Chapter 8 Empirical & Molecular Formulas. Types of Formulas The molecular formula of a compound...

Chapter 8

Empirical & Molecular Formulas

Types of Formulas

• The molecular formula of a compound indicates the actual number of each atom present in a compound.

• An empirical formula of a compound indicates the smallest whole number ratio of the elements of each atom present in a compound.

Empirical and Molecular Formulas

Substance Molecular Formula

Empirical Formula

Benzene C6H6 CH

Acetylene C2H2 CH

Glucose C6H12O6 CH2O

Water H2O H2O

Chemical Formulas of Chemical Formulas of CompoundsCompounds

• Chemical formulas give the relative numbers of Chemical formulas give the relative numbers of atoms or moles of each element in a compound - atoms or moles of each element in a compound - always a whole number ratio.always a whole number ratio.

• ExampleExample

COCO22 2 atoms of O for every 1 atom of C 2 atoms of O for every 1 atom of C

oror

2 moles of O for every 1 mole of C2 moles of O for every 1 mole of C

• If we know or can determine the relative number If we know or can determine the relative number of moles of each element in a compound, we can of moles of each element in a compound, we can determine a formula for the compound.determine a formula for the compound.

To Calculate an To Calculate an Empirical Empirical FormulaFormula

1.1. Determine the mass in grams of each Determine the mass in grams of each element present, if necessary.element present, if necessary.

2.2. Calculate the number of Calculate the number of molesmoles of of each each element.element.

3.3. Divide each answer from step 2 by the Divide each answer from step 2 by the smallest number of moles to obtain the smallest number of moles to obtain the simplest whole number ratio.simplest whole number ratio.

4.4. If whole numbers are not obtainedIf whole numbers are not obtained** in step 3), in step 3), multiply through by the smallest number multiply through by the smallest number that will give all whole numbersthat will give all whole numbers

** Be careful! Do not round off numbers prematurelyBe careful! Do not round off numbers prematurely

A sample of a brown gas, a major air pollutant, is found to contain A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance.2.34g N and 5.34g O. Determine a formula for this substance.

A sample of a brown gas, a major air pollutant, is found to contain A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance.2.34g N and 5.34g O. Determine a formula for this substance.

2.34g N

5.34g O

A sample of a brown gas, a major air pollutant, is found to contain A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance.2.34g N and 5.34g O. Determine a formula for this substance.

2.34g N

14.0g

mol= 0.167 mol N

5.34g O

16.0g

mol= 0.334 mol O

A sample of a brown gas, a major air pollutant, is found to contain A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance.2.34g N and 5.34g O. Determine a formula for this substance.

2.34g N

14.0g

mol= 0.167 mol N

5.34g O

16.0g

mol= 0.334 mol O

N0.167O0.334

A sample of a brown gas, a major air pollutant, is found to contain A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance.2.34g N and 5.34g O. Determine a formula for this substance.

2.34g N

14.0g

mol= 0.167 mol N

5.34g O

16.0g

mol= 0.334 mol O

N0.167O0.334

0.167 mol N

0.167 0.167

0.334 mol O= 1 mol N = 2 mol O

A sample of a brown gas, a major air pollutant, is found to contain A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance.2.34g N and 5.34g O. Determine a formula for this substance.

2.34g N

14.0g

mol= 0.167 mol N

5.34g O

16.0g

mol= 0.334 mol O

N0.167O0.334

0.167 mol N

0.167 0.167

0.334 mol O= 1 mol N = 2 mol O

NONO22

To Calculate a To Calculate a Molecular Molecular FormulaFormula

1.1. Determine the empirical formula and Determine the empirical formula and calculate its mass.calculate its mass.

2.2. Divide the molecular mass by the mass of Divide the molecular mass by the mass of the empirical formula you determined in step the empirical formula you determined in step 1. Your answer should be rounded to the 1. Your answer should be rounded to the nearest whole number.nearest whole number.

3.3. Multiply the empirical formula by the whole Multiply the empirical formula by the whole number answer found in step 2number answer found in step 2..

A colorless liquid, used in rocket engines has a molar A colorless liquid, used in rocket engines has a molar mass of 92.0 g/mol. What is the mass of 92.0 g/mol. What is the molecular formulamolecular formula of the of the

compound if the empirical formula of NOcompound if the empirical formula of NO22??

A colorless liquid, used in rocket engines has a A colorless liquid, used in rocket engines has a molar mass of 92.0 g/mol. What is the molar mass of 92.0 g/mol. What is the molecular formulamolecular formula

of the compound if the empirical formula of NOof the compound if the empirical formula of NO22??

NO2 = 14.0 + 2(16.0) = 46.0 g/mol

A colorless liquid, used in rocket engines has a A colorless liquid, used in rocket engines has a molar mass of 92.0 g/mol. What is the molar mass of 92.0 g/mol. What is the molecular molecular

formulaformula of the compound if the empirical formula of NOof the compound if the empirical formula of NO22??

NO2 = 14.0 + 2(16.0) = 46.0 g/mol

92.0 g/mol46.0 g/mol

= 2

A colorless liquid, used in rocket engines has a A colorless liquid, used in rocket engines has a molar mass of 92.0 g/mol. What is the molar mass of 92.0 g/mol. What is the molecular formulamolecular formula

of the compound if the empirical formula of NOof the compound if the empirical formula of NO22??

NO2 = 14.0 + 2(16.0) = 46.0 g/mol

92.0 g/mol46.0 g/mol

= 2

2 x NO2 = N2O4

What is the empirical formula of a compound What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S?that contains 53.73% Fe and 46.27% S?

What is the empirical formula of a compound What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S?that contains 53.73% Fe and 46.27% S?

53.73g Fe 46.27g S

What is the empirical formula of a compound What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S?that contains 53.73% Fe and 46.27% S?

53.73g Fe

55.8g

mol= 0.963 mol Fe

46.27g S

32.1g

mol= 1.44 mol S

What is the empirical formula of a compound What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S?that contains 53.73% Fe and 46.27% S?

53.73g Fe

55.8g

mol= 0.963 mol Fe

46.27g S

32.1g

mol= 1.44 mol S

0.963 mol Fe

0.963 mol 0.963 mol

1.44 mol S= 1 Fe = 1.5 S

What is the empirical formula of a compound What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S?that contains 53.73% Fe and 46.27% S?

53.73g Fe

55.8g

mol= 0.963 mol Fe

46.27g S

32.1g

mol= 1.44 mol S

0.963 mol Fe

0.963 mol 0.963 mol

1.44 mol S= 1 Fe = 1.5 S

FeS1.5 x 2 =

Fe2S3

What is the molecular formula of a substance with the empirical formula Fe2S3 and a gram

formula mass of 624 g/mol?

What is the molecular formula of a substance with the empirical formula Fe2S3 and a gram

formula mass of 624 g/mol?

FeFe66SS99

What is the molecular formula of a substance with the empirical formula Fe2S3 and a gram

formula molecular mass of 624 g/mol?

2Fe + 3S = 2(55.8) + 3(32.1) = 207.9 g/mol2Fe + 3S = 2(55.8) + 3(32.1) = 207.9 g/mol

624 g/mol 624 g/mol 207.9 g/mol207.9 g/mol

3 x Fe3 x Fe22SS33 = =

FeFe66SS99

≈ 3

What are the empirical and molecular formulas for What are the empirical and molecular formulas for Ibuprofen if the molar mass is 206 g/mole and the Ibuprofen if the molar mass is 206 g/mole and the

percent composition of 75.7% C, 8.80% H, and 15.5% O?percent composition of 75.7% C, 8.80% H, and 15.5% O?

What are the empirical and molecular formulas for What are the empirical and molecular formulas for Ibuprofen if the molar mass is 206 g/mole and the Ibuprofen if the molar mass is 206 g/mole and the

percent composition of 75.7% C, 8.80% H, and 15.5% O?percent composition of 75.7% C, 8.80% H, and 15.5% O?

75.7g C

8.80g H

15.5g O

What are the empirical and molecular formulas for What are the empirical and molecular formulas for Ibuprofen if the molar mass is 206 g/mole and the Ibuprofen if the molar mass is 206 g/mole and the

percent composition of 75.7% C, 8.80% H, and 15.5% O?percent composition of 75.7% C, 8.80% H, and 15.5% O?

75.7g C mol C12.0g C

= 6.31 mol C

8.80g H mol H1.0 g H

= 8.80 mol H

15.5g O mol O16.0g O

= 0.969 mol O

What are the empirical and molecular formulas for What are the empirical and molecular formulas for Ibuprofen if the molar mass is 206 g/mole and the Ibuprofen if the molar mass is 206 g/mole and the

percent composition of 75.7% C, 8.80% H, and 15.5% O?percent composition of 75.7% C, 8.80% H, and 15.5% O?

75.7g C mol C12.0g C

= 6.31 mol C

8.80g H mol H1.0 g H

= 8.80 mol H

15.5g O mol O16.0g O

= 0.969 mol O

6.31 mol C0.969 mol

= 6.5 C

8.80 mol H0.969 mol

= 9 H

0.969 mol H = 1 O0.969 mol

What are the empirical and molecular formulas for What are the empirical and molecular formulas for Ibuprofen if the molar mass is 206 g/mole and the Ibuprofen if the molar mass is 206 g/mole and the

percent composition of 75.7% C, 8.80% H, and 15.5% O?percent composition of 75.7% C, 8.80% H, and 15.5% O?

75.7g C mol C12.0g C

= 6.31 mol C

8.80g H mol H1.0 g H

= 8.80 mol H

15.5g O mol O16.0g O

= 0.969 mol O

6.31 mol C0.969 mol

= 6.5 C

8.80 mol H0.969 mol

= 9 H

0.969 mol H = 1 O0.969 mol

C6.5H9O x 2 = C13H18O2

What are the empirical and molecular formulas for What are the empirical and molecular formulas for Ibuprofen if the molar mass is 206 g/mole and the Ibuprofen if the molar mass is 206 g/mole and the

percent composition of 75.7% C, 8.80% H, and 15.5% O?percent composition of 75.7% C, 8.80% H, and 15.5% O?

C13H18O2

13(12.0) + 18(1.0) + 2(16.0) = 206g/mol

Therefore the molecular formula is also C13H18O2

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