EMPIRICAL FORMULA • The empirical formula represents the

smallest ratio of atoms present in a

compound.

• The molecular formula gives the total

number of atoms of each element present

in one molecule of a compound.

The empirical formula is the simplest

formula and the molecular formula is the

“true” formula.

EMPIRICAL FORMULA

Assume 100g sample Calculate mole ratio

Use Atomic Masses

Mass % of

elements

Grams of

each

element

Moles of

each

element

Empirical

Formula

% = mass

mass to moles

divide by smallest

multiply till whole!

Step 1: If given the % composition, assume a 100g

sample then convert % to grams.

Step 2: Use the atomic masses to convert grams to

moles.

Step 3: Divide the moles of each element by the

SMALLEST mole fraction.

Step 4: The results from step 3 should be a whole

number, if not, make it so by multiplying by a

common factor. Helpful Hint: if a # has a fraction:

.50 X by 2 .25 & .75 X by 4 .33 & .66 X by 3

EMPIRICAL FORMULA 1. Calculate the empirical formula from a sample

containing 43.4% Na, 11.3% C, and 45.3% O.

smallest

43.4% 43.4 g Na (1 mole / 23 g/mol) =1.887 moles Na

11.3% 11.3 g C (1 mole / 12 g/mol) = 0.9417 moles C

45.3% 45.3 g O (1 mole / 16 g/mol) = 2.831 moles O

1.887/0.9417 =2.00 Na

2.831/0.9417 = 3.00 O .

9417/.9417 = 1.00 C

Empirical Formula = Na2CO3

EMPIRICAL FORMULA 3. A compound was found to have a composition of 33.0 % Sr, 26.8

% Cl, and 40.2 % water. Calculate the empirical formula of this

hydrate.

smallest

33.0% 33.0 g Sr (1 mole/87.6 g/mol) = 0.3767 moles Sr

26.8% 26.8 g Cl (1 mole/35.45 g/mol) = 0.7560 moles Cl

40.2% 40.2 g H2O (1 mole/18.0g/mol) = 2.233 moles H2O

0.7560 / 0.3767 = 2 Cl 2.233 / 0.3767 = 5.9 = 6 H2O

Empirical Formula = SrCl2 . 6 H2O

EMPIRICAL FORMULA

& Molecular Formula 4. Propylene contains 14.3 % H, 85.7% C, and has a molar mass of

42.0 g/mol. What is its molecular formula?

smallest

14.3% 14.3 g H (1 mole/1.01 g/mol) = 14.19 moles H

85.7% 85.7 g C (1 mole/12.01 g/mol) = 7.142 moles C

14.19 / 7.142 = 1.987 = 2 H

Empirical Formula = CH2

Molar mass

empirical mass (42.0 g/mol / 14.0 g/mol) = 3

3 x CH2 becomes the molecular formula C3H6

PRACTICE PROBLEM #12 ______ 1. Which contains the larger number of MOLES of

atoms?

a) 125.0 g KCl b) 25.0 g CaSO4 c) 17.0 g of N2

______ 2. What is the empirical formula of the compound

whose composition is 39.7% K, 27.8% Mn, and 32.5% O?

______ 3. Determine the empirical formula of a compound

that contains 89.7 % bismuth and 10.3 % oxygen.

______ 4. Write the molecular formula for a compound

that contains 54.5 % C, 9.1% H, and 36.4 % O and has a

molar mass of 132 amu?

A

K2MnO4

Bi2O3

C6H12O3

GROUP STUDY PROBLEM #12 ______ 1. Which contains the larger number of MOLES of

atoms?

a) 125.0 g HBr b) 25.0 g C6H11O6 c) 17.0 g of Br2

______ 2. A sample of a compound weighing 4.18 g

contains 1.67 g of sulfur and the rest is oxygen. What is

the empirical formula?

______ 3. What is the empirical formula of the compound

whose composition is 28.7% K, 1.4% H, 22.8 % P, and

47.1% O?

______ 4. A compound contains 92.3% C and 7.7% H and

has a molar mass of 78.0 g/mol. Determine the

molecular formula.