Transcript chem chapter 09a notes

The Mole
Where Are We Going?
Count large objects by ones
(cars).
Count smaller objects by
dozens (eggs).
Count tiny objects by
hundreds (ream of paper).
Where Are We Going?
Smaller objects need larger
numbers to be sizeable.
Atoms and molecules are so
small that we must have a
HUGE number of them to
work with them.
Avogadro’s Number
2 is a pair.
12 is a dozen.
144 is a gross.
602,200,000,000,000,000,000,000
is Avogadro’s number.
It can be written 6.022 × 1023.
The Mole
A mole is the amount of
matter in Avogadro’s number
of particles (i.e., 6.022 × 1023
particles).
Use four significant digits for
calculations.
Why That Number?
That many particles weighs (in
grams) the exact weight (in u)
of one particle.
1 atom of H weighs 1.008 u.
1 mole of H weighs 1.008 g.
Molar Mass
The terms “gram-atomic
mass,” “gram-molecular
mass,” and “gram-formula
mass” are sometimes used.
We will use “molar mass”
(MM) for all these terms to
avoid confusion.
Molar Mass
Molar mass is the weight of
1 mole of the substance.
If the substance is more than
one element, add up the
molar mass of the elements.
Example 1
1 mole of H2SO4 has:
2 moles of H at 1 g each
1 mole of S at 32 g each
4 moles of O at 16 g each
(2)(1) + (1)(32) + (4)(16) = 98
So, H2SO4 has 98 g/mole.
Question
What is the MM of KMnO4?
1. 158 g/mole
2. 115 g/mole
3. 207 g/mole
4. 131 g/mole
5. 147 g/mole
1 mol
× molar mass
Mass
(grams)
Moles
molar mass
×
1 mol
1
mol
×
NA units
No. of
particles
N
units
A
×
1 mol
0.500 mole K atoms
4.00 g Cu
33.3 mg Au
70 g Al2(SO4)3
Types of Formulas
1. Empirical formula
2. Molecular formula
3. Structural formula
Empirical Formula
Shows only the ratio of atoms
Ex: CH2 for ethene
(really C2H4)
Molecular Formula
Shows the actual number of
atoms of each element in the
substance
Ex: C2H4 for ethene
Structural Formula
Not only has the exact number
of atoms, but also shows the
structure
Ex:
H
H
C=C
H
H
Percent Composition
Is always percent by weight,
unless stated to be by
volume
Deals with mass, not number
of particles
part
Equals
× 100%
whole
60.00 g H2O
53.28 g O
6.72 g H
Example 2
A laboratory analysis of a
30.00 g sample of Al2(SO4)3
showed that it contained
4.731 g of aluminum, 8.436 g
of sulfur, and 16.833 g of
oxygen. What is the percent
composition of this
compound?
Al:
Example 2
4.731 g Al
× 100% = 15.77% Al
30.00 g Al2(SO4)3
S:
8.436 g S
× 100% = 28.12% S
30.00 g Al2(SO4)3
O:
16.833 g O
× 100% = 56.11% O
30.00 g Al2(SO4)3
Formula-to-Percent
Problems
1. Find the number of grams
of each element in 1 mole
of the substance.
2. Find the molar mass of
substances.
Formula-to-Percent
Problems
3. Divide the number of grams
of each element by the
molar mass of substances.
4. Multiply by 100 to get
percent.
Example 3
Find the percent composition
of this compound: Al2(SO4)3.
2 mol of Al (26.98 g/mol) = 53.96 g
3 mol of S (32.07 g/mol) = 96.21 g
12 mol of O (16.00 g/mol) = 192.0 g
Total
342.2 g
Al:
Example 3
53.96 g Al
× 100% = 15.77% Al
342.2 g Al2(SO4)3
S:
96.21 g S
× 100% = 28.12% S
342.2 g Al2(SO4)3
O:
192.0 g O
× 100% = 56.11% O
342.2 g Al2(SO4)3
Example 4
How many grams of oxygen
would a 65.00 g sample of
Al2(SO4)3 contain?
65.00 g Al2(SO4)3
16.833 g O
30.00 g Al2(SO4)3
= 36.47 g O
Percent-to-Formula
Problems
1. Assume there is 100 g of
the substance.
2. Divide the number of grams
of each element by its
molar mass to find the
number of moles.
Percent-to-Formula
Problems
3. Divide the number of moles
of each element by the
smallest number of moles.
This gives the empirical
formula.
100.00 g unknown
75.00 g C
25.00 g H
Example 5
An unknown gas is 72.55% O
and 27.45% C by mass.
What is the empirical
formula?
Example 5
Percent Composition:
72.55% O
27.45% C
Example 5
Mass Composition:
In a 100 g sample of the gas,
there will be 72.55 g of O and
27.45 g of C.
Example 5
Mole Composition:
72.55 g O 1 mole O
16.00 g O
= 4.534 mole O
27.45 g C 1 mol C
12.01 g C
= 2.286 mol C
Example 5
Mole Ratio:
mol O : mol C = 4.534 : 2.286
reduced to lowest terms is
4.534 : 2.286
mol O : mol C = 2.286 2.286
= 1.983 : 1.000
Example 5
Empirical Formula:
For every 2 mol of oxygen,
there is 1 mol of carbon.
The empirical formula must
be CO2.
Example 6
A 5.000 g sample of an
unknown compound contains
1.844 g of N and 3.156 g of O.
Find the empirical formula.
Example 6
Mass Composition:
1.844 g N
3.156 g O
Example 6
Mole Composition of Sample:
1.844 g N 1 mol N
14.01 g N
= 0.1316 mol N
3.156 g O 1 mol O
16.00 g O
= 0.1973 mol O
Example 6
Mole Ratio:
mol N : mol O = 0.1316 : 0.1973
reduced to lowest terms is
0.1316 0.1973
mol N : mol O = 0.1316 : 0.1316
= 1.000 : 1.499
The ratio should be 1 : 1.5 (N : O).
Example 6
Empirical Formula:
2 3
1 : 1.5 = :
2 2
2 2 : 3 2
= 2:3
1 2 3 1
The empirical formula of the
compound is N2O3.
Example 7
Find the molecular formula of
caffeine, which has a molar
mass of 194.20 g/mol.
Percent composition:
5.170% H
16.49% O
28.86% N
49.48% C
Example 7
Mass composition:
In a 100 g sample of the
compound, there will be
5.170 g H
16.49 g O
28.86 g N
49.48 g C
Example 7
Mole Composition of Sample:
5.170 g H 1 mol H
1.008 g H
= 5.129 mol H
16.49 g O 1 mol O
16.00 g O
= 1.031 mol O
Example 7
Mole Composition of Sample:
28.86 g N 1 mol N
14.01 g N
= 2.060 mol N
49.48 g C
1 mol C
12.01 g C
= 4.120 mol C
Example 7
Mole
5.129
mol
H:
=
4.974
Ratio:
1.031 mol
1.031 mol
O: 1.031 mol = 1.000
2.060 mol
N: 1.031 mol = 1.998
4.120 mol
C: 1.031 mol = 3.996
Example 7
Molecular Formula:
empirical formula = C4H5N2O
molar mass of empirical
formula = 97.10 g/mol
molar mass of caffeine =
194.20 g/mol
194.20 g/mol
= 2.000
97.10 g/mol
Example 7
Molecular Formula:
C4H5N2O 2
C8 H10 N4O2