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Percentage Composition: is the percent mass of each element present in a compound. Percent Composition Can be calculated if given: the chemical formula OR masses of elements in compound By Chemical Formula % mass = molar mass of an element X 100% total molar mass of the compound Example: What is the % composition of CaCO3? CHEMICAL FORMULA Step 1: Find the molar mass of CaCO3 : Ca x 1 = 40.1 g/mol C x 1 = 12.0 g/mol O x 3 = 48.0 g/mol CaCO3 = 100.1 g/mol Step 2: Find the % composition: % Ca = 40.1 g/mol x 100 % = 40.1 % Ca 100.1 g/mol % C = 12.0 g/mol x 100% = 12.0 % C 100.1 g/mol % O = 48.0 g/mol x 100 % = 48.0 % O 100.1 g/mol Example: Calculate the percent composition of the compounds that is formed from this reaction: 29.0g of Ag combines completely with 4.30g of S. Masses of elements in compound STEP 1: find the total mass of the elements. 29.0g + 4.30g = 33.30g STEP 2: find the % composition Ag = 29.0g x 100% = 87.1% 33.30g S = 4.30g x 100% = 12.9% 33.30g Try These: 1) Find the percent composition of KMnO4. 2) Calculate the % composition of the compound that results from 9.03g Mg reacting completely with 3.48g N. ANSWERS: 1) K = 24.7% Mn = 34.8% O = 40.5% 2) Mg = 72.2% N = 27.8% 1) Do Problem 17 on page 131 2) Do Problem 18 & 19 on page 133 CHECK YOUR ANSWERS! Example 2 How much carbon is present in 15.2 g of carbon dioxide gas? % carbon = 12.0 g/mol x 100 % = 27.3 % 44.0 g/mol Xg carbon = 15.2 g of CO2 X 27.3 g C 100. g CO2 = 4.15 g of Carbon Percent Composition Can be used to: calculate the mass of elements in a compound determine the empirical formula of a compound determine the molecular formula of a compound Empirical Formula shows the simplest mole ratio of the elements. CO is a 1:1 ratio of carbon to oxygen H2O is a 2:1 ratio CO2 is a 1:2 ratio Empirical formulas can’t be reduced. Molecular Formula shows the actual number of atoms in a molecule. The molecular formula for hydrogen peroxide is H2O2. Its empirical formula would be HO. Often the molecular formula is the same as the empirical formula: H2O, CO2 Empirical? CH4O – yes, cannot be reduced further C 2H 6 – no, empirical would be CH3 C3H10O – yes C6H6O2 – no. What would empirical be? – C3H3O Calculating Empirical Formulas A chemist with an unknown compound can easily figure out its percent composition, but it is much more meaningful to know its formula. EXAMPLE: What is the empirical formula for a compound that is 25.9% nitrogen and 74.1% oxygen? Method 1. Write the mass (g) of each element in the compound. So….we assume that it is a 100g sample: 25.9% N = 25.9g 74.1% O = 74.1g 2. Convert the mass of each element to moles, by dividing by the molar mass. N = 25.9g = 1.85 mol 14.0g/mol O = 74.1g 16.0g/mol = 4.63 mol 3. Calculate the simplest whole number ratio by dividing the number of moles by the smallest number of moles. 1.85 1.85 : 4.63 = 1 : 2.5 1.85 (If the result is not within 0.1 of a whole number, multiply all numbers by a whole number) 2(1 : 2.5) = 2 : 5 4. Write the empirical formula using the numbers you obtained. N2 O5 NOTE: For inorganic compounds, write the most positive element first. For organic compounds, write C first, H second and all others alphabetically. A special present just for you…….. Page 135, Problems #20 & 21 Check your answers Molecular Formula Given the empirical formula and the gram formula mass (gfm) OR Given the percent composition and the gram formula mass (gfm) Example #1 Calculate the molecular formula for NaO having a gfm of 78g. Determine the efm (empirical formula mass). NaO = 23.0g + 16.0g = 39.0 Divide the efm into the gfm. 78.0 = 2 39.0 This is the conversion factor used to determine the molecular formula. Na2O2 Example #2 Find the molecular formula for a compound having a composition of 58.8% C, 9.8% H and 31.4% O and a gmm of 102g/mol. Determine the mass of each component. C = 102g/mol x 58.8% = 60.0g/mol H = 102g/mol x 9.8% = 10.0g/mol O = 102g/mol x 31.4% = 32.0g/mol convert to moles C = 60.0g/mol = 5 12.0g H = 10.0g/mol = 10 1.0g O = 32.0g/mol = 2 16.0g Use moles as subscripts for components of compound C5H10O2 Check the gmm of this compound…does it equal 102.0g/mol? 5(12.0) + 10(1.0) + 2(16.0) = 102.0g/mol YES! And Now….. Oh Yeah! And there’s more… Page 136, Problems #22 & 23 Now Try page 139, #41 44