Chapter Calculations with Chemical Formulas and Equations

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Chapter Calculations with Chemical Formulas and
Equations
Chemistry 1061 Principles of Chemistry I Andy
Aspaas, Instructor
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Molecular weight and formula weight

• Molecular weight sum of atomic weights for all
  atoms in a substance
• MW of H2O 2(1.008 amu) (16.00 amu) 18.016
  amu
• Formula weight same as above, but for any
  compound, molecule or not
• FW of NaCl (22.99 amu) (35.45 amu) 58.44 amu

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Mass and moles

• Mole quantity of a substance that contains as
  many molecules or formula units as the number of
  atoms in exactly 12 grams of carbon-12
• That number Avogadros number, 6.02 x 1023
• Simply a quantity like pair, dozen, gross
• Molar mass mass of 1 mole of a substance
• Equal to formula weight of that substance!
• One mole of oxygen has a mass of 16.00 g
• One mole of H2O has a mass of 18.016 g

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Mass and moles

• Mole calculations
• Use atomic, molecular, or formula weight as a
  conversion factor in dimensional analysis to
  convert between mass and moles, and vice-versa
• How many moles in 100.0 g Fe? 100g of H2O?
• What is the mass of 4.72 moles of Mg? 3.25 moles
  of glucose (C6H12O6)?

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Molecules and moles

• Use Avogadros number as a conversion factor to
  convert between number of particles and moles,
  and vice versa
• 6.02 x 1023 particles 1 mole
• Combine the two methods to convert mass to number
  of particles.

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Determining mass percentage from a formula

• Mass percentage mass of certain constituent
  divided by mass of whole, expressed as percentage
• Mass from formula
• Assume you have 1 mol of the substance, what mass
  do you have?
• Use formula to determine number of moles of each
  element present
• Convert moles of element to mass
• Divide mass of element by mass of compound and
  multiply by 100

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Elemental analysis

• Determines amount of C, H, and O in a compound
  that contains only those elements
• Compound is burned in an oven, amounts of CO2 and
  H2O given off are recorded
• Since 1 mol of CO2 contains 1 mol C atoms, the
  mass of carbon in the substance can be found
• 1 mol H2O contains 2 mol H atoms - mass of H can
  be found
• Mass of C and H found
• O mass is whats left over

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Determining empirical formulas from composition

• Empirical formula simplest formula of a
  substance that has all integer whole-number
  subscripts
• If given compositions of elements, assume you
  have 100.0 g of the substance
• 14.0 of an element becomes 14.0 g of that
  element
• Convert grams to moles for each element
• Divide all mole numbers by the smallest one
• If not all answers are integers (within
  experimental error), multiply to make them all
  the smallest possible integers

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Molecular formula from empirical formula

• Molecular formula is some multiple of the
  empirical (simplest) formula
• Empirical formulas of acetylene (C2H2) and
  benzene (C6H6) are both CH
• Molecular weight n x empirical formula weight
• n molecular weight / empirical formula weight
• Multiply all subscripts in empirical formula by n

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Stoiciometry

• Chemical equations can be interpreted as either
  individual molecules or numbers of moles
• N2(g) 3H2(g) ---gt 2NH3(g)
• 1 N2 molecule reacts with 3 H2 molecules to make
  2 NH3 molecules
• Or 1 mol N2 reacts with 3 mol H2 to make 2 mol
  NH3
• Use coefficients as conversion factors when
  converting moles of any constituent in a chemical
  reaction to moles of any other constituent

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Masses in chemical equations

• How many grams of a product will be yielded from
  a given mass of reactant?
• Do same process as before, but convert masses to
  moles first, and moles back to masses
• Mass NH3 --gt mol NH3 --gt mol H2 --gt mass H2

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Limiting reactant

• Often times, one reactant will be in excess of
  the other
• Number of moles of product is determined by
  starting moles of limiting reactant
• Determine number of moles of each reactant, and
  calculate how many moles of product each would
  make
• The reactant that makes the least amount of moles
  of product is the limiting reactant

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Theoretical yield

• Theoretical yield maximum amount of product that
  can be obtained by a reaction from given amounts
  of reactants
• In actuality, less product is often obtained
• Losses due to side-reactions
• Losses due to incomplete separations
• Percentage yield
• (actual yield / theoretical yield) x 100