Download POWERPOINT - Chapter 8

Chapter 8
Chemical
Composition
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Inc., or its affiliates. All Rights
Reserved.
.

Moles and Avogadro's number

Molar Mass

Convert between particles,moles, and mass

Mass percent of elements in compounds

Empirical formulas

Calculating molecular formulas



Nylon, aspartame, Kevlar (bulletproof vests), PVC,
Teflon
All originated in chemist's laboratory
Once they make it – they must determine what it is
◦ What is it's composition?
◦ What is it's chemical formula?
Recall that matter is composed of
atoms, molecules, and ions.
•
•
These particles are much, much smaller
than grains of sand, and an extremely
large number of them are in a small
sample of a substance.
Obviously, counting particles one by one
is not practical.
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Inc., or its affiliates. All Rights
Reserved.
.


Since atoms are so small, we need a
quantity with a larger amount
We use the mole (mol)
o
o
1 mole = 6.02 X 1023 representative particles
(atoms, molecules, formula units)
It is known as Avogadro’s number

We usually are dealing with three types
of representative particles
o
Molecules

o
Atoms

o
H2O and H2 are molecules, covalently bonded
Al and Na are atoms, not bonded
Formula units

CaCl2 and NaOH are formula units, ionicly
bonded
1. How many particles can be found in
10 grams of neon?
2. If you had a bottle that contained
5.69x1024 particles of water, how many
grams of water does the bottle hold?
3. How many particles are in 50 g of
Ca(OH)2?
10 g Ne X 1 mol
X
20.17 g Ne
6.02 x 1023 atoms Ne =
1 mole
= 2.98 x 1023 Ne atoms
5.69 x 1024 molecules H2O
X
________1
mol
6.02 x 1023 molecules H2O
= 170.32 g H2O
X
18.02 g H2O
1 mol
=
50
g Ca(OH)2
1 mol
6.02 x 1023 f.u. Ca(OH)2
X
X
=
74.10 g Ca(OH)2
1 mol
= 4.06 x 1023 f.u. Ca(OH)2
◦ The mole allows chemists to count the number
of representative particles in a substance.
◦ The atomic mass of an element expressed in
grams is the mass of a mole of the element.
◦ To calculate the molar mass of a compound,
find the number of grams of each element in
one mole of the compound. Then add the
masses of the elements in the compound.
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Inc., or its affiliates. All Rights
Reserved.
.
 Percent composition of a compound is
the relative amount of each element in the
compound.
Which of the following compounds contains
the highest percent of Iron?
Iron III Acetate, Iron II Hydroxide or Iron II
nitride.

Sometimes, percent composition can
include compounds called hydrates



Hydrates are compounds that bind
water molecules to their structure
BaCl2 • 6H2O
MgSO4 • 3H2O
Name -> barium chloride
hexahydrate
Name ->
Find the percentage of water in barium
chloride hexahydrate
H2O
MgSO4  7H2O
%H2O =
MM =24.31 + 32.06 + 4(16) + 7(18.02)
= 246.51 g/mol
7(18.02 g/mol)
246.51g/mol
= 51.17 % H2O
x 100

Find the percent composition of water in
potassium sulfate tetrahydrate.
K2SO4  4H2O
%H2O =
MM =2(39.10) + 32.06 + 4(16) + 4(18.02)
= 246.34 g/mol
4(18.02 g/mol)
246.34 g/mol
= 29.26 % H2O
x 100

Empirical formula is the simplest
whole number ratio between
elements in a compound
You are given % of an element. Change the % to
grams (the number does not change).
Example 1: You have 26.56% K, 35.41% Cr, and
38.03% O .
Therefore, you have: 26.56 g K, 35.41 g Cr, and
Now change Grams to Moles
26.56 g K x
1 mol K
39.10 g K
=
x
1 mol Cr
52 g Cr
=
0.6810 mol Cr
38.03 g O x
1 mol O
16 g O
=
2.3769 mol O
35.41 g Cr
0.6793 mol K

Divide each molar quantity by the smaller
number of moles to get 1 mol for the
element with the smaller number of moles.
0.6793 mol K
0.6793 mol
=
1
0.6810 mol Cr
0.6793 mol
=
1
2.3769 mol O
0.6793 mol
=
3.5

Multiply each part of the ratio by the smallest
whole number that will convert both
subscripts to whole numbers.

Multiply each part of the ratio by the smallest
whole number that will convert both
subscripts to whole numbers.
0.6793 mol K
0.6793 mol
=
0.6810 mol Cr
0.6793 mol
=
2.3769 mol O
0.6793 mol
=
1
x2
=
2
1
x2
=
2
3.5 x 2
Answer =
=
7
K2Cr2O7
1. Change % to grams
2. Change grams to moles
3. Divide all moles by the lowest number of
moles
4. If all the numbers are not whole
numbers, you must multiply the lowest
number possible to get a whole number.
Example 2: 63.52% Fe and 36.48% S Find
the Empirical Formula.
Calculate the empirical formula with the
following:
56.4% potassium, 8.7% carbon, 34.9% oxygen
Answer:
K2CO3

Molecular Formula – A multiple of an
empirical formula.
 Can also be the same as empirical
in some cases
Example: Empirical formula N2O – 2:1 ratio
Molecular Formula of N2O could be:
N4O2 – A multiple of 2
2(N2O) = N4O2
N6O3 – A multiple of 3
3(N2O) = N6O3
How to determine the multiple you need:
Molecular Formula Mass = Multiple
Empirical Formula Mass
Then multiply the multiple to the Empirical Formula

Example 1: Calculate the molecular
formula of a compound whose molar
mass is 60.0 g/mol and empirical
formula is CH4N.
First calculate the empirical formula mass.
Use the formula
Molecular Formula Mass = Multiple
Empirical Formula Mass
Multiply the formula subscripts by this multiple to
determine the molecular formula.
Example 2
 What is the molecular formula of a
compound with the empirical formula
CClN and a molar mass of 184.5 g/mol.

Find the molecular formula of
ethylene glycol, which is used as
antifreeze. The molar mass is 62.0
g/mole, and the empirical formula is
CH3O