AP Chemistry Compounds 2015-09-14 www.njctl.org Slide 3 / 163 - PDF document

Slide 1 / 163 Slide 2 / 163 AP Chemistry Compounds 2015-09-14 www.njctl.org Slide 3 / 163 Slide 4 / 163 Table of Contents: Compounds Pt. A Click on the topic to go to that section · Chemical Formulas Chemical Formulas · Metals & Alloys · Ionic Compounds · Covalent Compounds Return to Table of Contents Slide 5 / 163 Slide 6 / 163 The Mole The Mole Recall that 1 mole is defined as 6.022 x 10 23 units of a given Within 1 mole of a compound, there are often differing moles of substance. each element 1 mol of electrons = 6.022 x 10 23 electrons In 1 mole of aluminum nitrate, Al(NO 3 ) 3 1 mol of H 2 O molecules = 6.022 x 10 23 molecules of water = 1 mol of Al 3+ ions 1 mol of NaCl formula units = 6.022 x 10 23 formula units NaCl = 3 mol of NO 3- ions 1 mol of K atoms = 6.022 x 10 23 atoms of K = 3 mol of N atoms = 9 mol of O atoms

Slide 7 / 163 Slide 7 (Answer) / 163 1 How many moles of oxygen atoms are present in 2.0 1 How many moles of oxygen atoms are present in 2.0 moles of aluminum nitrate Al(NO 3 ) 3 ? moles of aluminum nitrate Al(NO 3 ) 3 ? Answer 18 moles [This object is a pull tab] Slide 8 / 163 Slide 8 (Answer) / 163 2 How many moles of oxygen atoms are in 2.0 moles of 2 How many moles of oxygen atoms are in 2.0 moles of sodium sulfate, Na 2 SO 4 ? sodium sulfate, Na 2 SO 4 ? Answer 8 moles [This object is a pull tab] Slide 9 / 163 Slide 9 (Answer) / 163 3 How many hydroxide ions are present in 1.2 x 10 24 3 How many hydroxide ions are present in 1.2 x 10 24 formula units of magnesium hydroxide: Mg(OH) 2 ? formula units of magnesium hydroxide: Mg(OH) 2 ? 2.4 x 10 24 hydroxide ions Answer [This object is a pull tab]

Slide 10 / 163 Slide 11 / 163 Molar Mass and Volume 4 What is the volume occupied @STP by 88 grams of carbon dioxide? Recall that the mass of 1 mol of a substance is called the molar mass and is measured in g/mol. This can be found on the periodic table. Molar mass of CaCl 2 = 110 g/mol Molar Mass of Ag = 108 g/mol Recall also that 1 mol of any gaseous substance will occupy 22.4 L of space at STP. 1 mol of Ar(g) = 22.4 L @STP 1 mol of H 2 (g) = 22.4 L @STP Slide 11 (Answer) / 163 Slide 12 / 163 4 What is the volume occupied @STP by 88 grams of 5 How many mL of methane gas (CH 4 ) are present @STP in a 100 gram sample of a gas that is 32% methane by carbon dioxide? mass? Answer 44.8 L [This object is a pull tab] Natural gas (methane) pipeline. Slide 12 (Answer) / 163 Slide 13 / 163 5 How many mL of methane gas (CH 4 ) are present @STP 6 Which of the following contains the most atoms of H? in a 100 gram sample of a gas that is 32% methane by mass? A 2 grams of H 2 gas B 16 grams of methane (CH 4 ) C 22.4 L of H 2 gas Answer 44,800 mL D 9 grams of water (H 2 O) Natural gas (methane) pipeline. [This object is a pull tab]

Slide 13 (Answer) / 163 Slide 14 / 163 6 Which of the following contains the most atoms of H? 7 How many moles of fluoride ions (F-) are present in a 79 gram sample of SnF 2 ? A 2 grams of H 2 gas A 0.5 moles B 16 grams of methane (CH 4 ) Answer B 78.5 moles B C 22.4 L of H 2 gas C 1 mole D 9 grams of water (H 2 O) D 1.5 moles [This object is a pull tab] Slide 14 (Answer) / 163 Slide 15 / 163 Chemical Formulas 7 How many moles of fluoride ions (F-) are present in a 79 gram sample of SnF 2 ? A chemical formula provides the ratio of atoms or moles of each element in a compound. A 0.5 moles B 78.5 moles H 2 O = 2 atoms H or 2 mol H Answer 1 atom O 1 mol O C 1 mole C D 1.5 moles Al(NO 3 ) 3 = 1 Al 3+ ion or 1 mol Al 3+ ions 3 NO 3- ions 3 mol NO 3- ions [This object is a pull tab] Slide 16 / 163 Slide 17 / 163 Calculating an Empirical Formula Empirical and Molecular Formulas To find an empirical formula: An empirical formula provides the simplest whole number ratio of atoms or moles of each element in a compound. 1. Determine the moles of each element within the compound then..... Compound "X" consists of 1.2 g C, 0.2 g H, and 1.6 g O Examples: H 2 O, NaCl, C 3 H 5 O = 0.1 mol C, 0.2 mol H, and 0.1 mol O A molecular formula represents the actual number of atoms or moles of each element in a compound. 2. Find the whole number ratio of these moles by dividing by smallest mole value! 0.1 mol C = 1 C 0.2 mol H = 2 H 0.1 mol O = 1 O Examples: H 2 O, C 3 H 5 O, C 6 H 12 O 6 0.1 mol 0.1 mol 0.1 mol Empirical formula = CH 2 O

Slide 18 / 163 Slide 19 / 163 Calculating a Molecular Formula Practice 8 Students type their answers here Given the following data, calculate the empirical formula of Determining the molecular formula of a compound is easy once phosphine gas. Phosphine gas is created by reacting solid the empirical formula and the molecular weight of the compound phosphorus with H 2 (g). are known. Mass of P(s) initial Mass of P(s) unreacted 1. Determine the ratio of the molecular weight to the empirical formula weight. 1.45 g 1.03 g MW of Compound "X" = 60 u Mass of H 2 (g) initial Mass of H 2 (g) unreacted Empirical formula weight of CH 2 O = 30 u 0.041 g 0.000 g Ratio = 60/30 = 2/1. The molecule is twice as heavy as the empirical formula. 2. Multiply each subscript of empirical formula by the ratio determined in step 1 CH 2 O x 2 = C 2 H 4 O 2 = Molecular Formula Slide 19 (Answer) / 163 Slide 20 / 163 Practice Students type their answers here Practice 8 9 Students type their answers here Given the following data, calculate the empirical formula of Black iron oxide (aka magnetite) is used as a contrast agent in phosphine gas. Phosphine gas is created by reacting solid MRI scans of human soft tissue. To determine the empirical phosphorus with H 2 (g). formula, a student reacted solid iron with O 2 (g). 0.42 g P reacted = 0.0135 mol P reacted Mass of P(s) initial Mass of P(s) unreacted Fe(s) reacted Mass of iron oxide obtained. Answer 0.041 g H 2 reacted = 0.0202 mol H 2 1.45 g 1.03 g 0.0202 mol H 2 = 0.0405 mol H 3.05 g 4.22 g 0.0405/0.0135 = 3 H 0.0135/0.0135 = 1 P Mass of H 2 (g) initial Mass of H 2 (g) unreacted What is the empirical formula? PH 3 = empirical formula 0.041 g 0.000 g [This object is a pull tab] Slide 20 (Answer) / 163 Slide 21 / 163 Students type their answers here Practice Students type their answers here Practice 9 10 Butane gas can be produced when solid carbon is reacted with Black iron oxide (aka magnetite) is used as a contrast agent in hydrogen gas. If 0.45 grams of carbon were found to react with MRI scans of human soft tissue. To determine the empirical 1.05 L of H 2 gas @STP, what is the molecular formula of butane 3.05 g Fe = 0.055 mol Fe formula, a student reacted solid iron with O 2 (g). given it has a molar mass of 58 g/mol. 4.22 - 3.05 = 1.17 g O 2 (g) Fe(s) reacted Mass of iron oxide obtained. Answer 1.17 g O 2 (g) = 0.037 mol O 2 3.05 g 4.22 g 0.037 mol O 2 = 0.073 mol O 0.055/0.055 = 1 Fe 0.073/0.055 = 1.33 mol O What is the empirical formula? x each by 3 to get a whole number ratio ... Fe 3 O 4 [This object is a pull tab]

Slide 21 (Answer) / 163 Slide 22 / 163 Students type their answers here Practice 10 11 Which of the following is NOT an empirical formula? Butane gas can be produced when solid carbon is reacted with hydrogen gas. If 0.45 grams of carbon were found to react with A Fe 2 O 3 1.05 L of H 2 gas @STP, what is the molecular formula of butane 0.45 g C = 0.0375 mol C given it has a molar mass of 58 g/mol. B H 2 NNH 2 1.05 L H 2 (g) = 0.0469 mol H 2 (g) 0.0469 mol H 2 (g) = 0.0938 mol H Answer C CH 3 OH 0.0375/0.0375 = 1 C x 2 = C 2 0.0938/0.0375 = 2.5 H x 2 = H 5 D CH 3 CH 2 Cl Empirical Formula = C 2 H 5 58/29 = 2 Molecular Formula = C 4 H 10 [This object is a pull tab] Slide 22 (Answer) / 163 Slide 23 / 163 11 Which of the following is NOT an empirical formula? 12 A compound used in airbags degrades into sodium and nitrogen gas (N 2 ) when ignited. If 3.36 L of N 2 (g) was produced @STP from an initial mass of the compound of A Fe 2 O 3 6.50 grams, what is the empirical formula? Answer B H 2 NNH 2 B A Na 2 N 3 C CH 3 OH B Na 3 N D CH 3 CH 2 Cl C NaN 3 [This object is a pull tab] D NaN Slide 23 (Answer) / 163 Slide 24 / 163 12 A compound used in airbags degrades into sodium and 13 A compound containing carbon, hydrogen, and chlorine is nitrogen gas (N 2 ) when ignited. If 3.36 L of N 2 (g) was 14.1% carbon by mass, 83.5% Cl, with the rest being produced @STP from an initial mass of the compound of hydrogen. What is the empirical formula? 6.50 grams, what is the empirical formula? A C 2 H 5 Cl A Na 2 N 3 Answer B CH 2 Cl 2 C B Na 3 N C C 2 H 6 Cl C NaN 3 D CH 3 Cl D NaN [This object is a pull tab]