Download Chapter 7 Chemical Quatities

Chapter 7
The Mole
Mole-Mass and Mole-Volume Relationships
Percent Composition and Chemical Formulas
Chapter 7.1 The Mole
What is a mole?
The Number of
Particles in a Mole
The Mass of a Mole
of an Element
The Mass of a Mole
of a Compound
What is a Mole?
A pair is how many?

2
A dozen is how
many?

12
Apples
If buy apples :

At a fruit stand, you buy them by count


At a supermarket, you buy them by mass


5 for $2.00
$1.95 per kilogram
At an orchard, you buy them by volume

$9.00 bushel
Dozen Apples
By count

12 apples
By mass

2.0kg of apples
By volume

.20 bushels of apples
A Mole
SI unit that
measures the
amount of a
substance
Mole
Can be related to



The number of
particles (the count)
The mass
The volume
The Number of Particles
in a Mole
1 Mole = 6.02 x1023
This is Avogadro’s Number
Representative Particle
Refers to the
species present in a
substance



Atoms - Fe
Molecules – H2O
Formula Units (Ions)
How many atoms?
CO
Consist of two atoms
So 2 times
Avogadro’s Number
2 x (6.02 x1023)atoms
The Mass of a Mole of
an Element
Gram Atomic Mass (gam)
Add up the atomic mass
of each atom
Carbon



Atomic Mass 12.0amu
Change to grams
12 grams of Carbon
How many atoms in a
Mole of Carbon?
The gam is equal to 6.02 x1023 atoms
The Mass of a Mole of a
Compound
Must know the formula
first
This will tell you the
number of atoms of
each element
Gram Molecular Mass

(gmm)
The Mass of a Mole of a
Compound
SO3
1 Sulfur atom, 3 Oxygen atoms
S = 32.06 amu, O = 15.9994
amu
S = 32.06g, O = 15.9994g
32.06g + (3 x 15.9994g)
gmm = 80.06g
Gram Formula Mass
Same as gmm but
for ionic compounds
Chapter 7.2 Mole-Mass
and Mole-Volume
Relationships
Molar Mass of a
Substance
Volume of a Mole of
a Gas
Mole Road Map
Molar Mass of a
Substance
gam -?

1 mole of atoms
gmm - ?

1 mole of a
molecular compound
gfm - ?

1 mole of an ionic
compound
Molar Mass of a
Substance
Molar Mass – the
mass in grams of
one mole of any
substance
Broader to refer to
all three

Atoms, molecular
compounds or ionic
compounds
What is the molar mass
of oxygen?
Molecular oxygen


O2
32.0g
Atomic oxygen


O
16.0g
Moles  Grams
9.45 mol of N2O3 is how many grams?
Dimensional Analysis

9.45 mol of N2O3 (76.0g N2O3 ) = 718.2g N2O3
(1.00 mol N2O3 )
718g N2O3
Moles  Grams
How many grams in each of the following?



3.32 mol K
.00452 mol C20H42
10.6 mol H2O
Grams  Moles
How many moles in 92.2g Fe2O3?
92.2g Fe2O3 (1 mol Fe2O3) = 0.5776 mol Fe2O3
(159.6g Fe2O3)
0.578 mol Fe2O3
Grams  Moles
How many moles in each of the following?



.373g B
27.4g TiO2
847 g (NH4) 2CO3
The Volume of a Mole of
Gas
STP – Standard Temperature and Pressure


Standard Temperature is 0oC (273K)
Standard Pressure is 101.3kPa (1 atmosphere (atm))
The Volume of a Mole of
Gas
Gasses are measured at STP
1 Mole of ANY gas at STP has a volume of 22.4L
1 Mole of any gas at STP also has 6.02 x1023 particles
What is the molar mass?
The density of a gas containing carbon and
oxygen is 1.964 g/L.
Use the relationship 1 mole of gas = 22.4L at STP
1.964g (22.4L) = 43.9936 g
L (1mol)
mol
What is the molar mass?
The density of a gas containing sulfur and
oxygen is 3.58 g/L at STP.
Mole Road Map
Ch 7.3 Percent
Composition and
Chemical Formulas
Percent Composition
Using Percent as a Conversion Factor
Calculating Empirical Formulas
Calculating Molecular Formulas
Percent Composition
The relative amounts of each element in a compound
The percent by mass of each element in a compound
Has the same number of percent values as elements
Percent Composition
The total of the percents must equal 100%

K2CrO4




K = 40.3%
Cr = 26.8%
O = 33.0%
Total 40.3% + 26.8% + 33.0% = 100.1%
% mass of element = grams of element x 100%
grams of compound
Percent Composition
An 8.20 gram piece of magnesium combines
completely with 5.40g of oxygen to form a
compound. What is the percent composition of
this compound?


% Mg = mass of Mg x 100%
grams of compound
= 8.20g x 100%
13.60g
= 60.3%
% O = mass of O x 100%
grams of compound
= 5.40g x 100%
13.60g
= 39.7%
Percent Composition
9.03g of Mg combines completely with 3.48g N
to form a compound. What is the percent
composition of this compound?


Mg 72.2%
N 27.8%
Percent Mass
% mass = grams of element in 1 mol compound x 100%
molar mass of compound
Calculate the percent composition of propane (C3H8).

%C = grams of C in 1 mol compound x 100%
molar mass of compound
= 36.0g x 100%
44.0g
= 81.8%

%H = grams of H in 1 mol compound x 100%
molar mass of compound
= 8.0g x 100%
44.0g
= 18.2%
Using percent as a
conversion factor
Calculate the mass of of carbon in 82.0g of
propane (C3H8).

82.0g C3H8 X (81.8g C) = 67.1g C
(100g C3H8)

Where did the 81.8g C come from?



Last page we calculated the percent mass of C and H in Propane
81.8% C and 18.2% H
This is the same as 81.8g C and 18.2g H.
5.08g of Li combines completely with 8.06g F
to form a compound. What is the percent
composition of this compound?
Calculate the percent composition of
ammonium peroxide.
Calculate the mass of of oxygen in 67.0g of
ammonium peroxide.
Calculating Empirical
Formulas
Empirical Formula – the lowest whole number
ratio of the atoms of elements in a compound
Empirical Formulas may or may not be the
same as molecular formulas.



Hydrogen Peroxide
Molecular Formula – H2O2
Empirical Formula HO
Calculating Empirical
Formulas
What is the molecular and empirical formula
for carbon dioxide?


Molecular CO2
Empirical CO2
What is the molecular and empirical formula
for dinitrogen tetrahydride?


Molecular N2H4
Empirical NH2
What is the empirical formula of a compound that is
25.9% nitrogen and 74.1% oxygen?









25.9g N (1 mol N) = 1.85 mol N
(14.0g N)
74.1g O (1 mol O) = 4.63 mol O
(16.0g O)
One you have the moles of each element, divide each by the
lowest number of moles.
1.85 mol N = 1 mol N
1.85 mol
4.63 mol O = 2.50 mol O
1.85 mol
This is not the empirical formula, since they are not whole
numbers. Multiply each by 2 to get whole numbers.
1 mol N x 2 = 2 mol N, 2.50 mol O = 5 mol O
The empirical formula is N2O5
A compound contains 42.9% C and 57.1% O.
What is the empirical formula?

CO
A compound contains 32.00% C, 42.66% O,
18.67% N and 6.67% H. What is the empirical
formula?

C2O2NH5
Calculating Molecular
Formulas
If you know the empirical formula and the
molar mass, you can determine the molecular
formula.
Determine the empirical formula mass(efm)
Then divide the molar mass by the efm.
Multiply the subscripts by this number.
Calculating Molecular Formulas
Calculate the molecular formula of the
compound whose molar mass is 60.0g and
the empirical formula is CH4N.
Empirical Formula Mass – 30.0g
Molar Mass – 60.0g
Molar mass / empirical formula mass
60.0g/ 30.0g = 2
CH4N x 2 = C2H8N2
Calculating Molecular Formulas