Stoichiometry Notes

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Stoichiometry Notes
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What is Stoichiometry?

• Stoichiometry is the accounting, or math, behind
  chemistry. Given enough information, one can use
  stoichiometry to calculate masses, moles, and
  percents within a chemical equation.

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Types of Formulas

• Empirical (Simplest) Formula- Formula whose
  subscripts represent the simplest whole number
  ratio of atoms in a molecule or the simplest
  whole number ratio of moles of each element in a
  mole of the compound.
• For example, CH2 says that there will be twice as
  many Hydrogens as there are carbons in the
  compound that has this simplest formula. It does
  not address how many exact numbers of Hydrogens
  and Carbons there will be in the compound.

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Types of Formulas

• Molecular Formula- A formula whose subscripts
  represent the absolute exact numbers of atoms of
  each element per molecule of the compound or the
  absolute number of moles of each element per mole
  of the compound. A molecular formula may be
  reducible to a simple formula if all its
  subscripts are divisible by a common denominator.
  

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Types of Formulas

• Structural Formula- Formula that not only gives
  via its subscripts the exact number of atoms of
  each element per molecule but it displays the way
  that the atoms are bonded together and the shape
  of the molecule is revealed.
• There are compounds that have the same empirical
  formula and even the same molecular formula and
  the only way that they can be distinguished is
  through their structural Formulas.

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Molar Ratio

• The molar ratio will assume a place of central
  importance in solving stoichiometry problems. The
  sources for these ratios are the coefficients of
  a balanced equation. A sample equation is
• 2 H2 O2 ---gt 2 H2O
• What is the molar ratio between H2 and O2?
• Answer two to one. So this ratio in fractional
  form is
• It is recommended that you write a one in the
  denominator of the ratio.

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Determining Empirical FormulaFrom the Mass of
Each Element

• We can determine the empirical (simplest) formula
  by using mass of each element in the compound
  data. A compound is composed of 7.20g of carbon,
  1.20g of hydrogen, and 9.60g of oxygen. Find the
  empirical formula for this compound.
• Convert the grams of each element to moles
  knowing that 1 mole of any element formula mass
  in grams
• 7.20 / 12 0.6 mole C
• 1.20 / 1.0 1.2 mols H
• 9.60 / 16 0.6 mols O

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Continued

• Divide each of the three mole figures by the
  lowest of the three in order to simplify the mole
  ratio since the definition of simple formula is
  the SIMPLEST whole number mole ratio.
• 0.6 mol C / 0.6 1.0 mol C
• 1.2 mol H / 0.6 2.0 mol H
• 0.6 mol O / 0.6 1.0 mol O
• If these mole figures are not whole numbers or
  within .2 of being to the next whole number then
  find the lowest common factor (ie 2,3,4) that
  when you multiply by that factor each of the mole
  figures determined in step 2 that will result in
  whole numbers since the definition mentioned in
  step 2 talks about WHOLE numbers. Use these mole
  figures arrived at in either step 2 or 3
  (whichever are whole number ratio) as the
  subscripts and use them after the respective
  symbol.
• CH2O would be the Empirical Formula

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Determining Mass From Formulas

• Formulas allow us to determine molecular mass of
  a compound. They can also lead to the mass
  composition of each element represented in the
  compound.
• Mass of X (Subscript of X in
  formula)(atomic mass of X)(100)
  Formula Mass
• The Formula Mass is the mass based upon the
  formula. It is determined by taking the subscript
  for each element represented in the formula and
  multiplying it by the atomic mass of the element
  to get a mathematical product.

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Example

• Determine the mass of Oxygen in Sulfuric Acid,
  H2SO4 - Multiply the subscript for Hydrogen in
  the formula by the atomic mass of Hydrogen
  (1.0)
• 2(1.0) 2.0
• Multiply the subscript for Sulfur by the atomic
  mass of Sulfur(32.1)
• 1(32.1) 32.1
• Multiply the subscript of Oxygen by the atomic
  mass of Oxygen
• 4(16.0) 64.0
• Add answers from steps 1,2, and 3 together for
  the formula mass
• 2.0 32.1 64.0 98.1 Formula Mass of H2SO4

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Continued

• We can determine the mass of Hydrogen in the
  above compound as well as the mass of Sulfur
  and Oxygen by using the equation for determining
  the mass introduced early in the lesson.
• mass H 2(1.0)(100) / 98.1 2.04 H
• mass S 1(32.1)(100)/ 98.1 32.72 S
• mass O 4(16.0)(100) / 98.1 65.24 O
• What this means in the above example is that
  65.24 of the total sample mass of Sulfuric Acid
  is due to the presence of Oxygen. Sulfur
  constitutes 32.72 of the total sample mass. In a
  100 gram sample of Sulfuric Acid 65.24 grams of
  Oxygen will be present 32.72 grams Sulfur will be
  present in that 100 gram sample , and 2.04 grams
  of Hydrogen are found in every 100 gram sample of
  Sulfuric Acid.

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Mass Percent

• Once the mass percent of each element is
  determined for a compound, then according to the
  Law of Constant Composition those mass
  percentages remain constant are fixed in value no
  matter what size sample you have.

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Determining Empirical Formulas From Mass
Composition

• One variation of the previous example is if the
  composition is given in mass percent. If this is
  the case then one assumes a 100 gram sample and
  then convert all percentage figures to gram
  figures. Then you would proceed with step 1.
  Assuming 100 gram sample allows us to use the
  same numbers as gram figures instead of
  percentage figures. The numerical percentages
  will remain fixed according to the Law of
  Constant Composition. Let's take the problem
  above where the combustion of the sample gave the
  mass The compound had the following mass
  composition
• 38.71 Carbon
• 9.71 Hydrogen
• 51.58 Oxygen

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Continued

• Determine the Empirical Formula for the compound.
  
• Assume 100 grams of sample
• so in a 100 gram sample there will be 38.71 grams
  of Carbon, 9.71 grams Hydrogen, and 51.58 grams
  Oxygen
• Convert grams of each element in that 100 gram
  sample to mols of the element dividing by the
  atomic mass of the element
• 38.71 grams C X 1 mol C / 12.0 grams C 3.23
  mols C
• 9.71 grams H X 1 mol H / 1.0 grams H 9.71 mols
  H
• 51.58 grams O X 1 mol O / 16.0 grams O 3.22
  mols O
• Get a simple ratio of mols by dividing each mol
  figure by the lowest of the three
• 3.23 mols C / 3.22 1.0 mol C
• 9.71 mol H / 3.22 3.0 mol H
• 3.22 mol O / 3.22 1.0 mol O
• If these are whole numbers they represent the
  subscripts in the Empirical Formula
• CH2O