Determining Empirical Formula from Mass % Data

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Determining Empirical Formula from Mass Data

• To convert the mass composition obtained from a
  combustion analysis into an empirical formula, we
  must convert the mass of each type of atom into
  the relative number of atoms.
• To do this, assume that we have 100g of sample
• The mass will then be in grams

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Determining Molecular Formulas

• Once we know the empirical formula, we need one
  more piece of information to determine the
  molecular formula The Molar Mass
• Say we know the empirical formula of a compound
  is C3H4O3.
• All we know about this compound at this point is
  the ratio of the 3 elements.
• We dont know the exact number of each type of
  atom in the molecule.
• Is the Molecular Formula C6H8O6, C12H16O12 or
  C18H24O18?

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G Molarity

• Hands down one of the most important concepts you
  need to master is you are going to stay in the
  sciences. Period.
• The Molar Concentration, c, of a solute in
  solution is the number of moles of solute divided
  by the volume of the solution (in liters).
• Also referred to as Molarity

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Molarity
The symbol M is used to denote the molarity of
the solution
1M NaCl 1 mole NaCl per liter of H2O
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G4 Dilutions

• Frequently in the laboratory, you will need to
  make dilutions from a stock solution.
• This involves taking a volume from the stock and
  bringing it to a new volume with solvent.
• In order to perform these dilutions, we can use
  the following equation
• c1V1 c2V2
• Where c1 Stock concentration
• V1 Volume removed from stock
• c2 Target conc of new soln
• V2 Volume of new solution

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Law of Conservation of Matter

• Matter can neither be created nor destroyed
  Antoine Lavoisier, 1774

If a complete chemical reaction has occurred, all
of the reactant atoms must be present in the
product(s)
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Law of Conservation of Matter
a)
b)

• Stoichiometry coefficients are necessary to
  balance the equation so that the Law of
  Conservation of Matter is not violated
• 6 molecules of Cl2 react with 1 molecule of P4
• 3 molecules of Cl2 react with 2 molecules of Fe

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Example of Using Stoichiometric Coefficients
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Balancing Chemical Reactions

• Lets look at Oxide Formation
• Metals/Nonmetals may react with oxygen to form an
  oxide with the formula MxOy
• Example 1 Iron reacts with oxygen to give Iron
  (III) Oxide

Fe (s) O2 (g) ? Fe2O3 (s)
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How do we solve it?
Fe (s) O2 (g) ? Fe2O3 (s)

• Step 1 Look at the product. There are 3 atoms
  of oxygen in the product, but we start with an
  even number of oxygen atoms.
• Lets convert the of oxygens in the product to
  an even number

Result Fe (s) O2 (g) ? 2Fe2O3 (s)
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How do we Solve It?
Fe (s) O2 (g) ? 2Fe2O3 (s)

• Then, balance the reactant side and make sure the
  number/type of atoms on each side balance.

Balanced Equation 4Fe (s) O2 (g) ? 2Fe2O3
(s)
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How do we Solve It?

• Example 2 Sulfur and oxygen react to form
  sulfur dioxide.

S (s) O2 (g) ? SO2 (g)

• Step 1 Look at the reaction. We lucked out!

Balanced Equation S (s) O2 (g) ? SO2 (g)
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How do we Solve It?

• Example 3 Phosphorus (P4) reacts with oxygen to
  give tetraphosphorus decaoxide.

P4 (s) O2 (g) ? P4O10 (s)

• Step 1 Look at the reaction. The phosphorus
  atoms are balanced, so lets balance the oxygens.

Balanced Equation P4 (s) 5O2 (g) ? P4O10 (g)
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How do we Solve It?

• Example 4 Combustion of Octane (C8H18).

C8H18 (l) O2 (g) ? CO2 (g) H2O (g)

• Step 1 Look at the reaction. Then
• Balance the Carbons

C8H18 (l) O2 (g) ? 8CO2 (g) H2O (g)
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How do we Solve It?
C8H18 (l) O2 (g) ? 8CO2 (g) H2O (g)

• Step 2 Balance the Hydrogens

C8H18 (l) O2 (g) ? 8CO2 (g) 9H2O (g)

• Step 3 Balance the Oxygens
• Problem! Odd number of oxygen atoms
• Solution Double EVERY coefficient (even those
  with a value of 1)

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How do we Solve It?
C8H18 (l) 12.5O2 (g) ? 8CO2 (g) 9H2O (g)

• Step 3 (contd) Balance the Oxygens

2C8H18 (l) 25O2 (g) ? 16CO2 (g) 18H2O (g)

• Step 4 Make sure everything checks out

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Review of Balancing Equations