5.5 Molecular FormulaPercent Composition

Review of

Empirical Formula

A compound is found to contain 77.9% I, 22.1 % O . Calculate the empirical formula.

=

=

= mol77.9 g I x 1 mole

g

= mol22.1 g O x 1 mole

g

Molecular Formula

• Actual number of atoms of each element in a molecule.

• Compare it to empirical formula, which is the simplest whole number ratio of atoms in a molecule.

Empirical Formula Molar Mass Molecular Formula Molecular Mass

C4H8NO 86.08 g/mol C12H24N3O3 258.24 g/mol

192.24 g/mol C14H24 96.12 g/molC7H12

360.4 g/molC20H40O572.08 g/mol C4H8O

78.06 g/molC6H613.01 g/molCH

192.24 96.12

= 2360.472.08

= 578.06

13.01= 6

A compound is found to contain 30.8 % C, 4.27 % H, 23.9 % N, and 41.0 % O. The molecular mass is 468.0 g/mole. Determine the empirical and molecular formulas.

= 2

= 1.504

= 5

117.0 g/mole 468.0 g/moleC3H5N2O3 C12H20N8O12

30.8 g C x 1 mol = 2.567 mol

12.0 g4.27 g H x 1 mol = 4.227 mol

1.01 g

23.9 g N x 1 mol = 1.707 mol14.0 g

41.0 g O x 1 mol = 2.563 mol 16.0 g

1.707 mol

1.707 mol

1.707 mol

1.707 mol

= 2.476

= 1

= 1.501

x 4 =

x 2 =

= 3

= 3

How to find the molecular formula

1- Find the empirical formula as you always do.

2- Calculate the molar mass of the empirical formula you got.

3- Divide the molecular mass by the molar mass:Molecular mass

Molar mass

4- Multiply the subscripts in the EF by this number in step(3)

This new formula will be your molecular formula!

TRY:A compound is found to contain 66.35 g C, 17.54 g H, 33.17 g N, and 63.22 g O. The molecular mass is 456.44 g/mole. Determine the empirical and molecular formulas. Hint: use 1.01 g/mol for H.

= 3

= 2.334

= 22

228.22 g/mole 456.44 g/moleC7H22N3O5C14H44N6O10

66.35 g C x 1 mol = 5.529 mol

12.0 g17.54 g H x 1 mol = 17.37 mol

1.01 g

33.17 g N x 1 mol = 2.369 mol14.0 g

63.22 g O x 1 mol = 3.9513 mol 16.0 g

2.369 mol

2.369 mol

2.369 mol

2.369 mol

= 7.332

= 1

= 1.668

x 2 =

x 3 =

= 7

= 5

Another type of questions

• You are not given the molecular mass.

• What to do?!

• Usually these questions are only about gases at STP.

A gas has the empirical formula CH2. If 0.85 L of the gas at STP has a mass of 1.59 g, what is the molecular formula?

What did we just learn?

• We use 22.4 L/mol to help us find the molecular mass

– For gases only

– At STP

• We use the density (g/L) to find molecular mass (g/mole)

Homework

Try in this order!

Page 95

#54, 50, 52.