Document related concepts
no text concepts found
Transcript
```Chemical Quantities
Unit 7
Chapter 10, Section 10.3
Percent Composition and Chemical
Formulas
Objectives
• When you complete this presentation, you will
be able to …
– describe how to calculate the percent by mass of
an element in a compound.
– interpret an empirical formula.
– distinguish between empirical and molecular
formulas.
Introduction
• Chemical formulas tell us about the number of
atoms in a compound.
• In general, there are two kinds of chemical
formulas.
– In molecular formulas, the total number of atoms
in the compound is used.
– In empirical formulas, the lowest whole number
ratio of atoms in the compound is used.
Molecular Formulas
• In molecular formulas, the total number of
atoms in the compound is used.
– For example, benzene has 6 carbon atoms and 6
hydrogen atoms in each molecule.
• Therefore, its molecular formula is C6H6.
Molecular Formulas
• In molecular formulas, the total number of
atoms in the compound is used.
– For example, acetic acid has 2 carbon atoms, 4
hydrogen atoms, and 2 oxygen atoms in each
molecule.
• Therefore, its molecular formula is C2H4O2.
Molecular Formulas
• In molecular formulas, the total number of
atoms in the compound is used.
– For example, propane gas has 3 carbon atoms and
8 hydrogen atoms in each molecule.
• Therefore, its molecular formula is C3H8.
Empirical Formulas
• In empirical formulas, the lowest whole
number ratio of atoms in the compound are
used.
– For example, benzene has 6 carbon atoms and 6
hydrogen atoms in each molecule.
• Therefore, its molecular formula is C6H6 and its
empirical formula is CH (we divide all subscripts by 6).
Empirical Formulas
• In empirical formulas, the lowest whole
number ratio of atoms in the compound are
used.
– For example, acetic acid has 2 carbon atoms, 6
hydrogen atoms, and 2 oxygen atoms in each
molecule.
• Therefore, its molecular formula is C2H4O2. and its
empirical formula is CH2O (we divide all subscripts by
2).
Empirical Formulas
• In empirical formulas, the lowest whole
number ratio of atoms in the compound are
used.
– For example, propane gas has 3 carbon atoms and
8 hydrogen atoms in each molecule.
• Therefore, its molecular formula is C3H8 and its
empirical formula is also C3H8 (there is no common
divisor for all subscripts).
Percent Composition
• If we know the mass of a compound and the
mass of one or more of the elements in the
compound, then we can find the percent
composition of those atoms in the compound.
% mass of an element =
mass of atoms × 100%
mass of compound
Percent Composition
• For example,
– 1.716 g of C, 0.577 g of H, and 2.286 g of O
combine together to form 4.579 g of a compound.
% mass C =
mass of C
1.716 g
× 100% =
× 100% = 37.48%
mass of compound
4.579 g
% mass H =
mass of H
0.577 g
× 100% =
× 100% = 12.60%
mass of compound
4.579 g
% mass O =
mass of O
2.286g
× 100% =
× 100% = 49.92%
mass of compound
4.579 g
Sample Problem 10.9
• When a 13.60 sample of a compound containing only
magnesium and oxygen is decomposed, 5.40 g of oxygen is
obtained. What is the percent composition of this compound?
Known:
mass of compound = 13.60 g
mass of O = 5.40 g
mass of Mg = 13.60 g – 5.40 g = 8.20 g
Unknown:
%Mg = ?%
%O = ?%
% Mg =
%O=
mass of Mg
8.20 g
× 100% =
× 100% = 60.3%
mass of compound
13.60 g
mass of O
5.40 g
× 100% =
× 100% = 39.7%
mass of compound
13.60 g
Percent Composition
• If we know the chemical formula of a
compound, then we can find the percent
composition of each of the atoms in the
compound.
• We use the molar mass of the compound and
the average atomic masses of the atoms in
the compound.
percent composition =
atomic mass of atoms
× 100%
molar mass of compound
Percent Composition
• For example:
– the percent composition of benzene, C6H6, is:
%C =
100%
atomic mass of C atoms
×
molar mass of C6H6
%H =
atomic mass of H atoms
6(1.01 g/mol)
× 100% =
× 100% = 7.76%
molar mass of C6H6
78.12 g/ mol
=
6(12.01
78.12
g/ mol
g/mol)
× 100% = 92.24%
Percent Composition
• For example:
– the percent composition of acetic acid, C2H3O2, is:
%C =
100%
atomic mass of C atoms
×
molar mass of C2H4O2
%H =
atomic mass of H atoms
4(1.01 g/mol)
× 100% =
× 100% = 6.73%
molar mass of C2H4O2
60.06 g/ mol
%O =
atomic mass of O atoms
× 100% =
molar mass of C2H4O2
=
2(12.01
60.06
g/ mol
g/mol)
2(16.00
60.06
g/ mol
g/mol)
× 100% = 39.99%
× 100% = 53.28%
Sample Problem 10.10
• Propane (C3H8), the fuel commonly used in gas grills, is one of
the compounds obtained from petroleum. Calculate the
percent composition of propane
Known:
molar mass of C3H8 = 44.0 g/mol
mass of C = 3 × 12.0 g/mol = 36.0
g/mol
Unknown:
% Mg =
%O=
mass of H = 8 × 1.0 g/mol = 8.0 g/mol
%C = ?%
%H = ?%
mass of C
36.0 g/mol
× 100% =
× 100% = 81.8%
molar mass of C3H8
44.0 g/mol
mass of H
8.0 g/mol
× 100% =
× 100% = 18%
molar mass of C3H8
44.0 g/mol
Percent Composition
Practice Problems: find the percent composition of …
1. C6H14
%C = [(6×12.01)/(86.20)]×100% = 83.60%
%H = [(14×1.01)/(86.20)]×100% = 16.40%
2. NaCl
%Na = [(22.99)/(58.44)]×100% = 39.34%
%Cl = [(35.45)/(58.44)]×100% = 60.66%
3. KNO3
4. CuSO4
5. FeCO3
%K = [(39.10)/(101.11)]×100% = 38.67%
%N = [(14.01)/(101.11)]×100% = 13.86%
%O = [(3×16.00)/(101.11)]×100% = 47.47%
%Cu = [(63.55)/(159.61)]×100% = 39.82%
%S = [(32.06)/(159.61)]×100% = 20.09%
%O = [(4×16.00)/(159.61)]×100% = 40.10%
%Fe = [(55.85)/(115.86)]×100% = 48.20%
%C = [(12.01)/(115.86)]×100% = 10.37%
%O = [(3×16.00)/(115.86)]×100% = 41.43%
Empirical Formulas
• In empirical formulas, the lowest whole
number ratios of atoms in a compound is
used.
– Benzene, C6H6, has an empirical formula of CH.
• A ratio of 6/6 = 1/1.
– Acetic acid, C2H4O2, has an empirical formula of
CH2O.
• A ratio of 2/4/2 = 1/2/1
– Propane, C3H8, has an empirical formula of C3H8.
• A ratio of 3/8 =3/8.
Empirical Formulas
• If we know the percent composition of a
compound, then we can find the empirical
formula of the compound.
– First, we assume that we have 100 g of the
compound and find the mass of each atom in the
compound.
– Second, we find the number of mols of each atom.
– Third, we find the lowest whole number ratio of
mols.
Empirical Formulas
• For example, we have a compound with 52.2%
C, 13.1% H, and 34.7% O.
– First, we assume that we have 100 g of the
compound and find the mass of each atom in the
compound.
• 52.2 g of C
• 13.1 g of H
• 34.7 g of O
Empirical Formulas
• For example, we have a compound with 52.2%
C, 13.1% H, and 34.7% O.
– Next, we we find the number of mols of each
atom.
• nC = mC/MC = 52.2 g/12.0 g/mol = 4.35 mol C
• nH = mH/MH = 13.1 g/1.01 g/mol = 13.0 mol H
• nO = mO/MO = 34.7 g/16.0 g/mol = 2.17 mol O
Empirical Formulas
• For example, we have a compound with 52.2%
C, 13.1% H, and 34.7% O.
– Finally, we find the lowest whole number ratio of
mols.
• nC/nO = 4.35 mol/2.17 mol = 2/1
• nH/nO = 13.0 mol/2.17 mol = 6/1
– This means that there is a ratio of 2:6:1 for C:H:O
• The empirical formula is C2H6O
Empirical Formulas
• For example, we have a compound with 44.9%
K, 18.4% S, and 36.7% O.
– First, we assume that we have 100 g of the
compound and find the mass of each atom in the
compound.
• 44.9 g of K
• 18.4 g of S
• 36.7 g of O
Empirical Formulas
• For example, we have a compound with 44.9%
K, 18.4% S, and 36.7% O.
– Next, we we find the number of mols of each
atom.
• nK = mK/MK = 44.9 g/39.1 g/mol = 1.15 mol K
• nS = mS/MS = 18.4 g/32.1 g/mol = 0.573 mol S
• nO = mO/MO = 36.7 g/16.0 g/mol = 2.29 mol O
Empirical Formulas
• For example, we have a compound with 44.9%
K, 18.4% S, and 36.7% O.
– Finally, we find the lowest whole number ratio of
mols.
• nK/nS = 1.15 mol/0.573 mol = 2/1
• nO/nS = 2.29 mol/0.573 mol = 4/1
– This means that there is a ratio of 2:1:4 for K:S:O
• The empirical formula is K2SO4
Sample Problem 10.11
• A compound is analyzed and found to contain25.9% nitrogen
and 74.1% oxygen. What is the empirical formula of the
compound?
– First, we assume that we have 100 g of the compound and
find the mass of each atom in the compound.
• mN = 25.9 g
• mO = 74.1 g
Sample Problem 10.11
• A compound is analyzed and found to contain25.9% nitrogen
and 74.1% oxygen. What is the empirical formula of the
compound?
– Next, we find the number of mols of each atom.
• nN = mN/MN = 25.9 g/14.0 g/mol = 1.85 mol
• nO = mO/MO = 74.1 g/16.0 g/mol = 4.63 mol
– Finally, we find the lowest whole number ratio of mols.
• nO/nN = 4.63 mol/1.85 mol = 2.5/1 = 5/2
– This means that there is a ratio of 5:2 for O:N
• The empirical formula is N2O5
Empirical Formulas
Practice problems: find the empirical formulas of compounds that have
the following percent compositions.
1. 11.21% H and 88.79% O
H2O
2. 92.24% C and 7.76% H
CH
3. 39.99% C, 6.73% H, and 53.28% O
CH2O
4. 24.27% C, 4.08%H, and 71.65% Cl
CH2Cl
5. 39.96% N, 14.40% H, and 45.64% O
NH5O
Molecular Formulas
• In molecular formulas, the total number of
atoms in the compound is used.
– Benzene has a molecular formula of C6H6.
– Acetic acid has a molecular formula of C2H4O2.
– Propane has a molecular formula of C3H8.
Molecular Formulas
• If we know the empirical formula of a compound,
then we can find the molecular formula of the
compound, if we know the molar mass of the
compound.
– First, we determine the empirical mass of the
compound.
– Second, we divide the molar mass by the empirical
mass to get our multiplier.
– Third, we multiply the subscripts of the empirical
formula by the multiplier to get the molecular
formula.
Molecular Formulas
• For example, we have a compound with a
molecular formula of CH2O and a molar mass
of 180.16 g/mol. What is the molecular
formula of the compound?
– First, we determine the empirical mass of the
compound.
EM = (1×12.01) + (2×1.01) + (1×16.00) = 30.03
Molecular Formulas
• For example, we have a compound with a
molecular formula of CH2O and a molar mass
of 180.16 g/mol. What is the molecular
formula of the compound?
– Next, we divide the molar mass by the empirical
mass to find the multiplier.
M/EM = (180.16)/(30.03) = 5.999333999 ≈ 6
Molecular Formulas
• For example, we have a compound with a
molecular formula of CH2O and a molar mass
of 180.16 g/mol. What is the molecular
formula of the compound?
– Finally, we multiply the subscripts of the empirical
formula by the multiplier to get the molecular
formula.
CH2O ⇒ C6H12O6
Sample Problem 10.12
• Calculate the molecular formula of a compound whose molar
mass is 60.0 g/mol and empirical formula is CH4N.
– First, we determine the empirical mass.
• EM = (1×12.0) + (4×1.0) + (1×14.0) = 30.0
– Next, divide molar mass by empirical mass.
• M/EM = (60.0)/(30.0) = 2
– Finally, multiply to get the molecular formula.
• CH4N ⇒ C2H8N2
Molecular Formulas
Practice problems: find the molecular formulas of compounds that
have the following empirical formulas and molar masses.
1. CH2O; M = 60.0 g/mol
C2H4O2
2. CH2; M = 42.1 g/mol
C3H6
3. NaCO2; M = 134.0 g/mol
Na2C2O4
4. CH2Cl; M = 98.96 g/mol
C2H4Cl2
5. NH5O; M = 35.06 g/mol
NH5O
```
Related documents