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Unit 08 – Moles and Stoichiometry. Molar Conversions. A. What is the Mole?. VERY. A large amount!!!!. A counting number (like a dozen) Avogadro’s number ( 6.02  10 23 particles ) (SI unit) 1 mol = molar mass. A. What is the Mole?. HOW LARGE IS IT???.

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a what is the mole
A. What is the Mole?

VERY

A large amount!!!!

  • A counting number (like a dozen)
  • Avogadro’s number (6.02  1023 particles) (SI unit)
  • 1 mol = molar mass
a what is the mole1
A. What is the Mole?

HOW LARGE IS IT???

  • 1 mole of pennies would cover the Earth 1/4 mile deep!
  • 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!
b molar mass
B. Molar Mass
  • Molar Mass- the mass of a mole of any element or compound (in grams)
      • Round to 2 decimal places
  • Also called:
      • Gram Formula mass – sum of the atomic masses of all the atoms in a formula of a compound
      • Formula weight
b molar mass examples
B. Molar Mass Examples
  • water
  • sodium chloride
  • H2O
  • (1.01g x 2) + 16.00g = 18.02 g
  • NaCl
  • 22.99g + 35.45g = 58.44 g
b molar mass examples1
B. Molar Mass Examples
  • sodium hydrogen carbonate
  • sucrose
  • NaHCO3
  • 22.99g + 1.01g + 12.01g + (16.00g x 3) = 84.01 g
  • C12H22O11
  • (12.01g x12) + (1.01g x 22) + (16.00g x11)= 342.34 g
c number of particles in a mole
C. Number of Particles in a Mole

1 mole = 6.02 × 10 23representative particles

(also called Avogadro’s Number)

Atom- rep. particle for most elements

Ions – if atom is charged

Molecule- rep. particle for covalent compounds and diatomic molecules “BrINCl HOF”

Formula unit- rep. particle for ionic compounds

What is a representative particle?

How the substance normally exists:

d volume of a mole of gas
D. Volume of a Mole of Gas
  • The Volume of a gas varies with a change in temperature or pressure.
  • Measured at standard temperature and pressure (STP)
      • 0°C at 1 atmosphere (atm)
  • 1 mole of any gas occupies a volume of 22.4L
the mole road map
The MoleRoad Map

Molecule

Atoms (ions)

Formula unit

e molar conversion examples
E. 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

e molar conversion examples1
E. Molar Conversion Examples
  • How many molecules are in 2.50 moles of C12H22O11?

6.02  1023

molecules

1 mol

2.50 mol

= 1.51  1024

molecules

C12H22O11

e molar conversion examples2
E. Molar Conversion Examples
  • Find the number of molecules of 12.00 L of O2 gas at STP.

6.02 x 1023 molecules

1 mol

1 mol

22.4 L

12.00 L

= 3.225 x 1023 molecules

a proportional relationships
A. Proportional Relationships

Ratio of eggs to cookies

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

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.

5 eggs

5 doz.

2 eggs

= 12.5 dozen cookies

a proportional relationships1
A. Proportional Relationships
  • Stoichiometry
    • mass relationships between substances in a chemical reaction
    • based on the mole ratio
  • Mole Ratio
    • indicated by coefficients in a balanced equation

2 Mg + O2 2 MgO

b stoichiometry steps
B. Stoichiometry Steps

1. Write a balanced equation.

2. Identify known & unknown.

3. Convert known to moles (IF NECESSARY) Line up conversion factors.

4. Use Mole ratio – from equation

5. Convert moles to unknown unit

(IF NECESSARY)

6. Calculate and write units.

slide17

C. Stoichiometry Problems

  • Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen.

N2 + 3H2→ 2NH3

a limiting reactants
A. 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
a limiting reactants1
A. 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
a limiting reactants2
A. Limiting Reactants

1. Write a balanced equation.

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

3. Smaller answer indicates:

  • limiting reactant
  • amount of product
a limiting reactants3
A. Limiting Reactants

Using the following equation identify the limiting reagent.

  • How many moles of ammonia (NH3) can be produced from the reaction of 28.2 L of nitrogen and 25.3 L of hydrogen?

N2 + 3H2 2NH3

28.2 L

? mol

25.3 L

a limiting reactants4
A. Limiting Reactants

N2 + 3H2 2NH3

28.2 L

? mol

25.3 L

28.2

L N2

1 mol

N2

22.4

L N2

2 mol

NH3

1 mol

N2

= 2.5 mol NH3

a limiting reactants5
A. Limiting Reactants

N2 + 3H2 2NH3

28.2 L

? mol

25.3 L

25.3

L H2

1 mol

H2

22.4 L H2

2 mol

NH3

3 mol

H2

= 0.753 mol NH3

a limiting reactants6
A. Limiting Reactants

N2: 2.5 mol NH3 H2: 0.753 mol NH3

Limiting reactant: H2

Excess reactant: N2

Product Formed: 0.753 mol NH3

limiting reactants
Limiting Reactants

Mg + 2HCl → MgCl2 + H2

How many grams of magnesium chloride are produced from the reaction of 2.08 mol of Mg and 2.08 mol of HCl?

b percent yield
B. Percent Yield

measured in lab

calculated on paper

b percent yield1
B. Percent Yield
  • When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.

K2CO3 + 2HCl  2KCl + H2O + CO2

45.8 g

? g

actual: 46.3 g

b percent yield2
B. Percent Yield

K2CO3 + 2HCl  2KCl + H2O + CO2

Theoretical Yield:

45.8 g

? g

actual: 46.3 g

45.8 g

K2CO3

1 mol

K2CO3

138.21 g

K2CO3

2 mol

KCl

1 mol

K2CO3

74.55

g KCl

1 mol

KCl

= 49.4

g KCl

b percent yield3
B. Percent Yield

46.3 g

49.4 g

K2CO3 + 2HCl  2KCl + H2O + CO2

Theoretical Yield = 49.4 g KCl

45.8 g

49.4 g

actual: 46.3 g

 100 =

93.7%

% Yield =

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