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 8 / 163 1 How many moles of oxygen atoms are present in 2.0 2 How many moles of oxygen atoms are in 2.0 moles of moles of aluminum nitrate Al(NO 3 ) 3 ? sodium sulfate, Na 2 SO 4 ? Slide 9 / 163 Slide 10 / 163 Molar Mass and Volume 3 How many hydroxide ions are present in 1.2 x 10 24 formula units of magnesium hydroxide: Mg(OH) 2 ? 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 / 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? Natural gas (methane) pipeline.

Slide 13 / 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 ) B 78.5 moles C 22.4 L of H 2 gas C 1 mole D 9 grams of water (H 2 O) D 1.5 moles Slide 15 / 163 Slide 16 / 163 Chemical Formulas Empirical and Molecular Formulas A chemical formula provides the ratio of atoms or moles of each An empirical formula provides the simplest whole number ratio of element in a compound. atoms or moles of each element in a compound. H 2 O = 2 atoms H or 2 mol H Examples: H 2 O, NaCl, C 3 H 5 O 1 atom O 1 mol O A molecular formula represents the actual number of atoms or moles of each element in a compound. Al(NO 3 ) 3 = 1 Al 3+ ion or 1 mol Al 3+ ions 3 NO 3- ions 3 mol NO 3- ions Examples: H 2 O, C 3 H 5 O, C 6 H 12 O 6 Slide 17 / 163 Slide 18 / 163 Calculating an Empirical Formula Calculating a Molecular Formula Determining the molecular formula of a compound is easy once To find an empirical formula: the empirical formula and the molecular weight of the compound 1. Determine the moles of each element within the compound are known. then..... 1. Determine the ratio of the molecular weight to the empirical Compound "X" consists of 1.2 g C, 0.2 g H, and 1.6 g O formula weight. = 0.1 mol C, 0.2 mol H, and 0.1 mol O MW of Compound "X" = 60 u Empirical formula weight of CH 2 O = 30 u 2. Find the whole number ratio of these moles by dividing by Ratio = 60/30 = 2/1. smallest mole value! The molecule is twice as heavy as the empirical formula. 0.1 mol C = 1 C 0.2 mol H = 2 H 0.1 mol O = 1 O 0.1 mol 0.1 mol 0.1 mol 2. Multiply each subscript of empirical formula by the ratio determined in step 1 Empirical formula = CH 2 O CH 2 O x 2 = C 2 H 4 O 2 = Molecular Formula

Slide 19 / 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). Mass of P(s) initial Mass of P(s) unreacted Fe(s) reacted Mass of iron oxide obtained. 1.45 g 1.03 g 3.05 g 4.22 g Mass of H 2 (g) initial Mass of H 2 (g) unreacted What is the empirical formula? 0.041 g 0.000 g Slide 21 / 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 given it has a molar mass of 58 g/mol. B H 2 NNH 2 C CH 3 OH D CH 3 CH 2 Cl Slide 23 / 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 B CH 2 Cl 2 B Na 3 N C C 2 H 6 Cl C NaN 3 D CH 3 Cl D NaN

Slide 25 / 163 Slide 26 / 163 14 Hydrazine is a component of rocket fuel. It consists of Law of Definite Composition 87.5% N with the rest being hydrogen by mass. If the molecular weight of the compound is 32 grams/mol, what is the molecular formula? If a compound is pure, it will always consist of the same composition no matter where the sample was taken or the size of the sample . A NH 2 Example: calcium carbonate B NH 3 If it's calcium carbonate, it's guaranteed to be 40% calcium, 48% C N 2 H 6 oxygen, and 12% carbon by mass. D N 2 H 4 Slide 27 / 163 Slide 28 / 163 Law of Definite Composition Law of Definite Composition Some substances are not pure and do not obey the law of definite Example: calcium carbonate composition. These are called mixtures. Sea water Sample Location Size Analysis Composition Sample % mass Sample Size location composition 20.0 g Ca 40% calcium Eastern 85.3% O 1 50.0 g 24.0 g O 48% oxygen Pennsylvania 10.7 % H Atlantic 1 500.0 g 6.0 g C 12% carbon Ocean 1.6% Na 100.0 g Ca 40 % calcium 2.4 % Cl 2 Wyoming 250.0 g 120.0 g O 48% oxygen 79.5 % O 30.0 g C 12% carbon 10.0 % H Indian 2 330.0 g Ocean 4.2 % Na 6.3 % Cl Slide 29 / 163 Slide 30 / 163 Practice Practice 15 16 Students type their answers here Students type their answers here The law of definite composition can be used mathematically to If a sample of water was found to contain 34.1 grams of oxygen, see how much of a given substance is present in a sample. how many grams of hydrogen and water would be present? Water is known to be 88.9% oxygen and 11.1 % hydrogen by mass. How many grams of oxygen would be present in a 400 gram sample of pure water?

Slide 31 / 163 Slide 32 / 163 17 Hydrogen peroxide is known to be 94.1 % oxygen by 18 Calcium oxide is 71.4% calcium by mass with the rest mass with the rest being hydrogen. How many grams of being oxygen. How many grams of calcium would be hydrogen would be present in a pure 230. gram sample present if a sample of calcium oxide sample was found to of hydrogen peroxide? contain 12.3 grams of oxygen? Slide 33 / 163 Slide 34 / 163 19 Calcium carbonate is known to be 40% Ca, 48% O, and Law of Definite Composition and the 12% carbon by mass. When a 200 gram sample of what Classification of Matter is thought to be pure calcium carbonate is decomposed, 18 grams of carbon are found in the sample. Is this This law allows us to classify the different types of matter substance pure? MATTER Does the material obey the Law of Definite composition? Yes Yes No Mixture Pure Substance No Can the material be broken down into different elements with distinct properties? Yes No Compounds Elements Slide 35 / 163 Slide 36 / 163 20 Two samples of a material are taken and the composition 21 A certain material is found to vary in composition by of each sample is given below. Is this material a pure mass. What kind of matter is this? substance? A Element Sample A Sample B Yes B Compound 45 % Cu, 12% Si, 43 % O 34% Cu, 19% Si, 47 % O No C Mixture Answer D Pure Substance