Empirical Formulas and

Molecular Formulas

Ch 3.5

• Empirical Formulas are the simplest (lowest) whole number ratio of atoms in a molecule or ionic compound

• Molecular Formulas are true formulas.

• For example:

• C6H6 = CH

• H2O2 = HO

• C6H12O6 = CH2O

Empirical Formulas

…can be determined from % composition, here is the “process:”

1. % is the same as grams

2. Convert from grams to moles

3. Next divide by the smallest # of moles

4. …this gives the empirical formula

Empirical Formula of Eugenol a Component of Clove Oil?

…is 73.14% C, 7.37% H and 19.49 g O

Remember, % is the same as grams (g)

Empirical Formula of Eugenol, continued…

Next, convert to the central unit, the mole

73.14 g C x 1 mole = 6.09 mol C

12.01 g

7.37 g H x 1 mole H = 7.31 moles H

1.0079 g

19.49 g O x 1 mole O = 1.22 moles O

15.9994 g

Empirical Formula of Eugenol, continued…

Finally divide by the smallest # of moles

6.09 mol C 7.31 moles H 1.22 moles O

1.22 mol 1.22 mol 1.22 moles

C: 4.99 H: 5.99 O: 1.00

Or C: 5 atoms H: 6 atoms O: 1.00 atoms

Therefore C5H6O

is Eugenol’s empirical formula

Eugenol

Count the number of carbon atoms, hydrogen atoms and oxygen atoms, does this fit the empirical formula that we just derived? No, because we did not find the molecular formula…

Molecular Formulas

Molecular formulas are also known as the “true formula” of a molecule.

To derive this use amu:

Molecular Formula = True amu

empirical amu

True Formulas

• The molar mass of Eugenol is 164.2 g/mol, what’s the molecular formula of Eugenol?

• Use: True amu

empirical amu

164.2 g/mol = 164.2 g/mol = 2

C5H6O 82 g/mol

Therefore 2(C5H6O) = C10H12O2

• Complete Practice Problems #1-#18

• Quiz Next Class Period, THEN begin Lab!... Whew! Finally!!

Hydrated Compounds

Hydrates

• A compound that is hydrated is called a hydrate since they form solids that include water in their crystal structure.

CuSO4 • 5 H2O

When figuring the molar mass should you add the amu of water?

Yes, therefore

CuSO4 • 5 H2O has an amu of 249.7 g/mol.

The “dot” does NOT mean to multiple the amu masses.

• Notice the color difference of the anhydrous crystals & hydrated crystals

• Cobalt (II) Chloride

• Copper (II) Sulfate

Naming Hydrates

1. Name the “criss-cross” compound

2. Use number prefixes to indicate the number of waters

3. Example: Copper (II) Sulfate Pentahydrate

CuSO4 • 5 H2O

Number Prefixes

• Mono- = 1

• Di- = 2

• Tri- = 3

• Tetra- = 4

• Penta- = 5

• Hexa- = 6

• Hepta- = 7

• Octa- = 8

• Nona- = 9

• Deca- = 10

• By simply heating the solid, water can be driven from a hydrate to leave an anhydrous compound.

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Naming Hydrates

• To name hydrates:

1. Name the compound

2. Plus the word hydrate—use prefixes to indicate how many waters are associated with the compound

3. Example: Copper (II) Sulfate pentahydrate

4. To write their formulas

Write: the name of the compound • number of H2O

CuSO4 • 5 H2O

Calculate The Empirical Formula Of Ca(NO3)2 _____ H2O

Units of Hydration

• A student heats hydrated crystals of CuSO4, how many moles of water are associated with the crystals?

• Step 1: Find the mass of the crystals:

1.023 g of CuSO4 • x H2O

• Step 2: Subtract the dehydrated crystal mass from the initial crystal mass = mass of water

1.023 g of CuSO4 • x H2O – 0.654 g of CuSO4

= 0.369 g water

Units of Hydration Continued…

• Step 3: Determine the number of moles 0.369 g H2O x 1 mol

18.02 g = 0.0205 mol H2O

0.654 g CuSO4 x 1 mol/159.6 g = 0.00410 mol CuSO4

• Step 4: Determine the molar ratio (see above) 0.0205 mol H2O 0.00410 mol CuSO4

0.00410 mol CuSO4 0.00410 mol CuSO4

1 CuSO4 • 5 H2O

Combustion Formula of a Compound

Ch 3.5

• The AP Exam will most likely give you a combination of work to complete by using a combustion device which analyzes substances containing C and H. It is burned in excess O2 producing CO2 and H2O, these products are then collected and from here one can determine the %C in CO2 and %H in H2O…

Combustion Analysis CxHy(any hydrocarbon) + O2 H2O + CO2

• 0.1156 g of a compound is reacted with O2 & 0.1638 g of CO2 & 0.1676 g of H2O is collected.

• The unknown compound has C, H and N, what’s the empirical & molecular formula molar mass = 31.06 g/mol?

• Go in the order of C, H, O, (N)

• Remember %C in CO2 (part/whole x 100% = % comp.):

%C: 12.01 g C x 0.1638 g CO2 = 0.04470 g C

44.01 g CO2

% H: 2.016 g H(note 2 hydrogen)x0.1676 g H2O = 0.01875 g of H

18.02 g H2O

0.1156 g compound = 0.04470 g C + 0.01875 g H + ______ g N

= 0.05211 g N

0.04470gC x 1mol 0.01875 g H x 1mol 0.05211g N x 1mol

12.01 g 1.01 g 14.01 g

= 0.003722 mol C = 0.01860 mol H = 0.003719 mol N

Divide by Smallest number of moles = 0.003719 mol

1 C : 5 H : 1 N

CH5N empirical amu = 31.07

True (given) amu = 31.06

31.06/31.07 = 1 (CH5N ) = CH5N is true formula

Percent Yield

Theoretical Yield

• The amount of product formed is controlled by the limiting reactant—products stop forming when on reactant runs out.

• The amount of product calculated in this way is called the theoretical yield.

• This is the amount of product predicted from the amount of reactants used.

Actual Yield

• However, the amount of product predicted (the theoretical yield) is seldom obtained.

• One reason for this is the presence of side reactions (other reactions that consume one or more of the reactants or products).

• The actual yield of product, is the amount of product actually obtained.

Percent Yield

• The comparison of the product actually obtained and theoretically obtained is called the percent yield:

• Percent Yield = Actual Yield x 100%

Theoretical Yield