...CHAPTER 3 MASS RELATIONSHIPS IN CHEMICAL REACTIONS This chapter reviews the mole concept, balancing chemical equations, and stoichiometry. The topics covered in this chapter are: • Atomic mass and average atomic mass • A vogadro’ s number, mole, and molar mass • Percent composition calculations • Empirical and molecular formula determinations • Chemical equations, amount of reactant and product calculations • Limiting reagents and reaction yield calculations Take Note: It is absolutely essential that you master the mole concept to do well on the quantitative aspects of AP Chemistry!! When solving quantitative problems on the Free Response section of the AP exam, supporting work must be shown to receive credit. Using dimensional analysis is a very powerful technique in solving problems. Be sure to report your answer to the correct number of significant figures (see Chapter 1 in this review book). Atomic mass and average atomic mass Atomic mass is the mass of an atom in atomic mass units (amu). One amu is defined as 1/12 of one C-12 atom. The C-12 isotope has a mass of exactly 12.000 amu. The C-12 isotope provides the relative scale for the masses of the other elements. Average atomic mass is the value reported on the periodic table, which takes into account the various isotopes of an element and their respective frequencies. To calculate the average atomic mass of an element, add up the different masses of the isotopes (using amu) multiplied by each isotope’s abundance...
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...Stoichiometry ________________________________________ Stoichiometry is simply the math behind chemistry. Given enough information, one can use stoichiometry to calculate masses, moles, and percents within a chemical equation. ________________________________________ ________________________________________ What is a Chemical Equation? In chemistry, we use symbols to represent the various chemicals. Success in chemistry depends upon developing a strong familiarity with these basic symbols. For example, the symbol "C"represents an atom of carbon, and "H" represents an atom of hydrogen. To represent a molecule of table salt, sodium chloride, we would use the notation "NaCl", where "Na" represents sodium and "Cl" represents chlorine. We call chlorine "chloride" in this case because of its connection to sodium. You should have reviewed naming schemes, or nomenclature, in earlier readings. A chemical equation is an expression of a chemical process. For example: AgNO3(aq) + NaCl(aq) ---> AgCl(s) + NaNO3(aq) In this equation, AgNO3 is mixed with NaCl. The equation shows that the reactants (AgNO3 and NaCl) react through some process (--->) to form the products (AgCl and NaNO3). Since they undergo a chemical process, they are changed fundamentally. Often chemical equations are written showing the state that each substance is in. The (s) sign means that the compound is a solid. The (l) sign means the substance is a liquid. The (aq) sign stands for aqueous in water and means...
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...Determining the Limiting Reactant and Percent Yield in a Precipitation Reaction Objectives: • Observe the reaction between solutions of sodium carbonate and calcium chloride. • Determine which of the reactants is the limiting reactant and which is the excess reactant. • Determine the theoretical mass of precipitate that should form. • Compare the actual mass with the theoretical mass of precipitate and calculate the percent yield. Materials: Balance 0.70 M sodium carbonate solution, Na2CO3(aq) Graduated cylinder 0.50 M calcium chloride solution, CaCl2(aq) Beaker (250 mL) Wash Bottle (distilled H2O) Filter paper Funnel Iron ring Ring stand Procedure: Part I: The Precipitation Reaction (Day 1) 1. Obtain two clean, dry 25 mL graduated cylinders and one 250 mL beaker. 2. In one of the graduated cylinders, measure 25 mL of the Na2CO3 solution. In the other graduated cylinder, measure 25 mL of the CaCl2 solution. Record these volumes in your data table. 3. Pour the contents of both graduated cylinders into the 250 mL beaker and observe the results. Record these qualitative observations in your observations table. Allow the contents of the beaker to sit undisturbed for approximately 5 minutes to see what happens to the suspended solid particles. Meanwhile, proceed to step 4. 4. Obtain a piece of filter paper and put your initials and your partner’s initials on it using a pencil. Measure and record the mass of the filter paper, then use...
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...Title: Experiment 2 Determination of the valency of the magnesium Objective: -To study the quantitative relationship between the amount of reactants and products of a reaction. -A known starting mass of magnesium and the measured collection of hydrogen gas will be used to determine the reaction stoichiometry and the valency of magnesium. - To identify the unknown X value in the chemical equation between magnesium and hydrochloric acid - To determine the valency of magnesium Introduction: Stoichiometry is the study of the quantitative relationship between amounts of reactants and products of a reaction (that is, how many moles of A react with a given number of moles of B). In this section, a known starting mass of magnesium and the measured collection of hydrogen gas will be used to determine the reaction stoichiometry. Magnesium reacts with hydrochloric acid to form hydrogen gases. The aim of this experiment is to determine the value of x in the following equation: Mg + X HCI → MgCIX + H2 A known amount of magnesium is reacted with a large excess of HCl, and the volume of H2 evolved is measured. As HCl is in excess, all the magnesium will be consumed, and the yield of both MgClX and H2 is dependent on the amount of magnesium used. A comparison of the amount of hydrogen produced with the amount of magnesium consumed will enable the X value to be determined. Apparatus and Materials: Magnesium ribbon, 0.5M Hydrochloric acid, Burette (50 cm3), Pipette (25...
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...Braidon Berry Lab 4: Stoichiometry of a Precipitation Reaction Background: Stoichiometry is a challenging yet essential part of learning chemistry. By utilitizing the resources of this course, as well as the lab work. Learning this difficult section of the course can be done with hard work and sharp attention. Purpose: The purpose of this lab was to learn how to connect calculations with real life examples. Doing this lab while doing the calculations for the molarity and molar masses, made it more challenging, however it did help in seating the knowledge and connecting it to the course. Procedure: To start, the dry Calcium Chloride Dihydrate, was weighed out and mixed with a small amount of distilled water. After that step the molarity of the CaCl2 was determined and logged in the data sheet. Similar steps were followed in terms of measuring dry material, weighing it, and then mixing it with distilled water. This will be necessary in order to observe the reaction and calculate more. Then the water was now filtered in the paper filter, this filter would later be dried in the sunlight and weighed to measure the mass of the product. After the filtering was complete, the percent yield could be calculated, as well as the actual mass of the precipitate. Data: |Initial: CaCl2*2H20 (g) |1.0 g | |Initial: CaCl2*2H20 (moles) ...
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...Study Guide for Exame 2 CHAPTER 3: Stoichiometry * Stoichiometry – study of quantitative aspects of formulas and relations * The mole – SI unit for the amount of a substance. * The amount of matter that contains the same number of atoms as 12.0g of carbon -> 6.022 x 10^23 (Avogadro’s number) * Avogadro’s number – 6.022 x 10^23 * How to determine how many atoms of each element is in a compound: * (moles or grams)(6.022x10^23)(Number of atoms/1molecule) * Molar mass - Molar mass is the weight of one mole (or 6.022 x 1023 molecules) of any chemical compounds. * Mass % of an element in a compound: * ((Number of atoms of element)(atomic weight))/(Formula weight) * Empirical formula – Gives the lowest whole number ratio of atoms of each element in a compound (Grams)/(atomic weight) --- divide by lowest number on all * Molecular formula – gives actual whole number ratio of atoms of each element in each compound. (Molecular formula weight)/(Empirical formula weight) x compound * Formulas from analysis: * Structured formula – a formula that shows the atoms of a compound, their relative positions, and the bonds between them. * Isomers – compounds with the same molecular formula, but different properties and different arrangements of atoms * Writing chemical equations (symbols) : * + adding 2 or more chemicals together * -> Yields (Products) * (arrow forward and backward) reaction...
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...FHSC1114 PHYSICAL CHEMISTRY Revision for Topics 1 – 3 Topic 1 Principle of Chemistry 1. 2. The number of electron for an ion, X–, is 18. Calculate the number of proton and identify the ion. (Ans: 17 p+; Cl–) [Sep 2012] 20 X2+ and Y– ions have the same number of electrons as 10 Z isotope. Identify X2+ and Y– ions. (Ans: X2+= Mg2+ ; Y– = F–) [Dec 2015] 3. Bromine has two naturally occurring isotopes 79Br and 81Br with their masses are 78.92 a.m.u. and 80.92 a.m.u, respectively. Given that the average atomic mass for bromine is 79.90, find the relative percentage abundances of these isotopes. (Ans: 79Br: 51 %, 81Br: 49 %) [Apr 2014] 4. An element has two isotopes, P and Q. The ratio of percentage abundance of P to Q is 0.32. The isotopic mass of P and Q are 36.9695 and 34.9689, respectively. (a) Find the percentage abundance of isotope P and Q. (Ans: P: 24.24%, Q: 75.76%) (b) Determine the average atomic mass of the element. (Ans: 35.45 a.m.u.) [Sep 2012] 5. The number of atoms present in a sample of formic acid, HCOOH was found to be 1.736 × 1025. Determine the mass of formic acid in this sample. (Ans: 265.38 g) [Apr 2014] 6. Salicylic acid is a pain reliever which consists of 6.558 g of carbon atoms, 0.471 g of hydrogen and 3.744 g of oxygen atoms. Determine the empirical formula of salicylic acid. (Ans: C7H6O3) [Apr 2015] Topic 2 Chemical Bonding 1. Draw the Lewis structure for nitrogen triiodide, NI3. Calculate...
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...Lecturer : : : Centre for Foundation Studies Foundation In Science Year 1 Trimester 1 Unit Code Unit Title Session : : : FHSC1114 Physical Chemistry 2015/05 : Ms. Amelia Chiang, Ms. Azlina Banu, Ms. Farhanah, Ms.Gurpreet, Ms. Jamie, Ms. Lau Mei Chien, Ms. Lily Lee, Ms. Nabilah, Mr. Ng Sweet Kin, Ms. Phang Ying Ning, Ms. Precilla, Ms. Rachel Tham, Ms. Rajalakshmi, Mr. Sivabalan, Ms. Tan Lee Siew Tutorial 3: Chapter 3 Stoichiometry and Solution Concentration 1. Balance the following equations: (a) (b) 2. V2O5(s) + CaS(s) CaO(s) + V2S5(s) GaBr3(aq) + Na2SO3(aq) Ga2(SO3)3(aq) + NaBr(aq) 316.0 g of aluminum sulfide, Al2S3 reacts with 493.0 g of water, H2O. Given the unbalanced equation as below: Al2S3(s) + H2O(l) → Al(OH)3(s) + H2S(g) (a) Find the excess reactant. (Ans: H2O) (b) Find the mole of the excess reactant that remains after the reaction. (Ans: 14.742 mole) [Sep 2014] 3. Consider the reaction below: 2Al(s) + 3I2(s) 2AlI3(s) (a) Determine the limiting reagent and the theoretical yield of the product if 1.20 moles of aluminium and 2.40 moles of iodine are used. (Ans: 489.218 g) (b) Calculate the percentage yield of the product if 450 g of AlI3 is obtained. (Ans: 91.98%) 4. A salt solution is produced when 2.9 g of sodium chloride, NaCl dissolved in 200 ml of water. Calculate the molality (m) of the NaCl solution, given that the density of water is 1.00 g ml⁻1. (Ans: 0.25 m) [Apr...
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...Jing Tyng, Ms Jamie Anne, Ms Lau Mei Chien, Mr Ng Sweet Kin, Ms Phang Ying Ning, Ms Precilla, Ms Rachel Tham, Ms Rajalakshmi, Mr Tan Jun Bin, Ms Tan Lee Siew Tutorial 3 : Chapter 3 Stoichiometry and Solution Concentration 1. Balance the following equations: (a) Al(s) + NH4ClO4(s) → Al2O3(s) + AlCl3(s) + NO(g) + H2O(g) (b) GaBr3(aq) + Na2SO3(aq) → Ga2(SO3)3(aq) + NaBr(aq) 2. Ethanol, C2H5OH, is a liquid with a density of 0.789 g ml-1 at 25 °C. Calculate the molarity of ethanol solution made by dissolving 20.00 mL of ethanol at 25 °C in enough water to make 250.0 ml of solution. [Ans: 1.37 mol L-1] 3. Copper sulfate is widely used as a dietary supplement for animal feed. A lab technician prepares a “stock” solution of CuSO4 by dissolving 79.80 g of CuSO4 in enough water to make 500.0 mL of solution. (a) Determine the molarity of the CuSO4 “stock” solution prepared by the technician. [Ans: 1 mol L-1] (b) Calculate the volume of CuSO4 “stock” solution that should be diluted to give 2.5 L of [Ans: 0.25 L] 0.1 M CuSO4. 4. Aluminum is a limiting reactant in the reaction with sulfur gas to form aluminum sulfide. Initially, 1.18 mol of aluminum and 2.25 mol of sulfur are combined. 2Al(s) + 3S(s) → Al2S3(s) (a) Calculate the aluminum sulfide formed in moles. (b) Calculate the mass of the excess reactant after the reaction. [Ans: 0.59 mol] [Ans: 15.39 g] [December 2013] 1 5. A mixture of excess carbon disulphide, CS2, and 650 cm3 of 9.2 g dm-3 chlorine, Cl2, are passed through...
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... 1.2 p. 13-26 | Atoms, Molecules and Ions * Structure of the atom: subatomic particles, the nuclear atom, isotopes, mass number, atomic number, isotope symbol * Ions: cations, anions, isotope symbol for ions * Naming ionic compounds * Naming covalent compounds | TextChapter 22.2 p. 60-682.3 p. 69-71Chapter 33.1 to 3.5p. 91-111 | Chemical Composition * Atomic mass: relative atomic mass, isotopic mass and isotopic abundance * Percent composition * Moles and Particles * Molar mass * Empirical Formula and Molecular Formula | TextChapter 22.4 p 72-74Chapter 44.1 to 4.3p. 130 - 148 | Quantities in Chemical Reactions * The meaning of a balanced reaction * Mole to mole conversion * mass to mass conversion * limiting reactants * percent yield | TextChapter 66.1 to 6.5p. 217-234 | Gases * Behavior of gases * Factors that affect the properties of gases * volume and pressure * volume and temperature * volume, pressure and temperature * Gay-Lussac’s law * Avogadro’s hypothesis * Ideal Gas Law * Calculations using ideal gas law * Kinetic Molecular Theory of...
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...the back of the lecture hall. Students may leave cell phones/PDAs on the front bench of the lecture hall. Students who are caught using cell phones/PDAs during an exam will be ejected from the exam and will receive a score of zero. This violation will be reported to the Dean and the Vice-Provost who will then take the appropriate disciplinary action. Only pens/pencils and scientific calculators (non-programmable and non-graphing) are allowed to be with students during exams. Students are not allowed to take an exam in a lecture section in which they are not registered. Chapter 4 – sections 4.1 – 4.8 Given a balanced chemical equation, you should be able to calculate grams (or moles) of a product (or reactant) given the grams (or moles) of another product (or reactant). Remember: grams A ( moles A ( moles B ( grams B (p. 118). Practice Example 4.1...
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...Department of Medical Technology Faculty of Pharmacy University of Santo Tomas, España, Manila Organic Chemistry Laboratory Recrystallization of Acetanilide Experiment 4 Author: Janina Erika G. Sese Group 8 – 2C Medical Technology (A.Y. 2015-2016) Members: Kathleen Danielle Marie A. Robles, Amiel C. Sabangan, Hanz Jefry A. Saliendra, Andrea Betina M. Vega, Anna Denise Z. Yang ------------------------------------------------- Date Submitted: October 28, 2015 ------------------------------------------------- ABSTRACT Recrystallization is the simplest and most widely used operation for purifying organic solids that differ in their solubility at different temperatures. In this experiment, three test tubes with water, hexane and, ethanol, respectively, were used to dissolve pure acetanilide. These test tubes were placed in a water bath for one to five minutes, then, were cooled through running water for the selection of the best solvent to use in recrystallization. Crude acetanilide was obtained by mixing two milliliters aniline and 20 milliliters distilled water with three milliliters acetic anhydride, which was cooled with running water. It was then filtered and dried for the procurement of crude acetanilide. This was mixed with 20 milliliters of the chosen solvent and heated through water bath until the solid was dissolved. Next, it was immediately filtered while hot. The filtrate was collected and was cooled with running water leading to...
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...POINT LISAS CAMPUS Esperanza Road, Brechin Castle, Couva www.utt.edu.tt LAB 1 Decomposition reaction Aim: Determination of the number of moles of water molecules of crystallization present in hydrated Magnesium Sulphate (MgSO4.xH2O) Apparatus: Mass balance, test tube, test tube holder, heat-proof mat and bunsen burner. Reagents: Hydrated sodium carbonate. Theory: Chemical decomposition, analysis or breakdown is the separation of a chemical compound into elements or simple compounds. A more specific type of decomposition is thermal decomposition or thermolysis, which is caused by heat. ABA+B, the reaction is endothermic, since heat is required to break the chemical bonds. Most decomposition reaction require energy either in the form of heat, light or electricity. Absorption of energy causes the breaking of the bonds present in the reacting substance which decomposes to give the product. When a hydrated salt is heated it decomposes into a pure form of the salt and water. MgSO4.xH2O MgSO4 + H2O Procedure: Refer to Handout Results: A. Mass of test tube/g = 21.77 B. Mass of the tube and salt/g = 24.0 A table showing the mass of the test tube and salt after 3 consecutive heating: Heating | Mass of the test tube and salt/g | 1st | 23.96 | 2nd | 23.81 | 3rd | 23.81 | Calculations: G. Mass of anhydrous magnesium sulphate/g = F - A = 23.81 – 21.77= 2.04 H. Mass of water of crystallization evaporated/g...
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...AP Chemistry Summer Assignment Mr. Ronnenkamp Welcome to SPHS AP Chemistry!! You already have a background in chemistry from your General Chemistry class, but AP Chem is very different. Rather than memorizing how to do particular types of problems, you must really understand the chemistry and be able to apply it to different kinds of problems. AP Chemistry is a challenging course. To succeed, you must keep up with the assignments and be willing to spend time working through the material. Like all AP classes, AP Chem comes with a summer assignment. It is due the second day of class - August 25th - and will count as ½ of a test grade. ALL YOUR ANSWERS FOR THE SUMMER ASSIGNMENT SHOULD BE PUT ON ANOTHER SHEET OF PAPER!!!! We will then have a CLASS TEST on the SUMMER ASSIGNMENT ON 9/1/2015 (depending on schedule)!!! This class test will count as the second ½ of your first test grade. I check my e-mail frequently, so feel free to contact me if you are having problems doing the summer assignment. I can offer help via email. My e-mail address is: ronnenkampd@pcsb.org Please take the summer assignment seriously. Completing the summer assignment will allow you to enter AP Chemistry in August ready to succeed!!!! Each section of the summer assignment is referenced with web tutorials to help you if you have forgotten some of your General Chemistry, but feel free to use any good website. I will also be giving you access to an online textbook. ONLINE...
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...CHAPTER 14 CHEMICAL KINETICS PRACTICE EXAMPLES 1A (E) The rate of consumption for a reactant is expressed as the negative of the change in molarity divided by the time interval. The rate of reaction is expressed as the rate of consumption of a reactant or production of a product divided by its stoichiometric coefficient. A 0.3187 M 0.3629 M 1min rate of consumption of A = = = 8.93 105 M s 1 t 8.25 min 60 s rate of reaction = rate of consumption of A2 = 8.93 105 M s 1 4.46 105 M s 1 2 1B (E) We use the rate of reaction of A to determine the rate of formation of B, noting from the balanced equation that 3 moles of B form (+3 moles B) when 2 moles of A react (–2 moles A). (Recall that “M” means “moles per liter.”) 0.5522 M A 0.5684 M A 3moles B rate of B formation= 1.62 104 M s 1 60s 2 moles A 2.50 min 1min 2A (M) (a) The 2400-s tangent line intersects the 1200-s vertical line at 0.75 M and reaches 0 M at 3500 s. The slope of that tangent line is thus 0 M 0.75 M slope = = 3.3 104 M s 1 = instantaneous rate of reaction 3500 s 1200 s The instantaneous rate of reaction = 3.3 104 M s 1 . (b) At 2400 s, H 2 O 2 = 0.39 M. At 2450 s, H 2 O 2 = 0.39 M + rate t At 2450 s, H 2 O 2 = 0.39 M + 3.3 10 4 mol H 2 O 2 L1s 1 50s = 0.39 M 0.017 M = 0.37 M 2B (M) With only the data of Table 14.2 we can use only the reaction rate during the...
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