1. Chemical Quantities
For example – let’s say you want to buy some Bazooka Gum
You could buy it by the piece from the deli
You could buy it by the box from Shop Rite
You could buy it by mass from the candy store

2. Chemical Quantities So, the different ways to measure Bazooka gum
1 dozen = 12 pieces
1 dozen = 1.75 boxes
1 dozen = 168 grams

3. Chemical Quantities
Based on those conversions, I can determine how much Bazooka gum I have, regardless of how it is measured.
I have 95 pieces of Bazooka gum, how many grams to I have?

4. Chemical Quantities
I have 95 pieces of Bazooka gum, how many grams to I have?

5. Chemical Quantities
I have 95 pieces of Bazooka gum, how many grams to I have?
1 dozen Bazooka
95 Bazookas X =
7.9 dozen Bazooka
12 Bazookas
168 g Bazookas =
1327 g Bazooka
7.9 dozen Bazooka X
1 dozen Bazooka =
1.3 X 103 g Bazooka

6. Chemical Quantities
In chemistry, there are many ways to measure the amount of substance
The count (quantity)
The volume (how much space it takes up)
The mass (the amount of matter)
Knowing how the count, mass, and volume of an item relate to a common unit allows you to convert among these units.

7. Chemical Quantities In chemistry, we do not use a dozen.
Since atoms are so small, we need a quantity with a larger amount
We use the mole (mol)
1 mole = 6.02 X 1023 representative particles (atoms, molecules, formula units)
It is known as Avogadro’s number

8. Chemical Quantities
We usually are dealing with three types of representative particles
H2O and H2 are molecules
Covalently bonded
Al and Na are atoms
Not bonded
CaCl2 and NaOH are formula units
Ionicly bonded

9. Chemical Quantities
However, 1 mole of any representative particle is always 6.02 X 1023
Using this value we can convert between moles and atoms/ molecules/ formula units

10. Chemical Quantities
For example: If a sample contains mol of O2, how many representative particles are present?

11. Chemical Quantities
For example: If a sample contains 5.2 X 1021 atoms of aluminum, how many moles are present?

12. Chemical Quantities
We can determine the molar mass of representative particles
Molar mass is the mass of 1 mole of that substance
The atomic mass is equal to the molar mass for an element
1 mol Na = g Na

13. Chemical Quantities
For molecules and formula units, we must calculate the molar mass
Let’s calculate the molar mass of SO3
S = 1 X = 32.06
O = 3 X = 48.00
= g/mol

14. Chemical Quantities
We can use the molar mass of a substance to convert from moles to mass
Remember the molar mass is the mass of 1 mole of that substance
SO3 = g/mol =
80.07 g SO3
1 mol SO3

15. Chemical Quantities
Using this relationship, we can easily convert a mole amount to a mass amount
How many grams is there in a mole sample of SO3?

16. Chemical Quantities We can also convert a mass amount to a mole amount
How many moles is there in a 28.5 gram sample of SO3?

17. Chemical Quantities
We can also make conversion between moles and volume, if the conditions are STP
Standard temperature (0 °C) and pressure (1 atm)

18. Chemical Quantities
The volume of solid and liquid substances may vary when the substance is changed
However, gases are unique in that they have a constant volume, regardless of the substance, at STP.
1 mole of gas at STP = 22.4 L
We can use this new relationship to do conversions

19. Chemical Quantities
For example: How many liters is present when you have 2.30 mol of CO2 at STP.

20. Chemical Quantities
The mole is at the center of your chemical calculations.
To convert from one unit to another, you must use the mole as an intermediate step.

21. Chemical Quantities
Percent composition defines what percentage, by mass, is present of each element in a compound.
% by mass of element =
mass of element
mass of compound
× 100

22. Chemical Quantities
Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane.

23. Chemical Quantities
Sometimes, percent composition can include compounds called hydrates
Hydrates are compounds that bind water molecules to their structure
BaCl2  6H2O

24. Chemical Quantities
Find the percentage of water in barium chloride hexahydrate

25. Chemical Quantities
Empirical formula is the symbols of the elements that make up a compound in the smallest whole number ratio.

26. Chemical Quantities
The percent composition of a compound can be used to calculate the empirical formula of that compound.
The percent composition tells the ratio of masses of the elements in a compound.
The ratio of masses can be changed to ratio of moles by using conversion factors based on the molar mass of each element.
The mole ratio is then reduced to the lowest whole-number ratio to obtain the empirical formula of the compound.

27. Chemical Quantities
A compound is analyzed and found to contain 92.3% carbon and 7.7% hydrogen. What is the empirical formula of the compound?

28. Chemical Quantities
A compound is analyzed and found to contain 25.9 g nitrogen and 74.1 g oxygen. What is the empirical formula of the compound?

29. Chemical Quantities You can also solve empirical formulas for hydrates
A gram sample of a hydrate was found to contain 7.05 grams of water. If the anhydrous salt left was sodium sulfate, determine the formula of the hydrate.

30. Chemical Quantities
Sometimes empirical formula problems will include real lab data.
For example:
A reaction occurs between nitrogen and oxygen. The initial amount of nitrogen is g. After the reaction is complete the final mass is 138 g. What is the empirical formula?

31. Chemical Quantities
The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole-number multiple of its empirical formula.
Once you have determined the empirical formula of a compound, you can determine its molecular formula, if you know the compound’s molar mass.

32. Chemical Quantities
Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N.
First calculate the empirical formula mass.
efm of CH4N = 12.0 g/mol + 4(1.0 g/mol) g/mol
= 30.0 g/mol

33. Chemical Quantities
Divide the molar mass by the empirical formula mass.
molar mass
efm =
60.0 g/mol
30.0 g/mol
= 2
Multiply the formula subscripts by this value.
(CH4N) × 2 = C2H8N2