chemical quantities

Chemical Quantities

Unit 7Chapter 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 = × 100%mass of atomsmass 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 = × 100%mass of C

mass of compound = × 100%1.716 g4.579 g = 37.48%

% mass H = × 100%

mass of Hmass of compound = × 100%0.577 g

4.579 g = 12.60%

% mass O = × 100%mass of Omass of compound = × 100%2.286g

4.579 g = 49.92%

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?

% Mg = × 100%mass of Mgmass of compound = × 100%8.20 g

13.60 g = 60.3%

% O = × 100%mass of Omass of compound = × 100%5.40 g

13.60 g = 39.7%

Known: mass of compound = 13.60 gmass of O = 5.40 gmass of Mg = 13.60 g –

5.40 g = 8.20 g Unknown: %Mg = ?%

%O = ?%

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 = × 100%atomic mass of atomsmolar mass of compound

Percent Composition

• For example:– the percent composition of benzene, C6H6, is:

%C = × 100%

atomic mass of C atomsmolar mass of C6H6

= × 100%6(12.01 g/mol)78.12 g/ mol = 92.24%

%H = × 100%atomic mass of H atomsmolar mass of C6H6

= × 100%6(1.01 g/mol)78.12 g/ mol = 7.76%

Percent Composition

• For example:– the percent composition of acetic acid, C2H3O2, is:

%C = × 100%

atomic mass of C atomsmolar mass of C2H4O2

= × 100%2(12.01 g/mol)60.06 g/ mol = 39.99%

%H = × 100%atomic mass of H atomsmolar mass of C2H4O2

= × 100%4(1.01 g/mol)60.06 g/ mol = 6.73%

%O = × 100%atomic mass of O atomsmolar mass of C2H4O2

= × 100%2(16.00 g/mol)60.06 g/ mol = 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

% Mg = × 100%mass of Cmolar mass of C3H8

= × 100%36.0 g/mol44.0 g/mol = 81.8%

% O = × 100%mass of Hmolar mass of C3H8

= × 100%8.0 g/mol44.0 g/mol = 18%

Known: molar mass of C3H8 = 44.0 g/molmass of C = 3 × 12.0

g/mol = 36.0 g/molmass of H = 8 × 1.0

g/mol = 8.0 g/mol Unknown: %C = ?%%H = ?%

Percent Composition

Practice Problems: find the percent composition of …

1. C6H14

2. NaCl

3. KNO3

4. CuSO4

5. FeCO3

%C = [(6×12.01)/(86.20)]×100% = 83.60%%H = [(14×1.01)/(86.20)]×100% = 16.40%

%Na = [(22.99)/(58.44)]×100% = 39.34%%Cl = [(35.45)/(58.44)]×100% = 60.66%

%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 FormulasPractice problems: find the empirical formulas of compounds that have the following percent compositions.

1. 11.21% H and 88.79% O

2. 92.24% C and 7.76% H

3. 39.99% C, 6.73% H, and 53.28% O

4. 24.27% C, 4.08%H, and 71.65% Cl

5. 39.96% N, 14.40% H, and 45.64% O

H2O

CH

CH2O

CH2Cl

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 C⇒ 6H12O6

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 C⇒ 2H8N2

Molecular FormulasPractice problems: find the molecular formulas of compounds that have the following empirical formulas and molar masses.

1. CH2O; M = 60.0 g/mol

2. CH2; M = 42.1 g/mol

3. NaCO2; M = 134.0 g/mol

4. CH2Cl; M = 98.96 g/mol

5. NH5O; M = 35.06 g/mol

C2H4O2

C3H6

Na2C2O4

C2H4Cl2

NH5O