Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1. A chemical equation. (C4.4)


 Andrea Page
 7 years ago
 Views:
Transcription
1 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page Question What is a symbolic representation of a chemical reaction? What 3 things (values) is a mole of a chemical equal to? What is the average mass (in atomic mass units) of the naturally occurring isotopes of an element? What is the number named for Amadeo Avogadro and what is that number equal to? What is the symbol for atomic mass unit? The typical formula for oxygen is? The typical formula for fluorine is? The typical formula for hydrogen is? The typical formula for iron is? The typical formula for argon is? 7. What is an equality statement? Answer A chemical equation. (C4.4) There are 3 answers that you should know so far from the unit handout and your mole day project: (C4.4) a) 1 mole of any elemental particle (atoms, molecules, formula units, electrons, neutrons, protons, monoatomic and polyatomic ions, and so on) is equal to 6.02 x of that elemental particle. i.e x of an object = 1 mole of those same objects. b) 1 mole of any gas is equal to 22.4 liters of that gas but only so long as the gas is at standard temperature and pressure (STP) c) 1 mole of any chemical is equal to the sum of the atomic masses of all the atoms in the formula unit of that chemical with those masses written in units of grams Atomic mass. (C4.4) and of anything equals a mole of those things. (C4.4) u is the preferred symbol for atomic mass unit but amu is an older symbol. (C4.4) O 2. F 2. H 2. Fe. Ar. A mathematic statement showing that 2 values with different measurement units are equal to each other. Example: 22.4 L of gas = 1 mole of gas. (C4.4) More information page 1. Every student was required to complete a mole project in which this question was answered. This can also be found on the Mole Map on page 23 of the unit handout. page 3. page 3. page 3. Previous units beginning on page 7 but mentioned throughout.
2 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 2 Question 8. What is formula mass? 9. What is molecular mass? Where do we get the molar ratio (be very specific)? What is molar mass and how do we in this class calculate it? What is the maximum amount of product that can be produced from a given amount of limiting reactant? What do we in this class use an equality statement to construct? Answer Formula mass is the term primarily used for ionic substances. It is the sum of the atomic masses of all of the elements contained in one formula unit of a compound; typically an ionic compound. (C4.4) Molecular mass is the term used for molecular compounds. It is the sum of the atomic masses of all of the elements in the molecular formula of the substance. (C4.4) The numerical values in molar ratios come from the coefficients in a balanced chemical equation. (C4.4) Molar mass is the sum of the atomic masses of the elements in a formula unit, molecular formula, or other chemical formula written with units of grams. This calculation is equal to one mole of that formula unit. There is a required format for calculating molar mass which shows the elements in the formula, the atomic masses of those elements, a multiplicands which is numerically equal to the number of atoms of that element in the formula, the product of those atomic masses and multiplicand, the sum of the products, units of grams and the species for the formula and a equality statement which shows the mass equal to a mole of that same species. (C4.4) Theoretical yield. (C4.4) An equality statement is used to construct conversion units. (C4.4) More information page 6. page 6. beginning on page 13 but mentioned throughout. Throughout the unit handout; but there is a clear example on page 10. page 19. Throughout the unit handout; but there is a clear example on page 7.
3 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page Question What is a limiting reactant and how do we determine which reactant is the limiting reactant? 15. What is percent yield? 16. The clean surface of pure aluminum quickly oxidizes to form a coating of aluminum oxide (Al 2 O 3 ) that actually protects the aluminum under the surface from further corrosion. What is the percent composition of aluminum oxide? Answer A limiting reactant is the reactant that runs out 1 st in the chemical reaction, leaving some amount of the excess reactant. In this unit, we determine the limiting reactant by performing a masstomass calculation from the given amount of one of the reactants to a calculated amount of the 2 nd reactant. If the calculated mass of the 2 nd reactant is greater than the given mass of that reactant, than that same reactant is the limiting reactant (the given mass is the mass in the problem or the mass that you measure in the lab). If not, the 2 nd reactant is the excess reactant and the 1 st reactant is the limiting reactant. (C4.4) Percent yield is the actual yield divided by the theoretical yield multiplied by 100. (C 4.4) % Al and % O. (C4.5) More information page 16 and 17. page 20. page 10. Also, see the complete explanation for this calculation below.
4 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page Question You perform the following lab procedure: 2.00g of calcium chloride and 1.00g of aluminum sulfate are dissolved in separate containers with 100mL of water which are then mixed together. A white mixture forms and the mixture is allowed to sit, undisturbed, for 10 minutes. After sitting it is observed that the white material has settled to the bottom of the beaker and a clear liquid remains in the beaker. The clear liquid is decanted and the white material is filtered, rinsed, dried, and weighed. The amount of product recovered weighs 0.45g. What is the solid, white product, what is the percent yield, and what is the amount of excess reactant that remains in solution? So long as there is plenty of oxygen, burning of hydrocarbon fuels produces carbon dioxide. When the supply of oxygen is limited however, burning of hydrocarbon fuels produces poisonous carbon monoxide. Propose an explanation for this phenomenon. Answer Solid, white product is CaSO 4. % yield is 38%. Amount of excess reactant remaining is 1.03 g CaCl 2. (C4.5) More information pages 17 through 22. Also, see the complete explanation for this calculation below. Since the reaction with a hydrocarbon fuel and oxygen can produce water and EITHER carbon dioxide and carbon monoxide there must be a model that would explain why one or the other might be produced. One possible model (and it is by no means the ONLY reasonable model) is this: When the oxygen is limited, the carbon will combine with oxygen in a form that uses less oxygen. CO (carbon monoxide) has less oxygen per molecule or per mole than CO 2 (carbon dioxide).
5 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 5 Problem #16; complete explanation: What is the percent composition of aluminum oxide (Al 2 O 3 )? Percent composition simply means the percent mass of each element in a compound. Since we were using carbon dioxide as an example earlier, let s continue with that for now. The atomic mass of aluminum (elemental symbol Al) is u and the atomic mass of oxygen (elemental symbol O) is u (see the periodic table in your test references; these numbers are in the bottom rectangle for these elements). The sum of the atomic masses of both aluminum atoms in aluminum oxide is u. The sum of the atomic masses of the 3 oxygen atoms in aluminum oxide is u. The total atomic mass of all the atoms in the formula (the sum of the atomic masses) for aluminum oxide is u which must be rounded to u. To find the percent composition for aluminum you simply take the total mass of the 2 aluminum atoms and divide by the entire mass of aluminum oxide formula unit then multiply by 100. To find the percent composition for oxygen you simply take the total mass of the 3 oxygen atoms and divide by the entire mass of aluminum oxide formula unit then multiply by 100. The percent composition for aluminum oxide comes in 2 parts: 1) the percent mass of aluminum and 2) the percent mass of oxygen. The percent composition for any compound comes in as many parts as there are elements in the formula. Example: Percent composition of aluminum oxide Example: The required format for calculating percent composition begins with the same format required for calculating molar mass in this class (meaning, if you don t do it I count off points of tests, homework, and class work). So for carbon dioxide you begin with The required format for calculating molar mass and the beginning of percent composition in this class. Al: = Mass of all aluminum atoms O: = Mass of all oxygen atoms g Al 2 O g Al 2 O 3 = 1 mol Al 2 O 3 Mass of the aluminum oxide formula unit Equality statement showing that the calculated mass of aluminum oxide is equal to a mole of aluminum oxide. From the calculations above, you have the mass of all the aluminum atoms and the mass of all the oxygen atoms and the mass of the entire compound. % mass of aluminum in Al O = 2 3 % mass of oxygen in Al O = u u u u % % Al in Al O % % O in Al2O 3. Check: The sum of the percent mass of aluminum and the percent mass of oxygen is equal to or is very close to 100%: % % % % 2 3
6 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 6 Problem #17; complete explanation: What is the solid, white product, what is the percent yield, and what is the amount of excess reactant that remains in solution? 2.00g of calcium chloride and 1.00g of aluminum sulfate dissolved in separate containers of water, a white precipitate forms and is collected and weighed. The amount of product recovered weighs 0.45g. Step 1: Make sure the chemical equation is balanced. If you have an unbalanced chemical equation presented to you, then you need to balance the equation first. If you are only given the reactants you will have to predict the products and balance the product formulas (you cannot properly balance a chemical equation unless the formulas are balanced). o You may only be given the word formulas for the reactants, in which case you must be able to write the symbolic formula from the word formula and balance those formulas as well. o In order to predict products you must know how to classify chemical equations. Calcium has a 2+ charge and chloride has a 1 charge (see your periodic table of oxidation numbers in your test references). So the balanced formula for this compound is CaCl 2. Aluminum has a 3+ charge and sulfate has a 2 charge (see your common ions chart in your test references to find the formula and charge on the sulfate polyatomic ion). So the balanced formula for this compound is Al 2 (SO 4 ) 3. Review unit 4 if you don t recall how balance and write formulas. So the beginning of our equation is CaCl 2 + Al 2 (SO 4 ) 3. Both reactants are soluble (see solubility rules in test references), so CaCl 2 (aq) + Al 2 (SO 4 ) 3 (aq). Both formulas are in 2 parts, so the likelihood is that this will be a double replacement reaction. The predicted products in this unbalanced equation will be CaCl 2 (aq) + Al 2 (SO 4 ) 3 (aq) AlCl 3 (aq) + CaSO 4 (s). Using the solubility rules you should be able to determine that aluminum chloride is soluble and since this reaction occurred in a water (aqueous) environment it will be identified as aqueous (aq). Calcium sulfate is not soluble and will, therefore, be a solid(s) and must be identified as such. This answers the 1 st part of the question, What is the solid, white product? Answer: calcium sulfate or CaSO 4. To balance the equation (or just to show that the equation is already balanced) you MUST have an atom inventory. The result is 3CaCl 2 (aq) + Al 2 (SO 4 ) 3 (aq) 2AlCl 3 (aq) + 3CaSO 4 (s). Ca Cl Al S O Review unit 5 if you don t recall how predict products or balance equations.
7 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 7 Step 2: Find the limiting reactant (also called the limiting reagent or limiting factor). In this unit, this determination will always be a masstomass calculation. This masstomass calculation will start with one reactant and end with the other reactant. o The rule is that if the calculated amount (called a theoretical amount) of the reactant at the end of the masstomass calculation is greater than the amount given in the problem, then the species at the end of the calculation is the limiting reactant. o If the calculated amount (called a theoretical amount) of the reactant at the end of the masstomass calculation is smaller than the amount given in the problem, then the species at the end of the calculation is the excess reactant. o One reactant is virtually always a limiting reactant and the other reactant is the excess reactant. To find a limiting reactant you must know how to calculate the molar mass (or molar weight) of the reactants and products AND you must know how to determine a molar ratio. o Determine molar mass by writing out the formula, such as H 2 O for water, and look up the weight of each atom on the periodic table. For example, multiply the hydrogen atom weight by two and add it to the weight of oxygen. Use only the molar mass calculation format required and approved for this class. o The molar ratio comes from the coefficients and formulas in the balanced chemical equation. Perform a masstomass calculation from one reactant to another. To do this we need to calculate the molar mass of calcium chloride and aluminum sulfate: Ca: = Cl: = g CaCl 2 = 1 mol acl 2 Now that we have this equality statement ( g CaCl 2 = 1 mol CaCl 2 ) we can enter a molar conversion unit for calcium chloride to the calculation. Al: = S: = O: = g Al 2 (SO 4 ) 3 = 1 mol Al 2 (SO 4 ) 3 Now, using these equality statements and the molar ratio from the balanced equation we can construct the following masstomass calculation from the mass for calcium chloride that is given in the problem to the mass of aluminum sulfate needed to completely react with that amount of calcium chloride g CaCl2 1 mol CaCl 2 1 mol Al SO g Al SO g CaCl2 3 mol CaCl 2 1 mol Al SO g Al SO 2.06 g Al SO. This masstomass calculation demonstrates that you need 2.06 grams of aluminum sulfate to completely react with all of the 2.00 grams of calcium chloride that you were given. The problem is that you don t have need 2.06 grams of aluminum sulfate. You were only given 1.00 gram of
8 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 8 aluminum sulfate. Therefore, aluminum sulfate must be the limiting reactant. Every masstomass calculation that follows this one will start with the given amount (1.00 gram) of aluminum sulfate. Step 3: Calculate the theoretical yield or how much will be synthesized under perfect conditions. In this unit, this determination will always be a masstomass calculation. To do this we must perform a masstomass calculation from the given amount of limiting reactant to the product that we are seeking. The product we are seeking is clearly identified as the solid, white precipitate (read the original question), and that is calcium sulfate (CaSO 4 ). Rule for determining the limiting reactant: 1. Complete a masstomass calculation from one reactant to the other. 2. If the amount of reactant that you calculated is greater than the amount given to you in the problem the ending species is the limiting reactant. 3. If the amount calculated is less than the amount given to you in the problem the ending species is the excess reactant. To perform this masstomass calculation, we will need a molar mass calculation for calcium sulfate: Ca: = S: = O: = g CaSO 4 = 1 mol CaSO g Al SO 1 mol 1 Al SO g Al SO 3 mol CaSO4 1 mol Al SO g CaSO 4 1 mol CaSO g CaSO 1.19 g CaSO. 4 4 Step 4: Know the actual yield, or the amount of product truly synthesized in the original experiment. This involves no additional calculations. It is simply the amount measured at the end of the experiment or given to you in a word problem. The problem states that, The amount of product recovered weighs 0.45 g. This, then, is the actual yield. Step 5: Calculate percentage yield (mass of actual yield divided by mass of theoretical yield multiplied by 100 percent). Percent yield is the actual yield divided by the theoretical yield multiplied by 100. Actual Yield Theoretical Yield 100 = Percentage Yield Therefore, the percentage yield would be: 0.45 g CaSO g CaSO4 100 = % 37% yield in calcium sulfate. Step 6: Calculate the amount of excess reactant remaining after the reaction is complete.
9 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 9 Before we can calculate the amount of excess reactant remaining, we must first determine the amount of excess reactant used. This also involves a masstomass calculation. For this calculation we also start with the limiting reactant, but we calculate to the other reactant the excess reactant. This allows us to determine just how much of the excess reactant is actually used up in the reaction: 1.00 g Al SO 1 mol 1 Al SO g Al SO 3 mol CaCl2 1 mol Al SO g CaCl 2 1 mol CaCl g CaCl g CaCl. 2 2 To find the amount of excess reactant remaining, we simply take this amount (0.973 g CaCl 2 ) and subtract it from the amount of calcium chloride given in the problem: 2.00 g CaCl 2 given in the problem g CaCl 2 used up in the reaction g CaCl g CaCl 2 excess reactant remaining after the reaction is finished.
Chapter 3: Stoichiometry
Chapter 3: Stoichiometry Key Skills: Balance chemical equations Predict the products of simple combination, decomposition, and combustion reactions. Calculate formula weights Convert grams to moles and
More informationChemistry B11 Chapter 4 Chemical reactions
Chemistry B11 Chapter 4 Chemical reactions Chemical reactions are classified into five groups: A + B AB Synthesis reactions (Combination) H + O H O AB A + B Decomposition reactions (Analysis) NaCl Na +Cl
More informationChapter 3. Chemical Reactions and Reaction Stoichiometry. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT
Lecture Presentation Chapter 3 Chemical Reactions and Reaction James F. Kirby Quinnipiac University Hamden, CT The study of the mass relationships in chemistry Based on the Law of Conservation of Mass
More information1. How many hydrogen atoms are in 1.00 g of hydrogen?
MOLES AND CALCULATIONS USING THE MOLE CONCEPT INTRODUCTORY TERMS A. What is an amu? 1.66 x 1024 g B. We need a conversion to the macroscopic world. 1. How many hydrogen atoms are in 1.00 g of hydrogen?
More informationThe Mole Concept. The Mole. Masses of molecules
The Mole Concept Ron Robertson r2 c:\files\courses\111020\2010 final slides for web\mole concept.docx The Mole The mole is a unit of measurement equal to 6.022 x 10 23 things (to 4 sf) just like there
More informationIB Chemistry 1 Mole. One atom of C12 has a mass of 12 amu. One mole of C12 has a mass of 12 g. Grams we can use more easily.
The Mole Atomic mass units and atoms are not convenient units to work with. The concept of the mole was invented. This was the number of atoms of carbon12 that were needed to make 12 g of carbon. 1 mole
More informationCalculating Atoms, Ions, or Molecules Using Moles
TEKS REVIEW 8B Calculating Atoms, Ions, or Molecules Using Moles TEKS 8B READINESS Use the mole concept to calculate the number of atoms, ions, or molecules in a sample TEKS_TXT of material. Vocabulary
More informationChemical Calculations: The Mole Concept and Chemical Formulas. AW Atomic weight (mass of the atom of an element) was determined by relative weights.
1 Introduction to Chemistry Atomic Weights (Definitions) Chemical Calculations: The Mole Concept and Chemical Formulas AW Atomic weight (mass of the atom of an element) was determined by relative weights.
More informationStoichiometry. What is the atomic mass for carbon? For zinc?
Stoichiometry Atomic Mass (atomic weight) Atoms are so small, it is difficult to discuss how much they weigh in grams We use atomic mass units an atomic mass unit (AMU) is one twelfth the mass of the catbon12
More informationUnit 2: Quantities in Chemistry
Mass, Moles, & Molar Mass Relative quantities of isotopes in a natural occurring element (%) E.g. Carbon has 2 isotopes C12 and C13. Of Carbon s two isotopes, there is 98.9% C12 and 11.1% C13. Find
More informationChemical Equations & Stoichiometry
Chemical Equations & Stoichiometry Chapter Goals Balance equations for simple chemical reactions. Perform stoichiometry calculations using balanced chemical equations. Understand the meaning of the term
More informationChemical Calculations: Formula Masses, Moles, and Chemical Equations
Chemical Calculations: Formula Masses, Moles, and Chemical Equations Atomic Mass & Formula Mass Recall from Chapter Three that the average mass of an atom of a given element can be found on the periodic
More informationChem 31 Fall 2002. Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations. Writing and Balancing Chemical Equations
Chem 31 Fall 2002 Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Writing and Balancing Chemical Equations 1. Write Equation in Words you cannot write an equation unless you
More informationElement of same atomic number, but different atomic mass o Example: Hydrogen
Atomic mass: p + = protons; e  = electrons; n 0 = neutrons p + + n 0 = atomic mass o For carbon12, 6p + + 6n 0 = atomic mass of 12.0 o For chlorine35, 17p + + 18n 0 = atomic mass of 35.0 atomic mass
More informationIB Chemistry. DP Chemistry Review
DP Chemistry Review Topic 1: Quantitative chemistry 1.1 The mole concept and Avogadro s constant Assessment statement Apply the mole concept to substances. Determine the number of particles and the amount
More informationEXPERIMENT 7 Reaction Stoichiometry and Percent Yield
EXPERIMENT 7 Reaction Stoichiometry and Percent Yield INTRODUCTION Stoichiometry calculations are about calculating the amounts of substances that react and form in a chemical reaction. The word stoichiometry
More informationFormulas, Equations and Moles
Chapter 3 Formulas, Equations and Moles Interpreting Chemical Equations You can interpret a balanced chemical equation in many ways. On a microscopic level, two molecules of H 2 react with one molecule
More information= 16.00 amu. = 39.10 amu
Using Chemical Formulas Objective 1: Calculate the formula mass or molar mass of any given compound. The Formula Mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all
More informationMole Notes.notebook. October 29, 2014
1 2 How do chemists count atoms/formula units/molecules? How do we go from the atomic scale to the scale of everyday measurements (macroscopic scale)? The gateway is the mole! But before we get to the
More informationChapter 3! Stoichiometry: Calculations with Chemical Formulas and Equations. Stoichiometry
Chapter 3! : Calculations with Chemical Formulas and Equations Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2 H 2 O (g) Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2
More informationChem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses
Chem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses B. Calculations of moles C. Calculations of number of atoms from moles/molar masses 1. Avagadro
More informationName Date Class STOICHIOMETRY. SECTION 12.1 THE ARITHMETIC OF EQUATIONS (pages 353 358)
Name Date Class 1 STOICHIOMETRY SECTION 1.1 THE ARITHMETIC OF EQUATIONS (pages 353 358) This section explains how to calculate the amount of reactants required or product formed in a nonchemical process.
More informationIntroduction to Chemistry
1 Copyright ç 1996 Richard Hochstim. All rights reserved. Terms of use. Introduction to Chemistry In Chemistry the word weight is commonly used in place of the more proper term mass. 1.1 Atoms, Ions, and
More informationMoles. Moles. Moles. Moles. Balancing Eqns. Balancing. Balancing Eqns. Symbols Yields or Produces. Like a recipe:
Like a recipe: Balancing Eqns Reactants Products 2H 2 (g) + O 2 (g) 2H 2 O(l) coefficients subscripts Balancing Eqns Balancing Symbols (s) (l) (aq) (g) or Yields or Produces solid liquid (pure liquid)
More informationBalance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O
Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O Ans: 8 KClO 3 + C 12 H 22 O 11 8 KCl + 12 CO 2 + 11 H 2 O 3.2 Chemical Symbols at Different levels Chemical symbols represent
More informationMOLES AND MOLE CALCULATIONS
35 MOLES ND MOLE CLCULTIONS INTRODUCTION The purpose of this section is to present some methods for calculating both how much of each reactant is used in a chemical reaction, and how much of each product
More informationW1 WORKSHOP ON STOICHIOMETRY
INTRODUCTION W1 WORKSHOP ON STOICHIOMETRY These notes and exercises are designed to introduce you to the basic concepts required to understand a chemical formula or equation. Relative atomic masses of
More informationStudy Guide For Chapter 7
Name: Class: Date: ID: A Study Guide For Chapter 7 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The number of atoms in a mole of any pure substance
More informationChem 115 POGIL Worksheet  Week 4 Moles & Stoichiometry
Chem 115 POGIL Worksheet  Week 4 Moles & Stoichiometry Why? Chemists are concerned with mass relationships in chemical reactions, usually run on a macroscopic scale (grams, kilograms, etc.). To deal with
More informationThe Empirical Formula of a Compound
The Empirical Formula of a Compound Lab #5 Introduction A look at the mass relationships in chemistry reveals little order or sense. The ratio of the masses of the elements in a compound, while constant,
More informationPart One: Mass and Moles of Substance. Molecular Mass = sum of the Atomic Masses in a molecule
CHAPTER THREE: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS Part One: Mass and Moles of Substance A. Molecular Mass and Formula Mass. (Section 3.1) 1. Just as we can talk about mass of one atom of
More informationCalculations and Chemical Equations. Example: Hydrogen atomic weight = 1.008 amu Carbon atomic weight = 12.001 amu
Calculations and Chemical Equations Atomic mass: Mass of an atom of an element, expressed in atomic mass units Atomic mass unit (amu): 1.661 x 1024 g Atomic weight: Average mass of all isotopes of a given
More informationChapter 1: Moles and equations. Learning outcomes. you should be able to:
Chapter 1: Moles and equations 1 Learning outcomes you should be able to: define and use the terms: relative atomic mass, isotopic mass and formula mass based on the 12 C scale perform calculations, including
More informationFormulae, stoichiometry and the mole concept
3 Formulae, stoichiometry and the mole concept Content 3.1 Symbols, Formulae and Chemical equations 3.2 Concept of Relative Mass 3.3 Mole Concept and Stoichiometry Learning Outcomes Candidates should be
More informationStoichiometry Exploring a StudentFriendly Method of Problem Solving
Stoichiometry Exploring a StudentFriendly Method of Problem Solving Stoichiometry comes in two forms: composition and reaction. If the relationship in question is between the quantities of each element
More informationLiquid phase. Balance equation Moles A Stoic. coefficient. Aqueous phase
STOICHIOMETRY Objective The purpose of this exercise is to give you some practice on some Stoichiometry calculations. Discussion The molecular mass of a compound is the sum of the atomic masses of all
More informationOther Stoich Calculations A. mole mass (mass mole) calculations. GIVEN mol A x CE mol B. PT g A CE mol A MOLE MASS :
Chem. I Notes Ch. 12, part 2 Using Moles NOTE: Vocabulary terms are in boldface and underlined. Supporting details are in italics. 1 MOLE = 6.02 x 10 23 representative particles (representative particles
More informationCalculation of Molar Masses. Molar Mass. Solutions. Solutions
Molar Mass Molar mass = Mass in grams of one mole of any element, numerically equal to its atomic weight Molar mass of molecules can be determined from the chemical formula and molar masses of elements
More information20.2 Chemical Equations
All of the chemical changes you observed in the last Investigation were the result of chemical reactions. A chemical reaction involves a rearrangement of atoms in one or more reactants to form one or more
More informationWhat s in a Mole? Molar Mass
LESSON 10 What s in a Mole? Molar Mass OVERVIEW Key Ideas Lesson Type Lab: Groups of 4 Chemists compare moles of substances rather than masses because moles are a way of counting atoms. When considering
More informationCONSERVATION OF MASS During a chemical reaction, matter is neither created nor destroyed.  i. e. the number of atoms of each element remains constant
1 CHEMICAL REACTINS Example: Hydrogen + xygen Water H + H + +  Note there is not enough hydrogen to react with oxygen  It is necessary to balance equation. reactants products + H + H (balanced equation)
More informationStoichiometry. Lecture Examples Answer Key
Stoichiometry Lecture Examples Answer Key Ex. 1 Balance the following chemical equations: 3 NaBr + 1 H 3 PO 4 3 HBr + 1 Na 3 PO 4 2 C 3 H 5 N 3 O 9 6 CO 2 + 3 N 2 + 5 H 2 O + 9 O 2 2 Ca(OH) 2 + 2 SO 2
More informationStoichiometry Limiting Reagent Laboratory. Chemistry 118 Laboratory University of Massachusetts, Boston
Chemistry 118 Laboratory University of Massachusetts, Boston STOICHIOMETRY  LIMITING REAGENT 
More informationF321 MOLES. Example If 1 atom has a mass of 1.241 x 1023 g 1 mole of atoms will have a mass of 1.241 x 1023 g x 6.02 x 10 23 = 7.
Moles 1 MOLES The mole the standard unit of amount of a substance (mol) the number of particles in a mole is known as Avogadro s constant (N A ) Avogadro s constant has a value of 6.02 x 10 23 mol 1.
More informationChemical Composition. Introductory Chemistry: A Foundation FOURTH EDITION. Atomic Masses. Atomic Masses. Atomic Masses. Chapter 8
Introductory Chemistry: A Foundation FOURTH EDITION by Steven S. Zumdahl University of Illinois Chemical Composition Chapter 8 1 2 Atomic Masses Balanced equation tells us the relative numbers of molecules
More informationYIELD YIELD REACTANTS PRODUCTS
Balancing Chemical Equations A Chemical Equation: is a representation of a chemical reaction in terms of chemical formulas Example: 1. Word Description of a Chemical Reaction When methane gas (CH 4 ) burns
More informationstoichiometry = the numerical relationships between chemical amounts in a reaction.
1 REACTIONS AND YIELD ANSWERS stoichiometry = the numerical relationships between chemical amounts in a reaction. 2C 8 H 18 (l) + 25O 2 16CO 2 (g) + 18H 2 O(g) From the equation, 16 moles of CO 2 (a greenhouse
More informationThe Mole and Molar Mass
The Mole and Molar Mass 1 Molar mass is the mass of one mole of a substance. Molar mass is numerically equal to atomic mass, molecular mass, or formula mass. However the units of molar mass are g/mol.
More informationName Class Date. Section: Calculating Quantities in Reactions. Complete each statement below by writing the correct term or phrase.
Skills Worksheet Concept Review Section: Calculating Quantities in Reactions Complete each statement below by writing the correct term or phrase. 1. All stoichiometric calculations involving equations
More informationAtomic Masses. Chapter 3. Stoichiometry. Chemical Stoichiometry. Mass and Moles of a Substance. Average Atomic Mass
Atomic Masses Chapter 3 Stoichiometry 1 atomic mass unit (amu) = 1/12 of the mass of a 12 C atom so one 12 C atom has a mass of 12 amu (exact number). From mass spectrometry: 13 C/ 12 C = 1.0836129 amu
More informationChapter Three: STOICHIOMETRY
p70 Chapter Three: STOICHIOMETRY Contents p76 Stoichiometry  The study of quantities of materials consumed and produced in chemical reactions. p70 31 Counting by Weighing 32 Atomic Masses p78 Mass Mass
More informationCHEM 101/105 Numbers and mass / Counting and weighing Lect03
CHEM 101/105 Numbers and mass / Counting and weighing Lect03 Interpretation of Elemental Chemical Symbols, Chemical Formulas, and Chemical Equations Interpretation of an element's chemical symbol depends
More informationThe Mole Notes. There are many ways to or measure things. In Chemistry we also have special ways to count and measure things, one of which is the.
The Mole Notes I. Introduction There are many ways to or measure things. In Chemistry we also have special ways to count and measure things, one of which is the. A. The Mole (mol) Recall that atoms of
More informationChapter 5, Calculations and the Chemical Equation
1. How many iron atoms are present in one mole of iron? Ans. 6.02 1023 atoms 2. How many grams of sulfur are found in 0.150 mol of sulfur? [Use atomic weight: S, 32.06 amu] Ans. 4.81 g 3. How many moles
More information2 The Structure of Atoms
CHAPTER 4 2 The Structure of Atoms SECTION Atoms KEY IDEAS As you read this section, keep these questions in mind: What do atoms of the same element have in common? What are isotopes? How is an element
More informationStoichiometry Limiting Reagent Laboratory. Chemistry 118 Laboratory University of Massachusetts, Boston
Chemistry 118 Laboratory University of Massachusetts, Boston STOICHIOMETRY  LIMITING REAGENT 
More informationMatter. Atomic weight, Molecular weight and Mole
Matter Atomic weight, Molecular weight and Mole Atomic Mass Unit Chemists of the nineteenth century realized that, in order to measure the mass of an atomic particle, it was useless to use the standard
More informationCHEM 110: CHAPTER 3: STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS
1 CHEM 110: CHAPTER 3: STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS The Chemical Equation A chemical equation concisely shows the initial (reactants) and final (products) results of
More informationATOMS. Multiple Choice Questions
Chapter 3 ATOMS AND MOLECULES Multiple Choice Questions 1. Which of the following correctly represents 360 g of water? (i) 2 moles of H 2 0 (ii) 20 moles of water (iii) 6.022 10 23 molecules of water (iv)
More informationSample Problem: STOICHIOMETRY and percent yield calculations. How much H 2 O will be formed if 454 g of. decomposes? NH 4 NO 3 N 2 O + 2 H 2 O
STOICHIOMETRY and percent yield calculations 1 Steps for solving Stoichiometric Problems 2 Step 1 Write the balanced equation for the reaction. Step 2 Identify your known and unknown quantities. Step 3
More informationEXPERIMENT 12: Empirical Formula of a Compound
EXPERIMENT 12: Empirical Formula of a Compound INTRODUCTION Chemical formulas indicate the composition of compounds. A formula that gives only the simplest ratio of the relative number of atoms in a compound
More informationChemical Reactions in Water Ron Robertson
Chemical Reactions in Water Ron Robertson r2 f:\files\courses\111020\2010 possible slides for web\waterchemtrans.doc Properties of Compounds in Water Electrolytes and nonelectrolytes Water soluble compounds
More informationCP Chemistry Review for Stoichiometry Test
CP Chemistry Review for Stoichiometry Test Stoichiometry Problems (one given reactant): 1. Make sure you have a balanced chemical equation 2. Convert to moles of the known substance. (Use the periodic
More informationChapter 8 How to Do Chemical Calculations
Chapter 8 How to Do Chemical Calculations Chemistry is both a qualitative and a quantitative science. In the laboratory, it is important to be able to measure quantities of chemical substances and, as
More informationGetting the most from this book...4 About this book...5
Contents Getting the most from this book...4 About this book....5 Content Guidance Topic 1 Atomic structure and the periodic table...8 Topic 2 Bonding and structure...14 Topic 2A Bonding....14 Topic 2B
More informationHow To Calculate Mass In Chemical Reactions
We have used the mole concept to calculate mass relationships in chemical formulas Molar mass of ethanol (C 2 H 5 OH)? Molar mass = 2 x 12.011 + 6 x 1.008 + 1 x15.999 = 46.069 g/mol Mass percentage of
More informationChapter 1 The Atomic Nature of Matter
Chapter 1 The Atomic Nature of Matter 6. Substances that cannot be decomposed into two or more simpler substances by chemical means are called a. pure substances. b. compounds. c. molecules. d. elements.
More informationMoles Lab mole. 1 mole = 6.02 x 1023. This is also known as Avagadro's number Demo amu amu amu
Moles I. Lab: Rice Counting II. Counting atoms and molecules I. When doing reactions chemists need to count atoms and molecules. The problem of actually counting individual atoms and molecules comes from
More informationAS1 MOLES. oxygen molecules have the formula O 2 the relative mass will be 2 x 16 = 32 so the molar mass will be 32g mol 1
Moles 1 MOLES The mole the standard unit of amount of a substance the number of particles in a mole is known as Avogadro s constant (L) Avogadro s constant has a value of 6.023 x 10 23 mol 1. Example
More informationChapter 3 Stoichiometry
Chapter 3 Stoichiometry 31 Chapter 3 Stoichiometry In This Chapter As you have learned in previous chapters, much of chemistry involves using macroscopic measurements to deduce what happens between atoms
More informationTHE MOLE / COUNTING IN CHEMISTRY
1 THE MOLE / COUNTING IN CHEMISTRY ***A mole is 6.0 x 10 items.*** 1 mole = 6.0 x 10 items 1 mole = 60, 00, 000, 000, 000, 000, 000, 000 items Analogy #1 1 dozen = 1 items 18 eggs = 1.5 dz.  to convert
More informationUnit 9 Stoichiometry Notes (The Mole Continues)
Unit 9 Stoichiometry Notes (The Mole Continues) is a big word for a process that chemist s use to calculate amounts in reactions. It makes use of the coefficient ratio set up by balanced reaction equations
More informationMoles. Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations
Moles Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations Micro World atoms & molecules Macro World grams Atomic mass is the mass of an
More informationChapter 3 Mass Relationships in Chemical Reactions
Chapter 3 Mass Relationships in Chemical Reactions Student: 1. An atom of bromine has a mass about four times greater than that of an atom of neon. Which choice makes the correct comparison of the relative
More informationMultiple Choice Identify the letter of the choice that best completes the statement or answers the question.
Introduction to Chemistry Exam 2 Practice Problems 1 Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1.Atoms consist principally of what three
More informationTopic 4 National Chemistry Summary Notes. Formulae, Equations, Balancing Equations and The Mole
Topic 4 National Chemistry Summary Notes Formulae, Equations, Balancing Equations and The Mole LI 1 The chemical formula of a covalent molecular compound tells us the number of atoms of each element present
More informationCHAPTER 8: CHEMICAL COMPOSITION
CHAPTER 8: CHEMICAL COMPOSITION Active Learning: 14, 68, 12, 1825; EndofChapter Problems: 34, 982, 8485, 8792, 94104, 107109, 111, 113, 119, 125126 8.2 ATOMIC MASSES: COUNTING ATOMS BY WEIGHING
More informationStoichiometry. Unit Outline
3 Stoichiometry Unit Outline 3.1 The Mole and Molar Mass 3.2 Stoichiometry and Compound Formulas 3.3 Stoichiometry and Chemical Reactions 3.4 Stoichiometry and Limiting Reactants 3.5 Chemical Analysis
More informationMOLES, MOLECULES, FORMULAS. Part I: What Is a Mole And Why Are Chemists Interested in It?
NAME PARTNERS SECTION DATE_ MOLES, MOLECULES, FORMULAS This activity is designed to introduce a convenient unit used by chemists and to illustrate uses of the unit. Part I: What Is a Mole And Why Are Chemists
More informationTutorial 4 SOLUTION STOICHIOMETRY. Solution stoichiometry calculations involve chemical reactions taking place in solution.
T27 Tutorial 4 SOLUTION STOICHIOMETRY Solution stoichiometry calculations involve chemical reactions taking place in solution. Of the various methods of expressing solution concentration the most convenient
More informationExam 2 Chemistry 65 Summer 2015. Score:
Name: Exam 2 Chemistry 65 Summer 2015 Score: Instructions: Clearly circle the one best answer 1. Valence electrons are electrons located A) in the outermost energy level of an atom. B) in the nucleus of
More information2. The percent yield is the maximum amount of product that can be produced from the given amount of limiting reactant.
UNIT 6 stoichiometry practice test True/False Indicate whether the statement is true or false. moles F 1. The mole ratio is a comparison of how many grams of one substance are required to participate in
More informationHow much does a single atom weigh? Different elements weigh different amounts related to what makes them unique.
How much does a single atom weigh? Different elements weigh different amounts related to what makes them unique. What units do we use to define the weight of an atom? amu units of atomic weight. (atomic
More informationLecture Topics Atomic weight, Mole, Molecular Mass, Derivation of Formulas, Percent Composition
Mole Calculations Chemical Equations and Stoichiometry Lecture Topics Atomic weight, Mole, Molecular Mass, Derivation of Formulas, Percent Composition Chemical Equations and Problems Based on Miscellaneous
More informationMultiple Choice questions (one answer correct)
Mole Concept Multiple Choice questions (one answer correct) (1) Avogadro s number represents the number of atoms in (a) 12g of C 12 (b) 320g of sulphur (c) 32g of oxygen (d) 12.7g of iodine (2) The number
More informationChapter 6 Chemical Calculations
Chapter 6 Chemical Calculations 1 Submicroscopic Macroscopic 2 Chapter Outline 1. Formula Masses (Ch 6.1) 2. Percent Composition (supplemental material) 3. The Mole & Avogadro s Number (Ch 6.2) 4. Molar
More informationUnit 6 The Mole Concept
Chemistry Form 3 Page 62 Ms. R. Buttigieg Unit 6 The Mole Concept See Chemistry for You Chapter 28 pg. 352363 See GCSE Chemistry Chapter 5 pg. 7079 6.1 Relative atomic mass. The relative atomic mass
More informationThe Mole. 6.022 x 10 23
The Mole 6.022 x 10 23 Background: atomic masses Look at the atomic masses on the periodic table. What do these represent? E.g. the atomic mass of Carbon is 12.01 (atomic # is 6) We know there are 6 protons
More informationPART I: MULTIPLE CHOICE (30 multiple choice questions. Each multiple choice question is worth 2 points)
CHEMISTRY 12307 Midterm #1 Answer key October 14, 2010 Statistics: Average: 74 p (74%); Highest: 97 p (95%); Lowest: 33 p (33%) Number of students performing at or above average: 67 (57%) Number of students
More information1. What is the molecular formula of a compound with the empirical formula PO and a grammolecular mass of 284 grams?
Name: Tuesday, May 20, 2008 1. What is the molecular formula of a compound with the empirical formula PO and a grammolecular mass of 284 grams? 2 5 1. P2O 5 3. P10O4 2. P5O 2 4. P4O10 2. Which substance
More informationChapter 5. Chemical Reactions and Equations. Introduction. Chapter 5 Topics. 5.1 What is a Chemical Reaction
Introduction Chapter 5 Chemical Reactions and Equations Chemical reactions occur all around us. How do we make sense of these changes? What patterns can we find? 1 2 Copyright The McGrawHill Companies,
More informationAtomic mass is the mass of an atom in atomic mass units (amu)
Micro World atoms & molecules Laboratory scale measurements Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12 C weighs 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00
More informationDescription of the Mole Concept:
Description of the Mole Concept: Suppose you were sent into the store to buy 36 eggs. When you picked them up you would get 3 boxes, each containing 12 eggs. You just used a mathematical device, called
More informationTuesday, November 27, 2012 Expectations:
Tuesday, November 27, 2012 Expectations: Sit in assigned seat Get out Folder, Notebook, Periodic Table Have out: Spiral (notes), Learning Target Log (new) No Backpacks on tables Listen/Pay Attention Learning
More informationAppendix D. Reaction Stoichiometry D.1 INTRODUCTION
Appendix D Reaction Stoichiometry D.1 INTRODUCTION In Appendix A, the stoichiometry of elements and compounds was presented. There, the relationships among grams, moles and number of atoms and molecules
More information602X10 21 602,000,000,000, 000,000,000,000 6.02X10 23. Pre AP Chemistry Chemical Quan44es: The Mole. Diatomic Elements
Pre AP Chemistry Chemical Quan44es: The Mole Mole SI unit of measurement that measures the amount of substance. A substance exists as representa9ve par9cles. Representa9ve par9cles can be atoms, molecules,
More informationChemical Reactions 2 The Chemical Equation
Chemical Reactions 2 The Chemical Equation INFORMATION Chemical equations are symbolic devices used to represent actual chemical reactions. The left side of the equation, called the reactants, is separated
More informationChapter 4. Chemical Composition. Chapter 4 Topics H 2 S. 4.1 Mole Quantities. The Mole Scale. Molar Mass The Mass of 1 Mole
Chapter 4 Chemical Composition Chapter 4 Topics 1. Mole Quantities 2. Moles, Masses, and Particles 3. Determining Empirical Formulas 4. Chemical Composition of Solutions Copyright The McGrawHill Companies,
More information75.5. Translate chemical symbols and the chemical formulas of common substances to show the component parts of the substances including:
75.5 Translate chemical symbols and the chemical formulas of common substances to show the component parts of the substances including: NaCl [salt], H 2 O [water], C 6 H 12 O 6 [simple sugar], O 2 [oxygen
More informationSample Exercise 3.1 Interpreting and Balancing Chemical Equations
Sample Exercise 3.1 Interpreting and Balancing Chemical Equations The following diagram represents a chemical reaction in which the red spheres are oxygen atoms and the blue spheres are nitrogen atoms.
More information