Empirical and Molecular Formulas

Empirical and Molecular
Formulas
A pure compound always consists of the
same elements combined in the same
proportions by weight.
Therefore, we can express molecular
composition as PERCENT BY
WEIGHT
Ethanol, C2H6O
52.13% C
13.15% H
34.72% O





Percent Composition
Consider some of the family of nitrogenoxygen
compounds:
NO2, nitrogen dioxide and closely
related, NO, nitrogen monoxide (or
nitric oxide)
Structure of NO2




Percent Composition
Consider some of the family of nitrogenoxygen
compounds:
NO2, nitrogen dioxide and closely
related, NO, nitrogen monoxide (or
nitric oxide)
Structure of NO2
Chemistry of NO,
nitrogen monoxide



Percent Composition
Consider NO2, Molar mass = ?
What is the weight percent of N and of
O?
Wt. % O 2 (16.0 g O per mole)
46.0 g
x 100% 69.6%
Wt. % N =
14.0 g N
46.0 g NO2
• 100% = 30.4 %
What are the weight percentages of
N and O in NO?


Determining Formulas
In chemical analysis we determine the % by
weight of each element in a given amount of
pure compound and derive the
EMPIRICAL or SIMPLEST formula.
PROBLEM: A compound of B and H is
81.10% B. What is its empirical
formula?



A compound of B and H is 81.10% B. What is its
empirical formula?
• Because it contains only B and H, it
must contain 18.90% H.
• In 100.0 g of the compound there are
81.10 g of B and 18.90 g of H.
• Calculate the number of moles of each
constitutent.



A compound of B and H is 81.10% B. What is its
empirical formula?
Calculate the number of moles of each
element in 100.0 g of sample.
81.10 g B •
1 mol
10.81 g
= 7.502 mol B
18.90 g H •
1 mol
1.008 g
= 18.75 mol H



A compound of B and H is 81.10% B. What is its
empirical formula?
Now, recognize that atoms combine in
the ratio of small whole numbers.
1 atom B + 3 atoms H --> 1 molecule BH3
or
1 mol B atoms + 3 mol H atoms --->
1 mol BH3 molecules
Find the ratio of moles of elements in the
compound.


A compound of B and H is 81.10% B. What is its
empirical formula?
But we need a whole number ratio.
2.5 mol H/1.0 mol B = 5 mol H to 2 mol B
EMPIRICAL FORMULA = B2H5


A compound of B and H is 81.10% B.
Its empirical formula is B2H5. What is
its molecular formula?
Is the molecular formula B2H5, B4H10,
B6H15, B8H20, etc.?
B2H6 is one example of this class of compounds.





A compound of B and H is 81.10% B. Its empirical
formula is B2H5. What is its molecular formula?
We need to do an EXPERIMENT to find
the MOLAR MASS.
Here experiment gives 53.3 g/mol
Compare with the mass of B2H5
= 26.66 g/unit
Find the ratio of these masses.
53.3 g/mol
26.66 g/unit of B2H5
=
2 units of B2H5
1 mol
Molecular formula = B4H10


Determine the formula of a
compound of Sn and I using the
following data.
• Reaction of Sn and I2 is done using
excess Sn.
• Mass of Sn in the beginning = 1.056 g
• Mass of iodine (I2) used
= 1.947 g
• Mass of Sn remaining
= 0.601 g
• See p. 133






Tin and Iodine Compound

Find the mass of Sn that combined with
1.947 g I2.
Mass of Sn initially = 1.056 g
Mass of Sn recovered = 0.601 g
Mass of Sn used = 0.455 g
Find moles of Sn used:
0.455 g Sn •
1 mol
118.7 g
= 3.83 x 10-3 mol Sn



Tin and Iodine Compound
Now find the number of moles of I2 that
combined with 3.83 x 10-3 mol Sn. Mass
of I2 used was 1.947 g.
1.947 g I2 •
1 mol
253.81 g
= 7.671 x 10-3 mol I2
This is equivalent to 2 x 7.671 x 10-3
or 1.534 x 10-2 mol iodine atoms



Tin and Iodine Compound
Now find the ratio of number of moles of
moles of I and Sn that combined.
1.534 x 10-2 mol I
3.83 x 10-3 mol Sn
=
4.01 mol I
1.00 mol Sn
Empirical formula is SnI4