter paper, etc

• Reactions other than the intended reaction occur, producing other products

• Impurities in reactants

Alternate Reaction Paths

Combustion of natural gas (methane)

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

Know this reaction takes place if have enough O2 – get nice blue flame

Know if limit O2, get yellow flame and soot (carbon) production

Same reactants yield different products

Percent Yield

Percent yield is a way of expressing the efficiency of a reaction

Percent yield = 100 actual yield

theoretical yield

Theoretical yield computed using same procedures used in previous sections (stoichiometry)

% Percent Yield

Theoretical yield of chemical reaction = 25 g. % yield if actual yield = 22 g ?

A. 88% B. 44% C. 50% D. 97%

?

% yield = 100 22 g / 25 g = 88%

Percent Yield

In commercial production of chemicals, attempt to produce as close to 100% yield as possible

• Optimum use of materials• Minimizes or eliminates need to

separate product from unused reactants and from unwanted products

Percent Yield for Multi-Stage

Attempt to produce as close to 100% yield as possible

• Critical for multi-stage reaction processes where product of one reaction used as one of the reactants of a subsequent reaction

• A three-stage reaction with 90% yield from each stage gives only 73% overall yield (0.9 x 0.9 x 0.9 = 0.73)

>100% Percent Yield

Actual Yield: Amount of product actually obtained when from reaction

Obtained from a laboratory measurement – most common is measuring mass on a balance

Assumption made is that the only “stuff” you’re measuring is the product of interest

>100% Percent Yield

Possible to get more “stuff” you think is product than predicted if:

Haven’t completely separated intended product from other products in reaction or from leftover reactants and/or solvent (insufficient drying)

Product reacts with air or water vapor before it is weighed

Practice

Problems 28 - 30 page 387

Problem 35, page 388


Problems 14 -15, page 983

Chapter 11 – Stoichiometry 11.1What is Stoichiometry? 11.2 Stoichiometric Calculations...

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