Chem Beginning of Mole Homework Key

Empirical Formula Problems - Set I Name: Hour: 1. Determine the empirical formula of a compound consisting of: 55.3% K, 14.6% P, and 30.1% O. (K3PO4) 55.3g * (1mol / 39.1g) = 1.41mol 14.6g * (1mol / 30.97g) = 0.471mol 30.1g * (1mol / 16g) = 1.88mol 1.41mol / 0.471mol = 3 0.471mol / 0.471mol = 1 1.88mol / 0.471mol = 4 K3PO4 52.14% C, 13.13% H, and 34.73% O. (C2H6O) 52.14g * (1mol / 12.01g) = 4.34mol 13.13g * (1mol / 1.01g) = 13.01mol 34.73g * (1mol / 16g) = 2.17mol 4.34mol / 2.17mol = 2 13.01mol / 2.17mol = 6 2.17mol / 2.17mol = 1 C2H6O 47.3% Cu and 52.7% Cl. (CuCl2) 47.3g * (1mol / 63.55g) = 0.744mol 52.7g * (1mol / 35.45g) = 1.49mol 0.744mol / 0.744mol = 1 1.49mol / 0.744mol = 2 CuCl2 40.0% C, 6.73% H, and 53.3% O. (CH2O) 40.0g * (1mol / 12.01g) = 3.33mol 6.73g * (1mol / 1.01g) = 6.66mol 53.3g * (1mol / 16g) = 3.33mol 3.33mol / 3.33mol = 1 6.66mol / 3.33mol = 2 3.33mol / 3.33mol = 1 CH2O 2. In vanadium oxide, the mole ratio is calculated to be: 2.50 mol O / 1 mol V. What is the simplest formula of vanadium oxide? (V2O5) V2O5 3. Some phosphorus is heated in air to produce phosphorus oxide. The following data was collected: mass of crucible (reaction vessel) 25.34 g mass of crucible + phosphorus 27.19 g mass of crucible + phosphorus oxide   29.58 g Determine the empirical formula of phosphorus oxide. (P2O5) mass P: 27.19g - 25.34g = 1.85g mass O: 29.58g - 27.19g = 2.39g 1.85g * (1mol / 30.97g) = 0.0597mol 2.39g * (1mol / 16g) = 0.149mol 0.0597mol / 0.0597mol = 1 0.149mol / 0.0597mol = 2.5 P2O5 4. Phosphorus forms two different compounds with chlorine. One compound contains 22.5% P by mass, and the other contains 14.87%   P by mass. Determine the empirical formulas of the two compounds. (PCI3, PCIS) P: 22.5g * (1mol / 30.97g) = 0.727mol Cl: 77.5g * (1mol / 35.45g) = 2.19mol 0.727mol / 0.727mol = 1 2.19mol / 0.727mol = 3 PC13 P: 14.87g * (1mol / 30.97g) = 0.48mol Cl: 85.13g * (1mol / 35.45g) = 2.40mol 0.48mol / 0.48mol = 1 2.40mol / 0.48mol = 5 PCIS 5. An ace chemistry student carries out the following reaction in an attempt to determine the empirical formula of a copper oxide:   CuO(s) + H2(g) -> Cu(s) + H2O(l) Mass of empty test tube 24.25 g Mass of empty test tube + copper oxide 26.26g Mass of test tube + solid copper (after heating) 25.85 g Mass of copper oxide 2.01 g Mass of copper (in copper oxide) 1.66 g Mass of oxygen (in copper oxide) 0.35 g 1.66g Cu * (1mol / 63.55g) = 0.025mol 0.35g O * (1mol / 16g) = 0.022mol 0.025mol / 0.025mol = 1 0.022mol / 0.025mol = 0.88 CuO 6. An ace chemistry student heats a piece of iron (Fe) metal in a crucible. The reaction is: Fe(s) + O2(g) -> Fe2O3(s) mass of crucible 27.50 g mass of crucible + Fe 28.62 g mass of crucible + iron oxide 29.10 g mass of Fe 1.12 g mass of iron oxide 1.69 g moles of Fe 1.12g * (1mol / 55.85g) = 0.02mol mass of oxygen 0.48 g moles of oxygen 0.48g * (1mol / 16g) = 0.03mol 0.02mol / 0.02mol = 1 0.03mol / 0.02mol = 1.5 Fe2O3 Empirical + Molecular Formula Problems - Set II 1. What is the empirical formula of the following compounds: (a) CO₃ (b) SiO₂ (c) NH₄Cl 2. A 10.00 g sample of vitamin C was analyzed and found to contain 4.092 g of C, 0.458 g of H, and 5.450 g of O. Given that the molar mass of the compound is 176g, determine the molecular formula of vitamin C. (answer: C₆H₈O₆) 4.092g C * (1mol / 12.01g) = 0.341mol 0.458g H * (1mol / 1.01g) = 0.453mol 5.450g O * (1mol / 16g) = 0.341mol 0.341mol / 0.341mol = 1 0.453mol / 0.341mol = 1.33 0.341mol / 0.341mol = 1 CH₁.₃O₁ molar mass: 88g/mol 176g / 88g = 2 C₆H₈O₆ 3. Cyclobutane has the empirical formula CH₂. Its molar mass is 42g. What is its molecular formula? (answer: C₃H₆) 12 + 2(1) = 14 42g / 14g = 3 C₃H₆ 4. The amino acid Histidine has a molar mass of 154g. Is the molecular formula of Histidine CH₄NO or C₆H₈N₂O₂? 3(12) + 4(1) + 1(14) + 1(16) = 70g 6(12) + 8(1) + 2(14) + 2(16) = 154g C₆H₈N₂O₂ 5. The major air pollutant of coal-burning power plants is a colorless, pungent gaseous compound containing only sulfur and oxygen. Chemical analysis of a 1.078   g sample of this gas showed that it contained 0.540   g of S and 0.538 g of O. What is the empirical formula of this compound? (answer: SO₂) 0.540g S * (1mol / 32.07g) = 0.0168mol 0.538g O * (1mol / 16g) = 0.0336mol 0.0168mol / 0.0168mol = 1 0.0336mol / 0.0168mol = 2 SO₂ 6. Determine the empirical formula of a compound containing: 29.2% N, 8.3% H, 12.5% C, and 50.0% O. (answer: N₂H₉CO₃) 29.2g N * (1mol / 14.01g) = 2.08mol 8.3g H * (1mol / 1.01g) = 8.22mol 12.5g C * (1mol / 12.01g) = 1.04mol 50.0g O * (1mol / 16g) = 3.13mol 2.08mol / 1.04mol = 2 8.22mol / 1.04mol = 8 1.04mol / 1.04mol = 1 3.13mol / 1.04mol = 3 N₂H₉CO₃ 7. Benzene has the empirical formula CH. Its molar mass is 78g. What is its molecular formula? (answer: C₆H₆) 12 + 1 = 13 78g / 13g = 6 C₆H₆ Moles Problem Set I 1. Determine the molar mass of each of the following compounds: a. HCHO: 1.01 + 12.01 + 16 = 29.02 g/mol b. (NH₄)₂S: 2(14.01) + 8(1.01) + 32.07 = 68.17 g/mol c. Ca(NO₃)₂: 40.08 + 2(14.01) + 6(16) = 164.1 g/mol d. UF₆: 238.03 + 6(19) = 352.03 g/mol 2. Calculate the number of grams in 0.3 mol of each of the compounds listed in question one. a. 0.3 mol * 29.02 g/mol = 8.71 g b. 0.3 mol * 68.17 g/mol = 20.5 g c. 0.3 mol * 164.1 g/mol = 49.2 g d. 0.3 mol * 352.03 g/mol = 105.6 g 3. Calculate the number of moles contained in 100.0 g of each compound listed in question one. a. 100.0 g * (1 mol / 29.02 g) = 3.45 mol b. 100.0 g * (1 mol / 68.17 g) = 1.47 mol c. 100.0 g * (1 mol / 164.1 g) = 0.61 mol d. 100.0 g * (1 mol / 352.03 g) = 0.284 mol 4. In 3.01x10²⁴ molecules of water: a. How many moles of water are present? 3.01x10²⁴ molecules * (1 mol / 6.02x10²³ molecules) = 5 moles b. How many grams of water are present? 5 moles * 18.02 g/mol = 90.1 g 5. How many atoms are in 2.4 mol of Copper? 2.4 mol Cu * (6.02x10²³ atoms / 1 mol) = 1.44x10²⁴ atoms 6. Calculate the molar mass of: a. NaHCO₃: 22.99 + 1.01 + 12.01 + 3(16) = 84.01 g/mol b. Mg(NO₃)₂: 24.31 + 2(14.01) + 6(16) = 148.33 g/mol c. (NH₄)₃PO₄: 3(14.01) + 12(1.01) + 30.97 + 4(16) = 149.12 g/mol 7. Calculate the mass of: a. 0.948 mol NaHCO₃: 0.948 mol * 84.01 g/mol = 79.64 g b. 3.67 mol Mg(NO₃)₂: 3.67 mol * 148.33 g/mol = 544.37 g c. 2.94 mol (NH₄)₃PO₄: 2.94 mol * 149.12 g/mol = 143.81 g 8. Calculate the number of moles in: a. 9.67 g of NaHCO₃: 9.67 g * (1 mol / 84.01 g) = 0.115 mol b. 38.4 g of Mg(NO₃)₂: 38.4 g * (1 mol / 148.33 g) = 0.259 mol c. 0.754 g of (NH₄)₃PO₄: 0.754 g * (1 mol / 149.12 g) = 0.005 mol 9. Calculate the percent composition of: a. NaHCO₃: Na 22.99/84.01 = 27.4%, H 1.01/84.01 = 1.2%, C 12.01/84.01 = 14.3%, O 3(16)/84.01 = 57.1% b. Mg(NO₃)₂: Mg 24.31/148.33 = 16.4%, N 2(14.01)/148.33 = 18.9%, O 6(16)/148.33 = 64.7% c. (NH₄)₃PO₄: N 3(14.01)/149.12 = 28.2%, H 12(1.01)/149.12 = 8.13%, P 30.97/149.12 = 20.77%, O 4(16)/149.12 = 42.9% 10. Calculate the number of grams of oxygen in 25.0g of: a. NaHCO₃: 25.0 g * (0.571) = 14.28 g b. Mg(NO₃)₂: 25.0 g * (0.6472) = 16.18 g c. (NH₄)₃PO₄: 25.0 g * (0.4295) = 10.74 g MOLES PROBLEM SET II 1. Determine the molar mass of the following compounds: a. Fe₃(PO₄)₂: 3(55.85) + 2(30.97) + 8(16) = 357.49 g/mol b. Hemoglobin (C₂₉₅₂H₄₆₆₄O₈₃₂S₈Fe₄): 2952(12.01) + 4664(1.01) + 832(16) + 8(32.07) + 4(55.85) = 63956.12 g/mol 2. Tin (II) fluoride is added to toothpaste to prevent tooth decay. In 24 g of SnF₂, how many moles are present? 24g / 156.7g = 0.153 mol SnF₂ 3. How many moles of benzene (C₆H₆) are present in 15.6 g of benzene? 15.6 g / 78.12 g/mol = 0.2 mol C₆H₆ 4. How many grams of NH₄Cl are present in 0.4 mol of ammonium chloride? 0.4 mol * 53.59 g/mol = 21.44 g NH₄Cl 5. Define molar mass molar mass = mass (in grams) of one mole of a substance 6. Calculate the molar mass of a compound if 0.22 mol of it weighs 48 g. 0.22 mol / 48 g = 1 mol / x g x = 218 g/mol 7. Which sample contains the largest number of hydrogen atoms? a. 7.2 g H₂O b. 0.4 mol H₂O c. all d. 6.02x10²³ molecules H₂ less than 8. Which of the following has more atoms? 2.4 g of He or 10.4 g of Cr? Show your work. 2.4 g He * (1 mol / 4 g) * (6.02x10²³ atoms / 1 mol) = 3.61x10²³ atoms He 10.4 g Cr * (1 mol / 52 g) * (6.02x10²³ atoms / 1 mol) = 1.20x10²³ atoms Cr 9. Determine the molar mass of each of the following compounds a. H₂SO₂ b. Cu(NO₃)₂ 2(1.01) + 32.07 + 4(16) = 98.09 g/mol 63.55 + 2(14.01) + 6(16) = 187.57 g/mol 10. Calculate the number of grams contained in 0.42 mol of each of the compounds in question 9 a. 0.42 mol * 98.09 g/mol = 41.2 g b. 0.42 mol * 187.57 g/mol = 78.78 g