MOLES

• 1 mole = 6.02 × 1023
  • No units by itself (gets a unit after defining what it measures)
• 6.02 × 1023 is Avogadro’s number (Amedeo Avogadro)
• 1 mole is extremely large
  • One mole of seconds has not occurred - 0.000073% of the way since the beginning of the universe
  • Necessary because atoms, molecules, and formula units are infinitesimally small
• 1 mole of an element = atomic mass number in grams/mole (grams per mole)
  • 1 atom of C = 12.01 amu
    • When looking at the atomic mass number, round to 2 decimal places
  • 1 mole of C atoms = (6.02 × 1023) × 12.01 amus = 12.01 grams
    • Molar mass (mass of mole)
      • Units show if the mass is atomic or molar (amu or gram)
    • True for all particle masses in amu
      • 1 mole of a molecule = molecular mass number (mass of one molecule) in grams
        • Molecular mass = atomic mass of each part of the molecule added together
          • H2O average molecular mass = 1.01 + 1.01 + 16.00 = 18.02 amu
        • Multiply molecular mass by 6.02 × 1023 to get average molar mass

MOLE CONVERSIONS

• Unit conversions (ratio to use when converting) is on the arrows
• For multiple steps, go through moles (convert to moles to move directions or to different places in chemical equations)
• Particle terminology:
  • Molecule for molecular compounds (non metal compounds made up of atoms)
  • Formula units for ionic compounds/salts (metal + non metal compounds made up of ions)
• Molarity (M) is written as x M
  • The unit conversion is $\frac{\text{x\ moles}}{1\ L\ sol'n}$
  • The compound after the molarity is the solute, assume the solvent is water
    • Solute - solid that goes into the solvent and dissolves
    • Solvent - typically water, does the dissolving
    • Solution - solute and solvent combined
      • When writing the volume of the solution before conversions, write x mL/L sol’n (not of the solvent)
  • The grams of solute that is found after converting would not be added to initial volume of water (would be less than molarity, more diluted), put solute and add water to amount of solution (for creating solutions with a specific concentration)
  • To convert from volume of solution and grams of solute to molarity, divide moles of solute by liters of solution (may need to convert units first)

EXAMPLE PROBLEMS

4.7 g NaNO3 = ? ions Na+

4.7 g NaNO3 $\times (\frac{1\ mol\ Na\text{NO}{3}}{85.00\ g\ Na\text{NO}{3}}) \times (\frac{1\ mol\ \text{Na}^{+}}{1\ mol\ Na\text{NO}_{3}}) \times (\frac{\ 6.02 \times 10^{23}\text{ions\ }\text{Na}^{+}}{\ 1\ mol\ \text{Na}^{+}})$ = 3.3x1022 ion Na+

30. mL CO2 = ? mol CO2

31. mL CO2 $\times (\frac{1\ L\ \text{CO}{2}}{\ 1000\ mL\ \text{CO}{2}}) \times (\frac{1\ mol\ \ \text{CO}{2}}{22.4\ \ L\ \text{CO}{2}})$ = 0.0012 mol CO2

15 mL of 6 M NaCl = ? g Cl

15 mL sol’n $\times (\frac{1\ L\ sol'n}{\ 1000\ mL\ sol'n}) \times (\frac{6\ mol\ \ NaCl}{\ 1\ L\ sol'n}) \times (\frac{1\ mol\ Cl}{\ 1\ mol\ \ NaCl}) \times (\frac{35.45\ g\ Cl}{\ 1\ mol\ \ Cl})$ = 3.2 g Cl

Molarity of 15 L solution with 13 grams of LiBr

13 grams LiBr = ? mol LiBr

13 g LiBr $\times (\frac{1\ mol\ LiBr}{\ 86.84\ g\ \ LiBr})$ = 0.15 mol LiBr

$\frac{0.15\ mol\ LiBr}{15\ L\ sol'n}$x = 0.01 M

PERCENT COMPOSITION

• Always by mass
  • H2O - most of molecule is H but most of mass is O (H has a smaller atomic mass)
    • % H is not 67%, it is 11.2%
• To find the percent composition of a formula, divide the amount of the element (mass) by the total mass of the compound, then multiply by 100
  • Percent composition of C in calcium acetate Ca(C2H3O2)2:
    • $\frac{4(12.01)}{40.08\ + \ 4(12.01)\ + \ 6(1.01)\ + \ 4(16.00)} \times 100 = 30.37$
      1. Ca(C2H3O2)2 is 30.37% Carbon
• To get from percent composition to a formula, find the empirical formula, then find the molecular formula
  • Molecular formula - molar ratio
    • C6H12O6
    • C5H10O5
    • CH2O
  • Empirical formula - lowest whole number ratio of elements in a compound, can be the same as the molecular formula (especially common for salts) (not unique for substance - multiple compounds can have the same empirical formula)
    • C6H12O6, C5H10O5, CH2O empirical formula is CH2O
    • Empirical formula problem (start where given information is):
      1. Percent composition
         • For salts, determine the possible empirical formulas (consider different way an element can ionize and polyatomic ions)
      2. Grams of each element (multiple the percent by 100 - would be the amount of each element if 100 grams of the compound was found)
      3. Moles of each element
         • Do not round to sig figs to get intended ratio
         • Go to at least 5 decimal places
      4. Normalize, divide the moles of each element by the smallest number of moles of an element (don’t normalize the mass, normalize the moles)
      5. Make the number for each element found by normalizing a whole number
         • May need to multiply to get to a whole number (multiply the number found for each element by the same number to maintain the ratio)
           • 0.125 x 8
           • 0.67 x 3
           • 0.50 x 2
           • 0.20 x 5
           • 0.25 x 4
           • 0.33 x 3
         • The whole number is the number of atoms/ions of that element in that compound
  • To get from the empirical formula to the molecular formula, need molar mass (MM) in grams/mole

1. Divide the molar mass by the empirical formula mass (will equal an exact whole number)
   1. May be given a range instead of an exact number - find the whole number that multiplies to a number within the range
2. Multiply the whole number found by each of the subscripts to get the molecular formula (the whole number means that there is x times the mass of the empirical formula compared to the molecular formula mass)

EXAMPLE PROBLEM

71.47% Ca, 28.53% O

COMBUSTION ANALYSIS

• Use to determine the empirical and molecular formulas
• Usually for hydrocarbons (CxHy) - elements made up primarily of hydrogen and carbon