S. 5.5 – Empirical and Molecular Formulae.
Empirical Formula:
Empirical formula: A formula that gives the smallest possible ratios of atoms in a compound.
Ex. 1: C2H4O,
C4H8O2,
C8H16O4,
C16H36O8
Each proceeding formula contains twice as many H’s, C’s, or O’s as the previous formula;
thus, the empirical formula (simplest formula) is C2H4O.
Finding molecular formula given percentages of elements within the compound.
Steps:
1) Find arbitrary amount of mass of each element within the compound.
- Pick an arbitrary amount of compound, usually 100 g.
 % 
Arbitrary amount of element = 
  arbitrary mass.
 100 
2) Find the moles of each element within the compound.
- Convert grams into moles.
3) Find the ratio of atoms within the compound.
- Divide the all the produced mole values by the smallest produced mole value.
4) Convert ratio values into whole numbers.
- Multiply ratio values by the same value to turn the values into whole numbers.
- This is done if whole values were not produced in step 3.
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Ex. 1: What is the empirical formula of a compound consisting of 15.9 % B, 84.1 % F.
Assume 100.0 g of the compound is taken.
Ex. 2: What is the empirical formula of a compound consisting of 21.8 % Mg, 27.9 % P, 50.3 % O.
Assume 100.0 g of the compound is taken.
Mass of Mg = 21.8 g, Mass of P = 27.9 g, Mass of O = 50.3 g
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What value to use to create whole numbers?
- Look at the first two decimal values.
Thirds always end in 0.33 or 0.67 and are multiplied by 3 to create whole values.
Quarters always end in 0.25 or 0.75 and are multiplied by 4 to create whole values.
Halves end in 0.50 and are multiplied by 2 to create whole values.
Question:
- Pg. 93, # 46a-j.
Molecular Formula:
Molecular Formula: The formula with the actual atom values.
Ex. 1: Molecular formula is C4H8O2 = 4 C, 8 H, and 2 O
OH
H
C
H 3C
CH
C
H
OH
Where
Empirical formula is C2H4O = 2 C, 4 H, and 1 O
H
C
H3C
O
Two compounds are different.
Finding molecular formula.
- Determine the multiple of the molecular formula compared to the empirical formula.
- Multiply empirical formula by multiple.
Steps:
1) Determine the molar mass/empirical masses of the compound.
2) Let N = the whole number multiple of the empirical formula.
 molar mass 

3) Equation is: multiple = N = 
 emperical mass 


4) Multiply the Empirical formula atom values by N to produce the molecular formula.
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Ex. 1: A molecule has an empirical formula of CH2 and a molar mass of 54.06 g. What is the
molecular formula?
Empirical mass = 18.02 g

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Ex. 2: A gas has the empirical formula POF3. If 0.350 L of the gas at STP has a mass of 1.62
g, what is the molecular formula of the compound?
Empirical mass = 104.0 g.
mass
Density of gas =
=
volume
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Ex. 3: The empirical formula of a compound is SiH3. If 0.0275 mol of a compound has a
mass of 1.71 g, what is the compound’s molecular formula?
Empirical mass = 31.1 g.
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Questions.
- Pg. 95, # 47, 49, 51, 53, 55.
- Quiz next class on S. 5.3 – 5.5.
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