Chapter 3continued

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Chapter 3-continued
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Patterns of Chemical Reactivity

• Combination reactions
• Two simpler substances combine to form one new
  substances
• A B C
• Examples
• - Recall 4 Fe 3 O2 2 Fe2O3 from
  last time
• - CO2 H2O H2CO3
• - Reaction between solid calcium oxide and water
  to produce calcium hydroxide
• Species present as reactants and products
• Ca2 O2- OH- and H2O
• CaO H2O Ca(OH)2 (s) (balanced as
  written)

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• Decomposition reactions One substance breaks
  down to yield two or more substances
• Examples
• - Recall 2 KClO3 (s) 2 KCl (s) 3 O2
  (g)
• - 2 NaN3 (s) 2 Na(s) 3 N2 (g)
• - Decomposition of solid ammonium nitrate to
  form dinitrogen oxide and gaseous water
• species present NH4 , NO3- , N2O , and H2O
  
• NH4NO3 (s) N2O (g) H2O (g)

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Combustion Reactions Substance reacts with
oxygen in the air, producing a flame Ex
Hydrocarbons sufficient air carbon dioxide
and water Recall propane oxygen carbon
dioxide water C3H8 (l) 5 O2 (g) 3 CO2
(g) 4 H2O (g) Note If there is not
sufficient oxygen, the carbon containing
product is CO.
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Formula weights (molar masses)

• Formula weights of molecules and ion (or ionic
  compounds) can be calculated based on the masses
  of the atoms in one formula unit of the compound.
• Ex NaCl
• Na 22.99 amu
• Cl 35.453 amu
• NaCl 58.443 amu

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• By analogy with the atomic and molar mass of
  atoms, we know that the molar mass of NaCl is
  58.44 g (the sum of the masses of mole sodium
  ions, or 22.99 g, and one mole of chloride ions,
  or 35.45 g).
• The concept of a mole can also be used to relate
  the number of particles, the number of moles,
  and the mass of a given substance.

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• How many potassium ions are in 0.1 g of potassium
  phosphate?

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Composition from formulas

• Sometimes we are interested in the amount of one
  particular ion or atom within a molecule.
• For example, we might be interested in the mass
  of sodium in our peanut butter.
• If there is sodium present as added sodium
  chloride, the mass of the sodium chloride must be
  corrected for the fact that the sodium makes up
  only part of the total mass of the added salt.

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• For any given element X

What is the percent sodium (by weight) in sodium
chloride?
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• The above process is reversed if we want to find
  the empirical formula of a compound. In this
  case we are given a total mass and a percentage
  and we then calculate the mass present of the
  element of interest.
• For example, suppose we are posed with the
  following problem
• A compound contains 92.31 C and 7.69 H, what
  is the atom ratio in small whole numbers (ie,
  the empirical formula) for the compound?

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• To approach this problem, we first assume that we
  have a fixed mass of compound 100 g or 1 g
  makes the process easier.
• Lets use 100 g, that means we have 92.31 g
  carbon atoms and 7.69 g hydrogen atoms

Therefore, the CH mole ratio is 11
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• If we now have the molar mass from some other
  experiment (like mass spectrometry), we can find
  the molecular formula.
• Suppose for the compound with a CH ratio of 11,
  the molar mass is found to be 78.11 g. What is
  the molecular formula?

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• We know that the molar mass must be a multiple of
  the mass of one of the CH units, so

or the molecular formula must be C6H6.
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Quantitative Information from Balanced Equations

• Recall that the equation
• xA yB 6 zC
• gives the following information
• x molecules of A react with y molecules of B to
  yield
• z molecules of C
• If we multiply x,y, and z all by 6.022 1023
  (Avogadros number) then we see that
• x mol of A reacts with y mol of B to yield z mol
  of C

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• If we multiply the number of moles by the molar
  masses of A, B, and C, we then get the mass
  relationships between the reactants and products
• What are the weight relationships implied by the
  reaction
• 4 Fe 3 O2 2 Fe2O3?
• 4 mol Fe 4 55.8 g 223.2 g Fe
• 3 mol O2 3 32.00 g 96.00 g O2
• 2 mol Fe2O3 319.2 g Fe2O3
• So 223.2 g Fe reacts with 96.00 g O2 to yield
  319.2 g Fe2O3

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• The above implies that if I know one of the
  masses or mol values involved, I can find the
  others.

How many g of Fe and O2 would be needed to
produce 100.0 g of Fe2O3?
Or 69.9 g Fe
Or 30.1 g Oxygen
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• Limiting reagents
• What about cases where the masses are not in
  exactly the correct ratio called for in the
  equation???
• Look at our propane combustion reaction
• C3H8 (l) 5 O2 (g) 3 CO2 (g) 4 H2O
  (g)
• If we are burning the propane in air, and we have
  44.0 g of propane, how many g of oxygen will be
  consumed and how many g of carbon dioxide and
  water will be produced?
• We will run out of propane first. (It is our
  limiting reagent.) Therefore, if we find out how
  many moles of propane are involved, we then can
  easily calculate the other masses.

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• Another example
• I combine 10.00 g of hydrogen and 10.00 g of
  oxygen in a container. The reaction is initiated
  with a spark and the reaction proceeds to
  completion. (Note My container has VERY strong
  walls!) What is in the flask after the reaction
  takes place?

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• First, write the balanced equation
• 2 H2 O2 ? 2 H2O
• The amount of product will be determined by the
  amount present of the reactant that gets consumed
  first, and..since the ratio of moles determines
  the ratio in which the reactants are consumed, we
  first find the number of moles of each.
• 10.00 g H2 4.961 mol 10.00 g O2
  0.3125 mol (A 15.9 1 ratio)
• The reaction only requires a 2 1 ratio so
• no oxygen will be left (IT IS THE LIMITING
  REAGENT)
• 0.625 mol hydrogen will be consumed,
  leaving 4.346 mol unreacted
• 0.625 mol water will be produced
• Or the flask will contain

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Theoretical yield
Theoretical yield mass you would expect to
produce based on stoichiometry of reaction
Example Back to our iron(III) oxide example.
Suppose we started out with the 69.9 g iron and
30.1 g oxygen required to make 100.0 g iron(III)
oxide, but actually only made 73.0 g. Our yield
would be
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Reactions in solution

• Same mole ratio relationships are essential to
  solve the problem but to get moles must multiply
  volume x concentration
• Example 100.0 mL of solution containing a total
  of 40.0 g NaOH per liter is mixed with 100.0 mL
  of a solution containing 18.75 g of HCl per
  liter. What is in solution at the end of the
  reaction?
• NaOH solution contains 4.00 g or 0.10 mol of NaOH
• HCl solution contains 1.825 g or 0.050 mol of HCl
• Reaction is NaOH HCl ? NaCl H2O
• Half of the NaOH will react, so 2.00 g NaOH is
  left. We will have produced 0.050 mol or 2.92 g
  NaCl and 0.050 mol or 0.90 g water

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Units of concentration

• One type of homogeneous mixtures that we
  mentioned earlier was a solution. Consists of a
  solute (what is dissolved) and a solvent.
• Usually we will express concentration as
  mass/volume or mol/volume

So, volume in liters x molarity mol of
solute. Other useful relationships are mL x
molarity mmol mmol/mL molarity