Chapter 3

Micro World atoms & molecules
Macro World grams
Atomic mass is the mass of an atom in atomic mass units (amu)
By definition:
1 atom 12 C “weighs” 12 amu
On this scale
1 H = 1.008 amu
16 O = 16.00 amu
3.1

Natural lithium is:
7.42% 6 Li (6.015 amu)
92.58% 7 Li (7.016 amu)
Average atomic mass of lithium:
(7.42% x 6.015) + (92.58% x 7.016)
= 6.941 amu
100
3.1

The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12 C
1 mol = N
A
= 6.0221367 x 10 23
Avogadro’s number ( N
A
)
3.2

eggs
Molar mass is the mass of 1 mole of in grams atoms
1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g
1 12 C atom = 12.00 amu
1 mole 12 C atoms = 12.00 g 12 C
1 mole lithium atoms = 6.941 g of Li
For any element atomic mass (amu) = molar mass (grams)
3.2

C
One Mole of:
S
Hg
Cu
Fe
3.2

1 g = 6.022 x 10 23 amu
1 amu = 1.66 x 10 -24 g
M
= molar mass in g/mol
N
A
= Avogadro’s number
3.2

How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K = 6.022 x 10 23 atoms K
0.551 g K x
1 mol K
39.10 g K x
6.022 x 10 23 atoms K
=
1 mol K
8.49 x 10 21 atoms K
3.2

Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule.
SO
2
1S 32.07 amu
2O + 2 x 16.00 amu
SO
2
64.07 amu
For any molecule molecular mass (amu) = molar mass (grams)
1 molecule SO
2
1 mole SO
2
= 64.07 amu
= 64.07 g SO
2
3.3

How many H atoms are in 72.5 g of C
3
H
8
O ?
1 mol C
3
H
8
O = (3 x 12) + (8 x 1) + 16 = 60 g C
3
H
8
O
1 mol C
3
H
8
O molecules = 8 mol H atoms
1 mol H = 6.022 x 10 23 atoms H
72.5 g C
3
H
8
O x
1 mol C
3
H
8
O
60 g C
3
H
8
O x
8 mol H atoms
1 mol C
3
H
8
O x
6.022 x 10 23 H atoms
1 mol H atoms
=
5.82 x 10 24 atoms H
3.3

KE = 1/2 x m x v 2 v = (2 x KE/m) 1/2
F = q x v x B
3.4

Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound
%C =
2 x (12.01 g)
46.07 g
%H =
6 x (1.008 g)
46.07 g
%O =
1 x (16.00 g)
46.07 g x 100% = 52.14% x 100% = 13.13% x 100% = 34.73%
C
2
H
6
O
52.14% + 13.13% + 34.73% = 100.0%
3.5

• Empirical Formula
The formula of a compound that expresses the smallest whole number
ratio of the atoms present.
Ionic formula are always empirical formula
• Molecular Formula
The formula that states the actual number of each kind of atom found in one
molecule of the compound.

1. Determine the mass in grams of each element present, if necessary.
2. Calculate the number of moles of each element.
3. Divide each by the smallest number of moles to obtain the simplest whole number ratio.
4. If whole numbers are not obtained * in step 3), multiply through by the smallest number that will give all whole numbers
* Be careful! Do not round off numbers prematurely

A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O.
Determine a formula for this substance.
require mole ratios so convert grams to moles moles of N = 2.34g of N = 0.167 moles of N
14.01 g/mole moles of O = 5.34 g = 0.334 moles of O
16.00 g/mole 
Formula:
N
0.167
O
0.334
N O
0.167
0.334
0.167
0.167
NO
2

Calculation of the Molecular Formula
A compound has an empirical formula of
NO
2
. The colourless liquid, used in rocket engines has a molar mass of 92.0 g/mole.
What is the molecular formula of this substance?
empirical formula mass: 14.01+2 (16.00) =
46.01 g/mol n = molar mass = 92.0 g/mol emp. f. mass 46.01 g/mol n = 2
N
2
O
4

Empirical Formula from % Composition
A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H
What is the empirical formula of the substance?
Consider a sample size of 100 grams
This will contain: 60.80 grams of Na, 28.60 grams of B, and 10.60 grams H
Determine the number of moles of each
Determine the simplest whole number ratio

Combust 11.5 g ethanol
Collect 22.0 g CO
2 and 13.5 g H
2
O g CO
2 mol CO
2 mol C g C 6.0 g C = 0.5 mol C g H
2
O mol H
2
O mol H g of O = g of sample – (g of C + g of H) g H 1.5 g H = 1.5 mol H
4.0 g O = 0.25 mol O
Empirical formula C
0.5
H
1.5
O
0.25
Divide by smallest subscript (0.25)
Empirical formula C
2
H
6
O
3.6

1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity
4. Convert moles of sought quantity into desired units
3.8

• Molarity
– Moles solute / L solution
• Gases
– 22.4 L = 1 mole of ANY GAS at STP

Methanol burns in air according to the equation
2CH
3
OH + 3O
2
2CO
2
+ 4H
2
O
If 209 g of methanol are used up in the combustion, what mass of water is produced?
grams CH
3
OH moles CH
3
OH molar mass
CH
3
OH moles H coefficients chemical equation
2
O grams H molar mass
H
2
O
2
O
209 g CH
3
OH
1 mol CH
3
OH x
32.0 g CH
3
OH x
4 mol H
2
O
2 mol CH
3
OH x
18.0 g H
2
O
1 mol H
2
O
=
235 g H
2
O
3.8

Limiting Reagents
3.9

• Pick A Product
• Try ALL the reactants
• The lowest answer will be the correct answer
• The reactant that gives the lowest answer will be the limiting reactant

• 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced?
2 Al + 3 Cl
2
• Start with Al:

2 AlCl
3
10.0 g Al 1 mol Al 2 mol AlCl
3
133.5 g AlCl
3
= 49.4g AlCl
3
27.0 g Al 2 mol Al 1 mol AlCl
3
• Now Cl
2
:
35.0g Cl
2
1 mol Cl
2
2 mol AlCl
3
133.5 g AlCl
3
= 43.9g AlCl
3
71.0 g Cl
2
3 mol Cl
2
1 mol AlCl
3

• Convert one of the reactants to the other
REACTANT
• See if there is enough reactant “A” to use up the other reactants
• If there is less than the GIVEN amount, it is the limiting reactant
• Then, you can find the desired species

Do You Understand Limiting Reagents?
g Al
In one process, 124 g of Al are reacted with 601 g of Fe
2
O
3
2Al + Fe
2
O
3
Al
2
O
3
+ 2Fe
Calculate the mass of Al
2
O
3 formed.
g Fe
2
O
3 needed g Fe
2
O
3 mol Al mol Fe
2
O
3 mol Fe
2
O
3 needed
OR mol Al needed g Al needed
124 g Al x
1 mol Al
27.0 g Al x
1 mol Fe
2
O
3
2 mol Al x
160. g Fe
2
O
3
1 mol Fe
2
O
3
= 367 g Fe
2
O
3
Start with 124 g Al need 367 g Fe
2
O
3
Have more Fe
2
O
3
(601 g) so Al is limiting reagent
3.9

Use limiting reagent (Al) to calculate amount of product that can be formed.
g Al mol Al
2Al + Fe
2
O
3 mol Al
2
O
3
Al
2
O
3
+ 2Fe g Al
2
O
3
124 g Al x
1 mol Al
27.0 g Al x
1 mol Al
2
O
3
2 mol Al x
102. g Al
2
O
3
1 mol Al
2
O
3
= 234 g Al
2
O
3
3.9

• 10.0g of aluminum reacts with 35.0 grams of chlorine gas
2 Al + 3 Cl
2

2 AlCl
3
• We found that chlorine is the limiting reactant, and
43.8 g of aluminum chloride are produced.
35.0 g Cl
2
1 mol Cl
2
2 mol Al 27.0 g Al
71 g Cl
2
3 mol Cl
2
1 mol Al
= 8.8 g Al
USED!
10.0 g Al – 8.8 g Al = 1.2 g Al EXCESS
Given amount of excess reactant
Amount of excess reactant actually used
Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

Theoretical Yield is the amount of product that would result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained from a reaction.
% Yield =
Actual Yield
Theoretical Yield x 100
3.10

Chemistry In Action: Chemical Fertilizers
Plants need: N, P, K, Ca, S, & Mg
3H
2
( g ) + N
2
( g ) 2NH
3
( g )
NH
3
( aq ) + HNO
3
( aq ) NH
4
NO
3
( aq ) fluorapatite
2Ca
5
(PO
4
)
3
F ( s ) + 7H
2
SO
4
( aq )
3Ca(H
2
PO
4
)
2
( aq ) + 7CaSO
4
( aq ) + 2HF ( g )