Tips for Solving Empirical Formula Problems in Chemistry
Empirical Formula is the simplest whole number ratio between atoms in a compound. So basically it tells us
the elements in their most basic ratios within a chemical formula. The Molecular Formula is going to have the
same ratio but multiplied by some basic integer value (1,2,3,4,5…). In a typical problem you will be given some
information about the elements, which could be gram amounts or percent composition. Sometimes you also
have to determine the amount of one of the elements from the information given (ex. a sulfur oxide compound
contains 70% S… then there must be 30% O!) Follow the steps below to solve problems and consult the
example problem at the bottom of the page.
Step 1: Determine the elements given and write each separately followed by a colon with significant space
between (as seen below H: … C: …)
Step 2: Determine the amounts of each element given and write the amount directly after the colon for each
specific element. If the amounts are given in a percent; assume that you are dealing with a 100g sample
and the percent would be the mass in grams (30% C is 30g C in a 100g sample).
Step 3: Convert the grams of each element to moles (elements not molecules…Oxygen in this case is not
existing as a diatomic gas so its molar mass is 16.00g/mol, not 32g/mol)
Step 4: Divide each of the mole amounts by the smallest mole amount that you solved for. This will give you a
ratio of 1 to something. You may have a exact whole number ratio at this point, if not continue to step 5
Step 5: If necessary multiply each of the values by prime numbers to determine the whole number ratio
between atoms. (start with 2, if that doesn’t work try 3, or 5, or 7) If this doesn’t work you have a
mistake somewhere.
Step 6: Write the empirical formula with the ratio as the number of each element. Remember the most metallic
element always goes first. Hydrogen only goes first if the compound is identified as an acid.
Step 7: To solve for the Molecular Formula use the formula:
(Empirical Formula)n = Molecular Formula
n
MolecularFormulaMolarMass
EmpiricalFormulaMolarMass
Example Problem:
The simple sugar galactose is chemically composed of 6.73% Hydrogen, 39.99% Carbon, and 53.28% Oxygen.
Determine the Empirical and Molecular Formula for Galactose if its molar mass is 180.18g/mol.
H:
6.73gH
1
1mole
1.01g
C:
39.99gC
1
1mole
12.01g
3.33molesC /3.33=1mole C
O:
53.28gO 1mole
1
16.00 g
3.33molesO /3.33= 1mole O
180.18g / mol
30.03g / mol
6.66molesH /3.33 = 2moles H
Empirical Formula CH₂O
6 Molecular Formula = (CH₂O)₆ = C₆H₁₂O₆