Empirical Formulas

A. Empirical formula: The simplest form of a compound

                a. They show only the simplest ratios, not the actual atoms

                                i. The empirical formula for chlorine gas is Cl, not Cl₂

                                ii. dinitrogen tetraoxide

                                                -It does not equal N₂O₄

                                                -To write it properly you have to only use the ratio;
                                                 for every nitrogen there are two oxygen atoms

                                                -The proper empirical formula is NO₂

B.  Molecular formula: Give the actual number of atoms

Ex.

Molecular Formula	Empirical Formula
P4O10	P2O5
C10H22	C5H11
C6H18O3	C2H6O

C. To determine the empirical formula we need to know the ratio of each element

Atom	Mass	Molar Mass	Moles	Moles/ Smallest Mole	Ratio

Ex. A sample of an unknown  compound is found to contain 8.4 g  of "C", 2.1 g of "H, and 5.6 g of "O". Determine the empirical formula.

Atom	Mass	Molar Mass	Moles (mass) x ( 1 mole/molar mass)	Moles/ Smallest Mole	Ratio
C	8.4 g	12.0	(2.4g) x (1 mol/ 12.0) = 0.7 mol	(0.7 mol/ 0.35 mol) = 1	2
H	2.1 g	1.0	(2.1 g) x (1 mol / 1.0) = 2.1 mol	(2.1 mol/ 0.35) = 6	6
O	5.6 g	16.0	(5.6 g ) x (1 mol/ 16.0) = 0.35 mol	(0.35 mol/ 0.35 mol) = 1	1

Empirical Formula: C₂H₆O

Ex. Find the empirical formula of a compound, given that a 48.5 g sample of the compound contains 1.75g of carbon and 46.75 g of bromine.

Atom	Mass	Molar Mass	Moles (mass) x ( 1 mole/molar mass)	Moles/ Smallest Mole	Ratio
C	1.75 g	12.0	(1.75 g) x (1 mol/ 12.0) = 0.145 mol	(0.145 mol/  0.145 mol) = 1	1
Br	46.75 g	79.9	(46.75 g) x (1 mol/ 79.9) = 0.585 mol	(0.585 mol / 0.145 mol) = 4	4

Empirical Formula: CBr₄

                       

                a. The simplest ratio may be decimals. For certain decimals you need to multiply
                 EVERYTHING by a common number.

               

Decimal	Multiplying Coefficient
0.5	2
0.33 or 0.66	3
0.25 or 0.75	4
0.2, 0.4, 0.6, 0.8	5

Ex.  Determine the empirical formula of a compound that contains 2.16 g of aluminum, 3.85 g of sulfur, and 7.68 g of oxygen.

Atom	Mass	Molar Mass	Moles (mass) x ( 1 mole/molar mass)	Moles/ Smallest Mole	Ratio
Al	2.16 g	27.0	(2.16 g) x (1 mol/ 27.0) = 0.08 mol	(0.08 mol/ 0.08 mol) = 1	1( x 2) = 2
S	3.85 g	32.1	(3.8 g) x (1 mol/ 32.1) = 0.119 mol	(0.119 mol/ 0.08 mol) = 1.5	1.5 (x 2) = 3
O	7.68 g	16.0	(7.68 g) x (1 mol/ 16.0) = 0.48 mol	(0.48 mol/ 0.08 mol) = 6	6 (x 2) = 12

Empirical Formula: Al₂S₃O₁₂

_-Nicole