Empirical & Molecular Formulas

• Law of Constant Composition a compound contains elements in a certain fixed proportions (ratios), regardless of how the compound is prepared or where it is found in nature.

• If you have one molecule of methane gas, you will always have 1 carbon atoms and 4 hydrogen atoms.

Empirical Formula• Empirical Formula is the formula that gives the

lowest ratio of atoms in a compound. It does not necessarily tell you the exact number of each type of atom.

• Example 1: The percent composition of a compound is 69.9 % iron and 30.1% oxygen. What is the empirical formula of a compound?

Example 1: The percent composition of a compound is 69.9 % iron and 30.1% oxygen.

What is the empirical formula of a compound?

Step 1: List the given valuesFe=69.9% and O = 30.1%

Step 2: Calculate the mass (m) of each element in a 100g sample.

mFe= 69.9 x 100g = 69.9g 100

mO= 30.1 x 100g = 30.1g 100

Step 3: Convert Mass (m) into moles (n)

nFe= m/M = 69.9g/55.86g/mol = 1.25 mol Fe

nO= m/M = 30.1g/16.00g/mol = 1.88 mol OStep 4: State the Amount RationFe : nO

1.25mol : 1.88 mol

Step 5: Calculate lowest whole number ratio1.25mol : 1.88 mol1.25mol 1.25 mol1 : 1.52 : 3 Empirical Formula

is Fe2O3

When you don’t get a whole number, multiply entire ratio by 2, 3, 4 etc. until you get a whole number

Example 2: The percent composition of a compound is

21.6% sodium, 33.3% chlorine, and 45.1% oxygen. What is the

empirical formula of the compound?

Step 1: List the given values

Cl=33.3%, Na = 21.6% and O = 45.1%

Step 2: Calculate the mass (m) of each element in a 100g sample.

mCl= 33.3 x 100g = 33.3g Cl

100

mNa= 21.6 x 100g = 21.6g Na

100

mO= 45.1 x 100g = 45.1g O

100

Step 3: Convert Mass (m) into moles (n)

nCl= m/M = 33.3g/35.5g/mol = 0.94 mol Cl

nNa= m/M = 21.6g/23.0g/mol = 0.94 mol Na

nO= m/M = 45.1g/16.00g/mol = 2.82 mol OStep 4: State the Amount RationFe : nNa : nO

0.94mol : 0.94mol : 2.82 mol

Step 5: Calculate lowest whole number ratio0.94mol : 0.94mol : 2.82 mol0.94mol : 0.94mol : 0.94 mol1 : 1: 3 Empirical Formula

is NaClO3

Molecular Formula

• Molecular Formula of a compound tells you exact number of atoms in one molecule of a compound. This formula may be equal to the empirical formula or may be a multiple of this formula.

• To determine, you need:– The empirical formula– The molar mass of the compound

Molecular Formula- shows the actual number of atoms

Example: C6H12O6

Empirical Formula - shows the ratio between atoms

Example: CH2O

The empirical formula of a compound is CH3O and its molar mass is 93.12g/mol. What is the

molecular formula?Step 1: List given valuesEmpirical Formula=CH3O

Mcompound = 93.12 g/mol

Step 2: Determine the molar mass for the empirical formula, CH3O.

MEmpirical = 12.01g/mol + 3(1.01g/mol) + 16.00g/mol

= 31.04 g/mol

Step 3. Divide the molar mass by the empirical formula molar mass.

= = 3Step 4. Calculate Molecular Formula by

multiplying this number by the empirical formula.

Molecular formula = x (empirical formula)3 x CH3O

Therefore, the molecular formula is C3H9O3

Molecular formula molar massEmpirical formula molar mass

93.12 g/mol31.04 g/mol

Example 2: The percent composition of a compound is determined by a combustion and analyzer is a 40.03% carbon, 6.67% hydrogen,

& 53.30% oxygen. The molar mass is 180.18g/mol. What is the molecular formula

Step 1: List given valuesC= 40.03%, O=53.30%, H=6.67%Mcompound = 180.18 g/mol

Step 2: Calculate the mass of each element in a 100g sample

mC=40.03g mO=53.30g mH=6.67g

Step 3: Convert Mass (m) into moles (n)

nC= m/M = 40.03g/12.01g/mol = 3.33 mol C

nH= m/M = 6.67g/1.01g/mol = 6.60 mol H

nO= m/M = 53.30g/16.00g/mol = 3.33 mol O

Step 4: State the Amount RationC : nH : nO

3.33mol : 6.60mol : 3.33 mol

Step 5: Calculate lowest whole number ratio3.33mol : 6.60mol : 3.33 mol3.33mol : 3.33mol : 3.33 mol

1 : 2: 1Empirical Formula is CH2O

Step 6: Determine the molar mass for the empirical formula

MEmpirical = 12.01g/mol + 2(1.01g/mol) + 16.00g/mol

= 30.03 g/mol

Step 7. Divide the molar mass by the empirical formula molar mass.

= = 6Step 8. Calculate Molecular Formula by

multiplying this number by the empirical formula.

Molecular formula = x (empirical formula)6 x (CH2O)

Therefore, the molecular formula is C6H12O6

Molar massEmpirical formula molar mass

180.18 g/mol30.03 g/mol

Example 3: The percent composition of a compound is determined by a combustion and analyzer is a 32.0% carbon, 6.70% hydrogen, 42.6% oxygen & 18.7% nitrogen. The molar mass is 75.08g/mol. What is the molecular

formula?Calculate the mass of each element in a 100g samplemC=32.0g mO=42.6g mH=6.70g mN=18.7g

Convert Mass (m) into moles (n)

nC= m/M = 32.0g/12.01g/mol = 2.66 mol C

nH= m/M = 6.70g/1.01g/mol = 6.65 mol H

nO= m/M = 42.6g/16.00g/mol = 2.66 mol O

nN= m/M = 18.7g/14.01g/mol = 1.33 mol N

State the Amount RationC : nH : nO : nN

2.66mol : 6.65mol : 2.6 mol: 1.33mol

Step 5: Calculate lowest whole number ratio2.66mol : 6.65mol : 2.6 mol: 1.33mol1.33mol : 1.33mol : 1.33 mol: 1.33mol

2 : 5: 2: 1

Empirical Formula is C2H5O2N

Determine the molar mass for the empirical formula

MEmpirical = 75.08g

Divide the molar mass by the empirical formula molar mass.

=

= 1Calculate Molecular Formula by multiplying

this number by the empirical formula.Molecular formula = x (empirical formula)1 x (C2H5O2N)

Therefore, the molecular formula is C2H5O2N

Molar massEmpirical formula molar mass

75.08 g/mol75.08 g/mol