# Download Chem 11 Empirical and Molecular Formulas Empirical formula

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```Empirical and Molecular Formulas
Chem 11
Empirical formula: shows the lowest whole number ratio of elements in a compound; also called the
simplest formula
Molecular formula: shows the actual number of atoms that make up a molecule
Ex: Glucose
Molecular formula
Empirical formula
C6H12O6
CH2O
Analysis of an unknown sample to determine the percent composition of elements can be used to find
the empirical formula. However, there can be many compounds that have the same empirical formula:
Ex: Benzene, C6H6 and acetylene, C2H2 are very different substances but both have the empirical
formula CH.
How to find the empirical formula of a compound from the percent composition:
Example 1:
Calculate the empirical formula of a compound that is 25.9 % nitrogen and 74.1 % oxygen.
Solution 1:
1. Assume 100 g of sample and determine the mass of each element.
Mass of nitrogen = 25.9g
Mass of oxygen = 74.1g
2. Find the number of moles of each element.
nN =
25.9 g
= 1.85mol N
14.0067 g / mol
nO =
74.1g
= 4.63mol O
15.9994 g / mol
3. Write the ratio of moles of each element, then reduce this ratio by dividing through by the
lowest number of moles. If necessary, multiply each term by a factor to convert decimals to
whole numbers.
Ratio (divide by smallest number of moles)
N : O
1.85 4.63
:
1.85 1.85
1 : 2.5 (we don’t use fractions in chemical formulas, so we need to multiply to get
whole numbers)
2:5
Therefore our empirical formula would be N 2 O5
Try: Write the empirical formula for a substance that is 38.7% C, 16.3% H and 45.1 % N.
Again assume 100g sample to find mass of each element present:
38.7g C, 16.3g H, 45.1g N
Find moles of each (in order as listed in question):
nC =
38.7 g
= 3.222molC
12.0107 g / mol
nH =
16.3g
= 16.17molH
1.00794 g / mol
nN =
45.1g
= 3.220molN
14.0067 g / mol
Write ratios of each and divide by smallest number of moles. Multiply, if necessary, to get whole
numbers throughout:
3.222 16.17 3.220
:
:
3.220 3.220 3.220
Therefore our empirical formula would be CH 5 N
1: 5 :1
How to determine the molecular (actual) formula, given the empirical formula and the molar
mass:
Example 1:
A substance has the empirical formula CH2O, and a molar mass of 180 g/mol. What is the molecular
formula for this compound?
Solution 1:
1. Using the empirical formula, calculate the mass of one mole of the compound (this is molar
mass).
12.0107g/mol + (2 x 1.00794g/mol) + 15.9994g/mol = 30.02598g/mol
2. Divide the given molar mass by the calculated molar mass of the empirical compound. Then
multiply this factor by the empirical formula to obtain the molecular formula.
180 g / mol
=6
30.02598 g / mol
Try:
So 6 times the E.F. is the M.F which gives us C 6 H 12 O6 .
Empirical formula: C2HCl
What is the molecular formula?
molar mass = 181.5 g/mol
Molar mass = (2 x 12.0107g/mol) + 1.00794g/mol + 35.453g/mol = 60.48234g/mol
181.5 g / mol
=3
60.48234 g / mol
Therefore our M.F. is C 6 H 3Cl 3
```