Percent Composition, Empirical Formulas, Molecular Formulas

Percent Composition

So… Percent Composition Part _______ Percent = x 100% Whole Percent Composition – the percentage by mass of each element in a compound Part _______ Percent = x 100% Whole So… Percent composition of a compound or = molecule Mass of element in 1 mol ____________________ x 100% Mass of 1 mol

Percent Composition Molar Mass of KMnO4 Example: What is the percent composition of Potassium Permanganate (KMnO4)? Molar Mass of KMnO4

Percent Composition Molar Mass of KMnO4 = % K % Mn % O Example: What is the percent composition of Potassium Permanganate (KMnO4)? Molar Mass of KMnO4 = % K % Mn % O

Percent Composition Determine the percentage composition of sodium carbonate (Na2CO3)? Molar Mass Percent Composition % Na = % C = % O =

Using Percent as a converting factor

Finding mass of an element: Directions: Step #1: Find Molar mass of Compound Step #2: Find % of Specific Element. Step #3: Use % to find mass of element in sample.

Percent Composition Calculate the mass of nitrogen in 85.0 g of the amino acid lysine, C6H14N2O2.

Review - Percent Composition Determine the percent composition of ethanol (C2H5OH)? _______________________________________________ Determine the mass of carbon in a 198-g sample of sodium oxalate (Na2C2O4)?

Empirical and Molecular Formulas Determining Formulas

Formulas Percent composition allows you to calculate the simplest ratio among the atoms found in compound. Empirical Formula – formula of a compound that expresses lowest whole number ratio of atoms. Molecular Formula – actual formula of a compound showing the number of atoms present Examples: C4H10 - molecular C6H12O6 - molecular C2H5 - empirical - empirical CH2O

Formulas Is H2O2 an empirical or molecular formula? Molecular, it can be reduced to HO HO = empirical formula

Calculating Empirical Formula An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula. 1. Determine the number of grams of each element in the compound. 2. Convert masses to moles.

Calculating Empirical Formula An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula. 3. Find ratio by dividing each element by smallest amount of moles. 4. Multiply by common factor to get whole number. (cannot have fractions of atoms in compounds)

Calculating Empirical Formula A 4.550 g sample of cobalt reacts with 5.475 g chlorine to form a binary compound. Determine the empirical formula for this compound.

Calculating Empirical Formula When a 2.000 g sample of iron metal is heated in air, it reacts with oxygen to achieve a final mass of 2.573 g. Determine the empirical formula. Fe = 2.000 g O = 2.573 g – 2.000 g = 0.5730 g 1 mol Fe 2.000 g Fe = 0.03581 mol Fe 55.85 g Fe 1 mol O 0.573 g O = 0.03581 mol Fe 16.00 g 1 : 1 FeO 17

Calculating Empirical Formula A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g lead, 0.00672g of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for lead arsenate. 1 mol Pb 1.3813 g Pb = 0.006667 mol Pb 207.2 g Pb 0.00672 gH 1 mol H = 0.00667 mol H 1.008 g H 1 mol As 0.4995 g As = 0.006667 mol As 74.92 g As 1 mol O 0.4267g Fe = 0.02667 mol O 16.00 g O

Calculating Empirical Formula A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g lead, 0.00672g of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for lead arsenate. 0.006667 mol Pb = 1.000 mol Pb 0.006667 0.00667 mol H = 1.00 mol H PbHAsO4 0.006667 0.006667 mol As = 1.000 mol As 0.006667 0.02667 mol O = 4.000 mol O 0.006667

Calculating Empirical Formula from % Composition The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6.

Calculating Molecular Formula A white powder is analyzed and found to have an empirical formula of P2O5. The compound has a molar mass of 283.88g. What is the compound’s molecular formula? Step 3: Multiply Step 1: Molar Mass P = 2 x 30.97 g = 61.94g O = 5 x 16.00g = 80.00 g 141.94 g (P2O5)2 = P4O10 Step 2: Divide MM by Empirical Formula Mass 238.88 g = 2 141.94g

Calculating Molecular Formula A compound has an experimental molar mass of 78 g/mol. Its empirical formula is CH. What is its molecular formula? (CH)6 = C = 12.01 g H = 1.01 g 13.01 g C6H6 78 g/mol = 6 13.01 g/mol