The Mole

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The Mole StoichiometryUnits 8 9
combined!Jan 2009
2
From Moles to Molecules

• How many molecules in 2.5 moles?
• 2.5 mol x 6.022x1023 molecules
  1.505 x 1024
  
  1 mole
• There are 1.5 x 1024 molecules in 2.5 mol

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From Molecules to Moles

• How many moles in 9.7 x 1015 molecules?
• 9.7x1015 molecules x 1 mole
  6.022 x 1023 molecules
• 1.6 x 10-8 moles

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Molar Mass

• The mass in grams of 1 mole of a substance.
• For example H2O
• H 2 x 1.00794 2.01588 g
• O 1 x 15.9994 15.9994 g
• 18.01528g
• The molar mass of H2O 18.0153 g/mol

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Formula vs Molar Mass

• The only difference between a formula mass and a
  molar mass is the unit!
• Formula mass represents the mass of one unit of a
  compound and is measured in amu. (Small amount
  uses a small unit.)
• Molar mass represents the mass of one mole of
  substance and is measured in grams. (Larger
  quantity uses a larger mass.)

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Molar Volume

• One mole of any gas at standard temperature and
  pressure (STP 0oC and 1 atm) has a volume of
  22.4 L or 22.4 dm3.
• The volume of 1 mole of gas at STP 22.4 L

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Molar Volume Sample Problem

• A canister with a volume of 25.0 L contains how
  many moles of oxygen at STP?

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Multi-Step Conversions
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Empirical Formula

• A formula that gives the simplest whole-number
  ratio of the atoms of in a compound.
• Molecular Empirical
• Examples C6H12O6 CH2O
• C5H10O5 CH2O
• H2O2 HO

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Finding Empirical Formulas

• Assume you are working with a 100.0g sample so
  the mass of each element will be the same as the
  percent of that element (for simplicity).
• Multiply the mass of each element by the molar
  mass of the element.
• Find the whole number ratio of these calculated
  amounts by dividing each mole value by the
  smallest one calculated.
• Round to whole numbers and use the ratio to
  determine the empirical formula.

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Empirical Formula Sample

• A compound was found to contain 29.6 Calcium,
  23.7 Sulfur and 46.8 Oxygen. What is the
  empirical formula for the compound?

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29.6 Ca, 23.7 S, 46.8 O

• Assume 29.6g Ca, 23.7g S and 46.8g O.
• 29.6g x 1 mole 0.739 mol Ca
• 40.078g
• 23.7g x 1 mole 0.739 mole S
• 32.066
• 46.8 x 1 mole 2.93 mole O
• 15.9994
• 0.739 1 0.739 1 2.93 3.96
• 0.739 0.739 0.739

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29.6 Ca, 23.7 S, 46.8 O

• Round to whole number
• 1 1 4
• The empirical formula is
• CaSO4

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Molecular Empirical Formulas

• Ribose has a molar mass of 150g/mol and a
  chemical composition of 40.0C, 6.67 hydrogen
  and 53.3 oxygen. What is the molecular formula
  for ribose?

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Empirical Formula for Ribose

• 40.0g x 1 mole 3.33 mole C
• 12.011
• 6.67g x 1 mole 6.62 mole H
• 1.00794
• 53.3g x 1 mole 3.33 mole O
• 15.9994

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Empirical Formula for Ribose

• 3.33 1 6.62 1.99 3.33 1
• 3.33 3.33 3.33
• Empirical Formula for Ribose
• CH2O
• One more step!

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Molecular Formula for Ribose

• Covalent bonds can form in many different ratios.
  (Ionic only form in one set ratio based on
  charges)
• Use the empirical formula and the ratio of molar
  mass to empirical formula mass to determine the
  molecular formula.

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Molecular Formula for Ribose

• Empirical Formula CH2O
• Molar mass (from original problem) 150g/mole
• Empirical mass 30.0g/mole
• (1x12.011 2x1.00794 1x15.999430.0)
• 150 g/mole (Molar mass of cmpd) 5
• 30.0 g/mole (Empirical mass)

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Molecular Formula for Ribose

• Ratio of molar mass to empirical mass 5
• Empirical formula CH2O
• Multiply empirical formula by 5
• Molecular Formula of Ribose C5H10O5

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Stoichiometry

• The mathematical relationships between reactants
  and products in a chemical reaction.
• A balanced equation gives the first and most
  important mathematical relationships in a reaction

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Stoichiometry

• A Balanced Equation
• ___ AgNO3 ___ Mg ? ___ Mg(NO3)2 ___
  Ag

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Stoichiometry

• A Balanced Equation
• ___ AgNO3 ___ Mg ? ___ Mg(NO3)2 ___
  Ag
• tells you the ratio of reactants and products
  that must be maintained in order for the reaction
  to go to completion. For this single replacement
  reaction, you need to react magnesium with silver
  nitrate in a 12 ratio. If you maintain the 12
  ratio of Mg to AgNO3, you will produce magnesium
  nitrate and metallic silver in a 12 ratio.

2
2
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Stoichiometry

• 2 AgNO3 Mg ? Mg(NO3)2 2 Ag
• Does this mean you always have to react 2 moles
  of silver nitrate with 1 mole of magnesium and
  produce 1 mole of magnesium nitrate and 2 moles
  of silver?

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Stoichiometry

• 2 AgNO3 Mg ? Mg(NO3)2 2 Ag
• Does this mean you always have to react 2 moles
  of silver nitrate with 1 mole of magnesium and
  produce 1 mole of magnesium nitrate and 2 moles
  of silver?
• No! You must maintain the 2112 ratio

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Stoichiometry

• 2 AgNO3 Mg ? Mg(NO3)2 2 Ag
• Balancing
• Ratio
• We only have 0.2 mole AgNO3, can we run the
  reaction?

2 1 1
2
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Stoichiometry

• 2 AgNO3 Mg ? Mg(NO3)2 2 Ag
• Balancing
• Ratio
• We only have 0.2 mole AgNO3, can we run the
  reaction? Yes, we adjust the ratio to fit

2 1 1
2
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Stoichiometry

• 2 AgNO3 Mg ? Mg(NO3)2 2 Ag
• Balancing
• Ratio
• We only have 0.2 mole AgNO3, can we run the
  reaction? Yes, we adjust the ratio to fit
• Adjusted
• Ratio
• We maintain the 2112 ratio.

2 1 1
2
0.2 0.1 0.1
0.2
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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• How many moles of potassium iodide will be needed
  to react completely with 0.3 mole of lead
  nitrate?

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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• How many moles of potassium iodide will be needed
  to react completely with 0.3 mole of lead
  nitrate?
• Balancing Ratio 2 1 2 1
• Adjusted Ratio

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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• How many moles of potassium iodide will be needed
  to react completely with 0.3 mole of lead
  nitrate?
• Balancing Ratio 2 1 2 1
• Adjusted Ratio 0.3

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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• How many moles of potassium iodide will be needed
  to react completely with 0.3 mole of lead
  nitrate?
• Balancing Ratio 2 1 2 1
• Adjusted Ratio 0.6 0.3 0.6 0.3

(0.3 x 2) (0.3 x 2) (0.3 x 1)
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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• How many moles of potassium iodide will be needed
  to react completely with 0.3 mole of lead
  nitrate?
• Balancing Ratio 2 1 2 1
• Adjusted Ratio 0.6 0.3 0.6 0.3
• You would react 0.6 mol KI

(0.3 x 2) (0.3 x 2) (0.3 x 1)
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Stoichiometry

• Practice Problems
• N2 3 H2 ? 2 NH3
• How many moles of hydrogen are needed to
  react completely with two moles of nitrogen gas?

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Stoichiometry

• N2 3 H2 ? 2 NH3
• How many moles of hydrogen are needed to
  react completely with two moles of nitrogen gas?
• Bal. Ratio 1 3 2
• Adj. Ratio 2 6 4

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Stoichiometry

• N2 3 H2 ? 2 NH3
• How many moles of hydrogen are needed to
  react completely with two moles of nitrogen gas?
• Bal. Ratio 1 3 2
• Adj. Ratio 2 6 4
• It will take 6 moles
• of hydrogen gas.

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Stoichiometry

• Limiting Reactant
• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• How many moles of lead iodide will be produced if
  0.2 mol KI are reacted with 0.3 mole of lead
  nitrate?

37
Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• How many moles of lead iodide will be produced if
  0.2 mol KI are reacted with 0.3 mole of lead
  nitrate?
• Balancing Ratio 2 1 2 1
• Adjusted Ratio 0.2 0.3

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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• How many moles of lead iodide will be produced if
  0.2 mol KI are reacted with 0.3 mole of lead
  nitrate?
• Balancing Ratio 2 1 2 1
• Adjusted Ratio 0.2 0.3 wait this is not a
  21 ratio, we have too much Pb(NO3)2.

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Stoichiometry

• Think about the mechanics of a double replacement
  reaction. We have ions in solution colliding
  with each other to produce two new compounds.

You can only make as much lead iodide as you have
lead and iodide ions. When you run out of
iodide, you can not make any more lead iodide
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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• 2 1 2
  1
• 0.2 0.3

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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• 2 1 2
  1
• 0.2 0.3
• Needed 0.1
• Excess 0.2
• The reaction is limited by the amount of KI
  present so KI is called the limiting reactant and
  the product ratio is set using the amount of KI
  present

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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• 2 1 2
  1
• 0.2 0.3 0.2 0.1
  
• Needed 0.1
• Excess 0.2 mole Pb(NO3)2

You would produce 0.1 mol lead iodide.
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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• 2 1 2
  1
• 0.2 0.3 0.2 0.1
  
• Needed 0.1
• Excess 0.2

You would produce 0.1 mol lead iodide.
You will also have excess Pb2 and NO31- ions in
the product mixture
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Stoichiometry

• Practice Problems
• 2 AgNO3 Mg ? 2 Ag Mg(NO3)2
• a.) How many moles of magnesium and silver
  nitrate will be needed to produce 3.2 moles of
  silver?
• b.) What would happen to the amount of
  product if twice the required amount of magnesium
  was added to the reaction?

45
Stoichiometry

• 2 AgNO3 Mg ? 2 Ag Mg(NO3)2
• Bal. Ratio 2 1 2 1
• Adj. Ratio 3.2
  

46
Stoichiometry

• 2 AgNO3 Mg ? 2 Ag Mg(NO3)2
• Bal. Ratio 2 1 2 1
• Adj. Ratio 3.2 1.6 3.2 1.6
  

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Stoichiometry

• Practice Problems
• 2 AgNO3 Mg ? 2 Ag Mg(NO3)2
• a.) How many moles of magnesium and silver
  nitrate will be needed to produce 3.2 moles of
  silver?1.6 mol Mg 3.2 mol AgNO3
• b.) What would happen to the amount of
  product if twice the required amount of magnesium
  was added to the reaction?

48
Stoichiometry

• 2 AgNO3 Mg ? 2 Ag Mg(NO3)2
• Bal. Ratio 2 1 2 1
• Adj. Ratio 3.2 1.6 3.2 1.6
• 3.2
  

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Stoichiometry

• 2 AgNO3 Mg ? 2 Ag Mg(NO3)2
• Bal. Ratio 2 1 2 1
• Adj. Ratio 3.2 1.6 3.2 1.6
• 3.2
• The amount of product will not change because
  the reaction is limited by the amount of Ag1 and
  NO31 present.

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Stoichiometry

• Practice Problems
• 2 AgNO3 Mg ? 2 Ag Mg(NO3)2
• a.) How many moles of magnesium and silver
  nitrate will be needed to produce 3.2 moles of
  silver?1.6 mol Mg 3.2 mol AgNO3
• b.) What would happen to the amount of
  product if twice the required amount of magnesium
  was added to the reaction?
• There will be no change, reaction limited by
  AgNO3

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Stoichiometry

• Taking the math one step further

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Stoichiometry

• Taking the math one step further
• How many grams of lead nitrate will be needed to
  react completely with 5.00 g of potassium iodide?
• How much lead iodide will be produced?

53
Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• 5.00g _____g _____g
• First we must convert to moles by
• dividing by the molar mass.

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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• 5.00g _____g _____g
• /166.0
• 0.030mol

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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• 5.00g _____g _____g
• /166.0
• 0.030mol
• Adjust the ratio to the number of
• moles of KI

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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• 5.00g _____g _____g
• /166.0
• 0.030mol 0.015mol 0.030mol 0.015mol

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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• 5.00g _____g _____g
• /166.0
• 0.030mol 0.015mol 0.030mol 0.015mol
• Convert moles back to grams by
• multiplying by each molar mass

58
Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• 5.00g _____g _____g
• 0.030mol 0.015mol 0.030mol 0.015mol
• Convert moles back to grams by
• multiplying by each molar mass

x 331.2
x 461.0
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Stoichiometry

• 2 KI Pb(NO3)2 ? 2 KNO3 PbI2
• 5.00g 4.97g 3.03 g 6.92g
• 0.030mol 0.015mol 0.030mol 0.015mol

x 331.2 x 101.1 x
461.0
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Stoichiometry

• How many grams of lead nitrate will be needed to
  react completely with 5.00 g of potassium iodide?
  4.97 g Pb(NO3)2
• How much lead iodide will be produced? 6.92 g
  PbI2