The Mole and Molar Mass
• Moles are the currency of chemistry
• 1 mole = 6.02 x 1023 particles = Avogadro’s number
• Molar mass of an element = mass of one mole of that
element
• Molar mass = atomic mass of element in grams
- Average mass of one atom of C = 12.01 amu
- Average mass of one mole of C = 12.01 g
• Molar mass of a compound = sum of molar masses of each
atom in the compound
• Example: What is the molar mass of MgCl2?
Molar mass of Mg = 24.31 g, Molar mass of Cl = 35.45 g
Molar mass of MgCl2 = 24.31 g + (2 x 35.45 g) = 95.21 g
• *Note: use as many sig. figs. as given in the atomic
masses (don’t round molar masses to 1/10 of a gram)
Calculations using Molar Mass
• Molar mass is used to convert between moles and grams
of a substance
• Chemical equations are balanced with atoms, and moles
are an extension of that (they also balance)
• Quantities of reactants and products are measured by
mass (grams) in the lab, but we use moles to balance
equations
• Example 1: How many grams of H2O are in 0.100 mol
H2O?
Molar mass of H2O = (2 x 1.008 g) + 16.00 g = 18.02 g
Grams H2O = 0.100 mol x (18.02 g/1 mol) = 1.80 g H2O
• Example 2: How many moles are in 0.250 g of HCl?
Molar mass of HCl = 1.008 g + 35.45 g = 36.46 g
Mol HCl = 0.250 g x (1 mol/36.46 g) = 6.86 x 10-3 mol HCl
Percent Composition
• Molecular formula tells you how many moles of each
type of atom are in one mole of a compound
• Using molar mass, you can calculate the % by mass of
each element in a compound
• Example: Calculate the % composition by mass of CO2
Molar mass of CO2 = 12.01 g + (2 x 16.00 g) = 44.01 g
%C = (12.01 g/44.01 g) x 100% = 27.29%
%O = (2 x 16.00 g/44.01 g) x 100% = 72.71%
Empirical Formulas
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Empirical formulas give the lowest whole-number ratio
of atoms in a compound
Molecular formula = multiple of empirical formula
Example: empirical formula for glucose = CH2O
molecular formula = C6H12O6
Can calculate empirical formula from masses or from
% composition by mass
Example: Calculate empirical formula for compound
with 8.0 g C and 2.0 g H
8.0 g C x (1 mol/12.01 g) = 0.6661 mol C
2.0 g H x (1 mol/1.008 g) = 1.984 mol H
0.6661 mol C/0.6661 mol = 1.000 C = 1 C
1.984 mol H/0.6661 mol = 2.979 H = 3 H’s
Empirical formula = CH3
Mole Relationships and Mole Calculations
•
Atoms, moles and mass are conserved in a chemical
reaction (shown in balanced equation)
•
Using chemical formulas, molar masses and mole
ratios (from balanced chemical equation), amounts
of products and reactants can be calculated
•
Example: For C3H8 + 5O2  3CO2 + 4H2O
How many moles of O2 are needed to produce 2.0 moles
of H2O?
1. Have moles H2O, want moles O2
2. 5 mol O2 = 4 mol H2O
3. 2.0 mol H2O x (5 mol O2 /4 mol H2O) = 2.5 mol O2
Mole Relationships and Mass Calculations
Example: C3H8 + 5O2  3CO2 + 4H2O
How many grams of CO2 are produced when 10.0 g of C3H8
are consumed?
1. Have grams C3H8, want grams CO2
(g C3H8  mol C3H8  mol CO2  g CO2)
2. Molar mass C3H8 = (3 x 12.01 g) + (8 x 1.008 g) =
44.094 g
Mol C3H8 = 10.0g x (1 mol/44.094 g) = 0.2268 mol
3. Mol CO2 = 0.2268 mol C3H8 x (3 mol CO2/1 mol C3H8)
= 0.6804 mol
4. Molar mass CO2 = 12.01 g + (2 x 16.00 g) = 44.01 g
Grams CO2 = 0.6804 mol CO2 x (44.01 g/1 mol) = 29.9 g
Percent Yield
•
Theoretical (or ideal) yield is calculated from balanced
chemical equation and molar masses
• Actual yield = what you got (usually < ideal)
• % yield = (actual yield/theoretical yield) x 100%
Example:
2Fe + 3S  Fe2S3
What is the % yield if 1.00 g Fe makes 1.50 g Fe2S3?
1. Calculate theoretical yield:
(g Fe  mol Fe  mol Fe2S3  g Fe2S3)
1.00 g Fe x (1 mol/55.85 g) = 0.0179 mol Fe
0.0179 mol Fe x (1 mol Fe2S3/2 mol Fe) = 0.00900 mol
Fe2S3
Molar mass Fe2S3 = (2 x 55.85 g) + (3 x 32.06 g) = 207.88 g
0.00900 mol Fe2S3 x (207.88 g/1 mol) = 1.87 g Fe2S3
2. % yield = (1.50 g Fe2S3 /1.871 g Fe2S3) x 100% = 80.2%
Limiting Reactant
• Limiting Reactant = reactant that is completely used up
• Example: If 1.0 mol of Fe are reacted with 3.0 mol of S,
which is the limiting reactant?
- Since 2.0 mol Fe are required to react with 3.0 mol of S, then
1.0 mol of Fe can only react with 1.5 mol of S
- So, all the Fe is use up and 1.5 mol of S is leftover
- Fe is the limiting reactant
Equilibrium Constants
• For reversible reactions at equilibrium:
- Rate of forward reaction = rate of reverse reaction
- Amounts of reactants and products are constant
(but not necessarily equal)
• Ratio of products to reactants = equilibrium constant (Keq)
• For a reaction aA + bB  cC + dD
Keq = [C]c [D]d/[A]a [B]b, where [A] = concentration of A
• Example: for 2H2 + O2  2H2O
Keq = [H2O]2/[H2]2 [O2]
• If Keq > 1, equilibrium favors products
If Keq < 1, equilibrium favors reactants