Transcript TheMole revised

The
Mole
Standards
Students know the quantity one mole is set by defining
one mole of carbon 12 atoms to have a mass of exactly 12
grams.
Students know one mole equals 6.02 x 1023 particles
(atoms or molecules).
Students know how to determine the molar mass of a
molecule from its chemical formula and a table of atomic
masses and how to convert the mass of a molecular
substance to moles, number of particles, or volume of gas
at standard temperature and pressure.
The Mole
1 dozen = 12
1 gross = 144
1 ream = 500
1 mole = 6.02 x 1023
There are exactly 12 grams of
carbon-12 in one mole of carbon-12.
Avogadro’s Number
6.02 x 1023 is called “Avogadro’s Number” in
honor of the Italian chemist Amadeo Avogadro
(1776-1855).
I didn’t discover it. Its
just named after me!
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
6.9 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
6.94 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
6.9 g Li
(18.2)(6.02 x 1023)/6.9
6.02 x 1023 atoms Li
1 mol Li
= 1.58 x 1024 atoms Li
Calculating Formula Mass
Calculate the formula mass of carbon
dioxide, CO2.
12.0 g + 2(16.0 g) =
44.0 g
 One mole of CO2 (6.02 x 1023 molecules)
has a mass of 44.0 grams
Standard Molar Volume
Equal volumes of all gases at the same temperature
and pressure contain the same number of molecules.
- Amedeo Avogadro
At STP (Standard Temperature and Pressure):
1 mole of a gas occupies 22.4 liters of volume
The Mole Highway
Form Mass 1 mol
22.4 L
x
x
x
1 mol
Form Mass
1 mol
Mass (g)
x
1 mol
22.4 L
Vol (L)
6.02x1023
Rep Part
1 mol
x
x
6.02x1023
1 mol
Rep Part
Rep Part
Calculations with Moles:
Two-Step Problem
How many atoms of lithium are in 18.2 g of
lithium?
18.2 g Li 1 mol Li
6.02 x 1023 atoms Li
6.9 g Li
(18.2)(6.02 x 1023)/6.9
1 mol Li
= 1.58 x 1024 atoms Li
Calculating Percentage Composition
Calculate the percentage composition of magnesium
carbonate, MgCO3.
From previous slide:
24.3 g + 12.0 g + 3(16.0 g) = 84.3 g
 24.31 
Mg  
  100  28.83%
 84.32 
 12.01 
C 
  100  14.24%
 84.32 
 48.00 
O
  100  56.93%
 84.32 
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
Rules
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
Example
Adipic acid contains 49.32% C, 43.84% O, and
6.85% H by mass. What is the empirical formula
of adipic acid?
 49.32 g C 1 mol C   4.107 mol C
12.01 g C 
 6.85g H 1 mol H   6.78 mol H
1.01 g H 
 43.84 g O 1 mol O   2.74 mol O
16.00 g O 
Empirical Formula Determination
(part 2)
Divide each value of moles by the smallest of the
values.
4.107
mol
C
Carbon:
 1.50
2.74 mol O
6.78 mol H
Hydrogen:
 2.47
2.74 mol O
2.74 mol O
Oxygen:
 1.00
2.74 mol O
Empirical Formula Determination
(part 3)
Multiply each number by an integer to obtain all
whole numbers.
Carbon: 1.50
x 2
3
Hydrogen: 2.50
x 2
5
Oxygen: 1.00
x 2
2
Empirical formula: 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(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g
Finding the Molecular Formula
(continued)
2. Divide the molecular mass by the
mass given by the empirical formula.
3(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g
146
2
73
3. Multiply the empirical formula by this
number to get the molecular formula.
(C3H5O2) x 2 = C6H10O4