One of the interesting aspects of our
study of moles is that it will give you the ability to determine the empirical formula of
a compound formed in our laboratory, by calculating the molar ratio by which the elements
combine. It will also allow you to determine the empirical formula of a compound
after you determine a compounds percentage composition by mass. This lesson will go
over the math involved in these types of procedures.
I. Determining the empirical formula of a
compound from mass.
In an upcoming laboratory activity, you will
be asked to experimentally determine the empirical formula of a compound that you produce
from the process of burning magnesium. This synthesis reaction will produce
magnesium oxide. Your purpose will be to determine the empirical formula of the
magnesium oxide by analyzing the molar ratio by which the elements combine.
Ex. 1. What is the empirical formula for a compound if an
8.1 g sample contains 4.9 g of magnesium and 3.2 g of oxygen?
Solution: Remember that a molar
ratio is a ratio of the number of moles, not the mass. We are not looking for a
ratio of mass to mass, but of moles-moles. Our first step is to determine how many
moles of each element are found in this compound. We do that be dividing the mass of
the given element by its molar mass:
mass of that element in the sample
# of moles of an element = -------------------------------------
molar mass of the element
Given: mass of magnesium = 4.9 g
molar mass of magnesium =
24.3 g
mass of oxygen = 3.2 g
molar mass of elemental
oxygen = 16.0 g (note we use elemental, not diatomic oxygen)
4.9 g
# of moles of magnesium = -------------- = 0.20 moles
24.3 g/mole
3.2 g
# of moles of oxygen = --------------- = 0.20 moles
16.0 g/mole
This problem is easy because, as you can see, the elements are combining in a 0.20 to 0.20
ratio, or in simpler terms, a 1:1 ratio. This means that for every one atom of
magnesium, there is one atom of oxygen in the compound. This gives us an empirical
formula of MgO.
Answer: MgO
Now, let us try a problem that is a little bit harder.
Example 2. Calculate the empirical formula of
a compound
that is made from 1.67 g of cerium and 4.54 g of iodine.
Solution: We will solve the problem the same way as example
1, with one exception. After we find out the number of moles of each element
present, we will convert are results into a simple, whole number ratio.
mass of that element in the sample
# of moles of an element = -------------------------------------
molar mass of the element
Given: mass of cerium = 1.67 g
molar mass of cerium =
140 g
mass of iodine = 4.54 g
molar mass of iodine = 127 g
1.67 g
# of moles
of cerium = ---------- = 0.0119 moles
140 g/mole
4.54 g
# of moles
of iodine = ------------- = 0.0357 moles
127 g/mole
Now, to turn the number of moles into a simple whole number
ratio, divide the smaller number of moles (0.0119) into both values:
# of moles of Ce = 0.0119 moles
-------------- = 1
0.0119 moles
# of moles of I = 0.0357 moles
--------------- = 3
0.0119 moles
This means that for every one atom of Ce there are 3 atoms of I,
so our empirical formula must be CeI3.
Answer: CeI3
II. Determining empirical formula from percentage
composition.
This type of problem is essentially the same as
the problems described above, with one slight difference. In these problems you
start with a percentage composition instead of mass. However, if you assume that you
are studying a 100g sample, you can easily change percentages to grams. Then solve
the problems exactly as shown above.
Example 1. Determine the empirical formula of a compound that
contains 36.5% sodium, 25.4% sulfur, and 38.1% oxygen.
Solution: By assuming that we can study a 100g sample of the compound, we can
change % to grams. so:
Example 1. Determine the empirical formula of a compound that
contains 36.5g sodium, 25.4g sulfur, and 38.1g oxygen.
Now we can solve them by finding the molar ratio by which the elements combine.
mass of that element in the sample
# of moles of an element = -------------------------------------
molar mass of the element
Given: mass of sodium = 36.5 g
molar mass of sodium = 23.0 g
mass of sulfur = 25.4 g
molar mass of sulfur = 32.1 g
mass of oxygen = 38.1 g
molar mass of oxygen = 16.0 g
36.5 g
# of moles of sodium = ------------- = 1.59 moles
23.0g/mole
25.4g
# of moles of sulfur = --------------- = 0.791 moles
32.1 g/mole
38.1 g
# of moles of oxygen = --------------- = 2.38 moles
16.0 g/mole
Now, find the simplest whole number ratio by dividing the
smallest number of moles into all three values.
# of moles of Na = 1.59 moles
-------------- = 2.01
0.791 moles
# of moles of S = 0.791 moles
-------------- = 1
0.791 moles
# of moles of O = 2.38 moles
-------------- = 3.01
0.791 moles
The ratio shows that 2 atoms of sodium combine with 1 atom of
sulfur and 3 atoms of oxygen, so our answer is Na2SO3.
Answer: Na2SO3
III. Determining the molecular formula from the empirical
formula and the molecular mass.
In this variation you are given the
empirical formula and the molecular mass of a compound and asked to determine the
molecular mass. This is accomplished by dividing the molecular mass by the mass
shown in the empirical formula. The result is a whole number that is used as a
multiplier for all of the subscripts in the empirical formula.
Example 1. What is the molecular formula of a compound with
an empirical formula of CH2 and a molecular formula of 56.0 u?
Solution: Find the mass of the empirical formula (CH2):
C = 12.0 x 1 atom = 12.0 u
H = 1.01 x 2 atoms = 2.02 u
----------
14.0 u
next, divide that number into the molecular mass:
56.0 u
------- = 4
14.0 u
Now use that number, 4, as a multiplier for the subscripts in the empirical formula:
CH2 x 4 = C4H8
Answer = C4H8
Now you must practice this type of math with the links below.
Determining the
Empirical Formula Quizzes |
| |
Please forward all questions, comments and criticisms to Gregory L. Curran.
© Copyright 2004 Fordham Preparatory School, All Rights Reserved.
Last Modified February 07, 2008 |
