Empirical Formulas

How would a scientist determine the identity of a substance that they made?

Percent Composition Mass of element in compound X 100% Total mass of compound Repeat for each element in compound Will add up to 100

Empirical Formula The symbols for the elements combined in the compound, with subscripts showing the smallest whole-number ratio of the different atoms in the compound. Ionic = formula unit Covalent = reduced molecular formula

If you know the percent composition of a compound, then you can determine its empirical formula. We do this by first assuming we have a 100.0 g sample so that the percentage values can be converted to grams. We then convert the grams to moles (to account for atoms having different molar masses) To correct for our assumption of 100.0 g we then divide each number of moles by the smallest number.

Mass composition (in 100.0g) = Composition in moles = What is the empirical formula if a compound contains 32.4% Na, 22.6% S, & 45.0% O? Mass composition (in 100.0g) = 32.4 g Na, 22.6 g S, 45.0 g O Composition in moles = 32.4 g Na x (1 mol Na / 22.99 g Na) = 1.41 mol Na 22.6 g S x (1 mol S / 32.07 g S) = 0.705 mol S 45.0 g O x (1 mol O / 16.00 g O) = 2.81 mol O Smallest whole-number mole ratio 1.41 mol Na / .705 = 2.00 mol Na .705 mol S / .705 = 1.00 mol S 2.81 mol O / .705 = 3.99 mol O

Percent Composition Mass Composition Composition in moles Smallest whole-number mole ratio

A compound’s empirical formula can also be used to determine its molecular formula. x (empirical formula) = molecular formula x = whole-number factor

So… CH  C6H6 Molecular formula mass = 78.11 amu Determine the molecular formula of the compound with an empirical formula of CH and a mass of 78.11 amu. Molecular formula mass = 78.11 amu Empirical formula mass = C + H = 12.01 amu + 1.01 amu = 13.02 amu x (empirical formula) = molecular formula x (13.02 amu) = 78.11 amu x = 78.11 amu / 13.02 amu = 5.999 So… CH  C6H6

In the green workbook: Pg 67 #1 b Pg 77 #1 a & b Pg 79 #1 a & b