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 Table of Contents: Compounds Pt. A Click on the topic to go to that section · Chemical Formulas · Metals & Alloys · Ionic Compounds · Covalent Compounds

Slide 4 / 163 Chemical Formulas Return to Table of Contents Slide 5 / 163 The Mole Recall that 1 mole is defined as 6.022 x 10 23 units of a given substance. 1 mol of electrons = 6.022 x 10 23 electrons 1 mol of H 2 O molecules = 6.022 x 10 23 molecules of water 1 mol of NaCl formula units = 6.022 x 10 23 formula units NaCl 1 mol of K atoms = 6.022 x 10 23 atoms of K Slide 6 / 163 The Mole Within 1 mole of a compound, there are often differing moles of each element In 1 mole of aluminum nitrate, Al(NO 3 ) 3 = 1 mol of Al 3+ ions = 3 mol of NO 3- ions = 3 mol of N atoms = 9 mol of O atoms

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

Slide 10 / 163 Molar Mass and Volume 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 4 What is the volume occupied @STP by 88 grams of carbon dioxide? Slide 12 / 163 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 mass? Natural gas (methane) pipeline.

Slide 13 / 163 6 Which of the following contains the most atoms of H? A 2 grams of H 2 gas B 16 grams of methane (CH 4 ) C 22.4 L of H 2 gas D 9 grams of water (H 2 O) Slide 14 / 163 7 How many moles of fluoride ions (F-) are present in a 79 gram sample of SnF 2 ? A 0.5 moles B 78.5 moles C 1 mole D 1.5 moles Slide 15 / 163 Chemical Formulas A chemical formula provides the ratio of atoms or moles of each element in a compound. H 2 O = 2 atoms H or 2 mol H 1 atom O 1 mol O Al(NO 3 ) 3 = 1 Al 3+ ion or 1 mol Al 3+ ions 3 NO 3- ions 3 mol NO 3- ions

Slide 16 / 163 Empirical and Molecular Formulas An empirical formula provides the simplest whole number ratio of atoms or moles of each element in a compound. Examples: H 2 O, NaCl, C 3 H 5 O A molecular formula represents the actual number of atoms or moles of each element in a compound. Examples: H 2 O, C 3 H 5 O, C 6 H 12 O 6 Slide 17 / 163 Calculating an Empirical Formula To find an empirical formula: 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 = 0.1 mol C, 0.2 mol H, and 0.1 mol O 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 0.1 mol 0.1 mol 0.1 mol Empirical formula = CH 2 O Slide 18 / 163 Calculating a Molecular Formula Determining the molecular formula of a compound is easy once the empirical formula and the molecular weight of the compound are known. 1. Determine the ratio of the molecular weight to the empirical formula weight. MW of Compound "X" = 60 u Empirical formula weight of CH 2 O = 30 u 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 / 163 Practice 8 Students type their answers here Given the following data, calculate the empirical formula of phosphine gas. Phosphine gas is created by reacting solid phosphorus with H 2 (g). Mass of P(s) initial Mass of P(s) unreacted 1.45 g 1.03 g Mass of H 2 (g) initial Mass of H 2 (g) unreacted 0.041 g 0.000 g Slide 20 / 163 Students type their answers here Practice 9 Black iron oxide (aka magnetite) is used as a contrast agent in MRI scans of human soft tissue. To determine the empirical formula, a student reacted solid iron with O 2 (g). Fe(s) reacted Mass of iron oxide obtained. 3.05 g 4.22 g What is the empirical formula? Slide 21 / 163 Students type their answers here Practice 10 Butane gas can be produced when solid carbon is reacted with hydrogen gas. If 0.45 grams of carbon were found to react with 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.

Slide 22 / 163 11 Which of the following is NOT an empirical formula? A Fe 2 O 3 B H 2 NNH 2 C CH 3 OH D CH 3 CH 2 Cl Slide 23 / 163 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 6.50 grams, what is the empirical formula? A Na 2 N 3 B Na 3 N C NaN 3 D NaN Slide 24 / 163 13 A compound containing carbon, hydrogen, and chlorine is 14.1% carbon by mass, 83.5% Cl, with the rest being hydrogen. What is the empirical formula? A C 2 H 5 Cl B CH 2 Cl 2 C C 2 H 6 Cl D CH 3 Cl

Slide 25 / 163 14 Hydrazine is a component of rocket fuel. It consists of 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? A NH 2 B NH 3 C N 2 H 6 D N 2 H 4 Slide 26 / 163 Law of Definite Composition 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 . Example: calcium carbonate If it's calcium carbonate, it's guaranteed to be 40% calcium, 48% oxygen, and 12% carbon by mass. Slide 27 / 163 Law of Definite Composition Example: calcium carbonate Sample Location Size Analysis Composition 20.0 g Ca 40% calcium Eastern 24.0 g O 48% oxygen 1 50.0 g Pennsylvania 6.0 g C 12% carbon 100.0 g Ca 40 % calcium 2 Wyoming 250.0 g 120.0 g O 48% oxygen 30.0 g C 12% carbon

Slide 28 / 163 Law of Definite Composition Some substances are not pure and do not obey the law of definite composition. These are called mixtures. Sea water Sample % mass Sample Size location composition 85.3% O 10.7 % H Atlantic 1 500.0 g Ocean 1.6% Na 2.4 % Cl 79.5 % O 10.0 % H Indian 2 330.0 g Ocean 4.2 % Na 6.3 % Cl Slide 29 / 163 Practice 15 Students type their answers here The law of definite composition can be used mathematically to see how much of a given substance is present in a sample. 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 30 / 163 Practice 16 Students type their answers here If a sample of water was found to contain 34.1 grams of oxygen, how many grams of hydrogen and water would be present?

Slide 31 / 163 17 Hydrogen peroxide is known to be 94.1 % oxygen by mass with the rest being hydrogen. How many grams of hydrogen would be present in a pure 230. gram sample of hydrogen peroxide? Slide 32 / 163 18 Calcium oxide is 71.4% calcium by mass with the rest being oxygen. How many grams of calcium would be present if a sample of calcium oxide sample was found to contain 12.3 grams of oxygen? Slide 33 / 163 19 Calcium carbonate is known to be 40% Ca, 48% O, and 12% carbon by mass. When a 200 gram sample of what is thought to be pure calcium carbonate is decomposed, 18 grams of carbon are found in the sample. Is this substance pure? Yes No

Slide 34 / 163 Law of Definite Composition and the Classification of Matter This law allows us to classify the different types of matter MATTER Does the material obey the Law of Definite composition? Yes No Mixture Pure Substance Can the material be broken down into different elements with distinct properties? Yes No Compounds Elements Slide 35 / 163 20 Two samples of a material are taken and the composition of each sample is given below. Is this material a pure substance? Sample A Sample B Yes 45 % Cu, 12% Si, 43 % O 34% Cu, 19% Si, 47 % O No Answer Slide 36 / 163 21 A certain material is found to vary in composition by mass. What kind of matter is this? A Element B Compound C Mixture D Pure Substance

Slide 37 / 163 22 Which of the following is TRUE regarding a pure compound? A It will not obey the law of definite composition B It can be broken down into different elements C The amounts of each element by mass in the compound will not vary D It must contain the same equal mass % of each element in the compound Slide 38 / 163 Metals & Alloys Return to Table of Contents Slide 39 / 163 Introduction to Chemical Bonds A chemical bond is an attraction between atoms to form a compound that contain two or more atoms. The bond forms due to the electrostatic attraction between opposite charges. There are three basic types of bonds: metallic, ionic, and covalent.