Stoichiometry

Newport High School Academic Chemistry Mrs. Teates

Lesson 1 – Introduction to Stoichiometry

 Lesson Essential Questions: • What is stoichiometry and how is it used to describe reactions?

Vocabulary: stoichiometry, composition stoichiometry, reaction stoichiometry, mole, Avogadro’s number, molar mass

Stoichiometry Definition



Stoichiometry

• mass relationships between substances in a chemical reaction 

Composition stoichiometry

deals with the mass relationships of elements in compounds. 

Reaction stoichiometry

involves the mass relationships between reactants and products in a chemical reaction.

What is the Mole?

 A counting number (like a dozen) • Can be used to measure anything (DVDs, cars, eggs, molecules)  1 mol =

6.02



10 23

items

A amount!!!!

large

What is the Mole?



1 mole of hockey pucks would equal the mass of the moon!



1 mole of basketballs would fill a bag the size of the earth!

 1 mole of pennies would cover the Earth 1/4 mile deep!

Molar Mass

 Mass of 1 mole of an element or compound.

 Atomic mass tells the...

• atomic mass units per atom • (amu) grams per mole (g/mol)  Round to 2 decimal places

Molar Mass Examples

 carbon  aluminum  zinc

12.01 g/mol 26.98 g/mol 65.39 g/mol

Molar Mass Examples

 water  

H 2 O 2(1.01) + 16.00 = 18.02 g/mol

 sodium chloride

NaCl



22.99 + 35.45 = 58.44 g/mol

Molar Mass Examples

 sodium bicarbonate  

NaHCO 3 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol

 sucrose

C 12 H 22 O 11



12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol

Molar Conversions MASS IN GRAMS molar mass 6.02



10 23 (g/mol) MOLES NUMBER OF PARTICLES (particles/mol)

Molar Conversion Examples

 How many moles of carbon are in 26 g of carbon?

26 g C 1 mol C 12.01 g C = 2.2 mol C

Molar Conversion Examples

 How many molecules are in 2.50 moles of C 12 H 22 O 11 ?

2.50 mol 6.02



10 23 molecules = 1.51



1 mol 10 24 molecules C 12 H 22 O 11

Molar Conversion Examples

 Find the mass of 2.1  molecules of NaHCO 3 .

10 24

2.1



10 24 molecule s 1 mol 6.02



10 23 molecules 84.01 g 1 mol = 290 g NaHCO 3

Percentage Composition

 Calculate because it is useful to know the percentage by mass of an element in a compound

mass

_

of

_

element mass

_

of

_ _

in

_

sample sample

_

of

_

of

_

compound

* 100  _

compound percentage

_

compositio n

or

mass

_

of

_

element molar

_ _

in mass

_ _

of

1

mol

_

of

_

compound

* 100  _

compound percentage

_

compositio n

Sample Problems

 Sample Problem J on page 243  Practice problem #1

Mole Ratio



A mole ratio

is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction

Example:

3O 2 (

g

) 2Al 2 O 3 (

l

) → 4Al(

s

) + , ,

Mole Ratios:

4 mol Al 4 mol Al 2 mol Al 2 O 3 3 mol O 2 2 mol Al 2 O 3 3 mol O 2

Converting Between Amounts in Moles

Stoichiometry Calculations

Lesson 2 – Ideal Stoichiometric Calculations

 Lesson Essential Questions: • How are calculations used to describe chemical reactions?

Proportional Relationships

2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies.

 I have 5 eggs. How many cookies can I make?

Ratio of eggs to cookies 5 eggs 5 doz.

= 12.5 dozen cookies 2 eggs

Stoichiometry Steps

1. Write a balanced equation.

2. Identify known & unknown.

3. Line up conversion factors.

• • • • Mole ratio moles moles Molar mass Molarity moles  moles  Molar volume - moles  grams liters soln liters gas Core step in all stoichiometry problems!!

4. Check answer.

Stoichiometry Calculations

Conversion of Quantities in Moles

Conversions of Quantities in Moles Sample Problem A

In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction with lithium hydroxide, LiOH, according to the following chemical equation.

CO 2 (

g

) + 2LiOH(

s

) → Li 2 CO 3 (

s

) + H 2 O(

l

) How many moles of lithium hydroxide are required to react with 20 mol CO 2 , the average amount exhaled by a person each day?

Conversions of Quantities in Moles Sample Problem A Solution

CO 2 (

g

) + 2LiOH(

s

) → Li 2 CO 3 (

s

) + H 2 O(

l

)

Given:

amount of CO 2 = 20 mol

Unknown:

amount of LiOH (mol)

Solution:

2 2  2 CO 2  LiOH

Conversions of Amounts in Moles to Mass

Conversions of Amounts in Moles to Mass Sample Problem B

In photosynthesis, plants use energy from the sun to produce glucose, C 6 H 12 O 6 , and oxygen from the reaction of carbon dioxide and water. What mass, in grams, of glucose is produced when 3.00 mol of water react with carbon dioxide?

Conversions of Amounts in Moles to Mass Sample Problem B Solution Given:

amount of H 2 O = 3.00 mol

Unknown:

mass of C 6 H 12 O 6 produced (g)

Solution:

Balanced Equation: 2 6CO 2 (

g

) + 6H 2 O(

l

) → C 6 H 12 O 6 (

s

) + 6O 2 (

g

)

mol ratio

12 6

molar mass factor

12 6  6 2 6 12 6 12 2 6 2 12 6 6 12 6 12 6 6

=

6 90.1 g C 6 H 12 O 6

Conversions of Mass to Amounts in Moles

Conversions of Mass to Amounts in Moles Sample Problem D

The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.

NH 3 (

g

) + O 2 (

g

) → NO(

g

) + H 2 O(

g

) (unbalanced) The reaction is run using 824 g NH 3 oxygen.

and excess a. How many moles of NO are formed?

b. How many moles of H 2 O are formed?

Conversions of Mass to Amounts in Moles Sample Problem D Solution Given:

mass of NH 3 = 824 g

Unknown: a.

amount of NO produced (mol)

b.

amount of H 2 O produced (mol)

Solution:

Balanced Equation: 4NH 3 (

g

) + 5O 2 (

g

) → 4NO(

g

) + 6H 2 O(

g

)

a.

b.

3 3

molar mass factor mol ratio

3 3 3  3 3 2 3  2

a.

b.

Conversions of Mass to Amounts in Moles Sample Problem D Solution

3

molar mass factor mol ratio

3 3 N H 3  48.4

mol N O 3  3 NH 3 2 4 mol NH 3  72.5

mo 2

Mass-Mass to Calculations

Mass-Mass to Calculations Sample Problem E

Tin(II) fluoride, SnF 2 , is used in some toothpastes. It is made by the reaction of tin with hydrogen fluoride according to the following equation.

Sn(

s

) + 2HF(

g

) → SnF 2 (

s

) + H 2 (

g

) How many grams of SnF 2 are produced from the reaction of 30.00 g HF with Sn?

Mass-Mass to Calculations Sample Problem E Solution Given:

amount of HF = 30.00 g

Unknown:

mass of SnF 2 produced (g)

Solution:

g HF  g HF 

molar mass factor mol ratio molar mass factor

2 2 2  2 2 2 2 = 117.5 g SnF 2

More Practice for Homework

 Page 320 #4-16

Lesson 3 – Limiting Reactants and Percent Yield

 Lesson Essential Questions: • How does the limiting reactant affect the percentage yield and actual yield in a chemical reaction?

Vocabulary: limiting reactant, excess reactant, theoretical yield, actual yield, percent yield

Limiting Reactants



Available Ingredients

• 4 slices of bread • • 1 jar of peanut butter 1/2 jar of jelly 

Limiting Reactant

• bread 

Excess Reactants

• peanut butter and jelly

Limiting Reactants



Limiting Reactant

• • used up in a reaction determines the amount of product 

Excess Reactant

• added to ensure that the other reactant is completely used up • cheaper & easier to recycle

Limiting Reactants

1. Write a balanced equation.

2. Write your known and unknown.

3. For each reactant, calculate the amount of product formed.

4. Smaller answer indicates: • limiting reactant • amount of product

Limiting Reactant Practice

 Sample Problem F on page 313.

• Practice problem #1 on page 313.

 Sample Problem G on pages 314-315.

• Practice problems #1 & 2 on page 315.

 Homework pg. 321 #22-25

Percent Yield

measured in lab

% yield



actual yield theoretica l yield



100

calculated on paper

Percent Yield

 When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 + 2HCl  45.8 g 2KCl + H 2 O + CO 2 ? g actual: 46.3 g

Percent Yield

K 2 CO 3 + 2HCl  45.8 g 2KCl + H 2 O + CO 2 ? g actual: 46.3 g Theoretical Yield: 45.8 g K 2 CO 3 1 mol K 2 CO 3 2 mol KCl 138.21 g K 2 CO 3 1 mol K 2 CO 3 74.55

g KCl = 49.4

1 mol g KCl KCl

Percent Yield

K 2 CO 3 + 2HCl  45.8 g 2KCl + H 2 O + CO 2 49.4 g actual: 46.3 g Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g  100 = 93.7%

Percent Yield Practice

 Sample problem H on page 317.

• Practice problem #1 & 2 on page 318.

Homework

 Page 321 #22-30, 31-36, 38  Review for the test – Worksheet titled Stoichiometry Review Problems

Works Cited

 Modern Chemistry Textbook  www.nclark.net

 http://mrsj.exofire.net/chem/