Unit 08 – Moles Unit 08 – Moles and Stoichiometryand Stoichiometry

I. Molar Conversions

A. What is the Mole?A. What is the Mole?

A counting number (like a dozen)

Avogadro’s number (6.02 1023

particles) (SI unit)

1 mol = molar mass

A large amount!!!!

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

A. What is the Mole?A. What is the Mole?

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

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

B. Molar MassB. Molar Mass

o Molar Mass- the mass of a mole of any element or compound (in grams)oRound to 2 decimal places

o Also called:

oGram Formula mass – sum of the atomic masses of all the atoms in a formula of a compoundoFormula weight

B. Molar Mass ExamplesB. 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 ExamplesB. Molar Mass Examplessodium 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 C. Number of Particles in a MoleMole

What is a representative particle? How the substance normally exists:How the substance normally exists:

1 mole = 6.02 × 10 23 representative particles(also called Avogadro’s Number)(also called Avogadro’s Number)

1.Atom- rep. particle for most elements2. Ions – if atom is charged3.Molecule- rep. particle for covalent

compounds and diatomic molecules “BrINCl HOF”

4.Formula unit- rep. particle for ionic compounds

D. Volume of a Mole of D. Volume of a Mole of GasGasThe 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 The MoleMoleRoad Road MapMap

Atoms (ions)

Molecule

Formula unit

E. Molar Conversion E. Molar Conversion ExamplesExamples

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

26 g C 1 mol C12.01 g C

= 2.2 mol C

E. Molar Conversion E. Molar Conversion ExamplesExamples

How many molecules are in 2.50 moles of C12H22O11?

2.50 mol6.02 1023

molecules1 mol

= 1.51 1024

molecules C12H22O11

E. Molar Conversion E. Molar Conversion ExamplesExamples

Find the number of molecules of 12.00 L of O2 gas at STP.

12.00 L 1 mol

22.4 L

= 3.225 x 1023 molecules

6.02 x 1023

molecules

1 mol

II. Stoichiometric II. Stoichiometric CalculationsCalculations

A. Proportional A. Proportional RelationshipsRelationships

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

3/4 c. brown sugar1 tsp vanilla extract2 eggs2 c. chocolate chipsMakes 5 dozen cookies.

2 1/4 c. flour1 tsp. baking soda1 tsp. salt1 c. butter3/4 c. sugar

5 eggs 5 doz.2 eggs

= 12.5 dozen cookies

Ratio of eggs to cookies

A. Proportional A. Proportional RelationshipsRelationships

StoichiometryStoichiometry• mass relationships between

substances in a chemical reaction• based on the mole ratio

Mole RatioMole Ratio• indicated by coefficients in a

balanced equation

2 Mg + O2 Mg + O22 2 MgO 2 MgO

B. Stoichiometry StepsB. Stoichiometry Steps1. Write a balanced equation.2. Identify known & unknown.3. Convert known to moles (IF

NECESSARY) Line up conversion factors.

4. Use Mole ratio – from equation5. Convert moles to unknown unit

(IF NECESSARY)

6. Calculate and write units.

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

C. Stoichiometry C. Stoichiometry ProblemsProblems

III. Stoichiometry in the Real World

A. Limiting ReactantsA. Limiting Reactants

Available IngredientsAvailable Ingredients• 4 slices of bread• 1 jar of peanut butter• 1/2 jar of jelly

Limiting ReactantLimiting Reactant• bread

Excess ReactantsExcess Reactants• peanut butter and jelly

A. Limiting ReactantsA. Limiting Reactants

Limiting ReactantLimiting Reactant• used up in a reaction• determines the amount of product

Excess ReactantExcess Reactant• added to ensure that the other

reactant is completely used up• cheaper & easier to recycle

A. Limiting ReactantsA. 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 ReactantsA. Limiting Reactants

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

28.2 L ? mol25.3 L

N2 + 3H2 2NH3

Using the following equation identify the limiting reagent.

A. Limiting ReactantsA. Limiting Reactants

28.2L N2

1 molN2

22.4L N2

= 2.5 mol NH3

2 molNH3

1 molN2

28.2 L ? mol25.3 L

N2 + 3H2 2NH3

A. Limiting ReactantsA. Limiting Reactants

25.3L H2

1 molH2

22.4 L H2

= 0.753 mol NH3

2 molNH3

3 molH2

28.2 L ? mol25.3 L

N2 + 3H2 2NH3

A. Limiting ReactantsA. 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 ReactantsLimiting Reactants

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

Mg + 2HCl → MgCl2 + H2

B. Percent YieldB. Percent Yield

100yield ltheoretica

yield actualyield %

calculated on paper

measured in lab

B. Percent YieldB. Percent YieldWhen 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 YieldB. Percent Yield

45.8 gK2CO3

1 molK2CO3

138.21 gK2CO3

= 49.4g KCl

2 molKCl

1 molK2CO3

74.55g KCl

1 molKCl

K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g

actual: 46.3 gTheoretical Yield:

B. Percent YieldB. Percent Yield

Theoretical Yield = 49.4 g KCl

% Yield =46.3 g49.4 g

100 =93.7%

K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g 49.4 g

actual: 46.3 g