Moles and Formula Mass
The Mole
1 dozen =
12
1 gross =
144
1 ream =
1 mole =
500
6.022 x 1023
(on your reference sheet)
There are exactly 12 grams of carbon-12 in
one mole of carbon-12.
Avogadro’s Number
6.022 x 1023 is called “Avogadro’s Number” in honor of
the Italian chemist Amadeo Avogadro (1776-1855).
I didn’t discover it. It’s just
named after me!
Note: If I sat for a portrait and this is how it
came out, I’d be devastated.
Amadeo Avogadro
Calculations with Moles:
Converting moles to grams
How many grams of lithium are in 3.50 moles of
lithium?
3.50 mol Li
6.941 g Li
1 mol Li
= 45.1 g Li
Calculations with Moles:
Converting grams to moles
How many moles of lithium are in 18.2 grams of
lithium?
18.2 g Li
1 mol Li
6.941 g Li
= 2.62 mol Li
Calculations with Moles:
Using Avogadro’s Number
How many atoms of lithium are in 3.50 moles of
lithium?
3.50 mol
6.02 x 1023 atoms
1 mol
= 2.07 x 1024 atoms
Calculations with Moles:
Using Avogadro’s Number
How many atoms of lithium are in 18.2 g of
lithium?
18.2 g Li
1 mol Li
6.941 g Li
(18.2)(6.022 x 1023)/6.941
6.022 x 1023 atoms Li
1 mol Li
= 1.58 x 1024 atoms Li
Calculating Formula Mass
Calculate the formula mass of magnesium carbonate,
MgCO3.
24.31 g + 12.01 g + 3(16.00 g) =
84.32 g
Calculating Percentage Composition
Calculate the percentage composition of magnesium
carbonate, MgCO3.
From previous slide:
24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
100.00
Formulas
Empirical formula: the lowest whole number ratio of
atoms in a compound.
Molecular formula: the true number of atoms of
each element in the formula of a compound.
 molecular formula = (empirical formula)n [n =
integer]
 molecular formula = C6H6 = (CH)6
 empirical formula = CH
Formulas (continued)
Formulas for ionic compounds are ALWAYS
empirical (lowest whole number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3
Formulas (continued)
Formulas for molecular compounds MIGHT be
empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11
Empirical Formula Determination
1. Base calculation on 100 grams of compound.
2. Determine moles of each element in 100
grams of compound.
3. Divide each value of moles by the smallest of
the values.
4. Multiply each number by an integer to obtain
all whole numbers.
Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by
mass. What is the empirical formula of adipic acid?
(part 1)
Divide each % by the element’s mass
Empirical Formula Determination
(part 2)
Divide each value of moles by the smallest of the
values.
Carbon:
Hydrogen:
Oxygen:
Empirical Formula Determination
(part 3)
Multiply each number by an integer to obtain all
whole numbers.
Carbon: 1.50
x 2
Hydrogen: 2.50
x 2
3
Empirical formula:
Oxygen: 1.00
x 2
5
2
C3H5O2
Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2.
The molecular mass of adipic acid is 146 g/mol.
What is the molecular formula of adipic acid?
1. Find the formula mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2.
The molecular mass of adipic acid is 146 g/mol.
What is the molecular formula of adipic acid?
2. Divide the molecular mass by the mass
given by the emipirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2.
The molecular mass of adipic acid is 146 g/mol.
What is the molecular formula of adipic acid?
3. Multiply the empirical formula by this number
to get the molecular formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
(C3H5O2) x 2 =
C6H10O4
Combustion Analysis
•
Device used to determine the mass percent
of each element in a compound.
20
Ex) Combustion Analysis
• A 2.04g sample containing C, H, & O
underwent combustion analysis.
• 4.48g of CO2 and 2.4g of H2O were produced.
• Find the empirical formula.
Process
• Use stoichiometry to find the grams of carbon
and hydrogen produced.
• Subtract the total mass from the sum of the
masses of carbon and hydrogen to find the mass
of oxygen.
• Find moles of C, H, & O.
Find masses of C, H, & O
• Mass of C
• Mass of H
• Mass of O
Find moles of C, H, & O
Divide # moles of each element by
smallest # calculated
Isopropyl alcohol, a substance sold
as rubbing alcohol, is composed of
C, H, and O. Combustion of 0.255g
of isopropyl alcohol produces
0.561g CO2 and 0.306g H2O.
Determine the empirical formula of
isopropyl alcohol.
• When 0.1156g of a compound that contains only
the elements carbon, oxygen, and hydrogen
undergoes combustion analysis, it is found to
produce 0.1638g of CO2 and 0.1676g of H2O.
Determine the empirical formula of this compound.