Unformatted text preview: 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
composition
PERCENT Percent Composition Percent Composition Consider some of the family of nitrogenoxygen compounds:
NO2, nitrogen dioxide and closely
related, NO, nitrogen monoxide (or
nitric oxide) Consider some of the family of nitrogenoxygen compounds:
NO2, nitrogen dioxide and closely
related, NO, nitrogen monoxide (or
nitric oxide) WEIGHT Ethanol, C2H6O
52.13% C
13.15% H
13.15%
34.72% O
34.72% Percent Composition
Consider NO2, Molar mass = ?
What is the weight percent of N and of
O?
Wt. % N = 14.0 g N
• 1 00% = 30.4 %
46.0 g NO 2 Wt. % O = 2 ( 16.0 g O per mole) x 1 00% = 6 9.6%
46.0 g What are the weight percentages of
N and O in NO? Structure of NO2 Structure of NO2
Chemistry of NO,
nitrogen monoxide 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.
SIMPLEST formula. PROBLEM : A compound of B and H is
81.10% B. What is its empirical
formula? Page 1 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. 1 mol
81.10 g B •
= 7 .502 mol B
10.81 g
18.90 g H • 1 mol
= 1 8.75 mol H
1.008 g A compound of B and H is 81.10% B.
Its empirical formula is B2H5. What is
Its empirical
its molecular formula?
its molecular
Is the molecular formula B 2H5, B4H 10,
B6H15, B8 H20, etc.? B2H6 A compound of B and H is 81.10% B. What is its
empirical formula? Now, recognize that atoms combine in
Now,
atoms the ratio of small whole numbers.
1 atom B + 3 atoms H > 1 molecule BH 3
or
1 mol B atoms + 3 mol H atoms >
1 mol BH 3 molecules
Find the ratio of moles of elements in the
compound. A compound of B and H is 81.10% B. Its empirical
formula is B2H 5. What is its molecular formula? We need to do an EXPERIMENT to find
We
EXPERIMENT
the MOLAR MASS.
Here experiment gives 53.3 g/ mol
Compare with the mass of B 2H5
= 26.66 g/unit
Find the ratio of these masses.
2 units of B2H5
53.3 g/mol
=
1 mol
26.66 g/unit of B2H5 B2H6 is one example of this class of compounds. Molecular formula = B 4H10 Page 2 A compound of B and H is 81.10% B. What is its
empirical formula? Take the ratio of moles of B and H. Always
Take
Always
divide by the smaller number.
divide 18.75 mol H
2.499 mol H
2.5 mol H
=
=
1.000 mol B
1.0 mol B
7.502 mol B
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 Determine the formula of a
Determine the formula of a
compound of Sn and II using the
compound of Sn and using the
following data.
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
used
= 1.947 g
• Mass of Sn remaining
= 0.601 g
0.601
• See p. 133 Tin and Iodine Compound Tin and Iodine Compound 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: Now find the number of moles of I 2 that
combined with 3.83 x 10 3 mol Sn . Mass
of I2 used was 1.947 g. Now find the ratio of number of moles of
moles of I and Sn that combined. 0.455 g Sn • 1 mol
= 3.83 x 103 mol Sn
118.7 g 1.947 g I2 • 1 mol
= 7.671 x 103 mol I2
253.81 g This is equivalent to 2 x 7.671 x 10 3
or 1.534 x 102 mol iodine atoms Page 3 1.534 x 10 2 mol I
3.83 x 103 mol Sn = 4.01 mol I
1.00 mol Sn Empirical formula is SnI4
Empirical
SnI ...
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 Spring '08
 Staff
 Chemistry, Mole, Nitric oxide, mol, Nitrogen dioxide

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