Sample Problem: STOICHIOMETRY and percent yield calculations. How much H 2 O will be formed if 454 g of. decomposes? NH 4 NO 3 N 2 O + 2 H 2 O

1 1

STOICHIOMETRY

STOICHIOMETRY

and percent yield calculations

and percent yield calculations

Reactants: Zn + I

Reactants: Zn + I₂₂ Product: Zn IProduct: Zn I₂₂

2

Steps for solving Stoichiometric Problems 2

Step 1

Write the balanced equation for the reaction.

Step 2

Identify your known and unknown quantities.

Step 3

Convert from grams of known to moles of known Step 4

Convert from moles of known to moles of unknown Step 5

Convert from moles of unknown to grams of unknown

3

Sample Problem:

3

Sample Problem:

Sample Problem:

How much H

How much H _{2 2} O O

will be

will be

formed if

formed if

454 g of

454 g of

NH NH _{4 4} NO NO _{3 3}

decomposes ?

decomposes ?

4

SAMPLE PROBLEM:

4

If 454 g of NH

₄

NO

₃

decomposes, how

much H

₂

O is formed?

SAMPLE PROBLEM:

SAMPLE PROBLEM:

If 454 g of NH

If 454 g of NH

₄₄

NO NO

₃₃

decomposes, how decomposes, how

much H

much H

₂₂

O is formed? O is formed?

STEP 1

STEP 1

Write the balanced

Write the balanced

chemical equation

chemical equation

NH NH

₄₄

NO NO

₃₃

  N N

₂₂

O + 2 H O + 2 H

₂₂

O O

5

SAMPLE PROBLEM:

5

If 454 g of NH

₄

NO

₃

decomposes, how

much H

₂

O is formed?

SAMPLE PROBLEM:

SAMPLE PROBLEM:

If 454 g of NH

If 454 g of NH

₄₄

NO NO

₃₃

decomposes, how decomposes, how

much H

much H

₂₂

O is formed? O is formed?

STEP 2

STEP 2

Identify your

known and

unknown quantities.

NH

NH

₄₄

NO NO

₃₃

  N N

₂₂

O + 2 H O + 2 H

₂₂

O O

6 6

NH

₄

NO

₃

 N

₂

O + 2 H

₂

O

NH NH

₄₄

NO NO

₃₃

  N N

₂₂

O + 2 H O + 2 H

₂₂

O O

STEP 3

STEP 3

Convert from grams of known

to moles of known

454 g • 1 mol

80.04 g = 5.68 mol NH4NO3 Use the

Use the

molar mass molar mass

as a

as a

CONVERSION FACTOR CONVERSION FACTOR

::

7 7

NH

₄

NO

₃

--> N

₂

O + 2 H

₂

O

NH NH

₄₄

NO NO

₃₃

--> N -- > N

₂₂

O + 2 H O + 2 H

₂₂

O O

Use the

Use the mole ratio mole ratio

2 mol H₂O produced 1 mol NH₄NO₃ used

STEP 4

STEP 4

Convert from moles of known to moles of unknown

1 mol NH

1 mol NH

₄₄

NO NO

₃₃

: : 2 mol H 2 mol H

₂₂

O O

as a

as a CONVERSION FACTOR CONVERSION FACTOR ^{: :}

8 8

NH

₄

NO

₃

--> N

₂

O + 2 H

₂

O

NH NH

₄₄

NO NO

₃₃

--> N -- > N

₂₂

O + 2 H O + 2 H

₂₂

O O

= 11.4 mol H

= 11.4 mol H

₂₂

O produced O produced

5.68 mol NH₄NO₃ • 2 mol H₂O produced 1 mol NH₄NO₃ used

STEP 4

STEP 4

Convert from moles of known to moles of unknown

Use the

Use the

mole ratio mole ratio

as a

as a

CONVERSION FACTOR CONVERSION FACTOR

::

9 9

NH

₄

NO

₃

 N

₂

O + 2 H

₂

O

NH NH

₄₄

NO NO

₃₃

  N N

₂₂

O + 2 H O + 2 H

₂₂

O O

11.4 mol H₂O • 18.02 g

1 mol = 204 g H₂O

STEP 5

STEP 5

Convert from moles of unknown

to grams of unknown

ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!ALWAYS FOLLOW THESE STEPS IN ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!

SOLVING STOICHIOMETRY PROBLEMS!

Use the

Use the

molar mass molar mass

as a

as a

CONVERSION FACTOR CONVERSION FACTOR

::

10

GENERAL PLAN FOR

10

STOICHIOMETRY

CALCULATIONS

GENERAL PLAN FOR

GENERAL PLAN FOR

STOICHIOMETRY

STOICHIOMETRY

CALCULATIONS

CALCULATIONS

Mass known

Mole Ratio Moles

known

Moles unknown

Mass unknown

Molar Mass known

Molar Mass unknown

11 11

OUR FINAL ANSWER WAS:

OUR FINAL ANSWER WAS:

SAMPLE PROBLEM:

If 454 g of NH

₄

NO

₃

decomposes, how

much H

₂

O is formed?

SAMPLE PROBLEM:

SAMPLE PROBLEM:

If 454 g of NH

If 454 g of NH

₄₄

NO NO

₃₃

decomposes, how decomposes, how

much H

much H

₂₂

O is formed? O is formed?

This is called the

This is called the THEORETICAL YIELD…

THEORETICAL YIELD

…

……it is how much water we it is how much water we theoreticallytheoreticallyshould produce.should produce.

= 204 g H

= 204 g H

₂₂

O produced O produced

12

Theoretical Yield:

12

Theoretical Yield:

• • a.k.a. a.k.a. “ “predicted yield predicted yield” ”

•

• is is calculated calculated (by (by stoichometry stoichometry) )

•

• The amount of product we The amount of product we “ “should should” ” get. get.

• •aka aka “ “Experimental Yield Experimental Yield” ”

• •is is measured measured in the lab in the lab

•

•What you actually did produce What you actually did produce

• •Always less than the theoretical yield. Always less than the theoretical yield.

Actual Yield:

Actual Yield:

13 13

Percent Yield =

Percent Yield =

Actual Yield

Actual Yield

Theoretical Yield

Theoretical Yield x 100 x 100

•

• Always less than 100 % Always less than 100 %

•

• Ratio of actual production Ratio of actual production

to theoretical production.

to theoretical production.

•

• Not the same as percent error! Not the same as percent error!

14 14

NH

₄

NO

₃

--> N

₂

O + 2 H

₂

O

NH NH

₄₄

NO NO

₃₃

--> N -- > N

₂₂

O + 2 H O + 2 H

₂₂

O O

Sample Problem

Sample Problem Con Con’ ’t t: :

Our theoretical yield was 204 g H

Our theoretical yield was 204 g H

₂₂

O O

If Johnny Q. Chemistry did the reaction and only

If Johnny Q. Chemistry did the reaction and only

collected 186 g H

collected 186 g H

₂₂

O, the percent yield would be: O, the percent yield would be:

186 g H

186 g H

₂₂

O O

204 g H

204 g H

₂₂

O O x 100 = x 100 = 91% 91%

15

Stoichiometry can be used

15

Stoichiometry can be used

to Determine a Formula:

to Determine a Formula:

The mass of CO

The mass of CO₂₂and Hand H₂₂O produced can be used to determine O produced can be used to determine the amount of C and H present in the formula.

the amount of C and H present in the formula.

4.0 g

4.0 g CC_xxHH_yy+ 4.5 g O+ 4.5 g O₂₂---> ---> 5.0 g CO5.0 g CO₂₂+ + 3.5 g H3.5 g H₂₂OO If we burn an hydrocarbon fuel with an unknown formula, If we burn an hydrocarbon fuel with an unknown formula,

““CC_xxHH_yy”, using a known quantity of oxygen gas:”, using a known quantity of oxygen gas:

16

Determining the Formula of a

16

Determining the Formula of a

Hydrocarbon by Combustion

Hydrocarbon by Combustion

CCR, page 138 CCR, page 138

17

Reactions Involving a

17

LIMITING REACTANT Reactions Involving a Reactions Involving a

LIMITING REACTANT

LIMITING REACTANT

•• Usually, there is not enough of one reagent to Usually, there is not enough of one reagent to use up the other reagent completely.

use up the other reagent completely.

•

• The reagent in short supply The reagent in short supply

LIMITS LIMITS

the the quantity of product that can be formed.

quantity of product that can be formed.

•

• It is called the It is called the ““limiting reagentlimiting reagent””

18 18

Rxn 1: Balloon inflates fully, some Zn left Rxn 1: Balloon inflates fully, some Zn left

*

* More than enough Zn to use up the 0.100 mol HClMore than enough Zn to use up the 0.100 mol HCl Rxn 2: Balloon inflates fully, no Zn left

Rxn 2: Balloon inflates fully, no Zn left

* Right amount of each (

* Right amount of each (HClHCland Zn)and Zn)

Rxn 3: Balloon does not inflate fully, no Zn left.

Rxn 3: Balloon does not inflate fully, no Zn left.

* Not enough Zn to use up 0.100 mol

* Not enough Zn to use up 0.100 mol HClHCl

LIMITING REACTANTS

LIMITING REACTANTS

LIMITING REACTANTS

React solid Zn with 0.100 React solid Zn with 0.100 mol

mol HClHCl(aq(aq)) Zn + 2

Zn + 2 HClHCl---> ZnCl---> ZnCl₂₂+ H+ H₂₂

1 2 3

(See CD Screen 4.8) (See CD Screen 4.8)

19 19

Rxn 1

Rxn 1 Rxn 2Rxn 2 Rxn 3Rxn 3 mass Zn (g)

mass Zn (g) ~ 7.0~ 7.0 3.273.27 1.311.31 mol Zn

mol Zn 0.1000.100 0.0500.050 0.0200.020 mol mol HClHCl 0.1000.100 0.1000.100 0.1000.100 mol mol HClHCl/mol Zn/mol Zn 1.00/11.00/1 2.00/12.00/1 5.00/15.00/1 Lim Reactant

Lim Reactant LR = LR = HClHCl no LRno LR LR = ZnLR = Zn

LIMITING REACTANTS

LIMITING REACTANTS

LIMITING REACTANTS

React solid Zn with 0.100 React solid Zn with 0.100 mol HClmol HCl(aq(aq))

Zn + 2

Zn + 2 HClHCl------> ZnCl> ZnCl₂₂+ H+ H_{2 2}

20 20

LIMITING REACTANTS

LIMITING REACTANTS

LIMITING REACTANTS

Demo of limiting reactants on Screen 4.7 Demo of limiting reactants on Screen 4.7

21 21

454 g of NH

₄

NO

₃

--> N

₂

O + 2 H

₂

O

454 g of NH

454 g of NH

₄₄

NO NO

₃₃

-- --> N > N

₂₂

O + 2 H O + 2 H

₂₂

O O

% yield = actual yield

theoretical yield • 100%

STEP 6

STEP 6 Calculate the percent yield

Calculate the percent yield

% yield = 131 g

250. g • 100% = 52.4%

22 22

Chemical Equations

Chemical Equations

Chemical Equations

•• Because the same atoms Because the same atoms are present in a reaction are present in a reaction at the beginning and at at the beginning and at the end, the amount of the end, the amount of matter in a system does matter in a system does not change.

not change.

•

• The The Law of the Law of the Conservation of Conservation of Matter

Matter Demo of conservation of matter, See Screen 4.3.

23

LIMITING REACTANTS

23

LIMITING REACTANTS

Reactantseactants ProductsProducts

2 NO(g) + O

₂

(g) 2 NO

₂

(g)

Limiting reactant = ___________

Limiting reactant = ___________

Excess reactant = ____________

Excess reactant = ____________

24 24

Reaction to be Studied

Reaction to be Studied

Reaction to be Studied

2 Al + 3 Cl

2 Al + 3 Cl _{2 2} --- ---> Al > Al _{2 2} Cl Cl _{6 6}

25 25

PROBLEM: Mix 5.40 g of Al with 8.10 g

of Cl

₂

. What mass of Al

₂

Cl

₆

can form?

PROBLEM:

PROBLEM: Mix 5.40 g of Al with 8.10 g Mix 5.40 g of Al with 8.10 g

of Cl

of Cl

₂₂

. What mass of Al . What mass of Al

₂₂

Cl Cl

₆₆

can form? can form?

Mass reactant

Stoichiometric factor Moles

reactant

Moles product

Mass product

26 26

Step 1 of LR problem:

compare actual mole ratio

of reactants to

theoretical mole ratio.

Step 1 of LR problem:

Step 1 of LR problem:

compare

compare actual actual mole ratio mole ratio

of reactants to

of reactants to

theoretical

theoretical mole ratio. mole ratio.

27 27

2 Al + 3 Cl

2 Al + 3 Cl _{2 2} --- --- > Al > Al _{2 2} Cl Cl _{6 6}

Reactants must be in the mole ratio

Reactants must be in the mole ratio

Step 1 of LR problem:

compare actual mole ratio of

reactants to theoretical

mole ratio.

Step 1 of LR problem:

compare actual mole ratio of

reactants to theoretical

mole ratio.

mol Cl ₂

mol Al = 3

2

28

Deciding on the Limiting

28

Reactant

Deciding on the Limiting

Deciding on the Limiting

Reactant

Reactant

If If

There is not enough Al to use up all

There is not enough Al to use up all

the Cl

the Cl

₂₂

2 Al + 3 Cl

2 Al + 3 Cl

₂₂

---> Al --- > Al

₂₂

Cl Cl

₆₆

mol Cl ₂

mol Al > 3

2

Lim reag Lim reag = Al = Al

29 29

If If

There is not enough Cl

There is not enough Cl

₂₂

to use to use

up all the Al

up all the Al

2 Al + 3 Cl

2 Al + 3 Cl

₂₂

---> Al --- > Al

₂₂

Cl Cl

₆₆

mol Cl ₂

mol Al < 3

2

Lim reag Lim reag = Cl = Cl _{2 2}

Deciding on the Limiting

Reactant

Deciding on the Limiting

Deciding on the Limiting

Reactant

Reactant

30 30

We have 5.40 g of Al and 8.10 g of Cl We have 5.40 g of Al and 8.10 g of Cl₂₂

Step 2 of LR problem:

Calculate moles of each reactant

Step 2 of LR problem:

Step 2 of LR problem:

Calculate moles of each reactant

Calculate moles of each reactant

5.40 g Al • 1 mol

27.0 g = 0.200 mol Al

8.10 g Cl₂ • 1 mol

70.9 g = 0.114 mol Cl₂

31 31

Find mole ratio of reactants

Find mole ratio of reactants

Find mole ratio of reactants

This This

would be 3/2, or 1.5/1, if would be 3/2, or 1.5/1, if reactants are present in the reactants are present in the exact stoichiometric ratio.

exact stoichiometric ratio.

Limiting reagent is Limiting reagent is

Cl Cl _{2 2}

mol Cl ₂

mol Al = 0.114 mol

0.200 mol = 0.57

2 Al + 3 Cl

2 Al + 3 Cl

₂₂

--- --- > Al > Al

₂₂

Cl Cl

₆₆

32

Mix 5.40 g of Al with 8.10 g of Cl

32

Mix 5.40 g of Al with 8.10 g of Cl

₂₂

. .

What mass of Al

What mass of Al

₂₂

Cl Cl

₆₆

can form? can form?

Limiting reactant = Cl

₂

Base all calcs. on Cl

₂

Limiting reactant = Cl

Limiting reactant = Cl

₂₂

Base all

Base all calcs calcs. on Cl . on Cl

₂₂

moles Cl₂

moles Al₂Cl₆ grams

Cl₂

grams Al₂Cl₆

1 mol Al₂Cl₆ 3 mol Cl₂ 2 Al + 3 Cl

2 Al + 3 Cl₂₂ ---> Al---> Al₂₂ClCl₆₆

33

CALCULATIONS: calculate mass of

33

Al

₂

Cl

₆

expected.

CALCULATIONS: calculate mass of

CALCULATIONS: calculate mass of

Al Al

₂₂

Cl Cl

₆₆

expected. expected.

Step 1:

Step 1: Calculate moles of AlCalculate moles of Al₂₂ClCl₆₆ expected based on LR.

expected based on LR.

0.114 mol Cl₂ • 1 mol Al₂Cl₆

3 mol Cl₂ = 0.0380 mol Al₂Cl₆

0.0380 mol Al₂Cl₆ • 266.4 g Al₂Cl₆

mol = 10.1 g Al₂Cl₆ Step 2:

Step 2: Calculate mass of AlCalculate mass of Al₂₂ClCl₆₆expected expected based on LR.

based on LR.

34 34

•

• Cl Cl

₂₂

was the limiting reactant. was the limiting reactant.

• • Therefore, Al was present Therefore, Al was present

in excess. But how much?

in excess. But how much?

• • First find how much Al was required. First find how much Al was required.

• • Then find how much Al Then find how much Al is in excess. is in excess.

How much of which reactant will

remain when reaction is complete? How much of which reactant will How much of which reactant will

remain when reaction is complete?

remain when reaction is complete?

35 35

2 Al + 3

2 Al + 3 Cl Cl

₂₂ productsproducts

0.200 mol

0.200 mol

0.200 mol 0.114 mol = LR 0.114 mol = LR 0.114 mol = LR

Calculating Excess Al

Calculating Excess Al

Calculating Excess Al

Excess Al = Al available

Excess Al = Al available --Al requiredAl required 0.114 mol Cl₂ • 2 mol Al

3 mol Cl₂ = 0.0760 mol Al req' d

= 0.200 mol

= 0.200 mol -- 0.0760 mol 0.0760 mol

=

= 0.124 mol Al in excess0.124 mol Al in excess

36

Using Stoichiometry to

36

Using Stoichiometry to

Determine a Formula

Determine a Formula

First, recognize that all C in CO

First, recognize that all C in CO

₂₂

and all H in H and all H in H

₂₂

O O

is from

is from C C

_xx

H H

_yy

. .

CC_xxHH_yy+ some oxygen + some oxygen ------> >

0.379 g CO

0.379 g CO₂₂+ 0.1035 g H+ 0.1035 g H₂₂OO

Puddle of CxHy

0.115 g

0.379 g CO 0.379 g CO₂₂

+O2

+O₂ 0.1035 g H₂O

1 H₂O molecule forms for each 2 H atoms in C_xH_y 1 CO₂molecule forms for

each C atom in C_xH_y

37

Using Stoichiometry to

37

Using Stoichiometry to

Determine a Formula

Determine a Formula

First, recognize that all C in CO

First, recognize that all C in CO

₂₂

and all H and all H

in H

in H

₂₂

O is from O is from C C

_xx

H H

_yy

. .

1. Calculate amount of C in CO

1. Calculate amount of C in CO

₂₂

8.61 x 10

8.61 x 10

^{-3 -3}

mol CO mol CO

₂₂

--> 8.61 x 10 -- > 8.61 x 10

^{-3 -3}

mol C mol C

2. Calculate amount of H in H

2. Calculate amount of H in H

₂₂

O O

5.744 x 10

5.744 x 10

^--33

mol H mol H

₂₂

O -- O -- >1.149 x 10 >1.149 x 10

^{-2 -2}

mol mol

H

H

C

C_xxHH_yy+ some oxygen + some oxygen ------> >

0.379 g CO

0.379 g CO₂₂+ 0.1035 g H+ 0.1035 g H₂₂OO

38

Using Stoichiometry to

38

Using Stoichiometry to

Determine a Formula

Determine a Formula

Now find ratio of mol H/mol C to find values of

Now find ratio of mol H/mol C to find values of

“

“x x” ” and and “ “y y” ” in C in C

_xx

H H

_yy

. .

1.149 x 10

1.149 x 10 ^{--2 2} mol H/

mol H / 8.61 x 10

8.61 x 10^{--3 3}

mol C

mol C

=

= 1.33 mol H

1.33 mol H

/ 1.00 mol C /

1.00 mol C

=

= 4 mol H

4 mol H

/ 3 mol C /

3 mol C

Empirical formula = C

Empirical formula = C₃₃

H

H₄₄

C

C_xxHH_yy+ some oxygen + some oxygen ------> >

0.379 g CO

0.379 g CO₂₂+ 0.1035 g H+ 0.1035 g H₂₂OO