ࡱ> !#  $bjbj 4.$lllll,$Y#:    """""""$Q$'"l"ll  #III l l "I"IIn hb! 0v ")#0Y# |'F' b!b!'l|"HI""? Y#' 2: Empirical Formula Worksheet A chemist finds that an unknown compound contains 50.5% S and 49.95 % O by weight. Calculate its empirical We will first determine the number of moles of each element based on a 100 g  EMBED Equation.3  divide by smallest (1.561mol) 1.561/1.561=1  EMBED Equation.3  3.122/1.561=2 empirical formula is:  EMBED Equation.3  A certain compound containing only phosphorus and oxygen is 56.3 % P (by mass) Determine the empirical formula of this compound. 100 g of compound contains 56.3 g P and 43.7 g O (56.3 g P) (1 mole P / 31.0 g) = 1.82 mole P /1.82 mole = 1 (43.7 g O) (1 mole O / 16.0 g) = 2.73 mole O/ 1.82 mole =1.5 Empirical formula: = P2O3 Determine the empirical formula for a compound which is 54.09% Ca, 43.18% O, and 2.73% H Assume 100 g of the substance, then 54.09 g Calcium, 43.18 g oxygen and 2.73g hydrogen. Ca = 54.09g C x (1 mol Ca/40g Ca) = 1.352; 1.352 mol Ca /1.352 mol = 1 O = 43.18 g O x (1 mol O /16 g O) = 2.699; 2.699 mol O /1.352 mol = 2 H = 2.73 g H x (1 mol H/1.01g H) = 2.73; 2.73 mol H /1.352 mol = 2 CaO2H2 = Ca(OH)2 A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Assume 100 g of the substance, then 72.2 g magnesium and 27.8 g nitrogen. for Mg: 72.2 g Mg x (1 mol Mg/24.3 g Mg) = 2.97 mol Mg 2.97 mol Mg / l.99 mol = 1.49 for N: 27.8 g N x (1 mol N/14.0 g N) = 1.99 mol N 1.99 mol N / l.99 mol = 1.00 for Mg: 2 x 1.49 = 2.98 (i.e., 3) for N: 2 x 1.00 = 2.00 Mg3N2 A 5.438 gram sample, was found to contain 2.549 grams of iron, 1.947 grams of oxygen, and 0.9424 grams of phosphorus. What is its empirical formula? 2.549 grams of Fe X 1 mole of Fe = 0.0456 moles of Fe 55.85 g Fe 1.947 grams of O X 1 mole of O = 0.1217 moles of O 16.00 g O 0.9424 grams of P X 1 mole of P = 0.0304 moles of P 30.97 g P Therefore the formula is Fe0.0456O0.1217P0.0304 = Fe1.5O4.0P1 =Fe3O8P2 A compound contains 63.11 % C, 12.36 % H, and 24.53 % N. What is its empirical formula? Get moles of each element, then determine the mole ratio. Assume 100 g of the compound, so that you have 63.11 g C, 12.36 g H, 24.53 g N. 1 extra digit shown in calculations below. 63.11 g C x (1 mol C / 12.011 g C) = 5.2543 mol C 12.36 g H x (1 mol H / 1.0079 g) = 12.263 mol H 24.53 g N x (1 mol N / 14.0067 g) = 1.7513 mol N Moles of N is the smallest number; get other ratios relative to N: 5.2543 mol C / 1.7513 mol N = 3.000 mol C / mol N 12.263 mol H / 1.7513 mol N = 7.002 mol H / mol N Therefore the empirical formula is C3H7N. (b) If the experimental molar mass is 114.2 g/mol, what is the molecular formula? C3H7N has a formula weight of 3*12.011 + 7*1.0079 + 14.0067 = 57.10 g/mol. This is exactly half of the experimental molar mass (114.2 / 57.10 = 2.00), so the molecular formula is C6H14N2. A certain carbohydrate compound (containing only C, H and 0) is 40.0% C, 6.72% H, and 53.3% O by mass. The experimentally determined molecular mass is 180 g/mol. What is the empirical and molecular formula for this carbohydrate? Choosing 100g as a convenient mass, we would have 40.0g of C, 6.72g H and 53.3g of O 40.0 g C * (1 mole/12.0g) = 3.33 moles C 6.72 g H * (1 mole/1.01g) = 6.65 moles H 53.3 g O * (1 mole/16.0g) = 3.33 moles O Calculating the relative stoichiometry by dividing by the smallest: (6.65/3.33) = 2.00 moles H, and 1.0 mole each for C and O This yields an empirical formula of CH2O This would have an atomic mass of (12.0 + 2.02 + 16.0) = 30 g/mol The actual molecular mass is 180 g/mol, or (180/30) = 6.0 times that of the empirical formula. Therefore, the chemical formula must be: C6H12O6 A certain hydrocarbon compound (containing only C and H) is 83.6% C and 16.4% H by mass. The experimentally determined molecular mass is 86.2 amu. What is the empirical and chemical formula for this hydrocarbon? (2 points) Choosing 100g as a convenient mass, we would have 83.6 g of C and 16.4 g of H 83.6 g C * (1 mole/12.01g) = 6.96 moles C 16.4 g H * (1 mole/1.01g) = 16.24 moles H Calculating the relative stoichiometry by dividing by the smallest: (6.96/6.96) = 1.00 moles C (16.24/6.96) = 2.33 moles H The value for H looks like 2 1/3, so multiply both molar amounts by 3 to give an empirical formula of C3H7 This would have an atomic mass of (3*12.01)+(7*1.01) = 43.1 amu The actual molecular mass is 86.2 amu, or (83.6/43.1) = 2.0 times that of the empirical formula. 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