EMPIRICAL FORMULAEMPIRICAL FORMULA• The empirical formulaempirical formula represents the

smallest ratio of atoms present in a compound.

• The molecular formulamolecular formula gives the total number of atoms of each element present in one molecule of a compound.

The empirical formula is the simplest formula The empirical formula is the simplest formula and the molecular formula is the “true” and the molecular formula is the “true”

formula.formula.

EMPIRICAL FORMULAEMPIRICAL FORMULA

Assume 100g sample Calculate mole ratio

Use Atomic Masses

Mass % of Mass % of elementselements

Grams of Grams of eacheach

elementelement

Moles of Moles of each each

elementelement

Empirical Empirical FormulaFormula

EMPIRICAL FORMULAEMPIRICAL FORMULAStep 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.

EMPIRICAL FORMULAEMPIRICAL FORMULA1. Calculate the empirical formula from a sample containing

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

smallestsmallest43.4% 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 = NaEmpirical Formula = Na22COCO33

EMPIRICAL FORMULAEMPIRICAL FORMULA2. When 8.00 g of calcium metal is heated in air, 11.20 g of

metal oxide is formed. Calculate the empirical formula..

According to the Law of Conservation of mass,

11.20 g Product - 8.00 g Ca = 3.20 g Oxygen (reactive part of air)

smallestsmallest 8.00 g Ca (1 mole / 40 g/mol) = 0.200 moles Ca

3.20 g O (1 mole / 16 g/mol) = 0.200 moles O

0.200 / 0.200 = 1

Empirical Formula = CaOEmpirical Formula = CaO

EMPIRICAL FORMULAEMPIRICAL FORMULA3. 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.

smallestsmallest33.0% 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 = SrClEmpirical Formula = SrCl22 . 6 H . 6 H22OO

EMPIRICAL FORMULA EMPIRICAL FORMULA & Molecular Formula& Molecular Formula

4. 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 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 = CHEmpirical Formula = CH22

Molar mass / empirical mass = multipierMolar mass / empirical mass = multipier

(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.