Empirical Formulas Empirical formulas: smallest whole number ratio of atoms present in a substance Molecular formula: actual number of each type of atom present in a molecule Many molecular compounds have different empirical and molecular formulas N2O4 (molecular) NO2 (empirical)

Empirical Formulas The empirical formula for a molecular compound can be found by dividing all subscripts in the molecular formula by the greatest common factor. Examples: C6H12O6 C4H10 Ca2SO4

Percent Composition Empirical formulas are generally obtained by determining the percent composition of a compound: Percent composition: the percentage of the mass contributed by each element in a substance % Element X = (# atoms of X)(AW) x 100% FW of compound

% C = (# atoms of C)(AW) x 100% Percent Compositon Example: Calculate the % composition of C2H4O2 (i.e find %C, %H, and %O). % C = (# atoms of C)(AW) x 100% FW of compound First, find the FW: FW = 2(12.0 amu) + 4(1.0 amu) + 2(16.0 amu) FW = 60.0 amu

Percent Composition Calculate the % composition for each element. % C = 2(12.0 amu) x 100% = 40.0% C 60.0 amu % H = 4(1.0 amu) x 100% = 6.7 % H % O = 2(16.0 amu) x 100% = 53.3 % O

Percent Composition N2O4 Example: Calculate the % nitrogen in dinitrogen tetroxide. N2O4 FW = 2(14.0 amu) + 4(16.0 amu) = 92.0 amu % N = 2(14.0 amu) x 100% = 30.4% 92.0 amu

Calculating Empirical Formulas Percent composition data is commonly used to determine the empirical formula of a compound: Four steps: percent to mass mass to mole divide by smallest multiple ‘til whole

Calculating Empirical Formulas Example: Calculate the empirical formula for a substance that contains 34.63% C, 3.875% H and 61.50% O. Step 1: % to mass If you assume 100.00 g of substance, then 34.63 % C  34.63 g C 3.875 % H  3.875 g H 61.50 % O  61.50 g O

Calculating Empirical Formulas Step 2: Mass to moles # moles C = 34.63 g C x 1 mole C = 2.883 mol C 12.01 g C # moles H = 3.875 g H x 1 mole H = 3.844 mol H 1.008 g H # moles O = 61.50 g O x 1 mole O = 3.844 mol O 16.00 g O

Calculating Empirical Formulas Step 3: Divide by smallest (this gives the molar ratio of the elements) C: 2.883 moles C = 1.000 2.883 moles C H: 3.844 mol H = 1.334 2.883 mol C O: 3.884 mol O = 1.334

Calculating Empirical Formulas Step 4: Multiply ‘til Whole (this step is necessary only when step 3 does not give whole number molar ratios) C: 2.883 moles C = 1.000 2.883 moles C H: 3.844 mol H = 1.334 2.883 mol C O: 3.884 mol O = 1.334 x 3 = 3.000 mol C x 3 = 4.000 mol H x 3 = 4.000 mol O

Calculating Empirical Formulas Use these numbers to write the empirical formula: 3.000 mol C 4.000 mol H C3H4O4 4.000 mol O

Calculating Empirical Formulas Example: An iron compound contains 69.943 % Fe and 30.057 % O. Calculate its empirical formula. Step 1: % to mass 69.943 % Fe  69.943 g Fe 30.057 % O  30.057 g O

Calculating Empirical Formulas Step 2: Mass to Moles mol Fe = 69.943 g Fe x mol Fe = 1.2524 mol Fe 55.845 g Fe mol O = 30.057 g O x mol O = 1.8786 mol O 15.9994 g O

Calculating Empirical Formulas Step 3: Divide by smallest O: 1.8786 mol O = 1.5000 1.2524 mol Fe Fe: 1.2524 mol Fe = 1.0000

Calculating Empirical Formulas Step 4: Multiply ‘til whole FeO1.5 doesn’t make sense. 1.5000 mol O 1.0000 mol Fe x 2 = 3.000 mol O Fe2O3 x 2 = 2.000 mol Fe

Using Empirical Formulas to Find Molecular Formulas Empirical formula for a molecule is the smallest whole number ratio of atoms in the molecule. The subscripts in the molecular formula must be some whole number multiple of the subscripts in the empirical formula: CH2O C2H4O2, C3H6O3, C4H8O4 X 2 X 3 X 4

Using Empirical Formulas to Find Molecular Formulas Steps: Find the empirical formula Calculate the formula weight for the empirical formula. MW = the whole number FW (empirical formula) ratio between MW and FW Multiply the subscripts in the empirical formula by the whole number ratio.

Using Empirical Formulas to Find Molecular Formulas Example: A certain substances has an empirical formula of CH2O. If its molecular weight is 180.0 amu, what is its molecular formula? Step 1: Empirical Formula = given = CH2O

Using Empirical Formulas to Find Molecular Formulas Step 2: FW = 1 (12.0 amu) + 2 (1.0 amu) + 1 (16.0 amu) = 30.0 amu Step 3: Whole number ratio = 180.0 amu = 6 30.0 amu Step 3: Molecular formula: C(1x6)H(2x6)O(1x6) = C6H12O6