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Mass Relationships in
Chemical Reactions
Micro World
atoms & molecules
Macro World
grams
Atomic mass is the mass of an atom in
atomic mass units (amu)
By definition:
1 atom 12C “weighs” 12 amu
On this scale
1H
= 1.008 amu
16O
= 16.00 amu
3.1
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (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 12C
1 mol = NA = 6.0221367 x 1023
Avogadro’s number (NA)
3.2
eggs
Molar mass is the mass of 1 mole of shoes in grams
marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
3.2
One Mole of:
S
C
Hg
Cu
Fe
3.2
Molecular mass (or molecular weight) is the
sum of the atomic masses (in amu) in a molecule.
For all practical purposes think of amu as grams
What is the molecular weight of SO2
1S
2O
+
1 X 32.07 g
2 x 16.00 g
SO2
1 molecule SO2
1 mole SO2
64.07 g/mole
= 64.07 amu
= 64.07 g SO2
3.3
Molecular weights (covalent compounds)
and formula weights (ionic compounds) are
found exactly the same way
Try a few
Find the formula weight for NaCl
Find the molecular weight for H2SO4
I
STOICHIOMETRY
A.
Relationship between balanced chemical
equations and mass or volume
1.
Mass - mass problems
a.
i.
The first thing to do is balanced the
chemical equations
this will allow you to find the mole
ratio between the reactants and products
2H2 +
O2 -->
2H2O
2 moles
1 mole
2 moles
4.04 g
32.0 g
36.0 g
ex.
How many grams of KCl are produced
when 200.0 g of KClO3 decomposes?
Oxygen is the other product
1st write the balanced equation
2nd find mole ratio
2moles
2moles
2KClO3 --->
2KCl
122.6g/m
200.0 g
3th write down given/asked
4th find mwt or fwt
3moles
+ 3 O2
74.55g/m 32.00g/m
x
5th solve the problem
a.
start with the given
b.
change grams of given to moles of given
using the molecular or formula weight
c.
change the moles of given to moles of
asked using the mole ratio
d.
change the moles of asked to grams of
asked using the molecular or formula weight
M.W
(200.0 g KClO3) (1 mole KClO3)
( 122.6 g KClO3 )
Given
M.W
(2 moles KCl) ( 74.55 g KCl)
(2 mole KClO3) (1 mole KCl)
mole ratio
Your answer will come out to be in g KCl
Remember your sig figs
243.2 g KCl
You try one
How many grams of NaOH is required to completely
neutralize 155 grams of HCl? The product of this
reaction is HOH and NaCl.
1 mole
NaOH
1 mole
+
Xg
HCl
1 mole
===>
HOH
+ NaCl
155 g HCl
Na = 1 x 23.0 g = 23.0 g
H = 1 x 1.01 g = 1.01 g
O = 1 x 16.0 g = 16.0 g
Cl =1 x 35.5 g = 35.5 g
H = 1.01 g
1 mole
= 1.01 g
170. g NaOH
40.0 g/mole
1 mole HCl
( 155 g HCl )(
36.5 g HCl
36.5 g/mole
1 mole NaOH
)(
1 mole HCl
40.0 g NaOH
)(1 mole NaOH )
Try another
How many grams of Potassium is required to produce
200. g of KOH when it is involved in a single
replacement reaction with water?
Hydrogen is the other product
2 moles
1 moles
+ 2 HOH ===> 2 KOH + H2
200. g KOH
2 moles
2 K
Xg
2 moles
39.1 g/mole
(
200.g KOH
)(
56.1 g/mole
1 mole KOH
2 moles K
)
(
)
(
56.1 g KOH
2 moles KOH
39.1g K
1 mole K
)
Methanol burns in air according to the equation
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the
combustion, what mass of water is produced?
grams CH3OH
moles CH3OH
molar mass
CH3OH
( 209 g CH3OH )
(
moles H2O
grams H2O
coefficients
chemical equation
molar mass
H2O
4 mol H2O
2 mol CH3OH
18.0 g H2O
1 mol H2O
1 mol CH3OH
32.0 g CH3OH
)(
)(
)
=
235 g H2O
3.8
1. If 120. grams of mercuric oxide are decomposed,
what mass of oxygen is produced?
2 mole
2 HgO
2 mole
==> 2 Hg
Hg = 1 x201 = 201g
O = 1 x 16.0 = 16.0 g
217 g/mole
O2
1 mole O2
2 moles HgO
(120. gHgO )( 217 g HgO ) (
= 8.85 g O2
+
x
O = 2 x 16.0 g = 32.0 g/mole
120.g
1 mole HgO
1mole
32.0 g O2
) ( 1 mole O )
2
2. When 70.0 grams of hydrogen peroxide are
decomposed, how many grams of oxygen are
released?
2 moles
2 moles
1 moles
2 H2O2 ===>2 H2O + O2
70.0 g
x
34.0 g/mole H2O2
32.0 g/mole O2
1 mole H2O2
1 mole O2
( 70.0 g H O ) ( 34.0 g H O )( 2 moles H O )(
2
2
2
= 32.9 g O2
2
2
2
32.0 g O2
1mole O2
)
3. If 40.00 grams of aluminum react with hydrochloric acid
(HCl), how many grams of hydrogen are produced?
2 moles Al
2 Al
+
40.00 g Al
6 moles HCl
2 moles AlCl3
3 moles H2
6 HCl ===> 2 AlCl3
+ 3 H2
xg
2.016 g/mole
26.98 g /mole Al
(
1 mole Al
40.00 g Al
3 moles H2
)( 26.98 g Al )( 2 moles Al ) (
2.016 g H2
1 mole H2
)
4. In the reduction of 84.0 grams of copper II oxide, how many
grams of hydrogen enter into the reaction?
1 mole
1 mole
CuO +
H 2 ===>
84.0 g CuO X g H2
1 mol 1mole
Cu + H2O
79.4 g/mole 2.00 g/mole H2
1 mole CuO
1 mole H2
( 84.0 g CuO)( 79.4 g CuO )(1 mole CuO )(
2.00 g H2
1 mole H2
)
5.
How many grams of potassium chlorate are required
in the preparation of 90.0 grams of oxygen?
KClO3 ===> KCl
+
O2
6. In the electrolysis of 144 grams of water, how many
grams of hydrogen are produced?
H2O ===> H2
+
O2
Mass Changes in Chemical Reactions
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
Limiting Reagents
6 red
green
leftused
overup
3.9
Method 1
• 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
Limiting Reactant: Method 1
• 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 Cl2  2 AlCl3
• Start with Al:
10.0 g Al
1 mol Al
27.0 g Al
2 mol AlCl3 133.5 g AlCl3
2 mol Al
1 mol AlCl3
= 49.4g AlCl3
• Now Cl2:
35.0g Cl2
1 mol Cl2
71.0 g Cl2
2 mol AlCl3 133.5 g AlCl3
3 mol Cl2
1 mol AlCl3
= 43.9g AlCl3
Method 2
• 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
5.0 g of H2 reacts with 10.0 g of O2. How many grams of H2O
is produced?
2 moles 1moles 2 moles
+ O2  2 H2O
2 H2
5.0 g
2.0 g/mole
( 5.0 g )
10.0 g
x
32.0g/mole 18 .0g/mole
1 mole H2
( 2.0g H ) (
2
( 10.0 g)
1 mole O2
1 mole O2
2mole H2
)(
2 mole H2O
32 g O2
1 mole O2
( 32.0g O )( 1mole O ) (
2
2
)
= 40. g O2
18.0 g H2O = 11.2 g H
2
1 mole H2
)
Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g
of Fe2O3
2Al + Fe2O3
Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
2Al + Fe2O3
g Al
mol Al
Al2O3 + 2Fe
mol Fe2O3 needed
g Fe2O3 needed
OR
g Fe2O3
( 124 g Al )(
mol Fe2O3
1 mol Al
27.0 g Al
)(
mol Al needed
1 mol Fe2O3
2 mol Al
)(
g Al needed
160. g Fe2O3
= 367 g Fe2O3
1 mol Fe2O3
)
3.9
Start with 124 g Al
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
Use limiting reagent (Al) to calculate amount
of product that can be formed.
g Al
mol Al
mol Al2O3
2Al + Fe2O3
(124 g Al )
1 mol Al
(27.0 g Al )(
g Al2O3
Al2O3 + 2Fe
1 mol Al2O3
2 mol Al
102. g Al2O3
)( 1 mol Al O )=
2
234 g Al2O3
3
3.9
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
x 100
Theoretical Yield
3.10
Percent composition of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
2 x (12.01 g)
x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H =
x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O =
x 100% = 34.73%
46.07 g
%C =
C2H6O
52.14% + 13.13% + 34.73% = 100.0%
3.5
You try two
Find percent composition of
H2O
H3PO4
To obtain an Empirical Formula
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
e.g. problem
Find empirical formula for a compound found to
contain
60.8 % Na
28.60 % B
10.60 % H
2.644 moles
60.80 % Na = ( 60.80 g Na) ( 1 mole Na )
= 1
=
)
( 23.00 g
2.644 moles
28.60 % B = (28.60g B) ( 1 mole B
( 10.81 g B
10.60 % H = (10.60g H) ( 1 mole H
( 1.008 g H
Na1 B1 H4
)
= 2.646 moles = 1
2.644 moles
)
)
)
= 10.52 moles
2.644 moles
= 4
Stop both 6th and 8th
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 g 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
You try one
A compound was found to contain 11 % Hydrogen
and 89 % O.
What is the empirical formula?
Types of Formulas
• 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.
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 =
moles of O =
(
2.34 g N
)(
1 mole N2
14.01g N2
) = 0.167 moles of N
0.167 moles of N
1 mole O2
0.334moles of O
2
0.167 moles of N
(5.34g O ) ( 16.0 g O ) =
NO2
Calculation of the Molecular Formula
From the previous problem, it was found a
compound has an empirical formula of NO2.
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 So the molecular formula is twice the
size of the empirical formula
NO2 = N2O4
8th period start here
6th period start here