Stoichiometry

1

Stoichiometry

Chapter 11Natural Approach to

Chemistry

Assignments:11.1 362/38cd, 39cd, 43acde, 40cd11.2 363/46-48; 363/49,50ab,5111.3 364/59-6211.4 365/64 and 66

2

3

Learning Objectives

• Apply the mole concept and the law of conservation of mass to calculate quantities of chemicals participating in reactions.

• Important terms: stoichiometry, percent yield, actual yield, theoretical yield, limiting reactant

4

Chemical equations tell stories…

… and stories can be put into different categories

Nonfiction

Science fiction

Adventure

Romance

History

Psychology

Children’s literature

…

Synthesis / Decomposition

Single / Double replacement

Precipitate reaction

Polymerization reaction

2CO(g) + O2(g) → 2CO2(g)

5


But what exactly do they tell us?

2CO(g) + O2(g) 2CO2(g)

They tell us what compounds we start with:

Carbon monoxide (CO) gas

Oxygen (O2) gas

what compounds are formed:

Carbon dioxide (CO2) gas

6


What else do they tell us?

2CO(g) + O2(g) 2CO2(g)

stoichiometry: the study of the amounts of substances involved in a chemical reaction.

2 CO molecules 2 CO2 molecules1 O2 molecules

They tell us how much of each compound is involved

7

2CO(g) + O2(g) 2CO2(g)

2 CO molecules2 dozen CO molecules2 moles CO molecules

2 x (6.023 x 1023) CO molecules

1 O2 molecules1 dozen O2 molecules1 mole O2 molecules

(1 x) 6.023 x 1023 O2 molecules

2 CO2 molecules2 dozen CO2 molecules2 moles CO2 molecules

2 x (6.023 x 1023) CO2 molecules

8

2CO(g) + O2(g) 2CO2(g)

2 moles

CO molecules

1 mole

O2 molecules

2 moles

CO2 molecules

Number of moles is

not conserved

Is that okay?

≠+

Yes, as long as the chemical equation is balanced! The coefficients are important!!!

9

2CO(g) + O2(g) 2CO2(g)

2 moles

CO molecules

1 mole

O2 molecules

2 moles

CO2 molecules

This chemical equation is balanced

The coefficients are correct

CoefficientsThese are important!

10

Coefficients are important

+ 3 eggs + ¼ cup oil + 1 cup water 1 batch cupcakes1 bagcake mix

1 14114

cup oil batch cupcakesor

batch cupcakes cup oilWrite as a ratio:

11

Write as a ratio:

Coefficients are important

+ 3 eggs + ¼ cup oil + 1 cup water 1 batch cupcakes1 bagcake mix

1 4

4 1

cup oil batches cupcakesor

batches cupcakes cup oil

12

1 2

2 1

mole glucose moles ethanol

moles ethanol mole glucose

C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g)

Fermentation of sugar (glucose) into alcohol:

1 moleglucose

2 molesethanol

2 molescarbon dioxide

Write as a ratio:

These are stoichiometric equivalents

13



1 moleglucose

2 molesethanol


Write as a ratio: 1 2

2 1



mole ratio: a ratio comparison between substances in a balanced equation. It is obtained from the coefficients in the balanced equation.

1411.1 Analyzing a Chemical Reaction



1 moleglucose

2 molesethanol


1 2

2 1



2

2

21

2 1

moles COmole glucose

moles CO mole glucose

2

2

22

2 2

moles COmoles ethanol

moles CO moles ethanol

mole ratios for

this chemical equation

Mole ratios

15

CO(g) + 2H2(g) CH3OH(l)

Mole ratiosConsider the following equation:

carbon monoxide

hydrogen methanol

If the reaction produces 5 moles of CH3OH, how many moles of H2 are consumed?

Asked: moles of H2

5 moles CH3OH x 2 moles H2 = 10 moles H2

1 mole CH3OH

16

A mixture of aluminum metal and chlorine gas reacts to form aluminum chloride (AlCl3): 2Al(s) + 3Cl2(g) → 2AlCl3(s). How many moles of aluminum chloride will form when 5 moles of chlorine gas react with excess aluminum metal?

Asked: moles AlCl3

Given: moles Cl2

5 mole Cl2 x 2 mole AlCl3 = 3.3 mole AlCl3

3 mole Cl2

17

Potassium + Hydrogen Phosphate

Finish the reaction in symbols and balance…

If 14.72 moles of hydrogen phosphate are consumed in the above reaction, how many moles of hydrogen are produced?

18

19

There is no scale that measures in moles!

How do you convert from moles to grams?

The mass of 1 mole of Al is not the same as the mass of 1 mole of Cl2.

How do you convert from grams of Al to grams of Cl2?

By using the molar mass (g/mole)

By using the molar mass (g/mole) and mole ratios

mass moles….

20

Process for calculating grams from grams given

21

If 45.0 g of calcium carbonate (CaCO3) decomposes in the reaction CaCO3(s) → CaO(s) + CO2(g), how many grams of CO2 are produced?

Asked: grams of CO2 Given: grams of CaCO3

Relationships: mole ratios molar mass of CaCO3 = 40.078 + 12.011 + (15.999 x 3) =

100.0 g/molemolar mass of CO2 = 12.011 + (15.999 x 2) = 44.01 g/mole

Strategy:

22

If 45.0 g of calcium carbonate (CaCO3) decomposes in the reaction CaCO3(s) → CaO(s) + CO2(g), how many grams of CO2 are produced?

0.45 mole CaC03 x 1 mole CO2 x 44.01 g CO2 = 19.8 g CO2

1 mole CaC03 1 mole C02

B B

23

CHAPTER 11

Stoichiometry

11.2 Percent Yield and Concentration

24

In theory, all 100 kernels should have popped.

Did you do something wrong?

+

100 kernels 82 popped 18 unpopped

25

+


In real life (and in the lab) things are often not perfect

In theory, all 100 kernels should have popped.

Did you do something wrong?No

26

+


100

82100

10082%

amount of corn poppedpercent yield

amount of kernels in the bag

percent yield

What you get to eat!

Percent yield

27

100actual yield

percent yieldtheoretical yield

100actual yield

theoreticalpercent yi

yieldeld

actual yield: the amount obtained in the lab in an actual experiment.

theoretical yield: the expected amount produced if everything reacted completely.

28

Percent yield in the labDecomposition of baking soda:

2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)

Heating

29



10.00 g 4.87 g

measured experimentally

-There is usually some human error, like not measuring exact amounts- carefully

-Maybe the heating time was not long enough; not all the Na2HCO3

- reacted

- Maybe Na2CO3 was not completely dry; some H2O(l) was measured too

- CO2 is a gas and does not get measured

Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?)

30


2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)10.00 g 4.87 g


Let’s calculate the percent yield

100actual yield


obtained in experiment

calculated

31

Percent yield in the lab

Decomposition of baking soda:2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)

10.00 g 4.87 g



4.810

70percent yield

theoretical y eld

g

i

calculated

32

Percent yield in the lab

Decomposition of baking soda:


10.00 g 4.87 g



4.87100

theoretica

gpercen

l yieldt yield

calculated

33

2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)10.00 g

This is a gram-to-gram conversion:

33 3

3

3

22.99 1.0079 12.011

110.00 0.1

84.01

19084.0

3(15.999

1

/)molar mass of NaHCO g mo

mole NaHCOg NaHCO moles NaHCO

g NaHCO

le

10.00g NaHCO3 1 mole NaHCO3 1 mole Na2CO3 1mole Na2CO3 = 84.01 g NaHCO3 2 mole NaHCO3 105.99 g Na2CO3

Answer: Mass of D = 6.306 g Na2CO3

2 3

2 32 3 2 3

2 3

105.990.

22.99 2 12.011 15.999 3

05950 6.3061

105.99 /

g Na COmoles Na CO g Na CO

mole Na CO

molar mass of Na CO g mole

34


0.05950 moles 6.306 g10.00 g 0.1190 moles

4.87 g

For 10.00 g of starting material (NaHCO3), the theoretical yield for Na2CO3 is 6.306 g.

The actual yield (measured) is 4.87 g.

35


For 10.00 g of starting material (NaHCO3), the theoretical yield for Na2CO3 is 6.306 g.

The actual yield (measured) is 4.87 g.

4.87 g

100

4.87100 77.2%

6.306

actual yieldpercent yield

theoretical yield

gpercent yield

g

36

Decomposition of baking soda: (We just looked at this.)


Reaction of solid zinc with hydrochloric acid:

Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq)50.0 mL of a

3.0 M solution

Convert to moles

Convert to moles

Reactions in solution

Stoichiometry with solutions

37

A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to:

Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess.

Solve:

Asked: grams of H2 producedGiven: 50.0 mL of 3.0 M HCl Reacting with excess zinc

Relationships: M = mole/LMole ratio: 2 moles HCl ~ 1 mole H2

Molar mass of H2 = 1.0079 x 2= 2.02 g/mole

Answer: 0.15 grams of H2 are produced

0.0500L HCl x 3.0mole HCl x 1 mol H2 x 2.02 g H2 = 0.15 g H2

L HCl 2 mol HCl 1 mol H2

38

39

Reaction of solid zinc with hydrochloric acid:

Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq)50.0 mL of a

3.0 M solution

Convert molarity to moles

Vinegar is 5% acetic acid by mass

Sometimes the concentration is written in mass percent

% 100mass of compound

mass of compoundtotal mass of solution

40

Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.)

Asked: grams of acetic acid in 120 mL of vinegarGiven: 120 mL of vinegar and 5% acetic acid by massRelationships: 120 mL = 120 g,

given a density of 1.0 g/mL

Solve:

% 100mass of acetic acid

massmass of solution

%

100

5%

100 120

mass mass of acetic acid

mass of solution

mass of acetic acid

g

Answer: 6.0 g of acetic acid.

0

0.05

.05

6

12

2

.

0

1

0

0

mass of acetic acid g

mass o

mass of acetic ac

f acetic acid

id

g

g

41

100actual yield


100actual yield

theoreticalpercent yi

yieldeld

% 100mass of compound

mass of compoundtotal mass of solution

Calculate using molar masses and mole ratios

Obtained from the experiment

Let’s Review:

42

Assignments 11.2:

Limiting Reactants

Ch 11.3

Suppose you want to make 2 ham & cheese sandwiches

Can you still make 2 ham & cheese sandwiches if you have 4 slices of bread, 4 slices of ham, and 1 slice of cheese?

No, you are limited by the cheese!You can only get 1 ham & cheese sandwich.

Limiting factor

limiting reactant: the reactant that “runs out” first in a chemical reaction.

excess reactant: the reactant that is remaining after the reaction is complete.

Limiting reactant

Excess reactant

Steps for Determining the Limiting Reactant Step 1 Step 2 Step 3

Convert both reactant masses to moles.

Multiply by the mole ratio from the balanced equation to find how much reactant is needed to use up all of the other reactant.

Compare the amounts of reactants. Compare what you have available to what you need.

This gives you the amount you have available to use.

This gives you the amount you need to consume all of the reactant.

If what you need is more than what you have, then this is the limiting reactant.

Sample Problem: 364/58.Iron can be produced from the following reaction:

Fe2O3 + 2Al 2Fe + Al2O3

a. If 100.0 g of Fe2O3 reacts with 30.0 g Al, which one will be used up first?

Sample Problem: 365/58.Iron can be produced from the following reaction:

Fe2O3 + 2Al 2Fe + Al2O3

b. For this next step, use the smaller mole answer from a. to find the needed amount

c. How much Fe can be produced?

1.112 mole Al 2 mole Fe 55.85 g Fe = 62.10 g Fe 2 mole Al 1 mole Fe

d. How much of the excessive reactant is remaining?Fe2O3 is the excessive reactant. mole Al mole Fe2O3 mass Fe2O3

Have – used = excess

1.112 mole Al 1 mole Fe2O3 159.70 g Fe2O3 = 88.795gFe2O3

2 mole Al 1 mole Fe2O3

100.0g – 88.80 g = 11.20 g remaining (excess)

Asgn: 364/59, 60

361/31,34,58-6211.3 Assignment

CHAPTER 14.4

Solving Stoichiometric Problems

Section 11.1 Analyzing a Chemical Reaction

Section 11.2 Percent Yield and Concentration

Section 11.3 Limiting Reactants

Section 11.4 Solving Stoichiometric Problems

Use what we’ve learned to answer these questions:

- What is the limiting reactant?

- What is the theoretical yield?

- What is the percent yield?

- How much excess reactant is left?

- How much reactant is used if it’s in a solution?











What is the limiting reactant?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the limiting reactant? Lithium is the only group 1 metal that is capable of reacting directly with nitrogen gas.

6Li(s) + N2(g) → 2Li3N(s)

Solve: 1) Moles of each reactant?2) Apply the mole ratio3) Compare what we have with what we need


When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the limiting reactant? Lithium is the only group 1 metal that is capable of reacting directly with nitrogen gas.

Asked: Limiting reactantGiven: 48.0 g of Li (have)

46.5 g of N2 (have)

6Li(s) + N2(g) → 2Li3N(s)


molar mass of Li = 6.941 g/molemolar mass of N2 = 28.01 g/molemole ratio: 6 moles Li ~ 1 mole N2


48.0 g of Li reacts with 46.5 g of N2

6Li(s) + N2(g) → 2Li3N(s)


2

148.0

6.941

146.5

28.01

6.92 ( )

1.66 ( )

moleg

g

moleg

moles Li have

moles N veg

ha

What is the limiting reactant?Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

6Li(s) + N2(g) → 2Li3N(s)?

2) Apply the mole ratio

How much N2 do we need to react with 6.92 moles Li?

We have: 6.92 moles Li; 1.66 moles N2

22 1.15 (6. 2

6)

19

mole Nmoles Li

molmoles N n

es Ld

iee

Do we have enough N2?

Yes, we have more than enough N2. That means we will run

out of Li before we run out of N2












How much lithium nitride (LiN3) can be produced from this reaction?

6Li(s) + N2(g) → 2Li3N(s)

What is the theoretical yield?

Asked: Amount of Li3N produced

Given: Li is the limiting reactant6.92 moles Li (have)

Relationships:molar mass of Li3N = 34.83 g/mole

mole ratio: 6 moles Li ~ 2 moles Li3N

3

3 6.941 14.007

34.83

molar mass of Li N

g g

mole mole

g

mole


How much lithium nitride (Li3N) can be produced from this reaction?

6Li(s) + N2(g) → 2Li3N(s)


From the last problem






How much lithium nitride (Li3N) can be produced from this reaction?

6Li(s) + N2(g) → 2Li3N(s)


Solve: 1) Find moles of Li3N

2) Convert moles to grams








6Li(s) + N2(g) → 2Li3N(s)


3

33

34.832.31 80.4

26. 2

6

926

.

1

31molemoles Li N

moles Limoles L

gmoles g Li N

mole

is Li N


6Li(s) + N2(g) → 2Li3N(s)








3

33

80.46

26.9

34.83

2 2.3

2.311

16

moles Li Nmoles Li mole

gmoles

mole

s Li Nmoles L

N

i

g Li


6Li(s) + N2(g) → 2Li3N(s)








3

33

80.46

26.92 2.31

6

34.832.31

1

moles Li Nmoles Li moles Li N

moles Li

gmoles

mi

oleg L N

Asked: 80.46 g of Li3N are produced












Calculate the percent yield of an experiment that actually produced 62.5 g of Li3N.

6Li(s) + N2(g) → 2Li3N(s)

What is the percent yield?


Calculate the percent yield of an experiment that actually produced 62.5 g of Li3N.

6Li(s) + N2(g) → 2Li3N(s)


From the last problem

100actual yield


Asked: Percent yield

Given: Theoretical yield: 80.46 g Li3N

Actual yield: 62.5 g Li3N

Relationships:

Solve: Use the percent yield formula


6Li(s) + N2(g) → 2Li3N(s)





Relationships:


100actual yield


62.5 ( )100

807

.46 (7. %

)7

g actualpercent yield

g theoretical


6Li(s) + N2(g) → 2Li3N(s)





Relationships:


100actual yield


62.5 ( )100

807

.46 (7. %

)7

g actualpercent yield

g theoretical

Answer: The percent yield in this particular experiment is 77.7%












How much of the excess reactant remains after the limiting reactant is completely consumed?

6Li(s) + N2(g) → 2Li3N(s)

What is left?

Asked: Amount of excess reactant leftGiven: N2 is the excess reactant

1.15 moles N2 (need)

1.66 moles N2 (have)

Relationships:Molar mass of N2 = 28.01 g/mole


How much of the excess reactant remains after the limiting reactant is completely consumed?

6Li(s) + N2(g) → 2Li3N(s)

What is left?

Solve: 1) How many moles N2 remain?



6Li(s) + N2(g) → 2Li3N(s)

What is left?

Asked: Amount of excess reactant left

Given: N2 is the excess reactant



Relationships:28.01 g/mole N2

Solve: 1) How many moles N2 remain?2) Convert moles to grams

2 2

2

1.66 ( ) 1.15 (

0.51 (

)

)

moles N hav

moles

e mo

N re

les N

main

nee

g

d

in


6Li(s) + N2(g) → 2Li3N(s)

What is left?

2 2

2

1.66 ( ) 1.15 ( )

0.51 ( )

moles N have moles N need

moles N remaining

2 2

28.010.51

114

gmoles N

moleg N









6Li(s) + N2(g) → 2Li3N(s)

What is left?

2 2

2

1.66 ( ) 1.15 ( )

0.51 ( )

moles N have moles N need

moles N remaining

2 2

28.010.51

114

gmoles N

moleg N








Answer: 14 g of N2 will remain at the end of the reaction.

365/64. Aluminum metal reacts with oxygen in the air to form a layer of aluminum oxide according to the equation:

4Al + 3O2 --> 2Al203(s)

A. If 28.0 g of Al reacts with excess oxygen in the air, what mass of aluminum oxide is formed?

B. Molar Mass Al203 = 2Al+3(0) = 2(27)+3(16) = 102 g/mole

28.0gAl 1mole Al 2 mole Al203 102 g Al203 = 52.9 g Al203

27g Al 4 mole Al 1 mole Al203


4Al + 3O2 --> 2Al203(s)

B How many moles of oxygen are consumed during this reaction?

28 g Al 1mole Al 3mole O2 = 0.778 mole O2

27 g Al 4 mole Al


4Al + 3O2 --> 2Al203(s)

C. If 250 g Al203 of is formed, how much Al reacted?

250 g Al203 1 mole Al203 4 mole Al 27.0 g Al

102 g Al203 2 mole Al203 1 mole Al

Answer: 132.4 g Al

365/66. Wine can spoil when ethanol is converted to acetic acid by oxidation:

C2H5OH + O2 --> CH3COOH + H2O

a. Determine the limiting reactant when 5.00 g of ethanol (C2H5OH) and 2.0 g of oxygen are sealed in a bottle.

2C + 6H + O = 24 + 6 + 16 = 46.1 g/mole 5.00g eth. 1 mole eth. = 0.108 moles ethanol 46.1 g eth 2.0 g oxygen 1 mole oxy. = 0.0625 moles oxygen 32.0g oxy.O2 is the limiting reactant.

We converted reactants to moles individually.



b. Calculate how much acetic acid will form in grams. Start with the oxygen because it is the limiting reactant.

0.0625 moles O2 1 mole CH3COOH 60 g CH3COOH

1 mole O2 1 mole CH3COOH

Answer: 3.75 g

Molar mass: 2C + 4H + 2(O) =24+4+32 = 60 g/mole



c. Calculate the amount of excess reactant remaining in grams.

Excess reactant is ethanol.Available less used: 0.109 moles – 0.0625 moles = 0.0465 moles0.0465 moles eth. 46 g eth. = 2.139 g eth.

1 mole eth.Amount of excess ethanol is 2.139 g











The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if 0.021 g of CuS precipitate is formed? The reaction is:

CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)

Reactants in solution



Asked: Concentration of CuSO4(aq)

Given: 1.0 L of solution is tested0.021 g CuS (formed)

Relationships:Molar mass of CuS = 95.61 g/moleMole ratio: 1 mole CuSO4 ~ 1 mole CuS

63.55 32.06 95.61

molar mass of CuS

g

mole






Relationships:Molar mass of CuS = 95.61 g/moleMole ratio: 1 mole CuSO4 ~ 1 mole CuS

Solve: 1) How many moles of CuS?2) How many moles of CuSO4?

3) What is the concentration of CuSO4?


The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is:





Relationships:95.61 g/mole CuS1 mole CuSO4 ~ 1 mole CuS



410.021

95.612.20 10

moleg CuS moles CuS

g









42.20 10 moles CuS

44

4 412.20 10

1

2.20 10

mole CuSOmoles CuS

mole Cu

moles C

S

uSO

Have:








42.20 10 moles CuS4

42.20 10 moles CuSO

Have:


4

44

42.20 10

2.20 1

1.0

0

moles of solutemolarity

L of solution

moles CuSO

M C

L of w

u

a er

SO

t









4

44

42.20 10

2.20 1

1.0

0

moles of solutemolarity

L of solution

moles CuSO

M C

L of w

u

a er

SO

t

Answer: The concentration of CuSO4 is

2.20 x 10-4 M.



Discussion: If the legal limit is 5.0 x 10-4 M of Cu2+, is the industrial plant following the environmental guidelines?



2.20 x 10-4 M.



Discussion: If the legal limit is 5.0 x 10-4 M of Cu2+, is the industrial plant following the environmental guidelines?

Yes, because 2.20 x 10-4 M is less than the legal limit.



2.20 x 10-4 M.

CuSO4(aq) → Cu2+(aq) + SO42–(aq)

42.20 10 M 42.20 10 M

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Stoichiometry