Slide 1 ___________________________________ 7 Quantitative Composition of Compounds
___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ Black pearls are composed of calcium carbonate, CaCO3.
The pearls can be measured by either weighing or counting.
Foundations of College Chemistry, 14th Ed.
Morris Hein and Susan Arena
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ ___________________________________ Chapter Outline
2 ___________________________________ 7.1 The Mole
___________________________________ 7.2 Molar Mass of Compounds
7.3 Percent Composition of Compounds
___________________________________ 7.4 Calculating Empirical Formulas
___________________________________ 7.5 Calculating the Molecular Formula from the Empirical Formula
___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ The Mole
3 ___________________________________ Individual atoms are tiny and have such a small mass,
more convenient units for atoms are needed to be
useful on the macroscale.
___________________________________ Analogy
Fruit in a supermarket is “counted” by weighing
the mass of fruit.
If the average mass for a piece of fruit is known,
the number of pieces of fruit can be calculated.
Example
___________________________________ ___________________________________ ___________________________________ If one orange has a mass of 186 g, then 75 oranges have what mass?
186 g
= 13,950 g = 13.95 kg
1 orange
Chemists count atoms in a similar way, by weighing.
75 oranges ×
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ Slide 4 ___________________________________ The Mole
The standard unit of measurement for chemistry.
1 mole = 6.022 x
1023
___________________________________ objects
The number represented by 1 mole, 6.022 x 1023,
is also called Avogadro’s number.
___________________________________ Such a large number is useful because even the smallest
amount of matter contains extremely large numbers of atoms.
___________________________________ The mole is similar to other common units of counting.
___________________________________ Example
___________________________________ 1 dozen = 12 objects
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 5 ___________________________________ The Mole
___________________________________ Moles can be used to describe elements, particles
or compounds.
___________________________________ Mole is often abbreviated as mol.
1 mol of atoms = 6.022 x 1023 atoms
___________________________________ 1 mol of molecules = 6.022 x 1023 molecules
___________________________________ 1 mol of electrons = 6.022 x 1023 electrons
Avogadro’s number can be used as a conversion factor.
1 mol
6.022 x 1023 objects
6.022 x
objects
1 mol
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 6 ___________________________________ 1023
___________________________________ ___________________________________ The Mole
___________________________________ How does the mol relate to masses of elements?
___________________________________ The atomic mass of 1 mol of any element is defined
as the amount of that substance that contains the same
number of particles as exactly 12 g of 12C.
___________________________________ ___________________________________ 1 mol of any element contains the same number of atoms,
but can vary greatly in the overall mass.
(Atoms of different elements have different masses)
___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Molar Mass
7 ___________________________________ Molar Mass: the atomic mass of an element or compound
(in grams) which contains Avogadro’s number of particles.
___________________________________ Molar masses are expressed to 4 significant figures in the text.
___________________________________ Determining Molar Mass
Convert atomic mass units on the periodic table to grams
and sum the masses of the total atoms present.
___________________________________ ___________________________________ Example CaF2
Molar Mass CaF2 = 40.08 g + 2(19.00) g = 78.08 g
Ca
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ 2F
___________________________________ Using the Mole and
Molar Mass Concepts
8 ___________________________________ We can use both the mol and molar mass as
conversion factors.
___________________________________ How many moles of lead does 15.0 g of Pb represent?
Solution Map
g Pb
___________________________________ mol Pb
The conversion factor relates g of Pb to moles of Pb.
___________________________________ 1 mol Pb
207.2 g Pb
or
207.2 g Pb
1 mol Pb
(Obtain molar mass from the periodic table.)
___________________________________ Calculate
1 mol Pb
= 7.24 x 10-2 mol Pb
15.0 g Pb ×
207.2 g Pb
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ ___________________________________ Using the Mole and
Molar Mass Concepts
9 ___________________________________ How many moles of mercury does 23.0 g of Hg represent?
a. 4.62 x 103 mol Hg
b. 1.15 x 10-1 mol Hg
c. 1.15 x 101 mol Hg
d. 4.62 x
10-3
mol Hg
___________________________________ Solution Map
g Hg
mol Hg
___________________________________ The conversion factor needed:
___________________________________ 1 mol Hg
200.6 g Hg
or
1 mol Hg
200.6 g Hg
___________________________________ Calculate
23.0 g Hg ×
1 mol Hg
= 1.15 x 10-1 mol Hg
200.6 g Hg
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ Slide ___________________________________ Using the Mole and
Molar Mass Concepts
10 ___________________________________ How many Au atoms are contained in 16.0 g of Au?
Solution Map
g Au
mol Au
___________________________________ atoms Au
Two conversion factors are needed:
___________________________________ 1 mol Au
6.022 x 1023 atoms Au
___________________________________ 1 mol Au
197.0 g Au
and
Calculate
16.0 g Au ×
___________________________________ 6.022 x 1023 atoms Au
1 mol Au
×
197.0 g Au
1 mol Au
___________________________________ = 4.89 x 1022 atoms Au
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Using the Mole and
Molar Mass Concepts
11 ___________________________________ How many Ti atoms are contained in 7.80 g of Ti?
g Ti
b. 2.25 x 1026 atoms Ti
c. 9.81 x
1022
d. 6.20 x
10-22
___________________________________ Solution Map
a. 2.71 x 10-25 atoms Ti
mol Ti
atoms Ti
___________________________________ Two conversion factors are needed:
atoms Ti
atoms Ti
1 mol Ti
1 mol Ti
and
6.022 x 1023 atoms Ti
47.87 g Ti
___________________________________ Calculate
___________________________________ 1 mol Ti
7.80 g Ti ×
47.87 g Ti
6.022 x 1023 atoms Ti
×
1 mol Ti
___________________________________ = 9.81 x 1022 atoms Ti
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Using the Mole and
Molar Mass Concepts
12 ___________________________________ What is the mass of 2.13 x 1018 atoms of Li?
Solution Map
atoms Li
mol Li
___________________________________ grams Li
___________________________________ Two conversion factors are needed:
1 mol Li
6.941 g Li
and
1 mol Li
6.022 x 1023 atoms Li
___________________________________ Calculate
2.13 x 1018
1 mol Li
atoms Li ×
6.022 x 1023 atoms Li
___________________________________ ×
6.941 g Li
1 mol Li
___________________________________ = 2.46 x 10-5 g Li
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Using the Mole and
Molar Mass Concepts
13 ___________________________________ What is the mass of 1.28 x 108 atoms of Ne?
a. 4.29 x 10-15 g Ne
b. 5.35 x 1032 g Ne
c. 3.06 x 10-17 g Ne
d. 1.11 x 1031 g Ne
Calculate
1.28 x
108
___________________________________ Solution Map
atoms Ne
mol Ne
grams Ne
___________________________________ Two conversion factors are needed:
1 mol Ne
1 mol Ne
and
20.18 g Ne
6.022 x 1023 atoms Ne
___________________________________ ___________________________________ 1 mol Ne
20.18 g Ne
atoms Ne ×
×
1 mol Ne
6.022 x 1023 atoms Ne
___________________________________ = 4.29 x 10-15 g Ne
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Using the Mole and
Molar Mass Concepts
14 ___________________________________ What is the mass of 1.05 mol of Ag?
Solution Map
mol Ag
___________________________________ grams Ag
___________________________________ One conversion factor is needed:
1 mol Ag
107.9 g Ag
___________________________________ Calculate
___________________________________ 1.05 mol Ag ×
107.9 g Ag
= 113. g Ag
1 mol Ag
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ ___________________________________ Using the Mole and
Molar Mass Concepts
15 ___________________________________ What is the mass of 8.21 mol of K?
___________________________________ Solution Map
a. 321. g K
mol K
b. 2.10 x 10-2 g K
___________________________________ grams K
One conversion factor is needed:
c. 113. g K
___________________________________ 1 mol K
39.10 g K
d. 1.11 x 1012 g K
___________________________________ Calculate
8.21 mol K ×
39.10 g K
1 mol K
___________________________________ = 321. g K
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Using the Mole and
Molar Mass Concepts
16 ___________________________________ How many hydrogen atoms are in
1.00 moles of H2 molecules?
Solution Map
mol H2
molecules H2
___________________________________ atoms H2
___________________________________ Two conversion factors are needed:
1 mol H2
6.022 x 1023 molecules H2
and
1 molecule H2
2 atoms H
___________________________________ Calculate
1.00 mol H2
___________________________________ 6.022 x 1023 molecules H2
2 atoms H
×
×
1 molecule H2
1 mol H2
___________________________________ = 1.20 x 1024 atoms H2
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Using the Mole and
Molar Mass Concepts
17 ___________________________________ How many sulfur atoms are in 2.27 mol of S8 molecules?
a. 1.33 x 10-23 atoms S8
b. 7.53 x 1022 atoms S8
d. 2.08 x 10-25 atoms S8
1.00 mol S8 ×
mol S8
molecules S8
atoms S8
___________________________________ Two conversion factors are needed:
c. 4.82 x 1024 atoms S8
Calculate
___________________________________ Solution Map
1 mol S8
1 molecule S8
and
6.022 x 1023 molecules S8
8 atoms S
___________________________________ ___________________________________ 6.022 x 1023 molecules S8
8 atoms S
×
1 molecule S8
1 mol S8
___________________________________ = 4.82 x 1024 atoms S8
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 18 ___________________________________ Molar Mass of Compounds
___________________________________ Much like an element,
molar mass can be defined for a compound.
___________________________________ Molar Mass (MM):
mass of one mole of the formula unit of a compound.
___________________________________ The molar mass of a compound is equal to the sum of
the molar masses of all the atoms in the molecule.
___________________________________ ___________________________________ Example H2O
Molar Mass = MMO + 2MMH = 16.00 g + 2(1.008 g) = 18.02 g
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ Slide ___________________________________ Molar Mass of Compounds
19 ___________________________________ What is the molar mass of aluminum hydroxide, Al(OH)3?
___________________________________ Al(OH)3
a. 43.99 g
Using the atomic masses of each element:
___________________________________ 1 Al = 1(26.98 g) = 26.98 g
___________________________________ b. 78.00 g
c. 75.99 g
3 O = 3(16.00 g) = 48.00 g
d. 46.00 g
3 H = 3(1.008 g) = 3.024 g
___________________________________ 78.00 g
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Molar Mass of Compounds
20 ___________________________________ The molar mass of a compound contains Avogadro’s
number of formula units/molecules.
H2
O
H 2O
2 x (6.022 x 1023)
H atoms
6.022 x 1023
O atoms
6.022 x 1023
H2O molecules
2 mol H atoms
1 mol O atoms
1 mol H2O molecules
2 x 1.008 g = 2.016 g H
16.00 g O
18.016 g H2O
___________________________________ ___________________________________ ___________________________________ Reminder: Pay close attention to whether the desired unit
involves atoms or formula units/molecules.
___________________________________ Example Cl2
___________________________________ Contains 2 mol of Cl atoms but only 1 mol of Cl2 molecules.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Molar Mass of Compounds
21 ___________________________________ As for elements, we can use both the mol and molar mass
of formula units/molecules as conversion factors.
___________________________________ How many moles of NaCl are there in 253 g of NaCl?
Solution Map
g NaCl
___________________________________ mol NaCl
To convert between g of NaCl and moles,
we must first calculate the molar mass of NaCl.
MM = 22.99 g + 35.45 g =
Calculate
253. g NaCl ×
1 mol NaCl
58.44 g NaCl
___________________________________ 58.44 g NaCl
___________________________________ = 4.33 mol NaCl
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Molar Mass of Compounds
22 ___________________________________ How many moles of TiCl4 are there in
12.5 g of titanium(IV) chloride?
a. 0.0659 mol TiCl4
___________________________________ Solution Map
b. 0.0321 mol TiCl4
g TiCl4
c. 2.37 x 103 mol TiCl4
d. 1.01 mol TiCl4
___________________________________ mol TiCl4
___________________________________ MMTiCl4 = 47.87 g + 4(35.45) g
= 189.7 g TiCl4
___________________________________ Calculate
12.5 g TiCl4 ×
1 mol TiCl4
= 0.0659 mol TiCl4
189.7 g TiCl4
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ ___________________________________ Molar Mass of Compounds
23 ___________________________________ What is the mass of 3.45 mol of Li2O?
Solution Map
mol Li2O
___________________________________ g Li2O
To convert between mol of Li2O and g,
we must first calculate the molar mass of Li2O.
___________________________________ MMLi2O = 2(6.941) g + 16.00 g =
___________________________________ 29.88 g Li2O
Calculate
3.45 mol Li2O ×
29.88 g Li2O
1 mol Li2O
___________________________________ = 103. g Li2O
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ ___________________________________ Molar Mass of Compounds
24 ___________________________________ What is the mass of 1.23 mol of PH3?
a. 3.62 x 10-2 g PH3
b. 1.22 x 10-6 g PH3
c. 39.33 g PH3
___________________________________ Solution Map
mol PH3
g PH3
___________________________________ MMPH3 = 30.97 g + 3(1.008) g
= 33.99 g PH3
d. 41.8 g PH3
___________________________________ Calculate
___________________________________ 1.23 mol PH3 ×
33.99 g PH3
1 mol PH3
= 41.8 g PH3
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ Slide ___________________________________ Molar Mass of Compounds
25 ___________________________________ How many molecules of H2S are present in 7.53 g of H2S?
How many atoms of H are present in the sample?
___________________________________ Solution Map g H2S
mol H2S
molecules H2S
atoms H
___________________________________ MMH2S = 2(1.008) g + 32.07 g = 34.09 g H2S
Calculate
1 mol H2S
7.53 g H2S ×
34.09 g H2S
×
___________________________________ 6.022 x 1023 molecules H2S
1 mol H2S
___________________________________ = 1.33 x 1023 molecules H2S
1.33 x 1023 molecules H2S ×
2 atoms H
= 2.66 x 1023 atoms H
1 molecule H2S
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 26 ___________________________________ ___________________________________ Molar Mass of Compounds
___________________________________ How many molecules of H2O2 are there in
0.759 g of the compound?
___________________________________ a. 4.29 x 10-23 molecules H2O2
b. 1.55 x 1025 molecules H2O2
c. 3.95 x 1024 molecules H2O2
g H2O2
Solution Map
mol H2O2
molecules H2O2
___________________________________ d. 1.34 x 1022 molecules H2O2
___________________________________ MMH2O2 = 2(1.008) g + 2(16.00) g = 34.02 g H2O2
0.759 g H2O2 ×
1 mol H2O2
6.022 x 1023 molecules H2O2
×
34.02 g H2O2
1 mol H2O2
___________________________________ ___________________________________ = 1.34 x 1022 molecules H2O2
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 27 ___________________________________ Percent Composition of Compounds
___________________________________ Percent = parts per 100 parts
___________________________________ Percent composition: mass percent of each element in
a compound
___________________________________ Molar mass: total mass (100%) of a compound
___________________________________ % composition is independent of sample size
% composition can be determined by:
1. Knowing the compound’s formula or
2. Using experimental data
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ ___________________________________ Slide 28 ___________________________________ Percent Composition from the Compound’s Formula
___________________________________ Two Step Strategy
___________________________________ 1. Calculate the molar mass of the compound.
2. Divide the total mass of each element by the
compound’s molar mass and multiply by 100.
% of the element =
Total element mass
Compound molar mass
___________________________________ ___________________________________ × 100
___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 29 ___________________________________ Percent Composition of Compounds
___________________________________ Calculate the percent composition of K2S.
Step 1 Calculate compound molar mass
___________________________________ MMK2S = 2(39.10) g + 32.07 g = 110.3 g
Step 2 Calculate % composition of each element.
%K=
%S=
2(39.10) g K
110.3 g
32.07 g S
110.3 g
___________________________________ × 100 = 70.90 % K
___________________________________ × 100 = 29.10 % S
___________________________________ Notice the sum of the percentages must equal 100%.
This provides another way of determining the %
composition of a specific element, if the other %s are known.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 30 ___________________________________ ___________________________________ Percent Composition of Compounds
___________________________________ Calculate the percent composition of O in H2O2.
a. 94.07 %
b. 5.93 %
c. 88.9 %
d. 11.1%
___________________________________ Step 1 Calculate molar mass
___________________________________ MMH2O2 = 2(1.008) g + 2(16.00)g = 34.02 g
___________________________________ Step 2 Calculate % composition O
%O=
2(16.00) g O
34.02 g
___________________________________ × 100 = 94.07 % O
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 31 ___________________________________ Percent Composition of Compounds
___________________________________ Calculate the % composition of K2CrO4.
Step 1 Calculate compound molar mass
___________________________________ MMK2CrO4 = 2(39.10) g + 52.00 g + 4(16.00) g = 194.2 g
___________________________________ Step 2 Calculate % composition
%K=
% Cr =
%O=
2(39.10) g K
194.2 g
52.00 g Cr
194.2 g
4(16.00) g O
194.2 g
× 100 = 40.27 % K
___________________________________ × 100 = 26.78 % Cr
___________________________________ × 100 = 32.95 % O
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 32 Percent Composition from
Experimental Data
___________________________________ Two Step Strategy
___________________________________ ___________________________________ 1. Calculate the mass of the compound formed.
2. Divide the mass of each element by the total mass and
multiply by 100.
% of the element =
Total element mass
Total compound mass
___________________________________ ___________________________________ ___________________________________ × 100
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 33 ___________________________________ Percent Composition from
Experimental Data
___________________________________ When heated in air, 1.63 g of Zn reacts with
0.40 g of oxygen to give ZnO. Calculate the percent
composition of the compound formed.
Step 1 Calculate the mass of the compound formed.
___________________________________ ___________________________________ Mass compound = MassZn + MassO = 1.63 g + 0.40 g
= 2.03 g compound
Step 2 Calculate % composition
% Zn =
%O=
1.63 g Zn
2.03 g
0.40 g O
2.03 g
___________________________________ × 100 = 80.3 % Zn
___________________________________ × 100 = 20. % O
100.3%
Total should be +/0.5% of 100
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ Slide ___________________________________ Percent Composition from
Experimental Data
34 ___________________________________ Aluminum chloride forms by reaction of
13.43 g of Al with 53.18 g of chlorine.
What is the percent composition of Cl in the compound?
___________________________________ ___________________________________ a. 53.2 % Step 1 Calculate the mass of the compound
formed.
b. 79.8 %
Mass compound = MassAl + MassCl
c. 20.2 %
= 13.43 g + 53.18 g = 66.61 g compound
d. 46.8 % Step 2 Calculate % composition
% Cl =
53.18 g Cl
66.61 g
35 ___________________________________ ___________________________________ × 100 = 79.8 %
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ ___________________________________ Empirical and Molecular Formula ___________________________________ Empirical Formula: smallest whole number ratio of atoms
in a compound
___________________________________ Molecular Formula: actual formula of a compound.
Represents the total number of atoms
in one formula unit of the compound.
___________________________________ Whole number multiple of the empirical formula
Example
Acetylene (C2H2)
and
___________________________________ Benzene (C6H6)
Both have the same empirical formula CH.
___________________________________ Each compound is a multiple of CH.
Acetylene C2H2 = (CH)2
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 36 ___________________________________ Benzene C6H6 = (CH)6
___________________________________ Empirical and Molecular Formula Formula
Composition
%C
%H
___________________________________ Molar Mass
(g/mol)
CH (empirical formula)
92.3
7.7
13.02
C2H2 (acetylene)
92.3
7.7
26.04 (2 x 13.02)
C6H6(benzene)
92.3
7.7
78.16 (6 x 13.02)
___________________________________ ___________________________________ Each compound has very different chemical and
physical properties even though they share
the same empirical formula.
___________________________________ Compounds with the same empirical formula have the
same percent composition.
___________________________________ Molar mass = molar mass of the empirical unit ×
multiple of the unit
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 37 ___________________________________ Calculating Empirical Formulas
___________________________________ To calculate an empirical formula, you need to know:
___________________________________ 1. The elements present in the compound
___________________________________ 2. The atomic masses of each element
(from the Periodic Table)
___________________________________ 3. The ratio (by mass or %) of the combined elements
___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 38 ___________________________________ Calculating Empirical Formulas
___________________________________ Strategy to Calculate an Empirical Formula:
1. Assume a starting mass of the compound (usually
___________________________________ 100.0 g) and express the mass of each element in grams.
2. Convert g of each element to mol using molar mass.
___________________________________ (These numbers may or may not be whole numbers.)
3. Divide each of the mole amounts from Step 2 by the
___________________________________ smallest mole amount. The new numbers are the
subscripts in the empirical formula.
___________________________________ Special Case:
If fractions are encountered, multiply by a common factor
to provide whole numbers for each subscript.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 39 ___________________________________ ___________________________________ Calculating Empirical Formulas
___________________________________ Calculate the empirical formula for a compound that
contains 11.19% H and 88.79% O.
___________________________________ Step 1 Find amounts of each element
In a 100.0 g sample, there are
___________________________________ 11.19 g H and 88.79 g O
___________________________________ Step 2 Convert g to moles using element molar masses
11.19 g H ×
1 mol H
1.008 g H
= 11.10 mol H
___________________________________ 88.79 g O ×
1 mol O
16.00 g O
= 5.549 mol O
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 40 ___________________________________ Calculating Empirical Formulas
___________________________________ Step 3 Convert to whole numbers by dividing by the
smallest mole amount.
11.10 mol H
5.549 mol O
5.549 mol O
5.549 mol O
___________________________________ ___________________________________ = 2.000
___________________________________ = 1.000
___________________________________ Empirical formula is H2O
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 41 ___________________________________ ___________________________________ Calculating Empirical Formulas
___________________________________ Calculate the empirical formula for a compound that
contains 56.68% K, 8.68% C and 34.73% O.
a. K3C2O3
b. K4C2O6
c. K2CO3
d. KCO2
In a 100.0 g sample, there are
Step 2 Convert g to moles
56.69 g K ×
×
34.73 g O ×
1 mol K
39.10 g K
1 mol C
12.01 g C
1 mol O
16.00 g O
___________________________________ = 1.447 mol K
___________________________________ = 0.723 mol C
___________________________________ = 2.171 mol O
© 2014 John Wiley & Sons, Inc. All rights reserved.
42 ___________________________________ 56.68 g K, 8.68 g C and 34.73 g O
8.68 g C
Slide ___________________________________ Step 1 Find amounts of each element
___________________________________ Calculating Empirical Formulas
___________________________________ Calculate the empirical formula for a compound that
contains 56.68% K, 8.68% C and 34.73% O.
___________________________________ Step 3 Convert to whole numbers by dividing by the
smallest mole amount.
a. K3C2O3
b. K4C2O6
c. K2CO3
d. KCO2
___________________________________ 1.447 mol K
= 2.000
0.723 mol
0.723 mol C
= 1.000
0.723 mol
___________________________________ 2.171 mol O
= 3.000
0.723 mol
___________________________________ ___________________________________ Empirical formula is: K2CO3
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ Calculating Empirical Formulas
43 ___________________________________ Calculate the empirical formula for a compound that
contains 2.233 g Fe and 1.926 g S?
a. FeS2
Step 1 Find amounts of each element
b.Fe3S2
Already provided in problem
___________________________________ 2.233 g Fe and 1.926 S
c. FeS
d. Fe2S3
___________________________________ Step 2 Convert g to moles
2.233 g Fe
×
1 mol Fe
55.85 g Fe
= 0.03998 mol Fe
1.926 g S
×
1 mol S
32.07 g S
= 0.06006 mol S
___________________________________ ___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide ___________________________________ ___________________________________ Calculating Empirical Formulas
44 ___________________________________ Step 3 Convert to whole numbers by dividing by the
smallest mole amount.
Common Fractions
0.03998 mol Fe
Decimal
Fraction
= 1.000 × 2 = 2.000
0.03998 mol
0.25
1/4
0.06006 mol S
0.03998 mol
= 1.502 × 2 = 3.000
Empirical formula is Fe2S3
0.33…
1/3
0.5
1/2
0.66….
2/3
0.75
3/4
___________________________________ ___________________________________ ___________________________________ To get a whole number, multiply the decimal by the
corresponding number in the denominator of the fraction.
Example After dividing, you get 0.75 (=3/4)
Multiply by the denominator 4(0.75) = 4(3/4) = 3
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 45 ___________________________________ ___________________________________ Calculating the Molecular Formula from the Empirical Formula
___________________________________ If molar mass is known, the molecular formula can be
calculated from the empirical formula.
___________________________________ ___________________________________ ___________________________________ Molecular formula is a multiple of the empirical formula.
Need to determine the value of n.
___________________________________ Solving for n
n =
Molar mass
Mass of empirical formula
___________________________________ = number of empirical units
in the molecular formula
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ Slide 46 ___________________________________ Calculating the Molecular Formula from the Empirical Formula
___________________________________ The molecular formula can be calculated from the
empirical formula if the compound’s molar mass is known.
___________________________________ Molecular formula = multiple of the empirical formula
___________________________________ (EF)n = MF
___________________________________ Determining the multiple n gives the molecular formula
n =
___________________________________ Molar mass
= number of empirical units
Mass of empirical formula
in the molecular formula
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 47 ___________________________________ ___________________________________ Calculating the Molecular Formula from the Empirical Formula
___________________________________ A compound with the empirical formula NH2
was found to have a molar mass of 32.05 g.
What is the molecular formula?
n =
___________________________________ Molar mass
Mass of empirical formula
n =
___________________________________ = number of empirical units
in the molecular formula
32.05
14.01 + 2(1.008)
___________________________________ = 2
___________________________________ ___________________________________ Molecular formula = (NH2)2 = N2H4
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 48 Calculating the Molecular Formula from the Empirical Formula
___________________________________ A compound with the empirical formula NO2 was found
to have a molar mass of 92.00 g. What is the molecular
formula?
___________________________________ a) NO2
b)N2O4
n =
92.00 g
14.01 + 2(16.00) g
___________________________________ ___________________________________ ___________________________________ = 2
c) N3O6
d) N4O8
___________________________________ Molecular formula = (NO2)2 = N2O4
___________________________________ © 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 49 ___________________________________ Calculating the Molecular Formula from the Empirical Formula
___________________________________ Propylene contains 14.3 % H and 85.7 % C and has a
molar mass of 42.08 g. What is its molecular formula?
___________________________________ Plan Calculate empirical formula and then determine the
molecular formula
___________________________________ Step 1 Find compound masses
In 100.0 g of compound, 14.3 g H and 85.7 g C
Step 2 Convert g to moles
1 mol H
= 14.2 mol H
14.3 g H ×
1.008 g H
___________________________________ 1 mol C
= 7.14 mol C
12.01 g C
___________________________________ 85.7 g C ×
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 50 ___________________________________ ___________________________________ Calculating the Molecular Formula from the Empirical Formula
___________________________________ Step 3 Convert to whole numbers by dividing by the
smallest mole amount.
14.2 mol H
= 1.99
7.14 mol
___________________________________ ___________________________________ Empirical formula = CH2
7.14 mol C
= 1.00
7.14 mol
___________________________________ With EF, calculate the molecular formula
n =
42.08 g
12.01 + 2(1.008) g
___________________________________ = 3
___________________________________ Molecular formula = (CH2)3 = C3H6
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 51 ___________________________________ Calculating the Molecular Formula from the Empirical Formula
___________________________________ Calculate the molecular formula for a compound that
contains 80.0% C and 20.0% H with a molar mass of 30.00 g.
a. CH3
b. CH2
c. C2H6
d. C2H4
___________________________________ Plan Calculate empirical and then
molecular formula
___________________________________ Step 1 Find compound masses
In 100.0 g of compound, 20.0 g H and 80.0 g C
___________________________________ Step 2 Convert g to moles
1 mol H
20.0 g H ×
1.008 g H
___________________________________ 80.0 g C
×
1 mol C
12.01 g C
= 19.8 mol H
___________________________________ = 6.66 mol C
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 52 ___________________________________ Calculating the Molecular Formula from the Empirical Formula
___________________________________ Step 3 Convert to whole numbers by dividing by the
smallest mole amount.
___________________________________ 19.8 mol H
= 2.97
6.66 mol
___________________________________ 6.66 mol C
= 1.00
6.66 mol
Empirical formula = CH3
From empirical formula, calculate the molecular formula
n =
30.00 g
12.01 + 3(1.008) g
___________________________________ ___________________________________ = 2
___________________________________ Molecular formula = (CH3)2 = C2H6
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 53 ___________________________________ Learning Objectives
___________________________________ 7.1 The Mole
___________________________________ Apply the concept of the mole, molar mass,
and Avogadro’s number to solve chemistry problems.
___________________________________ 7.2 Molar Mass of Compounds
Calculate the molar mass of a compound.
___________________________________ 7.3 Percent Composition of Compounds
___________________________________ Calculate the percent composition of a compound from
its chemical composition and from experimental data.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Slide 54 ___________________________________ ___________________________________ Learning Objectives
___________________________________ 7.4 Calculating Empirical Formulas
___________________________________ Determine the empirical formula for a compound
from its percent composition.
___________________________________ 7.5 Calculating the Molecular Formula
from the Empirical Formula
___________________________________ Compare an empirical formula to a molecular formula
and calculate a molecular formula from an
empirical formula, using the molar mass.
© 2014 John Wiley & Sons, Inc. All rights reserved.
___________________________________ ___________________________________