Section 6.2 Page 268 Empirical and Molecular Formulas

Section 6.2Section 6.2 Page Page 268268

Empirical and Molecular Empirical and Molecular FormulasFormulas

Empirical vs. Molecular Empirical vs. Molecular FormulaFormula

Empirical FormulaEmpirical Formula (E (Eff))

A formula that gives the simplest whole-number A formula that gives the simplest whole-number ratioratio of the atoms of each element in a of the atoms of each element in a compound.compound.

Empirical vs. Molecular Empirical vs. Molecular FormulaFormula

Empirical FormulaEmpirical Formula (E (Eff))

A formula that gives the simplest whole-number A formula that gives the simplest whole-number ratioratio of the atoms of each element in a of the atoms of each element in a compound.compound.

Molecular FormulaMolecular Formula (M (Mff))

A formula that gives the A formula that gives the actualactual number of atoms number of atoms of each element in a compound.of each element in a compound.

CHCH22OOCHCH33OOCH = COOCH = C22HH44OO22

CHCH33OOCHCH33OO

Molecular Molecular FormulaFormula

(actual)(actual)

Empirical FormulaEmpirical Formula

(simplest)(simplest)

HH22OO22 HOHO

CC66HH1212OO66 CHCH22OO

Steps to determining the Steps to determining the empirical formula.empirical formula.

Step 1.Step 1. Find mole amounts. Find mole amounts.

Steps to determining the Steps to determining the empirical formula.empirical formula.

Step 1.Step 1. Find mole amounts. Find mole amounts.

Step 2.Step 2. Divide each mole amount Divide each mole amount by the smallest mole amount.by the smallest mole amount.

Example 1:Example 1:

Determine the empirical formula for a Determine the empirical formula for a compound containing 2.128 g Cl and compound containing 2.128 g Cl and 1.203 g Ca.1.203 g Ca.

1.1. Find mole amounts.Find mole amounts.


nnClCl = = 2.128 g 2.128 g = 0.0600 mol Cl = 0.0600 mol Cl

35.45 g/mol35.45 g/mol



35.45 g/mol35.45 g/mol

nnCaCa = = 1.203 g 1.203 g = 0.0300 mol = 0.0300 mol CaCa

40.08 g/mol40.08 g/mol



35.45 g/mol35.45 g/mol

nnCaCa = = 1.203 g 1.203 g = = 0.03000.0300 mol mol CaCa

40.08 g/mol40.08 g/mol

2.2. Divide each mole by the Divide each mole by the smallest mole.smallest mole.


Cl = Cl = 0.0600 mol Cl0.0600 mol Cl = 2.00 mol Cl = 2.00 mol Cl 0.03000.0300

Ca = Ca = 0.0300 mol Ca0.0300 mol Ca = 1.00 mol = 1.00 mol CaCa

0.03000.0300


Cl = Cl = 0.0600 mol Cl0.0600 mol Cl = 2.00 mol Cl = 2.00 mol Cl 0.03000.0300

Ca = Ca = 0.0300 mol Ca0.0300 mol Ca = 1.00 mol Ca = 1.00 mol Ca 0.03000.0300

Ratio – 1 Ca : 2 ClRatio – 1 Ca : 2 Cl

Empirical Formula = CaClEmpirical Formula = CaCl22

Example 2:Example 2:A compound consists of 50.81% zinc, A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? oxygen. What is the empirical formula?

Example 2:Example 2:A compound consists of 50.81% zinc, A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? oxygen. What is the empirical formula?

• Always assume that you have a 100g sample.

• That way you can convert percent directly to grams...

• In this case you would have 50.81g of Zn, 16.04g of P, and 33.15g of O.

Rhyme

“Percent to mass

Mass to mole

Divide by small

Multiply ‘til whole”

A compound consists of 50.81% zinc, and A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? is the empirical formula?

Atom Mass Mole Mole Ratio

Whole # Ratio

Zn65.38g/mol

P30.97g/mol

O16g/mol



Whole # Ratio

Zn65.38g/mol50.81g

P30.97g/mol16.04g

O16g/mol33.15g



Whole # Ratio

Zn65.38g/mol50.81g nZn = 50.81g

65.38g/mol

= 0.777 mol

P30.97g/mol16.04g

O16g/mol33.15g



Whole # Ratio

Zn65.38g/mol50.81g nZn = 50.81g

65.38g/mol

= 0.777 mol

P30.97g/mol16.04g nP = 16.04g

30.97g/mol

= 0.518 mol

O16g/mol33.15g



Whole # Ratio

Zn65.38g/mol50.81g nZn = 50.81g

65.38g/mol

= 0.777 mol

P30.97g/mol16.04g nP = 16.04g

30.97g/mol

= 0.518 mol

O16g/mol33.15g nO = 33.15g

16g/mol = 2.072

mol



Whole # Ratio

Zn65.38g/mol50.81g nZn = 50.81g

65.38g/mol

= 0.777 mol

0.7770.518= 1.5

P30.97g/mol16.04g nP = 16.04g

30.97g/mol

= 0.518 mol

O16g/mol33.15g nO = 33.15g

16g/mol = 2.072

mol



Whole # Ratio

Zn65.38g/mol50.81g nZn = 50.81g

65.38g/mol

= 0.777 mol

0.7770.518= 1.5

P30.97g/mol16.04g nP = 16.04g

30.97g/mol

= 0.518 mol

0.5180.518= 1

O16g/mol33.15g nO = 33.15g

16g/mol = 2.072

mol



Whole # Ratio

Zn65.38g/mol50.81g nZn = 50.81g

65.38g/mol

= 0.777 mol

0.7770.518= 1.5

P30.97g/mol16.04g nP = 16.04g

30.97g/mol

= 0.518 mol

0.5180.518= 1

O16g/mol33.15g nO = 33.15g

16g/mol = 2.072

mol

2.0720.518= 4



Whole # Ratio

Zn65.38g/mol50.81g nZn = 50.81g

65.38g/mol

= 0.777 mol

0.7770.518= 1.5

1.5 x 2 = 3

P30.97g/mol16.04g nP = 16.04g

30.97g/mol

= 0.518 mol

0.5180.518= 1

O16g/mol33.15g nO = 33.15g

16g/mol = 2.072

mol

2.0720.518= 4



Whole # Ratio

Zn65.38g/mol50.81g nZn = 50.81g

65.38g/mol

= 0.777 mol

0.7770.518= 1.5

1.5 x 2 = 3

P30.97g/mol16.04g nP = 16.04g

30.97g/mol

= 0.518 mol

0.5180.518= 1

1 x 2 = 2

O16g/mol33.15g nO = 33.15g

16g/mol = 2.072

mol

2.0720.518= 4



Whole # Ratio

Zn65.38g/mol50.81g nZn = 50.81g

65.38g/mol

= 0.777 mol

0.7770.518= 1.5

1.5 x 2 = 3

P30.97g/mol16.04g nP = 16.04g

30.97g/mol

= 0.518 mol

0.5180.518= 1

1 x 2 = 2

O16g/mol33.15g nO = 33.15g

16g/mol = 2.072

mol

2.0720.518= 4

4 x 2 = 8



Whole # Ratio

Zn65.38g/mol50.81g nZn = 50.81g

65.38g/mol

= 0.777 mol

0.7770.518= 1.5

1.5 x 2 = 3

P30.97g/mol16.04g nP = 16.04g

30.97g/mol

= 0.518 mol

0.5180.518= 1

1 x 2 = 2

O16g/mol33.15g nO = 33.15g

16g/mol = 2.072

mol

2.0720.158= 4

4 x 2 = 8

Therefore empirical formula would be Zn3P2O8

Example 3:Example 3:The percentage composition of a The percentage composition of a compound is 48.63% carbon, 21.59% compound is 48.63% carbon, 21.59% oxygen, 18.90% nitrogen, and the rest oxygen, 18.90% nitrogen, and the rest hydrogen. Find the empirical formula hydrogen. Find the empirical formula for the compound.for the compound.

Molecular FormulaMolecular Formula

The molecular formula gives the The molecular formula gives the actualactual number of number of atoms of each element in a molecular compound.atoms of each element in a molecular compound.



StepsSteps

1. Find the empirical formula.1. Find the empirical formula.

2. Calculate the Empirical Formula Molar Mass. (M2. Calculate the Empirical Formula Molar Mass. (MEfEf))

3. Divide the Molecular Formula Molar Mass (M3. Divide the Molecular Formula Molar Mass (MMfMf) by ) by the “Mthe “MEfEf”.”.

4. Multiply empirical formula by factor.4. Multiply empirical formula by factor.



StepsSteps

1. Find the empirical formula.1. Find the empirical formula.

2. Calculate the Empirical Formula Molar Mass. (M2. Calculate the Empirical Formula Molar Mass. (MEfEf))

3. Divide the Molecular Formula Molar Mass (M3. Divide the Molecular Formula Molar Mass (MMfMf) by ) by the “Mthe “MEfEf”.”.

4. Multiply empirical formula by factor.4. Multiply empirical formula by factor.

Note: The molecular formula molar mass (MNote: The molecular formula molar mass (MMfMf) is always given ) is always given in the question. in the question.

Example 1:

Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH2O3.

Example 1:


Step 2. “MEf” = 62.03 g

Example 1:


Step 2. “MEf” = 62.03 g

Step 3. MMf/MEf = 124.06/62.03 = 2

Example 1:


Step 2. “MEf” = 62.03 g

Step 3. MMf/MEf = 124.06/62.03 = 2

Step 4. 2(CH2O3) = C2H4O6

Example 1:


Step 2. “MEf” = 62.03 g

Step 3. MMf/MEf = 124.06/62.03 = 2

Step 4. 2(CH2O3) = C2H4O6

Therefore the Molecular formula (Mf) is C2H4O6

Example 2:Example 2:

A compound containing carbon, A compound containing carbon, hydrogen, iodine and oxygen is found hydrogen, iodine and oxygen is found to be 18.0% carbon, 2.5% hydrogen, to be 18.0% carbon, 2.5% hydrogen, 63.5% iodine, and 16.0% O. The molar 63.5% iodine, and 16.0% O. The molar mass of this compound is known to be mass of this compound is known to be 400 g/mol. What is its molecular 400 g/mol. What is its molecular formula?formula?

1. Find the empirical formula.

2. Then you can find the molecular formula.

Mass Spectrometry Demo!Mass Spectrometry Demo!

Poison Identification.Poison Identification.

Acetone: C3H6O

Hydrogen cyanide: HCN

Ethylene Glycol: C2H6O2

Isopropanol: C3H8O

Methanol: CH4O

Sodium hypochlorite: NaOCl

Acetaminophen: C8H9NO2

Documents

Section 6.2 Page 268 Empirical and Molecular Formulas