Mass Relationships in
Chemical Reactions
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 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
1 g = 6.022 x 1023 amu
1 amu = 1.66 x 10-24 g
M = molar mass in g/mol
NA = Avogadro’s number
3.2
NMSI Renee Mc Cormick
Do You Understand Molar Mass?
How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K = 6.022 x 1023 atoms K
1 mol K
6.022 x 1023 atoms K
0.551 g K x
x
=
1 mol K
39.10 g K
8.49 x 1021 atoms K
3.2
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
1S
SO2
2O
SO2
32.07 amu
+ 2 x 16.00 amu
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
3.3
Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = 6.022 x 1023 atoms H
1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms
72.5 g C3H8O x
x
x
=
1 mol C3H8O
1 mol H atoms
60 g C3H8O
5.82 x 1024 atoms H
3.3
Mole fractions:
Xa= na_____ Where n is the number of moles
na+nb....
Use 1 mole to determine the ratio amounts:
What is the mole fraction of Na in Na2O.
Given the answer can you work backwards and
find the moles of O?
mass spectrometer—a device for measuring the mass of atoms
or molecules
 atoms or molecules are passed into a beam of high-speed
electrons
this knocks electrons OFF the atoms or molecules transforming
them into cations o apply an electric field
this accelerates the cations since they are repelled from the (+)
pole and attracted toward the (−) polesend the accelerated cations
into a magnetic field
an accelerated cation creates it’s OWN magnetic field which
perturbs the original magnetic fieldthis perturbation changes the
path of the cation
the amount of deflection is proportional to the mass; heavy cations
deflect little
• ions hit a detector plate where measurements can be obtained.
Heavy
Light
Heavy
Light
KE = 1/2 x m x v2
v = (2 x KE/m)1/2
F=qxvxB
3.4
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
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.
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
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
N 0.167 O 0.334 
Formula: N O
0.167
0.334
0.167
0.167
NO 2
Calculation of the Molecular Formula
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
N2O4
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
• Find the empirical formula using a equation.
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
g CO2
mol CO2
mol C
gC
6.0 g C = 0.5 mol C
g H2O
mol H2O
mol H
gH
1.5 g H = 1.5 mol H
g of O = g of sample – (g of C + g of H)
4.0 g O = 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
3.6
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
Other units
• Molarity
– Moles solute / L solution
• Gases
– 22.4 L = 1 mole of ANY GAS at STP
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 x
moles H2O
grams H2O
molar mass
coefficients
H2O
chemical equation
4 mol H2O
18.0 g H2O
1 mol CH3OH
=
x
x
32.0 g CH3OH
2 mol CH3OH
1 mol H2O
235 g H2O
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
This is the method I use. Pick
your favorite and use it.
• Limiting reagent
• To determine the limiting reagent requires
that you do two stoichiometry problems.
• Figure out how much product each reactant
makes.
• The one that makes the least is the limiting
reagent.
Limiting
Limiting
Reactant
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
• Limiting Reagents: shortcut • Limiting
reagent problems can be solved another way
(without using a chart)...
• • Do two separate calculations using both
given quantities. The smaller answer is
correct.
• Q-How many g NO are produced if
20gNH3 is burned in 30g O2 ?
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.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed
OR
g Fe2O3
124 g Al x
mol Fe2O3
1 mol Al
27.0 g Al
x
mol Al needed
1 mol Fe2O3
2 mol Al
Start with 124 g Al
160. g Fe2O3
=
x
1 mol Fe2O3
g Al needed
367 g Fe2O3
need 367 g Fe2O3
Have more Fe2O3 (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
mol Al2O3
2Al + Fe2O3
124 g Al x
1 mol Al
27.0 g Al
x
1 mol Al2O3
2 mol Al
g Al2O3
Al2O3 + 2Fe
102. g Al2O3
=
x
1 mol Al2O3
234 g Al2O3
3.9
Finding Excess Practice
• 10.0g of aluminum reacts with 35.0 grams of
chlorine gas
2 Al + 3 Cl2  2 AlCl3
• We found that chlorine is the limiting reactant, and
43.8 g of aluminum chloride are produced.
35.0 g Cl2 1 mol Cl2 2 mol Al
71 g Cl2
3 mol Cl2
27.0 g Al
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
x 100
Theoretical Yield
% Yield = what you got
x 100
what you could have got
3.10
Chemistry In Action: Chemical Fertilizers
Plants need: N, P, K, Ca, S, & Mg
3H2 (g) + N2 (g)
NH3 (aq) + HNO3 (aq)
2NH3 (g)
NH4NO3 (aq)
fluorapatite
2Ca5(PO4)3F (s) + 7H2SO4 (aq)
3Ca(H2PO4)2 (aq) + 7CaSO4 (aq) + 2HF (g)