Chapter 9 Chemical Quantities

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Chapter 9Chemical Quantities
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Formula Mass (Weight)Molecular Mass (Weight)

• Determine formula mass when given atomic masses
  and formula of a substance.
• (Also called molecular weight. When expressed in
  g, this is the molar mass, M)
• The formula mass is the sum of the atomic masses
  of the atoms in the element times their
  subscripts.

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For the Compound with FormulaAaBb

• M (AaBb) aA bB, where A and B are the atomic
  masses of A and B, respectively.
• Calculate the formula mass and molar mass of
  C6H12O6.
• M 6(12.0) 12(1.0) 6(16.0) 180
• 180 g (molar mass)

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The Mole

• The SI unit for the amount of substance.
• The amount of any substance containing the same
  number of formula units as 12.0000g of 612C.
• Formula units are ions, molecules, or atoms.
• One mole has 6.022 x 1023 formula units.
• 6.022 x 1023 is called Avogadros Number.

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The Mole

• Relate molar mass to atomic mass, Avogadros
  Number, and the mass of one atom.
• The molar mass is the mass of one mole of an
  element. One mole has 6.022 x 1023 atoms.
• The molar mass is also the atomic mass measured
  in grams. Thus, the atomic mass of Na is 22.99
  the molar mass is 22.99 g.

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Molar Mass of Monatomic Elements

• One atom of Na has a mass of 22.99 amu.
• One mole of Na has a mass of 22.99 g,
• or (22.99g/mol)/(6.022x1023atoms/mole
• 3.818x10-23g/atommass of one Na atom

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Composition Calculations

• Given Formula, calculate percent by mass of each
  element in a compound.
• Example Problem Calculate the mass percent for
  each element in isopropyl alcohol, C3H8O.
• 1. Calculate the molar mass
• C3 3 x 12.0 36.0
• H8 8 x 1.0 8.0
• O 1 x 16.0 16.0
• M 60.0

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Percent by Mass From Formula

• 2. For each element, multiply atomic mass by
  subscript
• C3 3 x 12.0 36.0
• H8 8 x 1.0 8.0
• O 1 x 16.0 16.0

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Percent by Mass From Formula

• 3. Divide each of the above numbers by M, and
  multiply by 100 to get
• C3 (36.0/60)x100 60 C
• H8 (8.0 /60) x100 13.3 H
• O (16.0 /60) x100 26.7 O

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Percent by Mass From Formula

• 4. Check. Percentages should add to 100.
• 60 C
• 13.3 H
• 26.7 O
• 100.0

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Moles and Mass

• Given two of the following, determine the third
  formula mass, mass, number of moles.
• Formulas
• Mol g/M
• g (mol)(M)
• M g/mol

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Example Problems

• Calculate the number of moles present in 50.0g of
  glucose (C6H12O6 , M 180 g/mol)
• mol g/M 50.0g/180g/mol

.278 mol
Calculate the mass of 0.233 mol of glucose.
g mol(M) 0.233 molx180g/mol
41.9 g
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Moles and Formula Units

• Relate the actual number of atoms or molecules to
  the number of moles.
• One mole is 6.022x1023 formula units (atoms,
  molecules, ions, etc.)
• formula units6.022x1023xmols
• mols(formula units)/6.022x1023

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Example Problems

• Calculate the number of moles present in 9.9x1022
  atoms of Na.
• Mole atoms/6.022x1023 atoms/mole
• 9.9x1022 atoms/6.022x1023 atoms/mol

1.64x10-1 0.16 mol
Calculate the number of atoms of Na present in
4.00 mol.
Atomsmolx6.022x10234.00molx6.022x1023atom/mol
2.41x1023atoms
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Molarity

• Molarity (M) is the moles of solute per Liters of
  solution.
• M mol/L
• mol MxL
• L mol/M

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Molarity Calculations

• Given two of the following, moles solute, volume
  solution, molarity calculate the third.
• Calculate the molarity of a solution prepared by
  dissolving 10.0g of NaCl in water to make a total
  volume of 250 mL.
• M mol/ L, L 250mlx(0.001L/mL) 0.25L, mol
  g/M, M 22.99 35.45 58.44g/mol, mol
  10.0/58.44 0.171mol
• M 0.171/0.250 0.684 M

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Preparing molar solutions.

• How would you prepare 250 mL of a 0.200M solution
  of NaCl?
• From the previous example, we have 0.25L, M(NaCl)
  58.44g/mol.
• Mol MxL 0.25x0.2 0.05 mol
• g molxM 0.05x58.44 2.92 g
• Add 2.92 g NaCl to a 250 mL volumetric flask,
  half filled with distilled water, mix until
  dissolved, dilute to the mark, mix.

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Preparing a Solution From a Solid
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Dilution Problem

• Calculate the volume of 2.50 M NaCl required to
  prepare 500. mL of 0.200 M NaCl.
• Mol NaCl required for 500mL, 0.2M NaCl
• mol MxL 0.200x0.500 0.100 mol
• L solution required
• L mol/M 0.100/2.50 0.0400 40.0 mL
• Add 40.0 mL 2.50M NaCl to a 500 mL volumetric
  flask. Dilute to mark. Mix.

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Alternative Approach

• Calculate the volume of 2.50 M NaCl required to
  prepare 500. mL of 0.200 M NaCl.
• Since the moles taken from the concentrated
  solution must equal the moles put into the dilute
  solution
• molcon moldil MconVconMdilVdil
• (2.5)(Vcon)(0.2)(500) Vcon (0.2)(500)/(2.5)
• 40 mL

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Given Molecular Formula, Determine the Empirical
Formula

• Determine the empirical formula of C6H12O6.
• The lowest common denominator of the subscripts
  is 6.
• Divide the subscripts by 6
• Empirical formula is CH2O.

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Determining Empirical Formulas(Simplest Formulas)

• Given percent or analytical data, determine the
  empirical formula.
• Sample Problem. Determine the empirical formula
  of a compound which contains 40.0 C, 6.7 H,
  53.3 O
• 1. Divide , g, etc. by atomic mass
• C 40.0/12.0 3.33
• H 6.7/1.00 6.7
• O 53.3/16.0 3.33

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Empirical Formula from

• 2. Divide each of the numbers from step 1 by the
  smallest of the numbers
• C 3.33/3.33 1
• H 6.7/3.33 2
• O 3.33/3.33 1

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3. Multiply by a common factor to get integers if
needed if .5 by 2 if .33 or .67 by 3 if .25
or .75 by 4 if .2, .4, or .6 by 5
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Empirical Formula from

• 4. Use the numbers from part 2 or 3 as subscripts
  in the formula
• C 1
• H 2
• O 1
• CH2O Empirical Formula
• The empirical mass is the sum of the masses of
  the elements in the empirical formula
• 12 2(1) 16 30

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Empirical Formulas and Molecular Formulas

• Given empirical formula and molecular mass,
  calculate the molecular formula.
• 1. Divide the molecular mass by the empirical
  mass.
• 2. Multiply each of the subscripts in the
  empirical by the number calculated above.

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Determining Molecular Formula

• Example problem Determine the molecular formula
  of a compound which has an empirical formula of
  CH2O and a molecular mass of 180.
• The empirical mass is 12 2(1) 16 30
• 180/30 6
• C1x6H2x6O1x6 C6H12O6