Transcript 7.5 Molecular Formulas

Chapter 7 Chemical
Quantities
7.5
Molecular Formulas
Brightly colored dishes are made of
melamine.
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Relating Molecular and
Empirical Formulas
A molecular formula
• is equal or a multiple of its empirical formula
• has a molar mass that is the product of the
empirical formula mass multiplied by a small
integer
molar mass
= a small integer
empirical mass
• is obtained by multiplying the subscripts in the
empirical formula by the same small integer
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Diagram of Molecular and
Empirical Formulas
A small integer links
• a molecular formula and its empirical formula
• a molar mass and its empirical formula mass
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Some Compounds with
Empirical Formula CH2O
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Calculating a Molecular Formula
from an Empirical Formula
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Finding the Molecular Formula
Determine the molecular formula of a compound that
has a molar mass of 78.11 g and an empirical
formula of CH.
STEP 1 Calculate the empirical formula mass.
Empirical formula mass of CH = 13.02 g
STEP 2 Divide the molar mass by the empirical
formula mass to obtain a small integer.
78.11 g = 5.999 ~ 6
13.02 g
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Finding the Molecular Formula
(continued)
STEP 3 Multiply the empirical formula by the
small integer to obtain the molecular formula.
Multiply each subscript in C1H1 by 6.
Molecular formula = C1x6 H1x6 = C6H6
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Learning Check
A compound has a molar mass of 176.1g and an
empirical formula of C3H4O3. What is its
molecular formula?
1) C3H4O3
2) C6H8O6
3) C9H12O9
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Solution
STEP 1 Calculate the empirical formula mass.
C3H4O3 = 88.06 g/EF
STEP 2 Divide the molar mass by the empirical
formula mass to obtain a small integer.
176.1 g (molar mass)
= 2
88.06 g (empirical formula mass)
STEP 3 Multiply the empirical formula by the
small integer to obtain the molecular formula.
molecular formula = 2 x empirical formula
C3x2H4x2O3x2
= C6H8O6
(2)
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Molecular Formula
A compound contains 24.27% C, 4.07% H, and
71.65% Cl. The molar mass is about 99 g. What
are the empirical and molecular formulas?
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Solution
STEP 1 Calculate the empirical formula mass.
24.27 g C x 1 mol C
12.01 g C
= 2.021 mol of C
4.07 g H
= 4.04 mol of H
x 1 mol H
1.008 g H
71.65 g Cl x 1 mol Cl
35.45 g Cl
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= 2.021 mol of Cl
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Solution (continued)
2.021 mol C
2.021
4.04 mol H
2.021
2.02 mol Cl
2.021
=
1 mol of C
=
2 mol of H
=
1 mol of Cl
Empirical formula = C1H2Cl1 = CH2Cl
Empirical formula mass (EM) CH2Cl = 49.48 g
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Solution (continued)
STEP 2 Divide the molar mass by the empirical
formula mass to obtain a small integer.
Molar mass
= 99 g = 2
Empirical formula mass
49.48 g
STEP 3 Multiply the empirical formula by the
small integer to obtain the molecular formula.
2 x (CH2Cl)
C1x2H2x2Cl1x2
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=
C2H4Cl2
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Learning Check
A compound is 27.4% S, 12.0% N, and 60.6
% Cl. If the compound has a molar mass of
351 g, what is the molecular formula?
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Solution
STEP 1 Calculate the empirical formula mass.
In 100 g, there are 27.4 g S, 12.0 g N, and 60.6 g Cl.
27.4 g S x 1 mol S = 0.854 mol of S
32.07 g S
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12.0 g N x
1 mol N = 0.857 mol of N
14.01 g N
60.6 g Cl x
1mol Cl = 1.71 mol of Cl
35.45 g Cl
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Solution (continued)
STEP 2 Divide the molar mass by the
empirical formula mass to obtain a small
integer.
0.854 mol S
= 1.00 mol of S
0.854
0.857 mol N
= 1.00 mol of N
0.854
1.71 mol Cl
= 2.00 mol of Cl
0.854
empirical formula
= SNCl2
empirical formula mass = 116.98 g
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Solution (continued)
STEP 3 Multiply the empirical formula by the
small integer to obtain the molecular
formula.
Molar mass
= 351 g
Empirical formula mass
116.98 g
=3
Molecular formula = (SNCl2)3 = S3N3Cl6
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Concept Map
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