Hein and Arena

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Calculations from Equations Chapter 9
Larry Emme
Chemeketa Community College
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A Short Review
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• The mole weight of an element is its atomic mass
  in grams.

• It contains 6.02 x 1023 atoms (Avogadros number)
  of the element.

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The mole weight of an element or compound is the
sum of the atomic masses of all its atoms.
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(No Transcript)
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Mole Weight
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1 mole 6.02 x 1023 molecule
1 mole 6.02 x 1023 formula units
1 mole 6.02 x 1023 atoms
1 mole 6.02 x 1023 ions
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• For calculations of mole-mass-volume
  relationships.

• The chemical equation must be balanced.

• The number in front of a formula represents the
  number of moles of the reactant or product.

The equation is balanced.
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Introduction to StoichiometryThe Mole-Ratio
Method
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• Stoichiometry The area of chemistry that deals
  with the quantitative relationships between
  reactants and products.

• Examples of questions asked
• How much product from a specific amount of
  reactant.
• How much reactant do you start with to get so
  much product.
• What if the reaction does not go to completion?
  What is the percent yield.
• What if the sample is impure? What is the percent
  purity.

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• Mole Ratio a ratio between the moles of any two
  substances involved in a chemical reaction.
• The coefficients used in mole ratio expressions
  are derived from the coefficients used in the
  balanced equation.

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Examples
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N2 3H2 ? 2NH3
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N2 3H2 ? 2NH3
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• The mole ratio is used to convert the number of
  moles of one substance to the corresponding
  number of moles of another substance in a
  stoichiometry problem.

• The mole ratio is used in the solution of every
  type of stoichiometry problem.

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The Mole Ratio Method

1. Convert the quantity of starting substance to
   moles (if it is not already moles)
2. Convert the moles of starting substance to moles
   of desired substance.
3. Convert the moles of desired substance to the
   units specified in the problem.

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Step 1 Determine the number of moles of starting
substance.

• Identify the starting substance from the data
  given in the problem statement. Convert the
  quantity of the starting substance to moles, if
  it is not already done.

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How many moles of NaCl are present in 292.215
grams of NaCl? The mole weight of NaCl 58.443
g.
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Step 2 Determine the mole ratio of the desired
substance to the starting substance.

• The number of moles of each substance in the
  balanced equation is indicated by the coefficient
  in front of each substance. Use these
  coefficients to set up the mole ratio.

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Step 2 Determine the mole ratio of the desired
substance to the starting substance.

• Multiply the number of moles of starting
  substance (from Step 1) by the mole ratio to
  obtain the number of moles of desired substance.

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In the following reaction how many moles of PbCl2
are formed if 5.000 moles of NaCl
react? 2NaCl(aq) Pb(NO3)2(aq) ? PbCl2(s)
2NaNO3(aq)
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Step 3. Calculate the desired substance in the
units specified in the problem.

• If the answer is to be in moles, the calculation
  is complete
• If units other than moles are wanted, multiply
  the moles of the desired substance (from Step 2)
  by the appropriate factor to convert moles to the
  units required.

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Step 3. Calculate the desired substance in the
units specified in the problem.
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Step 3. Calculate the desired substance in the
units specified in the problem.
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Step 3. Calculate the desired substance in the
units specified in the problem.
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Mole-Mole Calculations
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Phosphoric Acid

• Phosphoric acid (H3PO4) is one of the most widely
  produced industrial chemicals in the world.
• Most of the worlds phosphoric acid is produced
  by the wet process which involves the reaction of
  phosphate rock, Ca5(PO4)3,F with sulfuric acid
  (H2SO4).

Ca5(PO4)3F(s) 5H2SO4 ? 3H3PO4 HF 5CaSO4
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• Calculate the number of moles of phosphoric acid
  (H3PO4) formed by the reaction of 10 moles of
  sulfuric acid (H2SO4).
• Ca5(PO4)3F 5H2SO4 ? 3H3PO4 HF 5CaSO4

1 mol
5 mol
3 mol
1 mol
5 mol
Step 1 Moles starting substance 10.0 mol H2SO4
Step 2 The conversion needed is
moles H2SO4 ? moles H3PO4
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Calculate the number of moles of sulfuric acid
(H2SO4) that react when 10 moles of Ca5(PO4)3F
react. Ca5(PO4)3F 5H2SO4 ? 3H3PO4 HF 5CaSO4
1 mol
5 mol
3 mol
1 mol
5 mol
Step 1 The starting substance is 10.0 mol
Ca5(PO4)3F
Step 2 The conversion needed is
moles Ca5(PO4)3F ? moles H2SO4
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Mole-Mass Calculations
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• The object of this type of problem is to
  calculate the mass of one substance that reacts
  with or is produced from a given number of moles
  of another substance in a chemical reaction.

1. If the mass of the starting substance is given,
   we need to convert it to moles.

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1. We use the mole ratio to convert moles of
   starting substance to moles of desired substance.
   

1. We can then change moles of desired substance to
   mass of desired substance if called for by the
   problem.

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Examples
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• Calculate the number of moles of H2SO4 necessary
  to yield 784 g of H3PO4
• Ca5(PO4)3F 5H2SO4 ? 3H3PO4 HF 5CaSO4

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• Calculate the number of grams of H2 required to
  form 12.0 moles of NH3.
• N2 3H2 ? 2NH3

The conversion needed is
moles NH3 ? moles H2 ? grams H2
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Mass-Mass Calculations
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Solving mass-mass stoichiometry problems requires
all the steps of the mole-ratio method.

1. The mass of starting substance is converted to
   moles.
2. The mole ratio is then used to determine moles of
   desired substance.
3. The moles of desired substance are converted to
   mass of desired substance.

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Diagram to Successful Calculations
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Calculate the number of grams of NH3 formed by
the reaction of 112 grams of H2. N2 3H2 ? 2NH3

grams H2 ? moles H2 ? moles NH3 ? grams NH3
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Limiting-Reactant and Yield Calculations
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Limiting Reagent
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• The limiting reagent is one of the reactants in a
  chemical reaction.

• It is called the limiting reagent because the
  amount of it present is insufficient to react
  with the amounts of other reactants that are
  present.
• The limiting reagent limits the amount of product
  that can be formed.

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How many bicycles can be assembled from the parts
shown?
From eight wheels four bikes can be constructed.
From four frames four bikes can be constructed.
From three pedal assemblies three bikes can be
constructed.
The limiting part is the number of pedal
assemblies.
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H2 Cl2 ? 2HCl
4 molecules Cl2 can form 8 molecules HCl
Cl2 is the limiting reagent
3 molecules of H2 remainH2 is in excess
7 molecules H2 can form 14 molecules HCl
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Steps Used to Determine the Limiting Reagent
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1. Calculate the amount of product (moles or grams,
   as needed) formed from each reactant.

1. Determine which reactant is limiting. (The
   reactant that gives the least amount of product
   is the limiting reagent the other reactant is in
   excess.
2. Calculate the amount of the other reactant
   required to react with the limiting reagent, then
   subtract this amount from the starting quantity
   of the reactant. This gives the amount of the
   substance that remains unreacted.

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Examples
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How many moles of HCl can be produced by reacting
4.0 mol H2 and 3.5 mol Cl2? Which compound is
the limiting reagent?
H2 Cl2 ? 2HCl

• Step 1 Calculate the moles of HCl that can form
  from each reactant.

Step 2 Determine the limiting reagent.
The limiting reagent is Cl2 because it
produces less HCl than H2.
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How many grams of silver bromide (AgBr) can be
formed when solutions containing 50.0 g of MgBr2
and 100.0 g of AgNO3 are mixed together? How
many grams of the excess reactant remain
unreacted?
MgBr2(aq) 2AgNO3 (aq) ? 2AgBr(s) Mg(NO3)2(aq)

• Step 1 Calculate the grams of AgBr that can form
  from each reactant.

The conversion needed is g reactant ? mol
reactant ? mol AgBr ? g AgBr
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How many moles of silver bromide (AgBr) can be
formed when solutions containing 50.0 g of MgBr2
and 100.0 g of AgNO3 are mixed together? How
many grams of the excess reactant remain
unreacted?
MgBr2(aq) 2AgNO3 (aq) ? 2AgBr(s) Mg(NO3)2(aq)

• Step 2 Determine the limiting reagent.

The limiting reactant is MgBr2 because itforms
less AgBr.
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How many grams of the excess reactant (AgNO3)
remain unreacted?
MgBr2(aq) 2AgNO3 (aq) ? 2AgBr(s) Mg(NO3)2(aq)

• Step 3 Calculate the grams of unreacted AgNO3.
  First calculate the number of grams of AgNO3 that
  will react with 50 g of MgBr2.

The conversion needed is g MgBr2 ? mol MgBr2 ?
mol AgNO3 ? g AgNO3
The amount of MgBr2 that remains is
100.0 g AgNO3 -
92.3 g AgNO3
7.7 g AgNO3
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Percent Yield
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The quantities of products calculated from
equations represent the maximum yield (100) of
product according to the reaction represented by
the equation.
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Many reactions fail to give a 100 yield of
product.

• This occurs because of side reactions and the
  fact that many reactions are reversible.

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• The theoretical yield of a reaction is the
  calculated amount of product that can be obtained
  from a given amount of reactant.

• The actual yield is the amount of product finally
  obtained from a given amount of reactant.

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• The percent yield of a reaction is the ratio of
  the actual yield to the theoretical yield
  multiplied by 100.

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Silver bromide was prepared by reacting 200.0 g
of magnesium bromide and an adequate amount of
silver nitrate. Calculate the percent yield if
375.0 g of silver bromide was obtained from the
reaction
MgBr2(aq) 2AgNO3 (aq) ? 2AgBr(s) Mg(NO3)2(aq)
Step 1 Determine the theoretical yield by
calculating the grams of AgBr that can be formed.

The conversion needed is g MgBr2 ? mol MgBr2 ?
mol AgBr ? g AgBr
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Silver bromide was prepared by reacting 200.0 g
of magnesium bromide and an adequate amount of
silver nitrate. Calculate the percent yield if
375.0 g of silver bromide was obtained from the
reaction
MgBr2(aq) 2AgNO3 (aq) ? 2AgBr(s) Mg(NO3)2(aq)
Step 2 Calculate the percent yield.
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Percent Purity
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• The percent purity of a reaction is the ratio of
  the theoretical yield (calculated) to the sample
  weight multiplied by 100.

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A 5.00 gram sample of impure CaCO3 (limestone)
produces 1.89 grams of CO2 when heated. Calculate
the percent purity of CaCO3 according to
CaCO3 (s) ? CaO (s) CO2?
Step 1 Determine the actual grams (theoretical )
of CaCO3 in the original sample.
The conversion needed is g CO2 ? mol CO2 ? mol
CaCO3 ? g CaCO3
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A 5.00 gram sample of CaCO3 produces 1.89 grams
of CO2 when heated. Calculate the percent purity
according to
CaCO3 (s) ? CaO (s) CO2?
Step 2 Calculate the percent purity.
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484 gram of copper ore is oxidized by HNO3 to
yield 1000. grams of Cu(NO3)2. Find the percent
purity of copper in the ore according to
3 Cu(s) 8 HNO3(aq) ? 3 Cu(NO3)2(aq) 2 NO(g)
4 H2O(l)
The conversion needed is g Cu(NO3)2 ? mol
Cu(NO3)2 ? mol Cu ? g Cu
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The End