



The study of the quantitative relationships between reactants and products in a reaction


It is used to answer questions like; If I have this much reactant, how much product can I make?
If I want this much product, how much reactant do I need?
These questions have real life application, particularly in manufacturing .
It allows us to convert the mass of a substance to the number of particles (atoms, ions or molecules) it contains.
These numbers can be really large, so they are counted in groups
 Much like when we count a lot of pennies we stack them in 10’s and count by 10


Atoms are very tiny, so small that the grouping we use to count them must be very large

MOLE; the group (unit of measure) used to count atoms, molecules, formula units or ions of a substance

1 mole of a substance has a particular number of particles in it!

Much like 1 dozen always means 12; whether it is
12 eggs 12 oranges or 12 gold bars

The number of particles in a mole = 6.02 x 10 23 or
This is known as Avogadro’s Number
Using this, We can easily count the number of particles in all kinds of things !

There are 6.02 x 10 23 Carbon atoms in a mole of carbon
There are 6.02 x 10 23 CO
2 molecules in a mole of CO
2
There are 6.02 x 10 23 sodium ions in a mole of sodium
There are 6.02 x 10 23 marbles in a mole of marbles
That’s a lot of marbles!
The Size of a mole of a substance changes, the bigger the substance the more space a mole of the substance takes up, but the number of particles in a mole is always the same!


Chemicals do not come bundled in moles, like a dozen eggs comes in a 1 dozen or 1 ½ dozen package so we use the mole as a grouping unit.
The mass of 1 mole of a pure substance called it’s molar mass

If I want to produce 500g of methanol using the following equation, CO2 +3H2  CH3OH + H20 how many grams of CO2 and H2 do I need?

These are the questions stoichiometry answers !

If I want to produce 500g of methanol using the following equation;
CO
2
+3H
2
 CH
How many grams of CO
2
3
OH + H and H
2
2
0 do I need?
This equation relates the molecules of reactants and products, NOT THEIR MASSES!

1 molecule of CO molecule of CH
2
3
OH and 3 molecules of H
2 will make 1
We need to relate the masses to the number of molecules.

Remember; The average atomic masses of the elements are found on the
Periodic Table!

We can use the atomic masses on the PT to relate the mass of the compound to the mass of a mole!

Molecular mass: mass in atomic mass units of just one molecule
Formula Mass: mass in atomic mass units of one
formula unit of an ionic compound

Steps
1.
a) b)
Example: molar mass of Fe = 55.847 g
Example: molar mass of Pt = 195.08 g
2.
a)
If the element is a molecule, count the number of atoms in the molecule then multiply the atomic mass by the number of atoms.
Example: O
2
, the mass of O =16.0g There are 2 atoms of O in the O
2 molecule , 2 atoms X 16.0g =

Calculate the molar mass of each of the following:
1.
2.
3.
4.
5.
6.
N
2
Cl
2
Br
2
I
2
H
2
F
2

Calculate the molar mass of each of the following:
1.
2.
3.
4.
5.
6.
N
2
Cl
2
= 14.007g X 2 =28.014 g/mol
= 35.453g X 2 =70.906 g/mol
I
Br
2
2
= 79.904g X 2 =159.808 g/mol
= 126.904g X 2 =253.808 g/mol
H
2
F
2
= 1.008g X 2 =2.016 g/mol
= 18.998g X 2 =37.996 g/mol

Steps
1.
Count the number and type of atoms
2.
3.
Find the Atomic Mass of each atom type, on the periodic table. Write it in grams.
Multiply the mass times the # of Atoms. Then add the totals

1.
Count the number and type of atoms
Ethanol (C2H5OH)
Atom type
C
H
O
2.
3.
Amount of each atom
2
6
1
Find the Atomic Mass of each atom type, on the periodic table. Write it in grams.
Atom type Amount of atom Ave. Atomic Mass in g
C
H
O
2
6
1
12.0
1.00
16.0
Multiply The mass X the # of Atoms. Then add the totals.
Atom type Amount of atom Ave. Atomic Mass in g
C
H
O
2
6
1
12.0
1.00
16.0
Molar Mass Of Ethanol (C
2
H
5
OH)
Total
=24.0
=6.0
=16.0
= 46.0g/mole

Example: Calcium Chloride (CaCl
2
)
Atom
Types
Ca
Cl
Amount of
Atoms
1
2
Ave. Atomic
Mass in g
40.1
35.5
Mass of 1 mol of CaCl
2
(molar mass)
Total
40.1
71.0
111.1 g/mole

What is the molar mass of each of the following?
1.
2.
3.
4.
5.
6.
Fe
2
H
2
O
CO
2
O
3
NaCl
NH
3
BaI
2

Fe
2
O
3
= 55.85g X 2= 111.7 g
16.0g X 3 = 48.0g
= 159.7 g/mol
_______________________________________________
H
2
O = 1.01g X 2 = 2.02
16.0g X 1 = 16.0
= 18.02 g/mol
_______________________________________________
CO
2
= 12.01g X 1 = 12.01
16.0g X 2 = 32.0
= 44.01 g/mol
________________________________________________
NaCl = 22.99 gX1 = 22.99
35.45g X1 = 35.45
= 58.44 g/mol
________________________________________________
NH
3
=14.01g X 1 = 14.01
1.01g X 3 = 3.03
= 17.04 g/mol
________________________________________________
BaI
2
= 137.33g X 1 = 137.33
126.90g X 2 = 253.80
= 391.13 g/mol

If I want to produce 500g of ethanol using the following equation;
6CO
2
+17H
2
 3C
2
How many grams of CO
2
H
5
OH + 9H and H
2
The Molar Mass Of Ethanol (C
2
H
5
OH)
2
0 do I need?
= 46.0g/mole
 Now we need to find the number of atoms in the sample.
How many molecules of ethanol are in 500g?

Steps to finding the number of atoms in a given mass of a sample
1.
2.
Use PT to find the molar mass of the substance
Convert the mass of the substance to number of moles in the sample (convert using mass of one mole as conversion factor)
3.
4.
Use the number of atoms in a mole to find the number of atoms in the sample
Solve and check answer by canceling out units

The mass of an iron bar is 16.8g
. How many iron(Fe) atoms are in the sample?
Step 1: Use PT to find the molar mass of the substance : The molar mass of Fe =55.8g/mole
Step 2: Convert the given mass of the substance to number of moles in the sample: Fe =55.8g/mole
(16.8g Fe) (1 mol Fe) (6.022 X 10 23 Fe atoms)
(55.8g Fe) (1 mol Fe)
= 1.81 X 10
23 Fe atoms
Step 3: Use the number of atoms in a mole to find the number of atoms in the sample = 1.18 X 10 23

1.
25.0 g silicon, Si
2.
1.29 g chromium, Cr

( 25.0 g Si
1
) ( 1 mol Si ) ( 6.02 X 10 23 Si atoms
28.1g Si 1 mol Si
= 5.36 X10 23 atoms Si
)
( 1.29 g Cr
1
) ( 1 mol Cr ) ( 6.02 X 10 23 Cr atoms
52.0g Cr 1 mol Cr
= 1.49 X10 22 atoms Cr
)

1.
2.
3.
4.
98.3g mercury, Hg
45.6g gold, Au
10.7g lithium, Li
144.6g tungsten, W

1.
( 98.3 g Hg ) ( 1 mol Hg )( 6.02 X 10 23 Hg atoms
1 200.6g Hg 1 mol Hg
= 2.95 X10 23 atoms Hg
)
2.
( 45.6 g Au ) ( 1 mol Au )( 6.02 X 10 23 Au atoms
1 197.0g Au 1 mol Au
= 1.39 X10 23 atoms Au
)
3.
( 10.7 g Li ) ( 1 mol Li )( 6.02 X 10 23 Li atoms
1 6.94g Li 1 mol Li
= 9.28 X10 23 atoms Li
)
4.
( 144.6 g W ) ( 1 mol W )( 6.02 X 10 23 W atoms
1 183.8g W 1 mol W
= 4.738 X10 23 atoms W
)

Steps
1.
Use the PT to calculate the molar mass of one formula unit
2.
3.
Convert the given mass of the compound to the number of molecules in the sample (use the molar mass as the conversion factor)
Multiply the moles of the compound by the number of the formula units in a mole
(Avagadro’s number) and solve
4.
Check by evaluating the units

1.
Calculate the molar mass (Fe
2
O
3
)
2 Fe atoms 2X 55.8 = 111.6
3 O atoms 3 X 16.0 = +48.0 molar mass 159.6 g/mol
(given mass X 1 mole per molar mass X Form Units per 1 mole)
( 16.8 g Fe
2
O
3
1
) ( 1 mol Fe
2
O
3
159.6g Fe
2
O
3
)( 6.02 X 10 23 Fe
2
O
3 Formula units
1 mol Fe
2
O
3
)
= 6.34 X10 22 Fe
2
O
3
Formula units

1.
89.0g sodium oxide (Na
2
O)
2.
10.8g boron triflouride ( BF
3
)

1.
89.0g sodium oxide (Na
2
O)
Calculate the molar mass (Na
2
O)
2 Na atoms 2X 23.0 = 46.0
1 O atoms 1 X 16.0 = +16.0 molar mass 62.0 g/mol
(given mass X 1 mole per molar mass X molecules per 1 mole)
( 89.0 g Na
2
O
1
) ( 1 mol Na
2
O
62.0g Na
2
)( 6.02 X 10 23
Na
2
O Form Units
2
O
)
= 8.64 X10 23 Na
2
O Formula units

2.
10.8g boron trifloride ( BF
3
Calculate the molar mass (Na
2
)
O)
1 B atom 1X 10.8 = 10.8
3 F atoms 3 X 19.0 = +57.0 molar mass 67.8 g/mol
( given mass X 1 mole per molar mass X molecules per 1 mole)
( 10.8 g BF
3
1
) ( 1 mol BF
3
67.8g BF
3
)( 6.02 X 10 23
BF
3
1 mol BF
3
Form units )
= 9.59 X10 22 BF
3
Formula units

1.
2.

Calculate the number of moles in 6.84g sucrose (C
12
12 C atoms 12 X 12.0 = 144.0
H
22
O
11
)
22 H atoms 22 X 1.0 = 22.0
11 O atoms 11 X 16.0 = +176.0 molar mass 342.0 g/mol
(given mass/1) X (1 mole/ molar mass)
( 6.84 g sucrose
1
) ( 1 mol sucrose
= 2.0 X10 -02 moles of sucrose

1.
2.
3.
16.0g sulfur dioxide, SO
2
68.0g ammonia, NH
3
17.5g copper(II) oxide, CuO

1.
16.0g sulfur dioxide, SO
2
(16.0g/1) (1mole/64.1g ) = 0.250 mol SO
2
2.
68.0g ammonia, NH
3
( 68.0g/1) (1 mole/ 17.0g) = 4.00 mol NH
3
3.
17.5g copper(II) oxide, CuO
( 17.5g/1) (1 mole/ 79.1g) = 0.22 mol CuO

Steps:
1.
Find the molar mass of the compound
2.
Use the molar mass to convert the given number of moles to a mass (moles) X (g/mol)
3.
4.
Solve
Check using dimensional analysis (make sure units cancel and leaves only grams)

2.
3.
1.
2.
Find the molar mass of the compound (H
2
H - 2 atoms – 1.0 = 2.0
O)
O - 1 atom - 16.0 = 16.0 18.0 g/mol
Use the molar mass to convert the given number of moles to a mass (moles) X (g/mol)
(7.5 mol H
2
Solve : 7.5 X 18.0g H
2
O) ( 18.0 g H
2
( 1 mol H
2
O = 135 g H
2
O
O)
O)
Check using dimensional analysis (make sure units cancel and leaves only grams) “mol H other out, units are correct!
2
O” cancel each

1.
2.
3.
4.
3.52 mol Si
1.25 mol aspirin, C
9
H
8
O
4
0.550 mol F
2
2.35 mol Barium Iodide, BaI
2

1.
What mass of Si = 3.52 mol Si
(3.52 mol Si) (28.1g Si) = 98.9g Si
1 (1 mole Si)
2.
What mass of C
9
H
8
O
4
C -9 atoms – 12.0 – 108.0
H- 8 atoms – 1.0 8.0
O – 4 atoms – 16.0 - 64.0
180.0g/mol
= 1.25 mol aspirin, C
9
H
8
O
4
(1.25 mol C
9
H
8
O
4
) (180.0g C
1 (1 mole C
9
H
8
O
9
H
8
O
4
) = 225.0g C
9
H
4
)
8
O
4

3.
What mass of F
2
= 0.550 mol F
F- 2 atoms – 19.0 = 38.0 g/mol
2
(0.550 mol F
2
) (38.0 g F
2
) = 20.9g F
2
1 (1 mole F
2
)
4.
What mass of BaI
2
Ba-1 atom – 137.3 - 137.3
I – 2 atoms – 126.9 - 253.8
= 2.35 mol Barium Iodide, BaI
2
391.1g/mol
(2.35 mol BaI
2
) (391.1g BaI
2
) = 919.1g BaI
2
1 (1 mole BaI
2
)

Know:
1.
2.
What stoichiometry is
What a mole is
3.
1.
2.
3.
How to calculate molar mass of an element and of a compound
How to determine the number of atoms or formula units in a given mass of sample
How to determine the number of moles in a given mass of a sample
How to determine the mass of a given molar quantity

To Find molar mass (g/mol)
Review of Calculation Rules
(atomic mass of each atom) X (amount of each atom)
Then add together mass of all atoms
(g/mol)
To Find the # atoms in a given mass
To Find the # moles in a given mass
(given mass) X (1mole) /(molar mass(g)) X
(# atoms) /(1 mole)
(given mass) X (1mole)/(molar mass(g)) X
(#atom)/(1mole)
To Find the mass(g) of a given molar quantity
(#moles) X (grams/1 mole)  (from molar mass)




Balanced chemical equations relate moles of reactants to moles of products

Just like when baking, reactants have to be mixed in the proper proportions to make a certain amount of the desired product
Specific amounts of reactants produce specific amounts of product

Steps

You can not move directly from the mass of one substance to the mass of the second
1.
You MUST convert the given mass to moles first!
2.
3.
The coefficients of balanced reactions tell you
the NUMBER OF MOLES of each chemical in the reactant
Once you know the number of moles of any reactant or product use the coefficients in the equation to convert the moles of the other reactants and products

Ammonia gas is synthesized from nitrogen gas and hydrogen gas according to the balanced equation :
N
2
+ 3H
2
 2NH
3
How many grams of hydrogen gas are required for 3.75g of nitrogen gas to react completely? What mass of ammonia is formed?



Reactants and products are related in terms of moles
The amount of H
2 of N
2 needed depends on the moles of N
present in 3.75g and the ratio of moles of H
2 in the equation.
2 to moles
The amount of ammonia formed depends on the ratio of moles N
2 to moles of ammonia

1.
Convert the given mass to moles
Find the # of moles of N
2
(3.75g N
2
) (1 mol N
2
(28.0 g N
2
)
)
2.
using molar mass
The coefficients of balanced reactions tell you
the NUMBER OF MOLES of each chemical in the reactant
3.
Once you know the number of moles of any reactant or product use the coefficients in the equation to convert the moles of the other reactants and products