Stoichiometry
Chapter 3
Chemical Stoichiometry
Stoichiometry - The study of quantities
of materials consumed and produced in
chemical reactions.
Atomic Mass Unit
Atoms are so tiny that the gram is much too
large to be practical.
The mass of a single carbon atom is
1.99 x 10-23 g.
The atomic mass unit (amu) is used for atoms
and molecules.
The Mole
The number equal to the number of carbon
atoms in exactly 12 grams of pure 12C.
1 mole of anything = 6.022  1023 units of
that thing
Equal moles of
substances have equal
numbers of atoms,
molecules, ions,
etc.
The Mole
Substance Average Atomic Mass
(g)
Na
22.99
Cu
63.55
S
32.06
Al
26.98
# Moles
1
1
1
1
# Atoms
6.022 x 1023
6.022 x 1023
6.022 x 1023
6.022 x 1023
Avogadro’s number
equals
23
6.022  10 units
on reference table
Molar Mass
A substance’s molar mass ,molecular
weight, or gram formula mass(are all the
same thing) is the mass in grams of one
mole of the compound.
CO2 = 44.01 grams per mole
Practice old way
How many grams are in
5 moles of NaCl ?
Answer Slide
292.2
292g
Factor label method
(Dimensional analysis)
36 eggs is equivalent to how many dozen ?
How many eggs in 5 dozen?
Calculating Mass from Moles
How many grams in 4.86 mol CaCO3
CaCO3
1 Ca = 1 (40.08 g) = 40.08 g
1 C = 1 (12.01 g) = 12.01 g
3 O = 3 (16.00 g) = 48.00 g
100.09 g
Answer Slide
486g
Practice
How many grams are in
2 moles of KCl ?
Answer Slide
149.1g
149g
Practice
How many moles are in
5.0 g of HCl ?
Answer Slide
.137
Calculating Moles from Mass
How many moles are in 435 g of
Calcium Bromide?
Answer Slide
2.2 mol
2.18 mol
Calculating Number of Atoms
from Mass
How many atoms are in
10.00 grams of carbon?
Moles are the doorway
grams <---> moles <---> atoms
Answer Slide
5.01x1023 atoms
Calculating Mass Using
AMU’S
How much does 19 atoms of nitrogen weigh?
Answer Slide
4.42x10-22 g
Practice
How many molecules of CaF2 are in 121g of
Calcium fluoride ?
Answer Slide
9.32x1023
Challenge
How many atoms of oxygen are there in 75
grams of Al2O3?
Answer Slide
1.33x1024
Calculating Masses of
Reactants and Products
1.
2.
3.
4.
Balance the equation.
Convert mass to moles.
Set up mole ratios.
Use mole ratios to calculate moles of
desired product.
5. Convert moles to grams, if necessary.
Challenge
How many grams of Al2O3 are produced
when reacting 400 g of Al with excess O2?
4Al + 3O2
2Al2O3
Answer Slide
755.8g
756 g
Gram to Mole & Gram to Gram
__Al(s)
+ __I2(s) ---> __AlI3(s)
How many moles of aluminum iodide
can be produce from 35.0 g of
aluminum?
Answer Slide
1.30 mol AlI3
___N2 +___O2
___NO2
How many moles of NO2 can be produce
from 101.0 g of nitrogen and excess oxygen
Answer Slide
331.9g
Practice with factor label method
Green Review book P73
62-64, 68(show work for all)
Limiting Reactant
The limiting reactant is the reactant
that is consumed first, limiting the
amounts of products formed.
Demo NaHCO3 + CH3COOH →
CH3COONa + H2O + CO2(g)
Go to sandwich shop pp
Run pp
Solving a Stoichiometry Problem
Theoretical yield
1. Balance the equation.!!!!!!!!
2. Convert one mass to moles and then use
equation to determine mass needed .
Determine which reactant is limiting.
4. Use moles of limiting reactant and mole
ratios to find moles of desired product.
5. Convert from moles to grams.
Limiting Reactant Problem
If 56.0 g of Li reacts with 56.0 g of N2, how
many grams of Li3N can be produced?
__Li(s) + __N2(g) ---> __Li3N(s)
Answer Slide
Li Limiting
93.7g Li3N
Limiting Reactant Problem
6 Li(s) + N2(g) ---> 2 Li3N(s)
How many grams of nitrogen are left?
56.0g N2 given - 37.7 g used = 18.3 g excessN2
l
Ammonia is produced by the following
reaction
__N2 +__ H2  __NH3
What mass of ammonia can be
produced from a mixture of 100. g N2
and 500. g H2 ?
Answer Slide
N2 limiting
123.1
123g ammonia
Sodium Hydroxide reacts with phosphoric
acid to give sodium phosphate and water. If
17.80 g of sodium hydroxide with 15.40g of
phosphoric acid , how many grams of
sodium phosphate are formed?
Gfm
39.998 = NaOH
97.994 = H3PO4
163.94 =Na3PO4
Answer Slide
25.76 g of Na3PO4
Ditto/
answer
% Yield
Values calculated using stoichiometry are
always theoretical yields!
Values determined experimentally in the
laboratory are actual yields!
Limiting Reactant & % Yield
If 68.5 g of CO(g) is reacted with 8.60 g of H2(g),
what is the theoretical yield of methanol that
can be produced?
__H2(g) + __CO(g) ---> __CH3OH(l)
Answer Slide
Limiting Reactant & % Yield
2 H2(g) + CO(g) ---> CH3OH(l)
Since only 8.60 g of H2 were provided, the H2 is
the limiting reactant, and the CO is in excess.
(8.60 g H2)(1 mol/2.02 g)(1 mol CH3OH/2 mol H2)(32.0
g/1 mol) = 6.85g
Limiting Reactant & % Yield
2 H2(g) + CO(g) ---> CH3OH(l)
If in the laboratory only 3.57 g of CH3OH is
produced, what is the % yield?
actual _ yield
100 % 
% Yield 
theoretica l _ yield
3 . 57 g
100 % 
% Yield 
6 . 85 g
% Yield = 52.1 %
Practice Review Book
P.74
Q70,71
all parts show all work