% Composition & Empirical Formulas Monday, March 7th, 2016

% Composition = the mass % of each element in a compound Calculating Percent Composition (From Formula) *** Assume 1 mole *** Use Molar Mass For H2O % H = mass H x 100 % O = mass 0 x 100

In-Class Practice Determine the % Composition for: NaBr, Ca3N2, Aluminum Phosphate

Empirical Formulas The empirical formula of a compound is the simplest whole number ratio of the atoms present in the compound. The empirical formula can be found from the percent composition of the compound.

Empirical vs. Molecular Formulas Shows Simplest whole number ratio of atoms for a compound The number and kind of atoms that make up one molecule of a compound How Determined Mole ratio elements Mass ratio (element to compound) Ex: Glucose CH2O C6H12O6 Empirical = Simplest Ratio of atoms Molecular = Actual Ratio of atoms

Steps for Determining the Empirical Formula of a Compound A gaseous compound containing carbon and hydrogen was analyzed and found to consist of 83.65% carbon by mass. Determine the empirical formula of the compound. Step 1: Obtain the mass of each element present (in grams). Assume you have 100 g of the compound. 83.65% C = 83.65 g C (100.00 – 83.65) 16.35% H = 16.35 g H

Steps for Determining the Empirical Formula of a Compound Step 2: Determine the number of moles of each type of atom present.

Steps for Determining the Empirical Formula of a Compound Step 3. Divide the number of moles of each element by the smallest number of moles to convert the smallest number to 1. If all of the numbers so obtained are integers, these are the subscripts in the empirical formula. If one or more of these numbers are not integers, go on to step 4.

Steps for Determining the Empirical Formula of a Compound Step 4. Multiply the numbers you derived in step 3 by the smallest integer that will convert all of them to whole numbers. This set of whole numbers represents the subscripts in the empirical formula.

In-Class Practice Problem *** Assume 100.0 grams *** Use Molar Mass to Calculate Moles *** Determine simplest Mole Ratio 85.62% C 14.38% H

Extra Example #2: What is the Empirical Formula if the % composition is 43.2% K, 39.1% Cl, and some O? 1 mole K 1.10 1.10 43.2 g K -------------- = 1.10 moles K =1.0 mole K 39.098g K 1 mole Cl 1.10 1.10 39.1 g Cl -------------- = 1.10 moles Cl = 1.0 mole Cl 35.453 g Cl 1 mole O 1.10 1.10 17.7 g O ------------- = 1.10 moles O = 1.0 mole O 15.999g O The Empirical Formula is KClO (MW =90.550g/mole) If MW of the real formula is 90.550, what is the actual formula? (90.550)/(90.550) = 1 KClO x 1 = KClO

Extra Example #3 What is the Empirical Formula if the % composition is 40.0% C, 6.7% H, and 53.3% O? 1 mole C 3.33 3.33 40.0 g C -------------- = 3.33 moles C =1.0 mole C 12.011g C 1 mole H 6.64 3.33 6.7 g H -------------- = 6.64 moles H = 2.0 mole H 1.0079g H 1 mole O 3.33 3.33 53.3 g O ------------- = 3.33 moles O = 1.0 mole O 15.999g O The Empirical Formula is CH2O (MW =30.026) If MW of the real formula is 180.155, what is the actual formula? (180.155)/(30.026) = 6 CH2O x 6 = C6H12O6