Stoichiometry Chapter 11 (page 326). Essential Question!! Why do scientist compare mole ratios, and why is it important in chemical reactions??

Stoichiometry

Chapter 11(page 326)

Essential Question!!

• Why do scientist compare mole ratios, and why is it important in chemical reactions??

Vocabulary: Section 1

• Stoichiometry• Moles• Stoichiometric equivalent• Mole ratio• Gram to gram calculations

Chemical equations tell stories…

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

Chemical equations tell stories…

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

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 molecules

1 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

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?

≠+

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

2 C atoms

2 O atoms2 O atoms

2 C atoms

4 O atoms

Number of atoms is

conserved

This chemical equation is balanced

=

=+

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

Coefficients are important

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

Fermentation of sugar (glucose) into alcohol:

1 moleglucose

2 molesethanol

2 molescarbon dioxide

1 moles

3 moles

7.5 moles

glucose will yield

2 moles

6 moles

15 moles

ethanol/ carbon dioxide

x 7.5x 3

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

Fermentation of sugar (glucose) into alcohol:

1 moleglucose

2 molesethanol

2 molescarbon dioxide

Write as a ratio: 1 2

2 1

mole glucose moles ethanol

moles ethanol mole glucose

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

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

Mole ratios

Consider the following equation:

carbon monoxide

hydrogen methanol

3

3

11

1 1

molemole

m

CH

ol

OHCO

e mCH OH Oole C

Compare the reactant CO to the product CH3OH.

Stoichiometry1. Mass of substance A

2. Amount in mol of

Substance A

3. Amount in mol of

substance B

4. Mass of Substance B

Convert by dividing by the molar mass of A

Convert by mult. by molar mass of B

Convert using

the mol ratio

given in the chemical equation

B

A

Process for calculating grams from grams given

Introduction to Stoichiometry

1. mole --> molemoles of A X coeff B

coeff A= moles B

2. Mole--> massmoles of A X coeff B

coeff AX molar mass

of B = mass of B

Solution Route

3. mass --> molemass A FM A X

coeff Bcoeff A

= mol B

4. mass --> massmass AFM A X coeff B

coeff A X FM B = mass B

Method Mass - MassMethod Mass - Mass

mass Amass Acoeff A X FM A coeff A X FM A ==

mass Bmass B

coeff B X FM Bcoeff B X FM B

Method Mass - VolumeMethod Mass - Volume

mass Amass Acoeff A * FM A coeff A * FM A

== vol Bvol Bcoeff B * 22.4coeff B * 22.4

Method Volume - VolumeMethod Volume - Volume

Vol AVol Acoeff Acoeff A

** vol Bvol Bcoeff Bcoeff B

Assignment

• Take a new sheet of paper and fold it into three sections

• Write your name, the title of the chapter and the number

• On the first section from the sheet of paper, please write six things that you learned from your notes so far that could appear on your test.

1.Given the equation N2 + 3 H2 2NH3 If 4.00 mol of H2 react, how many mol of NH3 will be produced?

4.0 mol X23

= 2.66 mol

How many moles of sodium will react with water to produce 4.00 mol of Hydrogen?

2 Na + 2H2O 2 NaOH + H2

4 mol H2 X2

18 mol sodium=

How many moles of H2SO4 will react with 18.0 mol of Al?

2 Al + 3 H2SO4 Al2(SO4)3 + 3 H2

18.0 mol Al X 32

= 27.0 moles

What mass of KClO3 do you need to produce 0.50 mol of O2?

2 KClO3 2 KCl + 3 O2

0.50 mol X 23

= 0.33 mol KClO3

0.33 mol KClO3 X 122.6 g/mol

= 40.866 g

What mass of Zn metal do you need to produce 0.50 mol of ZnCl2?

Zn (s) + 2HCl ZnCl2(s) + H2(g)

0.50 mol X 11

= 0.5 Zn

0.5 mol Zn X 65.4 g/mol

= 32.7 g Zn

What mass of NH3 do you need to produce 0.25 mol of H2?

N2 + 3H2(g) 2 NH3

0.25 mol X 32

= 0.375 H2

0.375 mol H2 X 0.75 g/mol

= 0.75 g H2

Assignment

• Write a three dollar summary of this section (based on what you learned) and using three vocabulary words learned

• Answer questions # 1- 3 on page 360• Turn in completed work

• Honors Chemistry Homework:– Page 362 # 38 - 45


• Percent yield• Actual yield• Theoretical yield

In theory, all 100 kernels should have popped.

Did you do something wrong?

+

100 kernels 82 popped 18 unpopped

+

100 kernels 82 popped 18 unpopped

82100 82%

100

100amount of corn popped

percent yieldamount of kern

percent yie

els in the bag

ld

What you get to eat!

Percent yield

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.

Percent yield in the lab

Decomposition of baking soda:

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

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?)

Percent yield in the lab

Decomposition of baking soda:

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

10.00 g 4.87 g

measured experimentally

Let’s calculate the percent yield

100actual yield

percent yieldtheoretical yield

obtained in experiment

calculated

Conversions of Mass to Amounts In Moles

Example

Mass to Mass Calculations

Example

Assignment

• Write a three dollar summary of this section (based on what you learned)


• Honors Chemistry Homework– Page 363 # 46 - 57

Vocabulary: 3

• Limiting reactant• Excess reactant

4 slices of bread 4 slices of ham 2 slices of cheese2 ham & cheese

sandwich

Suppose you want to make 2 ham & cheese sandwiches

Recipe:

Suppose you want to make 2 ham & cheese sandwiches

Can you still make 2 ham & cheese sandwiches if you have4 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 Reactants

• A limiting reactant is the reactant that limits the amount of the other reactants that can combine to form product in a chemical reaction

• An excess reactant is the reactant that is left over after the other reactant runs out, or the reaction is completed

Excess reactant

For a chemical reaction:

Reactant A is in excess

so the reaction will stop when you

run out of reactant B.

Reactant B is the limiting

reactant.

The amount of product C will

depend on how much reactant B

is present.

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)150 g 60 g

1. What is the limiting reactant?

Step 1: Convert masses to moles

Step 2: Use mole ratios to find the limiting reactant





2 32 3 2 3

2 3

1150 0.94

159.7

160.0 2.22

26.98

mole Fe Og Fe O moles Fe O

g Fe O

mole Alg Al moles Al

g Al

2 32 3

2 32 3

20.94 1.88

1

12.22 1.11

2

moles Almoles Fe O moles Al

moles Fe O

mole Fe Omoles Al moles Fe O

moles Al

needed to react with all Fe2O3

needed to react with all Al

available Fe2O3

available Al





2 32 3

22

33

1150

159.7

16

0.94

2.0.026.

28

29

molesmole Fe O

g Fe Og Fe O

mole Al

Fe O

g Al mg A

ol s ll

e A

2 3

2 32 3

2 3

20.94

1

12.22

1.88

1.12

1

moles Almoles Al

moles Fe Omoles Fe O

mole Fe Omoles A ml

mololes

es Ae

lF O

needed to react with all Fe2O3


available Fe2O3

available Al



2 32 3

22

33

1150

159.7

160.0 2.22

0.

26.98

94 molesmole Fe O

g Fe Og Fe O

mole Al

Fe O

g Al moles Alg Al

2 3

2 32 3

2 3 1.11

20.94 1.88

1

12.22

2

moles Almoles Fe O moles Al

moles Fe O

mole Fe Omoles Al

momole

less Fe O

Al


available Fe2O3

There is not enough Fe2O3 available to react with all the Al, so Fe2O3 is the limiting reactant

Example # 1 & 2

N2(g) + 3H2(g) → 2NH3(s)

The Haber-Bosch process for the synthesis of ammonia:

N2(g) + 3H2(g) → 2NH3(s)

The Haber-Bosch process for the synthesis of ammonia:


• Excess reactant• Percent yield

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

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

Asked: Amount of excess reactant left

Given: N2 is the excess reactant

1.15 moles N2 (need)

1.66 moles N2 (have)

Relationships:

28.01 g/mole N2

Solve: 1) How many moles N2 remain?

2) Convert moles to grams

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

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:

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

Asked: Concentration of CuSO4(aq)

Given: 1.0 L of solution is tested

0.021 g CuS (formed)

Relationships:

95.61 g/mole CuS

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?

42.20 10 moles CuS4

42.20 10 moles CuSO

Have:

Reactants in solution

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

Assignment

• On the second section of that sheet of paper, please write six things that you learned from your notes so far that could appear on your test.

Percent Yield• Theoretical yield is the maximum amount of

product that can be produced from a given amount of reactant

• Actual yield is the measured amount of product obtained from a reaction

• Percent yield is the ratio of the actual yield to the theoretical yield

Percent Yield

Percent yield = Actual yield

theoretical yieldX 100 %

Assignment

• Write a three dollar summary of this section (based on what you learned) and using the vocabulary learned this section


• Honors Chemistry Homework– Page 364 # 58 - 63

Test:- Next Tuesday or Thursday depending on your class

• Homework requirement: page 360 # 1 - 8

• Make sure you have all work assigned between pages 360 and 363 completed and turned in by your test date.

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Stoichiometry Chapter 11 (page 326). Essential Question!! Why do scientist compare mole ratios, and why is it important in chemical reactions??