1. Quantitative Analysis
Balanced Equations
The Mole
Percent Composition
The Chemical Formula
Solutions
Stoichiometry
Limiting Reagents
Percent Yield

2. Relative Atomic Mass
The Atomic Mass for each element is found on the periodic table
Determined by Law of Definite Proportions
Atomic mass determined by comparing a standard mass
Originally Hydrogen
Current Standard is Carbon – 12

3. Carbon-12 Standard Carbon is a very common substance found everywhere
Assigning a value of 12 to C-12 allows all other elements to have value close to a whole number
One atomic mass unit = 1/12 the mass of a C-12 atom

4. Atoms in 12 grams of C-12 A defined amount of substance
Value = 6.02 x 1023
Estimated through experimentation as the number of atoms present in 12 grams of Carbon-12
12 g = 6.02 x 1023 atoms = 1 mole C-12

5. 1 Mole = 6.02 x 1023 things The mass of 1 Mole of an element =
The Mole
1 Mole =
6.02 x 1023 things
The mass of 1 Mole of an element =
the atomic mass of that element

6. Moles to Mass Given the number of moles of a compound or element
Find the Molar Mass of the substance on the Periodic Table
Moles x Molar Mass = Mass of substance

7. Mass to Moles Given the Mass of a compound or element
Find the Molar mass of the substance on the Periodic Table
Mass / Molar Mass = # of Moles

8. Moles to Particles Given the number of moles of a compound or element
Use Avogadro’s Constant
6.02 x 1023
Moles x 6.02 x 1023 = # of particles

9. Particles to Moles
Given the number of particles of a compound or element
Use Avogadro’s Constant
6.02 x 1023
number of particles / 6.02 x = # of Moles

10. Mass to Particles Given the Mass of a compound or element
Find the Molar Mass of the substance
Mass / Molar Mass = # of Moles
Take the number of moles
Use Avogadro’s Constant = 6.02 x 1023
Moles x 6.02 x 1023 = # of particles

11. Particles to Mass
Given the number of particles of a compound or element
Use Avogadro’s Constant = 6.02 x 1023
number of particles / Avogadro’s Constant = # of Moles
Take the number of moles
Find the Molar mass of the substance
Moles x Molar Mass = Mass of substance

12. Moles Mass (g) Particles  molar mass  6.02 x 1023 x molar mass

13. Percent Composition
A descriptive method of indicating how much there is of an element in a compound by comparing its mass to the mass of the compound
Ex. CO2 molar mass = 44g
C = 12g / 44g x 100 = 27%
O = (16gx2) / 44g x 100 = 73%

14. Empirical Molecular Formula Formula
vs.
Provides the lowest ratio of atoms of the elements in the compound
CH3O
C2H6O2CH3O
C4H12O4CH3O
Provides the actual ratio of atoms of each element in the compound
CH3O
C2H6O2
C4H12O4

15. Empirical Formulas From Percent Composition
Determine relative masses of each element (assume sample=100g)
Convert relative masses into moles
Divide each number of moles by the smallest number of moles
Express number of moles of each element as the smallest ratio

16. Example: C=85.7%;H=14.3% C = 85.7g H = 14.3g
C = 85.7g/12g = 7.14 moles
H = 14.3g/1g = 14.3 moles
C:H = 7.14:14.3 = 1:2
CH2

17. Molecular Formulas Using Molar Mass
Determine the Empirical Formula
Calculate the molar mass of the Empirical Formula
Compare with the actual molar mass of the unknown (will be given)
Actual molar mass
empirical formula molar mass
Multiply the empirical formula by (n)
= n

18. Concentrations in Solutions
Solutions are homogeneous mixtures
Solutions have definite percent compositions
Solutions are made up of:
Solute = substance in smaller amount
Solvent = substance in larger amount
Concentration(c) =
Amount of Solute
Amount of Solvent

19. Concentrations in Solutions
Dilute Solutions have low solute and high solvent ratios
Concentrated Solutions have high solute and low solvent ratios

20. Solution Concentration
Percentage by volume
Percentage by weight
Vsolute (smaller volume of two substances)
Vsolution (larger volume of two substances)
V/V =
X 100
msolute (mass of substance)
Vsolution (volume of solution)
W/V =
X 100

21. Solution Concentration
Molar Concentration: Molarity
Moles of solute per litre of solution
Grams of solute/molar mass
Volume of solution in litres
Cmol =
Mol L
Unit =

22. Solution Concentration
Parts per Million (ppm)
Used to describe very low concentrations in solutions 1:
1 g : ml
1 g : L
1 mg: 1 L
Normally expressed as mg/L

23. Diluting Solutions
Many lab activities include creating and using a series of concentrations of a particular solution.
Stock Solution provides a concentrated version of a solution that can quickly be diluted for other purposes.
Laboratory solutions are dilutions of the stock solution

24. civi = cfvf Diluting Solutions Need to know three of the following:
concentration of the stock solution (ci)
final concentration of the solution (cf)
volume of stock solution (vi)
volume of the lab solution (vf)
Relationship:
civi = cfvf

25. Balanced Chemical Reactions
Count the number of atoms of each element on the reactant side and the product side
Add coefficients in front of molecular formulas to balance
Coefficients affect the entire molecule
Don’t alter chemical formulas
Don’t add additional compounds or elements to balance

26. Balance the Reaction
Ba(OH)2(aq) + Na2SO4(aq) -> BaSO4(s) + NaOH(aq)
Left Side Right Side
Ba= Ba= 1
OH= OH = 1
Na= Na = 1
SO4= SO4= 1
The equation is not balanced.

27. Balancing the Reaction
Ba(OH)2(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaOH(aq)
Left Side Right Side
Ba= Ba= 1
OH= OH = 2
Na= Na = 2
SO4= SO4= 1
The equation is now balanced.

28. Mole Ratios in Balanced Equations
2H2 + O2  2H2O
Moles in the equation
Two moles of hydrogen
One mole of oxygen
Two moles of water
Mole Ratios present
H2:O2 = 2:1
H2:H2O = 2:2
O2:H2O = 1:2

29. Mole Ratio Applied 2H2 + O2  2H2O
If 4 moles of H2 are used you need 2 moles of O2
If 10 moles of O2 are used you get 20 moles of H2O
If 40 moles of water was made 40 moles of hydrogen and 20 moles of oxygen are needed

30. Calculating Masses of reactants and products
Given mass of substance
Convert mass of substance to moles
Find mole ratio of given to required substance
Calculate moles of required
Convert moles of required to mass of required

31. Limiting Reagents Goal: To discover which reactant will run out first.
The mole ratio is used to determine the required amounts
The reactant left over it is labeled the excess reagent
The reactant completely used up is labeled the limiting reagent

32. Yield of a Chemical Reaction
Yield is the amount of product produced in a reaction
Theoretical yield is the amount of product produced