Percent Composition Percentage composition of a compound gives the relative amount of each element present. % = mass element x 100 mass compound The sum of percentages of each element in a compound should be = 100

Calculate the percent composition for carbon dioxide (CO2) Calculate the molar mass for each element and the compound C: 1 x 12.0 = 12.0 O: 2 x 16.0 = 32.0 + 44.0 g/mol %C = 12 x 100 =27.3 % 44 % O = 32 x 100 = 72.7% Sum of percentages = 100 Round percentages to the tenth.

Calculate the percent composition for Al(OH)3 aluminum hydroxide Calculate the molar mass for each element and the compound Al: 1 x 27.0 = 27.0 H: 3 x 1.0 = 3.0 O: 3x 16.0 = 48.0 + 78.0 g/mol %Al = 27 x 100 = 34.6% 78 % H = 3 x 100 = 3.8% %O = 48 x 100 = 61.5%

Percent Composition What is the percent carbon in C5H8NO4 (the glutamic acid used to make MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? a) 8.22 %C b) 24.3 %C c) 41.1 %C

Empirical and molecular formulas If the identities of the elements of a compound are known and you also know its percentage composition, you can find the formula of the compound.

Types of Formulas Empirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. Ionic compounds formula are always empirical formula Molecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

Empirical Formula from % Composition A compound analyzed and found to contain 25.9% nitrogen and 74.1% O. What is the empirical formula of the compound? Since percent means parts per 100, you can assume 100.g of the compound contains 25.9g N and 74.1 g O.

Empirical Formula from % Composition A compound analyzed and found to contain 25.9% nitrogen and 74.1% O. What is the empirical formula of the compound? Convert those values to moles: 25.9 g N 1 mol N = 1.85 mol N 14.0 g N 74.1 g O 1 mol O = 4.63 mol O 16.0 g O

Empirical Formula from % Composition A compound analyzed and found to contain 25.9% nitrogen and 74.1% O. What is the empirical formula of the compound? Divide each molar quantity by the smaller number of moles. 1.85 mol N = 1 1.85 4.63 mol O = 2.50 mol O

Empirical Formula from % Composition A compound analyzed and found to contain 25.9% nitrogen and 74.1% O. What is the empirical formula of the compound? The subscript for O is still NOT a whole number. Multiply both numbers by the smallest whole number that will convert both subscripts to whole numbers 1 mol N x 2 = 2 mol N 2.50 mol O x 2 = 5 mol O Empirical Formula: N2O5

A sample of a brown gas, a major air pollutant, is found to contain 2 A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine the empirical formula for the compound. convert grams to moles moles of N = 2.34g of N 1 mol = 0.167 moles of N 14.0 g moles of O = 5.34 g 1 mol = 0.334 moles of O 16.0 g divide by smaller mole value: 0.167 mol N = 1 mol N 0.334 mol O = 2 mol O 0.167 0.167 Empirical formula : NO2

Molecular Formula Some elements combine in more than one way. The molecular formula of a compound is either its empirical formula or a whole-number multiple of its empirical formula.

Calculate molar mass of empirical formula: Molar mass CH4N: 30.0 g/mol Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH4N. Calculate molar mass of empirical formula: Molar mass CH4N: 30.0 g/mol Divide molar mass molecular formula by molar mass empirical formula: 60.0 g/mol = 2 30.0 g/mol Multiply the subscripts by this value. Molecular formula: C2H8N2

Calculate empirical formula: C5H10O2 The compound methyl butanoate smells like apples. Its percent composition is 58.8% C, 9.8%H, and 31.4% O. Its molar mass is 102 g/mol. What is its empirical formula? Its molecular formula? Calculate empirical formula: C5H10O2 Calculate molar mass empirical formula: 102 g/mol Divide 102/102= 1 Multiply subscripts