• Mass, volume, length, density… all these are measurements of matter.
• What is matter made of though? atoms
• If you measured the density of an object to be 2.04 g/mL, you have a certain number of atoms all joined together in a certain amount of space.
• So how many atoms are in there?

Why is it important to know HOW MANY atoms there are in a substance?
Chemistry is just like cooking!!! If you are making cookies and your recipe calls for 3 eggs in order to make 24 cookies, what happens if you only have
1 egg? Will you make 24 cookies?
No, quantity matters 
Same would be true if you were making table salt
(NaCl), except it’s a lot more dangerous to make.
https://www.youtube.com/watch?v=d2geiGKFveE

It’s a beauty mark…
It’s a small furry garden pest…
No, wait… its how we count ATOMS!

• Problem: How do you count something as small as an atom?
• Solution: The Mole
• A mole is a quantity (certain amount) of atoms. Due to atoms being so tiny, they are hard to individually measure. Scientists have to measure atoms in large quantities.

• Specific quantity (amount) of atoms in a substance. Basically, a word that represents a specific number (ex: pair = 2)
• 1 mole = 6.02 x 10 23 representative particles
(a lot of very tiny particles)
• This value is called Avogadro’s
Number

What are representative particles?
• smallest particle that retains chemical and physical properties
• 3 types depending on the compound:
• Atoms: Single element
• Molecules: covalent compound
• Formula Units: Ionic Compounds or ions

• Used to count groups of small particles for the purpose of mixing them with other particles in the lab.
• Ex: If 2.5 moles of sodium were used in an experiment, how many atoms of sodium were there?
1 mole = 6.02x10
23
2.5 mole = ___________

Convert: change units (cancel Unit) by using a conversion factor
Conversion Factor: equality using different units

19

• In a lab, the exact number of particles are needed before you can mix them with anything else.
• Problem – No way to count the particles (too small) and no piece of equipment that measures in moles. How do you measure a mole?
• Solution – Using a balance to measure the mass of a mole
(molar mass)

The periodic table contains the molar mass of all elements
Molar Mass = mass of 1 mole (6.02x10
23 )of its atoms.
6.02x10
23 atoms Carbon = 1 mole carbon = 12.01 grams Carbon

FORMULAS - review
2

How to calculate molar mass
1. Identify the # of atoms of each element
2. Multiply # atoms by the atomic mass of that element. (round to 2
#’s after the decimal)
3. Add them all together
4. Grams (g) is the unit

Using Molar Mass as a conversion factor
The periodic table contains the molar mass of all elements.
Molar Mass = mass of 1 mole (6.02x10
23 )of its atoms.
6.02x10
23 atoms Carbon = 1 mole carbon = 12.01 grams Carbon
• If 2.5 moles of carbon were used in an experiment, how many grams of carbon must you weigh out?

Using molar mass
• How many grams are in 15.7 mole MgCl
2

Why would it be difficult to measure the mass of helium?
(how would it stay on the balance)
Scientist have discovered that 1 mole of any gas has a constant volume at STP.
(standard temperature & pressure)
1 mole = 22.4 L of any gas
This value is called Molar Volume

Using molar volume as a conversion factor
• How many moles are in 12.95 L of oxygen gas?
• How many liters are in 0.758 moles of nitrogen gas?

What does a mole tell us?
1. Number of particles – Avagadro’s number
2. Mass particles in solid or liquid –
Molar mass
3. Volume of particles in a gas –
Molar Volume

What is a percentage mean?
• How is a percent calculated?
• What is the percentage of girls in the class?
• What is the percentage of students in
Mrs. Johnson’s class if she has 10 girls and 12 boys?

Percent Composition
• The percent BY MASS of each element in a compound – divide the element’s total mass (part) by the molar mass (whole) then multiple by 100 to get the percent.
(PART)
(WHOLE)
• Ex: % composition of MgCl
2
(PART) (WHOLE)
Mg – 1 (24.31) = 24.31 / 95.21 x 100 = 25.53% Mg
Cl – 2 (35.45) = 70.90 / 95.21 x 100 = 74.47% Cl
Molar mass = 95.21 g/mol

Practice Calculate the % comp of KMnO
4
:
K – 1 (39.10) = 39.10 / 158.04 x 100 = 24.74% K
Mn – 1 (54.94) = 54.94 / 158.04 x 100 = 34.76% Mn
O – 4 (16.00) = 64.00 / 158.04 x 100 = 40.50% O molar mass KMnO
4
= 158.04

Calculating the mass of an element in a sample
1. Find the % comp of the element in the compound
2. Change the % to a decimal (move decimal 2 times to the left or divide by 100)
3. Multiply that decimal by the amount (g) of the sample.
Ex: Calculate the mass of chlorine in 203.5 grams of MgCl
2. (use the % we found earlier)
74.47% Cl = .7447 x 203.5 = 151.5 grams Cl

Practice:
Calculate the amount of oxygen in 15.75 grams of water.
H – 2(1.01) = 2.02
O – 1(16.00) = 16.00
/ 18.02 x100 = 88.79% O
Molar mass H
2
O = 18.02
88.79% O = .8879 x 15.75 = 13.98 grams O
Complete % comp worksheet

Empirical Formula
The lowest whole number ratio
(subscripts) of elements in a compound.
Cannot be reduced!!!
not empirical empirical
Ex: C
6
H
12
O
6
 CH
2
O

Molecular Formula
Actual number of atoms in a chemical compound molecular
EX: C
12
H
24
O
12
Molecular Formulas can be reduced to Empirical Formulas molecular empirical
EX: C
12
H
24
O
12
 CH
2
O
Different molecular formulas can have similar empirical formulas
EX: molecular empirical
N
3
O
9

NO
3
N
12
O
36


PRACTICE:
1. Identify each as empirical (can’t be reduced) or molecular (can be reduced)
2. If its molecular – write the empirical
1. C
2
H
4
2. NO
3 molecular empirical
3. S
9
Cl
12
4. C
3
Cl
9
5. N
4
S
9 molecular molecular empirical
CH
S
3
2
Cl
4
CCl
3
- empirical
- empirical
- empirical

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Finding Empirical Formula from Percent Composition
• Ex: A compound was found to be 54.53% Carbon, 9.15%
Hydrogen, and 36.32% Oxygen. Find its Empirical Formula.
Steps:
1. Assume a 100g sample (change %  g)
2. ÷ by the molar mass of that element to find moles (sig fig it!)
3. Identify the lowest # of moles and ÷ them all by that number
4. Round each to the nearest whole #
5. The resulting whole #are the subscripts for that element in the empirical formula

Calculating Empirical Formula
63.5% Silver 8.2% Nitrogen 28.3% Oxygen
63.5 g Ag
107.87
8.2 g N
14.01
28.3 g O
16.00
.589 mole Ag
.589
1
.59 mole N
.589
1
AgNO
3
1.77 mole O
.589
3

Calculating Empirical Formula (special)
60.00%C 4.48%H 35.53%O
60.00g C
12.01
4.48g H
1.01
35.53g O
16.00
4.996 mole C
2.221
2.249
x4
9
4.44 mole H
2.221
2 x4
8
2.221 mole O
2.221
1 x4
4
C
9
H
8
O
4

Calculating Molecular Formula
1.Find the empirical formula
2.Calculate the molar mass of your empirical formula
3.Identify the molar mass of your molecular
(GIVEN in the problem every time!)
4.Divide the molecular mass / empirical mass
5.Round to the nearest whole #
6.Multiply the whole # by the subscripts in the Empirical formula

Practice
If a compound has an empirical formula of
NO
3 and a molecular mass of 186g – what is the molecular formula?
Empirical formula: NO
3 molar mass:
62.01g
Molecular mass (given) empirical mass
186g
62.01
3 x NO
3
= N
3
O
9

Calculating Empirical Formula (special)
60.00%C 4.48%H 35.53%O
60.00g C
12.01
4.48g H
1.01
35.53g O
16.00
4.996 mole C
2.221
2.249
x4
9
4.44 mole H
2.221
2 x4
8
2.221 mole N
2.221
1 x4
4
C
9
H
8
O
4