4.6 MOLECULAR FORMULAS

1. Determine the percent composition of all elements.

2. Convert this information into an empirical formula

3. Find the true number of atoms/ elements in the compound (Molecular Formula)

2

3 Steps for determining Chemical Formulas

4.6 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

• Empirical Formula - shows the ratio between atoms

Example: CH2O• Molecular Formula- shows the actual number of atoms

Example: C6H12O6

Steps to Determine Molecular Formula1. List the given values (if the empirical formula is

not given, you must determine that before moving

on).

2. Determine the molar mass for the empirical

formula.

3. Divide the Molecular molar mass by the

empirical formula molar mass.

4. Calculate Molecular Formula by multiplying this

number by the empirical formula.

Example 1: 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 Molecular molar mass by the empirical formula molar mass.

Step 4. Calculate Molecular Formula by multiplying this number by the empirical formula.

Molecular formula = x (empirical formula)3 x CH3OTherefore, 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 values

C= 40.03%, O=53.30%, H=6.67%Mcompound = 180.18 g/mol

The empirical formula is not given so you must calculate it now. a. Calculate the mass of each element in a 100g sample

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

b. Convert Mass (m) into moles (n)

nC= 40.03g x 1mol/ 12.01g = 3.33 mol C

nH= 6.67g x 1 mol/ 1.01g = 6.60 mol H

nO= 53.30g x 1 mol/ 16.00g = 3.33 mol O

c. State the Amount RationC : nH : nO

3.33mol : 6.60mol : 3.33 mol

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

1 : 2: 1Empirical Formula is CH2O

Step 2: Determine the molar mass for the empirical formula

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

= 30.03 g/mol

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

Step 4. 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 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?

1. List given values C = 32.0% H = 6.70% O = 42.6% N = 18.7%Mformula = 75.08g/mol

*** Empirical Formula is not given so you must now calculate it!

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= 32.0g x 1mol/ 12.01g = 2.66 mol C

nH= 6.70g x 1mol/ 1.01g = 6.65 mol H

nO= 42.6g x 1mol/ 16.00g = 2.66 mol O

nN= 18.7g x 1mol/14.01g = 1.33 mol N

State the Amount RationC : nH : nO : nN

2.66mol : 6.65mol : 2.6 mol : 1.33mol

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

2:5:2:1Empirical Formula is C2H5O2N

2. Determine the molar mass for the empirical formula

MEmpirical = 75.08g

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

=

4. Calculate 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 mass Empirical formula molar mass

75.08 g/mol75.08 g/mol

Assignment• Proceed with the assignment that follows

Molecular Formula Check Point

1. The empirical formula of a compound is CH2. Its molecular mass is 70g/mol. What is its molecular formula?

2. A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60.0g/mol. What is its molecular formula?

Molecular Formula Check Point #2

1. A class of compounds called sodium metaphosphates were used as additives to detergents to improve cleaning ability. One of them has a molecular mass of612g. Analysis shows the composition to be 22.5% Na, 30.4% P, and 47.1 % O. Determine the molecular formula of this compound