Download Unit 5.6 Empirical Formula

Unit 5.6 Empirical Formula
What’s occurring in this picture? In the early days of
chemistry, there were few tools for the detailed study of
compounds. Much of the information regarding the
composition of compounds came from the elemental analysis
of inorganic materials. The “new” field of organic chemistry
(the study of carbon compounds) faced the challenge of not
being able to characterize a compound completely. The
relative amounts of elements could be determined, but so
many of these materials had carbon, hydrogen, oxygen, and
possibly nitrogen in simple ratios. We did not know exactly
how many of these atoms were actually in a specific
Determining Empirical Formulas
molecule.
An empirical formula is one that shows the lowest whole-number ratio of the elements in a compound.
Because the structure of ionic compounds is an extended three-dimensional network of positive and
negative ions, all formulas of ionic compounds are empirical. However, we can also consider the empirical
formula of a molecular compound. Ethene is a small hydrocarbon compound with the formula
C2H4 (see diagram below). While C2H4 is its molecular formula and represents its true molecular
structure, it has an empirical formula of CH2. The simplest ratio of carbon to hydrogen in ethene is 1:2.
There are two ways to view that ratio. Considering one molecule of ethene, the ratio is 1 carbon atom
for every 2 atoms of hydrogen. Considering one mole of ethene, the ratio is 1 mole of carbon for every 2
moles of hydrogen. So the subscripts in a formula represent the mole ratio of the elements in that
formula.
Ball and stick model of ethane, C2H4.
Carbon is shown as black balls and
Hydrogen as white balls.
In a procedure called elemental analysis, an unknown compounds can be analyzed in the laboratory in
order to determine the percentages of each element contained within it. These percentages can be
transformed into the mole ratio of the elements, which leads to the empirical formula. The steps to be
taken are outlined below.
Steps used in determining Empirical Formula
1.
Assume use of a 100 g sample of the compound so that the given percentages can be directly
converted into grams.
2.
Use each element’s molar mass to convert the grams of each element to moles.
Moles =
3.
Grams
Molar Mass
In order to find a whole-number ratio, divide the moles of each element by whichever of the
moles from step 2 is the smallest.
4.
If all the moles at this point are whole numbers (or very close), the empirical formula can be
written with the moles as the subscript of each element.
5.
In some cases, one or more of the moles calculated in step 3 will not be whole numbers. Multiply
each of the moles by the smallest whole number that will convert each into a whole number.
6.
Write the empirical formula.
Sample Problem One: Determining the Empirical Formula of a Compound
A compound of iron and oxygen is analyzed and found to contain 69.94% iron and 30.06% oxygen. Find
the empirical formula of the compound.
Step 1: List the known quantities.
Known

% of Fe = 69.94%

% of O = 30.06%
Unknown

Empirical formula = Fe?O?
Step 2: Calculate.
1. Assume a 100 g sample.
69.94 g Fe
30.06 g O
2. Convert to moles.
69.94 g Fe
x
1 mol Fe
=
1.252 mol Fe
=
1.879 mol O
55.85 g Fe
30.06 g O
x
1 mol O
16.00 g O
3. Divide both moles by the smallest of the results.
1.252 mol Fe
=
1.000 mol Fe
=
1.501 mol O
1.252 mol
1.879 mol O
1.252 mol
4/5. Since the moles of O, is still not a whole number, both moles can be multiplied by 2,
resulting in both mole values becoming whole numbers.
(1 mol Fe) x (2) = 2.000 mol Fe
2 mol Fe
(1.501 mol O) x (2) = 3.002 mol O
3 mol O
With true experimental data, the numbers will rarely be exactly
whole number. Rather they will tend to be close to whole numbers
and rounding is necessary.
6.These mole values become the subscripts in the empirical formula. The empirical formula of
the compound is Fe2O3.
Step 3: Think about your result.
The subscripts are whole numbers and represent the mole ratio of the elements in the compound. The
compound is the ionic compound iron (III) oxide.
Glucose
Sucrose
How can you determine the differences between these two molecules?
Above we see two carbohydrates: glucose and sucrose. Sucrose is almost exactly twice the size of
glucose, although their empirical formulas are very similar. Some people could distinguish them on the
basis of taste, but it’s not a good idea to go around tasting chemicals. The best way is to determine the
molecular weights – this approach allows you to easily tell which compound is which.
Molecular formulas give the kind and number of atoms of each element present in a molecular compound.
In many cases, the molecular formula is the same as the empirical formula. The molecular formula of
methane is CH4 and because it contains only one carbon atom and four hydrogen, that is also its empirical
formula. Sometimes, however, the molecular formula is a simple whole-number multiple of the
empirical formula. Acetic acid is an organic acid that is the main component of vinegar. Its molecular
formula is C2H4O2. Glucose is a simple sugar that cells use as a primary source of energy. Its molecular
formula is C6H12O6. The structures of both molecules are shown in the figure below. They are very
different compounds, yet both have the same empirical formula of CH2O.
Acetic Acid has the
Glucose has the molecular
molecular formula of C2H4O2.
formula C6H12O6
Both acetic acid and glucose have the empirical formula of CH2O.
Empirical formulas can be determined from the percent composition of a compound as shown earlier. In
order to determine its molecular formula, it is necessary to know the molar mass of the compound.
Chemists use an instrument called a mass spectrometer to determine the molar mass of compounds.
In order to go from the empirical formula to the molecular formula, follow these steps:
1.
Calculate the empirical formula mass (EFM), which is simply the molar mass represented by the
empirical formula.
2.
Divide the molar mass of the compound by the empirical formula mass. The result should be a
whole number or very close to a whole number.
3.
Multiply all the subscripts in the empirical formula by the whole number found in step 2. The
result is the molecular formula.
Sample Problem Two: Determining the Molecular Formula of a Compound
The empirical formula of a compound of boron and hydrogen is BH3. Its molar mass is 27.70 g/mol.
Determine the molecular formula of the compound.
Step 1: List the known quantities.
Known

empirical formula = BH3

molar mass = 27.70 g/mol
Unknown

molecular formula = ?
Step 2: Calculate.
1. The empirical formula molar mass (EFM)
(10.81 g B) + (3 x 1.01 g H) = 13.84 g/mol
2.
molar mass of compound
empirical formula molar mass
3. (BH3) x (2)
=
27.70 g/mol
=
2
13.84 g/mol
= B2H6 is molecular formula for the compound.
Step 3: Think about your result.
(2 x 10.81 g B) + (6 x 1.01 g H) = 27.68 g/mol is the molar mass of the molecular formula.
The molar mass of the molecular formula matches the molar mass of the compound.
Sample Problem Three: Determining both the Empirical and Molecular Formula of a Compound
During strenuous exercise, lactic acid builds up in muscles resulting in the “burn” that is felt in muscles.
It has been determined that lactic acid consists of 40.0 % C, 6.67 % H, and 53.3 % O, and has a molar
mass of 90.0 g/mol. Determine both the empirical formula and molecular formula for lactic acid.
Step 1: List the known quantities.
Known

compound is 40.0 % C, 6.67 % H, and 53.3 % O

molar mass = 90.0 g/mol
Unknown

empirical formula = ?

molecular formula = ?
Step 2: Calculate Empirical Formula.
1. Assume a 100 g sample of lactic acid.
40.0 g C
6.67 g H
53.3 g O
2. Convert to moles.
40.0 g C
x
1 mol C
=
3.33 mol C
=
6.60 mol H
12.01 g C
6.67 g H
x
1 mol H
1.01 g H
53.3 g O
x
1 mol O
=
3.33 mol O
16.0 g O
3. Divide both moles by the smallest of the results.
3.33 mol C
=
1.00 mol C
3.33 mol
6.60 mol H
=
1.98 mol H
3.33 mol
3.33 mol O
3.33
=
1.00 mol O
mol
4,5,6. Both C and O mol values are whole number. 1.98 mol of H is close enough to be rounded to
2.00 mol H. These mole values become the subscripts in the empirical formula. The empirical
formula of the compound is CH2O.
Step 3: Calculate Molecular Formula.
1. The empirical formula molar mass (EFM)
(12.0 g C) + (2 x 1.01 g H) + (16.0 g O) = 30.02 g/mol
2.
molar mass of compound
empirical formula molar mass
3. (CH2O) x (3)
=
90.0 g/mol
=
2.998 or rounded to 3
30.02 g/mol
= C3H6O3 is molecular formula for the compound
Step 3: Think about your result.
(3 x 12.0 g C) + (6 x 1.01 g H) + (3 x 16.0 g O) = 90.06 g/mol is the molar mass of the molecular formula.
The molar mass of the molecular formula matches the molar mass of the compound.
Summary

A process is described for the calculation of the empirical formula for a compound based on the
percent composition of that compound.

A procedure is described that allows the calculation of the exact molecular formula for a
compound.
Review
1.
What is an empirical formula?
2.
What does an empirical formula tell you?
3.
What does it not tell you?
4.
What is the difference between an empirical formula and a molecular formula?
5.
In addition to the elemental analysis (to determine the empirical formula), what do you need to
know to calculate the molecular formula?
6.
What does the empirical formula mass tell you when compared to the molar mass of the
compound?
Answers
1.
An empirical formula is one that shows the lowest whole-number ratio of the elements in a
compound.
2.
The ratio of each kind of atom in the molecule, or the ratio of the moles of each kind of atom.
3.
It does not tell you if it is the true molecular formula.
4.
The empirical formula tells you the ratio of each kind of atom in a molecule, the molecular
formula tell the exact number of each kind of atom in a molecule.
5.
The true molar mass of the compound must be known to determine the molecular formula.
6.
It will tell the number to multiply the empirical formula by to obtain the molecular formula.