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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???. - PowerPoint PPT Presentation
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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