# Calculating Empirical Formulas ```Calculating Empirical Formulas
– Lowest whole-number ratio of the atoms of
the elements in a compound
• C6H12O6 (glucose)
• The ratio that glucose normally has for
carbon:hydrogen:oxygen is 6:12:6.
• The lowest ratio that glucose has for
carbon:hydrogen:oxygen is 1:2:1 (each
number can be divided by the smallest
number in the ratio which is 6).
• The empirical formula for glucose is
CH2O since this is the lowest wholenumber ratio of atoms for that
compound.
– May or may not be the same as the normal
molecular formula of a compound
• Next - Calculating empirical formulas
What is the empirical formula of a
compound that is 25.9% nitrogen and
74.1% oxygen?
• If 25.9% of the compound is nitrogen and
74.1% of the compound is oxygen, then a
compound with a mass of 100 g has 25.9 g of
nitrogen and 74.1 g of oxygen.
• To calculate the empirical formula, we need to
relate the moles of each atom in the
compound, so we need to convert the
masses of the elements to moles.
25.9 g N
1 mol N
mol N 

 1.85 mol N
1
14.0067 g N
74.1 g O
1 mol O
mol O 

 4.63 mol O
1
15.9994 g O
This would mean that the ratio of nitrogen to oxygen
is N1.85O4.63.
We can divide each number in the ratio of N1.85O4.63 by
1.85 to get N1O2.50.
Since we cannot have 2.50 atoms of
oxygen, we must multiply through each number by 2 to
even it out, getting N2O5 as our empirical formula.
• In calculating empirical formulas, remember
that the number of atoms is a whole number.
If the number of atoms for an element is close
to a whole number (i.e., 2.1 or 2.2 or 2.8, or
2.9), you can usually round up or down to get
a whole number.
• If you should get a number of atoms closer to
2.33 or 2.5, multiply each number in the
formula by a number that gets that to a whole
number. For example, if you calculated 2.33,
you would multiply this by 3 to get a value of
7 for that number.
Empirical vs. Molecular
Formula
Calculating Molecular Formulas
• Although sometimes a molecular
formula may be the same as a
molecule’s empirical formula, like in
carbon dioxide (CO2), we have seen
that the empirical formula for glucose is
not the same as its molecular formula.
• One can determine the molecular
formula of a compound by knowing its
empirical formula and its mass.
• Next - Example
Calculate the molecular formula of the compound
whose molar mass is 180.1583 g and empirical formula
is CH2O.
We know that the molecular formula will have a molar
mass of 180.1583 g. We also know, by calculating the
gmm of CH2O, that CH2O has an empirical formula
mass (efm) = 30.0264 g CH2O.
Now, in order to figure out what we must multiply each
number in the empirical formula by, we must figure out by
what number we must multiply the empirical formula mass
to get the molecular formula mass.
To get from 30.0264 to 180.1583, 180.1583
30.0264
6
• Therefore, we must multiply each number of
atoms in CH2O by 6 to get the molecular
formula of C6H12O6.
recalculating the molar mass of C6H12O6.
• gmm C6H12O6 = 6 x 12.0111 g + 12 x 1.00794
g + 6 x 15.9994 g = 180.1583 g C6H12O6
• This agrees with the molar mass we were
given, so the molecular formula we calculated
is correct.
Example (toughy)
• 1.00 g of menthol on combustion yields
1.161 g of H2O and 2.818 g of CO2.
What is the empirical formula?
• Solution:
– 1.161 g H2O  0.1298 g H = 0.1288 mol H
– 2.818 g CO2  0.7690 g C = 0.0640 mol C
– difference = 0.101 g O = 0.00632 mol O
– ratios: H/O = 20.4 and C/O =10.1
– Therefore, C10H20O
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