Empirical Formulae and Percent Composition

Empirical Formulae and Percent Composition 1. Calculate the mass percentage of each element in the following compounds: a. Rh(CN)4 b. Co(NO2)2 2. Calculate the mass proportions, expressed as percentage composition, for each of the elements in the compounds below: a. Ca2P2O7 b. Be2SiO4 3. Analysis of a 15.0 g sample of brass reveals it contains 13.0 g of copper. a) What is the mass percentage of copper in the sample? b) How many grams of brass would yield 95.0 g of pure copper? c) The only other component of this alloy is zinc. How many grams of zinc would be needed to prepare 600. g of this type of brass? 4. Calcium, nitrogen and oxygen were found to be present in a compound in the following mass percentages: 24.4% Ca, 17.1% N and 58.5% oxygen. Determine the empirical formula of the compound. 5. A compound of carbon and fluorine is found to contain 17.4% carbon and 82.6% fluorine. a) How many moles of carbon atoms are there in a 100. g sample? b) How many moles of fluorine atoms are there in a 200. g sample? c) What is the simplest formula for this compound? 6. Calculate the empirical formula for each compound: a) 82.6% C, 17.4% H b) 7.20% P, 92.8% Br c) 38.4% Mn, 16.8% C, 44.8% O d) 43.7% Zr, 10.3% B, 46.0% O 7. The composition of 2,4,6-trinitrotoluene (TNT) is 37.0% C, 2.22% H, 18.5% N, and 42.3% O. The molar mass of TNT is 227 g/mole. What is its molecular formula? 8. A 13.4 g sample of a compound contains 5.36 g of carbon, 0.902 g of hydrogen and 7.14 g of oxygen. Determine the empirical formula of the compound. Empirical Formulas from chemical reactions 1) A 36 g sample of iron oxide is composed of 78% iron. What is the empirical formula of this iron oxide? 2) 15.53 g of rust (FexOy) is decomposed to produce 10.87 g iron. What is the empirical formula of rust? 3) 72.06 g of carbon react with hydrogen gas to produce 88.06 g of propane. What is the empirical formula of propane? 4) 58.5 g C reacts with hydrogen gas producing 73.0 g ethane. What is the formula of ethane? 5) 20.16 g of Vanadium Oxide decomposes to give 8.88 g of Oxygen. What is the formula of this Vanadium Oxide? Answers to Empirical Formulae and Percent Composition 1. a) Molar mass of Rh(CN)4 = 206.9 g/mole % MassRh = ((1 x 102.9 g/mole) / 206.9 g/mole) x 100% = 49.7% Rh % MassC = (4 x 12.0 g/mole) / 206.9 g/mole x 100% = 23.2% C % MassN = (4 x 14.0 g/mole) / 206.9 g/mole x 100% = 27.1% N Check! Σ % Mass = 100% ???49.7% + 23.2% + 27.1% = 100.0% Good! b) Co(NO2)2: 39.0% Co, 18.6% N, 42.4% O 2. a) Ca2P2O7: 31.6% Ca, 24.4% P, 44.1% O b) Be2SiO4: 16.4% Be, 25.5% Si, 58.1% O 3. a) % MassCu = (massCu / massBrass) x 100 % = (13.0 g / 15.0 g) x 100%=% MassCu = 86.7% b) ? gBrass = massCu / % massCu = 95.0 g / 86.7% = 110. gBrass c) MassZn = % massZn x massBrass = (100.0% - 86.7%) (600. g) = 79.8 gZn 4. In 100.0 g of compound, we have 24.4 g Ca, 17.1 g N, and 58.5 g O. ? moles Ca = (1 mole / 40.0 g) (24.4 g) = 0.610 moles Ca ? moles N = (1 mole / 14.0 g) (17.1 g) = 1.22 moles N ? moles O = (1 mole / 16.0 g) (58.5 g) = 3.66 moles O Mole ratio for Ca = 0.610 moles / 0.610 moles = 1.00 Mole ratio for N = 1.22 moles / 0.610 moles = 2.00 Mole ratio for O = 3.66 moles / 0.610 moles = 6.00 The formula is CaN2O6, which we recognise as Ca(NO3)2 5. a) 1.45 moles C b) 8.69 moles F c) CF3 6. a) C2H5 b) PBr5 c) MnC2O4 d) Zr(BO3)2 7. Empirical formula: C7H5N3O6 Molecular formula: C7H5N3O6 8. Empirical formula: CH2O Answers to Empirical Formulas from chemical reactions 1) 2) 3) 4) 5) FeO Fe2O3 C3H8 C2H6 V2 O 5