Determining Empirical Formula from Mass % Data

• To convert the mass % composition obtained from a combustion analysis into an empirical formula, we must convert the mass % of each type of atom into the relative number of atoms.– To do this, assume that we have 100g of

sample– The mass % will then be in grams

Determining Molecular Formulas

• Once we know the empirical formula, we need one more piece of information to determine the molecular formula: The Molar Mass

• Say we know the empirical formula of a compound is C3H4O3.– All we know about this compound at this point is the

ratio of the 3 elements. – We don’t know the exact number of each type of atom

in the molecule.– Is the Molecular Formula C6H8O6, C12H16O12 or

C18H24O18?

G: Molarity

• Hands down one of the most important concepts you need to master is you are going to stay in the sciences. Period.

• The Molar Concentration, c, of a solute in solution is the number of moles of solute divided by the volume of the solution (in liters).– Also referred to as Molarity

Molarity

The symbol M is used to denote the molarity of the solution

1M NaCl = 1 mole NaCl per liter of H2O

G4: Dilutions• Frequently in the laboratory, you will need to make dilutions

from a stock solution.• This involves taking a volume from the stock and bringing it

to a new volume with solvent.• In order to perform these dilutions, we can use the following

equation:

c1V1 = c2V2

Where: c1 = Stock concentration

V1 = Volume removed from stock

c2 = Target conc of new sol’n

V2 = Volume of new solution

Law of Conservation of Matter

• “Matter can neither be created nor destroyed” – Antoine Lavoisier, 1774

If a complete chemical reaction has occurred, all of the reactant atoms must be present in the product(s)

Law of Conservation of Matter

a)

b)

•Stoichiometry coefficients are necessary to balance the equation so that the Law of Conservation of Matter is not violated

•6 molecules of Cl2 react with 1 molecule of P4

•3 molecules of Cl2 react with 2 molecules of Fe

Example of Using Stoichiometric Coefficients

Balancing Chemical Reactions

• Let’s look at Oxide Formation

• Metals/Nonmetals may react with oxygen to form an oxide with the formula MxOy

• Example 1: Iron reacts with oxygen to give Iron (III) Oxide

Fe (s) + O2 (g) → Fe2O3 (s)

How do we solve it?

• Step 1: Look at the product. There are 3 atoms of oxygen in the product, but we start with an even number of oxygen atoms.– Let’s convert the # of oxygens in the product

to an even number

Fe (s) + O2 (g) → Fe2O3 (s)

Result: Fe (s) + O2 (g) → 2Fe2O3 (s)

How do we Solve It?

• Then, balance the reactant side and make sure the number/type of atoms on each side balance.

Fe (s) + O2 (g) → 2Fe2O3 (s)

Balanced Equation: 4Fe (s) + O2 (g) → 2Fe2O3 (s)

How do we Solve It?

• Example 2: Sulfur and oxygen react to form sulfur dioxide.

S (s) + O2 (g) → SO2 (g)

• Step 1: Look at the reaction. We lucked out!

Balanced Equation: S (s) + O2 (g) → SO2 (g)

How do we Solve It?

• Example 3: Phosphorus (P4) reacts with oxygen to give tetraphosphorus decaoxide.

P4 (s) + O2 (g) → P4O10 (s)

• Step 1: Look at the reaction. The phosphorus atoms are balanced, so let’s balance the oxygens.

Balanced Equation: P4 (s) + 5O2 (g) → P4O10 (g)

How do we Solve It?

• Example 4: Combustion of Octane (C8H18).

C8H18 (l) + O2 (g) → CO2 (g) + H2O (g)

• Step 1: Look at the reaction. Then: – Balance the Carbons

C8H18 (l) + O2 (g) → 8CO2 (g) + H2O (g)

How do we Solve It?

• Step 2: Balance the Hydrogens

C8H18 (l) + O2 (g) → 8CO2 (g) + 9H2O (g)

C8H18 (l) + O2 (g) → 8CO2 (g) + H2O (g)

• Step 3: Balance the Oxygens– Problem! Odd number of oxygen atoms– Solution: Double EVERY coefficient (even those with a value of ‘1’)

How do we Solve It?

• Step 3 (cont’d): Balance the Oxygens

2C8H18 (l) + 25O2 (g) → 16CO2 (g) + 18H2O (g)

C8H18 (l) + 12.5O2 (g) → 8CO2 (g) + 9H2O (g)

• Step 4: Make sure everything checks out

Review of Balancing Equations