Topic 1:Stoichiometry 12'5 hours

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Topic 1Stoichiometry(12.5 hours)

• 1.1 The mole concept and Avogardos number
• 1.2 Formulas
• 1.3 Chemical equations
• 1.4 Mass and gas vol. relationships in chemical
  reactions

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1.1 The mole concept and Avogadros constant1.2
Formulas

• 1.1.1 Apply the mole concept to substances
• 1.1.2 Determine the number of particles and the
  amount of substance (in moles)
• 1.2.1 Define the terms relative atomic mass (Ar)
  and relative molecular mass (Mr)
• 1.2.2 Calculate the mass of one of mole of
  species from its formula
• 1.2.3 Solve problems involving the relationships
  between the amount of substance in moles, mass,
  and molar mass

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Quick Review

• Atoms are the smallest unit
• The same atoms make up elements (Na or Cl2)
• Different whole combination of atoms make up
  compounds (NaCl) or molecules (CH4)

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Atomic mass

• Atoms contain different numbers of protons and
  neutrons, so they have different masses
  (isotopes)
• Carbon-12 (6 p, 6 no)has a mass 12 times greater
  than hydrogen-1(1 p, 0 no)
• Its impossible to measure the mass of individual
  atoms on a balance, because it is so small
• Chemists developed their own unit, called the
  mole

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The Mole

• Similar to a dozen, except instead of 12, its
  602,000,000,000,000,000,000,000
• 6.02 X 1023 mol-1(in scientific notation)
• It applies to all kinds of particles atoms,
  particles, molecules, ions, electrons, formula
  units depending the way the question is asked, so
  be careful.
• This number is named in honor of Amedeo Avogadro
  (1776 1856)

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Just how big is Avogadro's constant?

• The earth is estimated to be 4.54 billion years
  old, that is approximately 1.43 x 1017 s. Still
  much smaller than a mole.
• If you were to count every grain of sand in the
  Sahara desert, assuming its 4000 km by 1500 km,
  and its depth is 10 m, and assuming that 1 cm3
  contains 1000 grains of sand. It would contain 6
  x 1022 grains of sand. Only one tenth the size
  of a mole!!!

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Relative atomic mass (Ar)

• As the weighted mean mass of all the naturally
  occurring isotopes of an element relative to one
  twelfth of the mass of a carbon-12 atom.
• Not whole numbers
• No mass unit
• Also applies to relative molecular mass (Mr)

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Molar Mass (or atomic mass)

• The Mass of 1 mole (in grams)
• Equal to the numerical value of the average
  atomic mass (get from periodic table), or add the
  atoms together for a molecule
• 1 mole of C atoms 12.0 g
• 1 mole of Mg atoms 24.3 g
• 1 mole of O2 molecules 32.0 g

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Molar Mass of Compounds

• The molar mass (MM) of a compound is determined
  the same way, except now you add up all the
  atomic masses for the molecule (or compound)
• Ex. Molar mass of CaCl2
• Avg. Atomic mass of Calcium 40.08g
• Avg. Atomic mass of Chlorine 35.45g
• Molar Mass of calcium chloride 40.08 g/mol Ca
  (2 X 35.45) g/mol Cl ? 110.98 g/mol CaCl2

20 Ca  40.08
17Cl 35.45
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Atoms or Molecules
Divide by 6.02 X 1023
Multiply by 6.02 X 1023
Moles
Multiply by atomic/molar mass from periodic table
Divide by atomic/molar mass from periodic table
Mass (grams)
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Practice

• Calculate the Molar Mass of calcium phosphate
• Formula
• Masses elements
• Ca 3 Cas X 40.1
• P 2 Ps X 31.0
• O 8 Os X 16.0
• Molar Mass

Ca3(PO4)2
120.3 g
62.0 g
128.0 g
120.3g 62.0g 128.0g
310.3 g/mol
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Calculations

• molar mass
  Avogadros number Grams
  Moles particles
• Everything must go through Moles!!!

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Atoms/Molecules and Grams

• How many moles of Cu are present in 35.4 g of Cu?
  How many atoms?
• 35.4 g x 1 mol 0.557 mol Cu
• 63.55 g
• 0.557 mol Cu x 6.02x1023 atoms 3.35x1023 atoms
• 1 mol

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On your own!

• How many moles of Fe are present in 102.4 g of
  Fe? How many atoms?

102.4 g x 1 mol 1.83 mol Fe 55.85 g 1.83 mol
Fe x 6.02x1023 atoms 1.10x1024 atoms
1 mol
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Work backwards!

• What is the mass (in grams) of 1.20x1024
  molecules of glucose (C6H12O6)?
• 1.20x1024 molec. X 1 mol 1.99 mol
• 6.02x1023 molec.
• 1.99 mol x 180.12 g 358.4 g
• 1 mol
• From periodic table

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This is as tricky as it gets!

• How many atoms of carbon are found in 2.6g of
  glucose (C6H12O6)?

2.6 g x 1 mole 0.014 mol C6H12O6
180.12 g 0.014 mol x 6.02 x 1023 molecules 8.4
x 1021 molecules mol 8.4 x 1021 molecules x 6
atoms of C 5.06 x 1022 atoms 1
molecule C6H12O6
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1.3 Chemical Equations

• 1.3.1 Deduce chemical equations when all
  reactants and products are given
• 1.3.2 Identify the mole ratio of any two species
  in a chemical equation
• 1.3.3 Apply the state symbols (s), (l), (g) and
  (aq)

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Chemical equations (review)

• To deduce the products of an equation we must
  know the type of reaction occuring.
• Single displacement
• Double displacement
• Combustion
• Synthesis
• Decomposition

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Single displacement

• either the metal changes position with the metal
  ion
• AgNO3 (aq) Cu(s) ? CuNO3(aq) Ag(s)
• or the non-metal changes position with the
  non-metal ion.
• F2(g) NaCl(aq) ? NaF(aq) Cl2(g)

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Double displacement

• The cations (positively charged ions) change
  position with each other.
• NaOH HCl ? HOH NaCl
• Fe(OH)2 2KBr ? 2KOH FeBr2
• Balancing an equation means that both sides must
  have the same number and type of atoms present.
• Coefficients are red (indicating total number of
  compounds), subscripts are blue (indicating
  number of atoms).

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Organic Combustion

• A hydrocarbon reacting with oxygen to produce
  carbon dioxide and water
• CH4 (g) 2O2 (g)? 2H2O (g) CO2 (g)
• 2C2H6 (g) 7O2 (g)? 6H2O (g) 4CO2 (g)
• The coefficient is also related the mole concept.
  For the 1st reaction, we would say that one mole
  of methane reacts with two moles of oxygen to
  produce two moles of water and one mole of carbon
  dioxide gas.

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Synthesis

• The combination of two or more particles to
  produce one compound.
• N2 (g) H2 (g) ? NH3 (l)
• Use (g) for gas, (l) for liquid, (s) for solids
  and (aq) if it dissolves in water/aqueous.

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Decomposition

• A single large compound will be broken down into
  more simple parts.
• H2O (l) ? H2 (g) O2 (g)

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1.4 Mass relationships in chemical equations

• 1.4.1 Calculate theoretical yields from chemical
  equations
• 1.4.2 Determine the limiting reactant the
  reactant in excess when quantities of reacting
  substances are given.
• 1.4.3 Solve problems involving theoretical,
  experimental and percentage yield.

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Chemistry Recipes

• Looking at a reaction tells us how much of
  something you need to react with something else
  to get a product
• Be sure you have a balanced reaction before you
  start!
• Example 2 Na Cl2 ? 2 NaCl
• This reaction tells us that by mixing 2 moles of
  sodium with 1 mole of chlorine we will get 2
  moles of sodium chloride
• What if we wanted 4 moles of NaCl? 10 moles? 50
  moles?

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Practice

• Write the balanced reaction for hydrogen gas
  reacting with oxygen gas.
• 2 H2 O2 ? 2 H2O
• How many moles of reactants are needed?
• What if we wanted 4 moles of water?
• What if we had 3 moles of oxygen, how much
  hydrogen would we need to react and how much
  water would we get?
• What if we had 50 moles of hydrogen, how much
  oxygen would we need and how much water produced?

2 mol H2 1 mol O2
4 mol H22 mol O2
6 mol H2, 6 mol H2O
25 mol O2, 50 mol H2O
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Mole Ratios

• These mole ratios can be used to calculate the
  moles of one chemical from the given amount of a
  different chemical
• Example How many moles of chlorine is needed to
  react with 5 moles of sodium (without any sodium
  left over)?
• 2 Na Cl2 ? 2 NaCl

5 moles Na 1 mol Cl2 2 mol Na
2.5 moles Cl2
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Mole-Mole Conversions

• How many moles of sodium chloride will be
  produced if you react 2.6 moles of chlorine gas
  with an excess (more than you need) of sodium
  metal?
• 2 Na Cl2 ? 2 NaCl

2.6 moles Cl2 2 mol NaCl 1 mol Cl2
5.2 moles NaCl
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You practice

• Aluminum reacts with oxygen to produce aluminum
  oxide. If I have 2.6 mol of Al, how much Al2O3
  would I produce?
• 4Al 3O2 ? 2Al2O3
• 2.6 mol Al x 2 mol Al2O3 1.3 mol Al2O3
• 4 mol Al

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Mole-Mass Conversions

• Most of the time in chemistry, the amounts are
  given in grams instead of moles
• We still go through moles and use the mole ratio,
  but now we also use molar mass to get to grams
• Example How many grams of chlorine are required
  to react completely with 5.00 moles of sodium to
  produce sodium chloride?
• 2 Na Cl2 ? 2 NaCl

5.00 moles Na 1 mol Cl2 70.90g Cl2
2 mol Na 1 mol Cl2
177g Cl2
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You Practice

• Calculate the mass in grams of Iodine required to
  react completely with 0.50 moles of aluminum.
• 2 Al 3 I2 ? 2 AlI3

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Mass-Mole

• We can also start with mass and convert to moles
  of product or another reactant
• We use molar mass and the mole ratio to get to
  moles of the compound of interest
• Calculate the number of moles of ethane (C2H6)
  needed to produce 10.0 g of water
• 2 C2H6 7 O2 ? 4 CO2 6 H20

10.0 g H2O 1 mol H2O 2 mol C2H6
18.0 g H2O 6 mol H20
0.185 mol C2H6
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Mass-Mass Conversions

• Most often we are given a starting mass and want
  to find out the mass of a product we will get
  (called theoretical yield) or how much of another
  reactant we need to completely react with it (no
  leftover ingredients!)
• Now we must go from grams to moles, mole ratio,
  and back to grams of compound we are interested in

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Mass-Mass Conversion

• Ex. Calculate how many grams of ammonia are
  produced when you react 2.00g of nitrogen with
  excess hydrogen.
• N2 3 H2 ? 2 NH3

2.00g N2 1 mol N2 2 mol NH3 17.06g NH3
28.02g N2 1 mol N2 1 mol
NH3
2.4 g NH3
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Practice

• How many grams of calcium nitride are produced
  when 2.00 g of calcium reacts with an excess of
  nitrogen?
• __Ca __N2 ? __Ca3N2

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Grilled Cheese Sandwich
Bread Cheese ? Cheese Melt
2 B C ? B2C
100 bread 30 slices ? sandwiches
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Limiting Reactants
aluminum chlorine gas ? aluminum
chloride
Al(s) Cl2(g) ?
AlCl3
2 Al(s) 3 Cl2(g) ? 2
AlCl3
100 g 100 g ? g
A. 200 g B. 125 g C. 667 g D. ???
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Limiting Reactant

• Most of the time in chemistry we have more of one
  reactant than we need to completely use up other
  reactant.
• That reactant is said to be in excess (there is
  too much).
• The other reactant limits how much product we
  get. Once it runs out, the reaction
  s. This is called the limiting reactant.

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Limiting Reactant

• To find the correct answer, we have to try all of
  the reactants. We have to calculate how much of
  a product we can get from each of the reactants
  to determine which reactant is the limiting one.
• The lower amount of a product is the correct
  answer.
• The reactant that makes the least amount of
  product is the limiting reactant. Once you
  determine the limiting reactant, you should
  ALWAYS start with it!
• Be sure to pick a product! You cant compare to
  see which is greater and which is lower unless
  the product is the same!

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Limiting Reactant Example
LimitingReactant

• 10.0g of aluminum reacts with 35.0 grams of
  chlorine gas to produce aluminum chloride. Which
  reactant is limiting, which is in excess, and how
  much product is produced?
• 2 Al 3 Cl2 ? 2 AlCl3
• Start with Al
• Now Cl2

10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g
AlCl3 27.0 g Al 2 mol Al
1 mol AlCl3
49.4g AlCl3
35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g
AlCl3 71.0 g Cl2 3 mol Cl2
1 mol AlCl3
43.9g AlCl3
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LR Example Continued

• We get 49.4g of aluminum chloride from the given
  amount of aluminum, but only 43.9g of aluminum
  chloride from the given amount of chlorine.
  Therefore, chlorine is the limiting reactant.
  Once the 35.0g of chlorine is used up, the
  reaction comes to a complete .

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Limiting Reactant Practice

• 15.0 g of potassium reacts with 15.0 g of iodine.
  Calculate which reactant is limiting and how
  much product is made.

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Finding the Amount of Excess

• By calculating the amount of the excess reactant
  needed to completely react with the limiting
  reactant, we can subtract that amount from the
  given amount to find the amount of excess.
• Can we find the amount of excess potassium in the
  previous problem?

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Finding Excess Practice

• 15.0 g of potassium reacts with 15.0 g of iodine.
  2 K I2 ? 2 KI
• We found that Iodine is the limiting reactant,
  and 19.6 g of potassium iodide are produced.

15.0 g I2 1 mol I2 2 mol K 39.1 g K
254 g I2 1 mol I2 1
mol K
4.62 g K USED!
15.0 g K 4.62 g K 10.38 g K EXCESS
Given amount of excess reactant
Amount of excess reactant actually used
Note that we started with the limiting reactant!
Once you determine the LR, you should only start
with it!
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Limiting Reactant Recap

• You can recognize a limiting reactant problem
  because there is MORE THAN ONE GIVEN AMOUNT.
• Convert ALL of the reactants to the SAME product
  (pick any product you choose.)
• The lowest answer is the correct answer.
• The reactant that gave you the lowest answer is
  the LIMITING REACTANT.
• The other reactant(s) are in EXCESS.
• To find the amount of excess, subtract the amount
  used from the given amount.
• If you have to find more than one product, be
  sure to start with the limiting reactant. You
  dont have to determine which is the LR over and
  over again!

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Percent Yields

• Theoretical yield max amount of a product that
  is formed in a reaction.
• Actual yield amount of product that is actually
  obtained in a reaction
• Usually less than theoretical. Why?

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Why?

• Theoretical has assumed that all of limiting
  reagent has completely reacted.
• Many reactions do not go to completion
• Unexpected competing side reactions limit the
  formation of products.
• Some reactants are lost during the separation
  process (remember in the lab, pouring off water,
  leaving silver behind!)
• Impure reactants
• Faulty measuring
• Poor experimental design or technique

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How to calculate?

• Percent Yield Actual Yield x 100
• Theoretical yield

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Practice together

• 20g of HBrO3 is reacted with excess HBr.
• What is the theoretical yield of Br2?
• What is the percent yield, if 47.3g is produced?
• HBrO3 5HBr ? 3H2O 3 Br2

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Practice alone

• When 35g of Ba(NO3)2 is reacted with excess
  Na2SO4, 29.8g of BaSO4 is recovered.
• Ba(NO3)2 Na2SO4 ? BaSO4 2NaNO3
• Calculate the theoretical yield of BaSO4
• Calculate the percent yield of BaSO4