Percent Composition, Empirical Formulas, Molecular Formulas

Formula Mass The sum of atomic masses of all atoms in its formula Important role in nearly all chemical calculations Can be calculated for compounds and diatomic elements The atoms present in one formula unit of a substance Can be calculated for both ionic and molecular substances

Calculating Formula Mass Calculate the formula mass of calcium chloride Write the formula from the name given Ca2+ (from group II) and Cl- (from group VII) Formula is CaCl2 due to charge balance Formula mass: Sum of the atomic masses of atoms in the formula (1 Ca atom + 2 Cl atoms) Write the name of the compound by combining Ca2+ and Cl- 1 amu Ca + 2 amu Cl is the formula mass of CaCl2 = 40.08 amu = 70.90 amu Formula mass of CaCl2

Practice Complete # 1-3, 10-12

Percent Composition 1 pizza 2 slices pineapple 6 slices pepperoni 4 slices cheese What % of the pizza slices are pineapple? pepperoni? just cheese? (2/12) x 100 = 17% pineapple, (6/12) x 100 = 50% pepperoni, (4/12) x 100 = 33% cheese

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 Molar Mass of KMnO4 K = 1(39.1) = 39.1 Example: What is the percent composition of Potassium Permanganate (KMnO4)? Molar Mass of KMnO4 K = 1(39.1) = 39.1 Mn = 1(54.9) = 54.9 O = 4(16.0) = 64.0 MM = 158 g

Percent Composition Molar Mass of KMnO4 = 158 g 39.1 g K % K x 100 = Example: What is the percent composition of Potassium Permanganate (KMnO4)? Molar Mass of KMnO4 = 158 g 39.1 g K % K x 100 = 24.7 % 158 g 54.9 g Mn 34.8 % x 100 = % Mn 158 g K = 1(39.10) = 39.1 64.0 g O x 100 = 40.5 % Mn = 1(54.94) = 54.9 % O 158 g O = 4(16.00) = 64.0 MM = 158

Percent Composition Determine the percentage composition of sodium carbonate (Na2CO3)? Molar Mass Percent Composition 46.0 g x 100% = 43.4 % Na = 2(23.00) = 46.0 C = 1(12.01) = 12.0 O = 3(16.00) = 48.0 MM= 106 g % Na = 106 g 12.0 g x 100% = 11.3 % % C = 106 g 48.0 g x 100% = 45.3 % % O = 106 g

Percent Composition Determine the percentage composition of ethanol (C2H5OH)? % C = 52.13%, % H = 13.15%, % O = 34.72% _______________________________________________ Determine the percentage composition of sodium oxalate (Na2C2O4)? % Na = 34.31%, % C = 17.93%, % O = 47.76%

Percent Composition Calculate the mass of bromine in 50.0 g of Potassium bromide. 1. Molar Mass of KBr K = 1(39.10) = 39.10 Br =1(79.90) =79.90 MM = 119.0 2. 79.90 g ___________ = 0.6714 119.0 g 3. 0.6714 x 50.0g = 33.6 g Br

Percent Composition Calculate the mass of nitrogen in 85.0 mg of the amino acid lysine, C6H14N2O2. 1. Molar Mass of C6H14N2O2 C = 6(12.01) = 72.06 H =14(1.01) = 14.14 N = 2(14.01) = 28.02 O = 2(16.00) = 32.00 MM = 146.2 2. 28.02 g ___________ = 0.192 146.2 g 3. 0.192 x 85.0 mg = 16.3 mg N

Formulas Percent composition allow 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: C6H12O6 - molecular C4H10 - molecular C2H5 - empirical CH2O - empirical

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

Relating Empirical and Molecular Formulas n represents a whole number multiplier from 1 to as large as necessary Calculate the mass of the empirical formula Divide the given molecular mass by the calculated empirical mass Answer is a whole number multiplier

Relating Empirical and Molecular Formulas Multiply each subscript in the empirical formula by the whole number multiplier to get the molecular formula

Calculate Empirical Formula from Percent Composition Lactic acid has a molar mass of 90.08 g. It has an empirical formula of CH2O. What is the molecular formula of lactic acid? Lactic acid builds up in muscles after aerobic exercise Since percent represents parts of one element per 100 parts of the total compound

Determination of the Molecular Formula Obtain the value of n (whole number multiplier) Multiply the empirical formula by the multiplier Molecular formula = n х empirical formula Molecular formula = 3 (CH2O) C3H6O3