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South Pasadena • Honors Chemistry Name 2 • Elements and Compounds Period 2.5 PROBLEMS 1. Write the balanced formation equation for the following compounds. Indicate the states. (a) N2O (g) 2 N2 (g) + O2 (g) → 2 N2O (g) (b) NaCl (s) 2 Na (s) + Cl2 (g) → 2 NaCl (s) (c) PBr3 (ℓ) OR 2 P (s) + 3 Br2 (ℓ) → 2 PBr3 (ℓ) P4 (s) + 6 Br2 (ℓ) → 4 PBr3 (ℓ) (d) SO3 (g) OR 2 S (s) + 3 O2 (g) → 2 SO3 (g) S8 (s) + 12 O2 (g) → 8 SO3 (g) 2. Write the balanced decomposition equation for the following compounds. Indicate the states. (a) Al2O3 (s) 2 Al2O3 (s) → 4 Al (s) + 3 O2 (g) (b) CaO (s) 2 CaO (s) → 2 Ca (s) + O2 (g) (c) OF2 (g) 2 OF2 (g) → O2 (g) + 2 F2 (g) (d) HI (g) 2 HI (g) → H2 (g) + I2 (s) 3. A sample of the common amino acid lysine was found to have 226.4 mg of carbon, 44.3 mg of hydrogen, 88.0 mg of nitrogen, and 100.5 mg of oxygen. (a) Find the % composition of lysine. Total mass = 226.4 mg C + 44.3 mg H + 88.0 mg N + 100.5 mg O = 459.2 mg 226.4 mg C %C = × 100% = 49.30% C 459.2 mg 44.3 mg H %H = × 100% = 9.65% H 459.2 mg 88.0 mg N %N = × 100% = 19.2% N 459.2 mg 100.5 mg O %O = × 100% = 21.89% O 459.2 mg (b) Find the empirical compound for lysine. Assume 100 g sample. 1 mol C nC = 49.30 g C 12.01 g C = 4.105 mol C 1 mol H nH = 9.65 g H 1.008 g H = 9.57 mol H 1 mol N nN = 19.2 g N 14.01 g N = 1.37 mol N 1 mol O nO = 21.89 g O 16.00 g O = 1.368 mol O – Date CHEMICAL ANALYSIS Empirical Formula: C4.105 H 9.57 N 1.37 O1.368 1.368 1.368 1.368 1.368 C3H7NO (c) If the molar mass of lysine is 146.19 g/mol, find the molecular formula for lysine. Molar Mass of Lysine = Empirical Formula Mass 146.19 g/mol = 3(12.01) + 7(1.008) + 1(14.01) + 1(16.00) 146.19 g/mol 2 = 73.10 g/mol 1 Molecular Formula = (Empirical Formula) × 2 = (C3H7NO) × 2 = C6H14N2O2 4. 0.85 mol of a sample of benzene, composed of only carbon and hydrogen, has a mass of 66.4 g. (a) What is the molar mass of benzene? Molar Mass = total mass 66.4 g = total moles 0.85 mol = 78 g/mol (b) How any molecules of benzene are in this sample? 66.4 g benzene 1 mol benzene 78 g benzene 6.022 × 10 molecules benzene 1 mol benzene 23 = 5.1 × 1023 molecules benzene (c) If benzene contains 7.74% H, find the empirical formula and molecular formula of the compound. 1 mol C nC = 92.26 g C 12.01 g C = 7.681 mol C 1 mol H nH = 7.74 g H = 7.68 mol H 1.008 g H Empirical Formula: C7.681 H7.68 CH 7.68 7.68 Molecular Formula: Molar Mass = Empirical Formula Mass 78 g/mol 78 g/mol 6 = = 1(12.01) + 1(1.008) 13.02 g/mol 1 Molecular Formula = (Empirical Formula) × 6 = (CH) × 6 = C6H6 (d) If the density of benzene is 0.876 g/mL, find the volume of this sample. 66.4 g C6H6 1 mL C6H6 0.876 g C6H6 = 75.8 mL C6H6(Note: do not use 1 mol = 22.4 L since it’s not a gas at STP) 5. A compound was analyzed and found to contain 53.30% carbon, 11.19% hydrogen, and 35.51% oxygen by mass. (a) Calculate the empirical formula of the compound. Assume 100 g sample. 1 mol C nC = 53.30 g C 12.01 g C = 4.438 mol C 1 mol H nH = 11.19 g H = 11.10 mol H 1.008 g H 1 mol O nO = 35.51 g O 16.00 g O = 2.219 mol O Empirical Formula: C4.438 H11.10 O2.219 2.219 2.219 P + Se → PxSey nP: 45.2 g P 1 mol P 30.97 g P = 1.46 mol P nSe: (131.6 g – 45.2 g) = 86.4 g Se Empirical Formula: 1 mol Se 78.96 g Se = 1.09 mol Se P1.46 Se1.09 P1.33Se 1.09 1.09 P4Se3 8. An 80.50 g sample of an unknown oxide of chromium (CrxOy) is heated until all of the oxygen is vaporized, leaving 55.08 g of solid chromium metal. The molar mass of this unknown compound is between 150 and 160 g/mol. (a) Find the % composition of each element of the unknown compound. 2.219 C2H5O (b) If the molar mass of the compound is 90.12 g/mol, what is the molecular formula of the compound? Molar Mass = Empirical Formula Mass 90.12 g/mol 90.12 g/mol = 2(12.01) + 6(1.008) + 1(16.00) 46.07 g/mol 2 = 1 Molecular Formula = (Empirical Formula) × 2 = (C2H5O) × 2 = C4H10O2 6. A 0.77 g sample of nitrogen reacts with chlorine to form 6.61 g of the chloride. Determine the empirical formula of nitrogen chloride. N2 + Cl2 NxCly 1 mol N 0.77 g N 14.01 g N = 0.0550 mol N 1 mol Cl nCl: (6.61 g – 0.77 g) = 5.84 g Cl 35.45 g Cl = 0.165 mol Cl Empirical Formula: N0.0550 Cl 0.165 NCl3 (b) Find the empirical formula and molecular formula of the unknown compound. (c) Write the balanced chemical equation for this decomposition. Include states. nN: 0.0550 (d) Find the volume at STP of the oxygen gas that is vaporized. 0.0550 7. A 45.2 g sample of phosphorous reacts with selenium to form 131.6 g of the selenide. Find the empirical formula of phosphorous selenide. (e) Find the number of chromium atoms in this sample. 9. Tartaric acid is the white, powdery substance that coats tart candies such as Sour Patch Kids. Combustion analysis of a 12.01 g sample of tartaric acid – which contains only carbon, hydrogen, and oxygen – produced 14.08 g CO2 and 4.32 g H2O. Determine the empirical formula of tartaric acid. CxHyOz (s) + O2 (g) → CO2 (g) + H2O (ℓ) (a) Determine the empirical formula of tartaric acid. CxHyOz (s) + O2 (g) → CO2 (g) + H2O (ℓ) MM: 1(12.01) + 2(16.00) = 44.01 g/mol 2(1.008) + 1(16.00) = 18.016 g/mol Carbon: 14.08 g CO2 1 mol CO2 1 mol C 44.01 g CO2 1 mol CO2 12.01 g C = 0.3199 mol C 1 mol C = 3.843 g C Hydrogen: 1 mol H2O 2 mol H 4.32 g H2O 18.02 g H2O 1 mol H2O 1.008 g H = 0.480 mol H 1 mol H = 0.483 g H Oxygen: 12.01 g CxHyOz – (3.843 g C + 0.483 g H) = 7.68 g O 1 mol O 7.68 g O 16.00 g O = 0.480 mol O Empirical Formula: C0.3199 H 0.480 O 0.480 0.3199 CH1.5O1.5 0.3199 0.3199 C2H3O3 (b) If the molar mass of tartaric acid is 150 g/mol, find its molecular formula. Molecular Formula: Molar Mass Empirical Formula Mass 150 g/mol = 2(12.01) + 3(1.008) + 3(16.00) 150 g/mol 2 = = 75.04 g/mol 1 Molecular Formula = (Empirical Formula) × 2 = (C2H3O3) × 2 = C4H6O6