1 Chapter 6 Chemical Calculations 1 Submicroscopic Macroscopic
2 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 Mass (Ch 6.3) 5. Chemical Formula & Mole Calculations (Ch 6.4 & 6.5) 6. Empirical & Molecular Formula (supplemental material & Lab Exp 8) 7. Chemical Equations & Stoichiometric Calculations (Ch 6.6, 6.7, 6.8) 8. Heat of Reaction (supplemental material) 9. Percent Yield (supplemental material) 10. Limiting Reagent (supplemental material)
3 3 1. Formula Masses (Ch 6.1) Molecular mass/weight sum of atomic masses of the atoms in a molecule. Example: M.W. = 46.0 amu C: 2 x 12.0 amu = 24.0 amu H: 6 x 1.0 amu = 6.0 amu O: 1 x 16.0 amu = 16.0 amu 46.0 amu
4 4 Formula mass/weight sum of atomic masses of atoms in an ionic substance (formula unit, NOT a molecule). Example: Ammonium sulfide F.W. = 68.1 amu N: 2 x 14.0 amu = 28.0 amu H: 8 x 1.0 amu = 8.0 amu S: 1 x 32.1 amu = 32.1 amu 68.1 amu Due to its offensive smell, ammonium sulfide it is the active ingredient in a variety of foul pranks, including the common stink bomb.
5 5 2. Percent Composition (supplemental material) % Composition = # of g of each element in 100 g of a compound = mass of element x 100 total mass Example: Ammonium sulfide (NH 4 ) 2 S F.W. = 68.1 amu % N = 2 x 14.0 amu N x 100 = % N 68.1 amu (NH 4 ) 2 S % H = 8 x 1.0 amu H x 100 = % H 68.1 amu (NH 4 ) 2 S % S = 32.1 amu S x 100 = % S 68.1 amu (NH 4 ) 2 S
6 6 3. The Mole & Avogadro s Number (Ch 6.2) 1 dozen = 12 objects 1 ream = 500 sheets 1 mole = objects x Avogadro s Number
7 7 Sample problem: How many He atoms are in 2.55 moles of He? information x = information given factor sought 2.55 mole He x 6.02 x He atoms = He atoms 1 mole He
8 8 4. Molar Mass (Ch 6.3) 1 mole 12 C = exactly 12 g 12 C demo sample Molar mass of 12 C = We now have three conversion factors to relate moles, number of atoms, and mass of Carbon: 6.02 x C atoms 12 g C 1 mole 1 mole 6.02 x C atoms 12 g C
9 9 Figure mole of S, Zn, C, Mg, Pb, Si, Cu, Hg (start counterclockwise from yellow S and Hg in center) 12 C is our standard and when we compare it to other elements, we find that there are Avogadro s number of atoms of any element in a sample whose mass in grams is numerically equal to its atomic weight. Mg = amu 1 mole Mg = g Mg = 6.02 x atoms Mg Pb = amu 1 mol Pb = g Pb = 6.02 x atoms Pb
10 10 Now we can do the same for ionic compounds as well as for molecules because the molar mass is the mass (in grams) of a substance that is numerically equal to the substance s formula mass. Ammonium sulfide (NH 4 ) 2 S formula mass = 68.1 amu 68.1 g (NH 4 ) 2 S = 1 mole (molar mass or formula weight, F.W.: ) 68.1 g (NH 4 ) 2 S = 6.02 x formula units of (NH 4 ) 2 S Carbon dioxide CO 2 formula mass = amu g CO 2 = 1 mole (molar mass or molecular weight, M.W.: ) g CO 2 = 6.02 x molecules of CO 2
11 11 Sample Problem: If 7.50 moles of ammonia, NH 3, are required for a certain experiment, what mass of ammonia is needed? Formula mass = 3 x 1.0 (H) (N) = 17.0 amu 1 mole NH 3 (molar mass) = 17.0 g NH moles NH 3 x 17.0 g NH 3 = g NH 3 1 mole NH 3
12 12 5. Chemical Formula & Mole Calculations (Ch 6.4 & 6.5) The subscripts in a chemical formula give the number of moles of atoms present in 1 mole of the substance: Example: Ammonium sulfide (NH 4 ) 2 S For N: 2 moles of N atoms or 1 mole (NH 4 ) 2 S 1 mole (NH 4 ) 2 S formula units moles N For H: 8 moles of H atoms or 1 mole (NH 4 ) 2 S 1 mole (NH 4 ) 2 S formula units moles H For S: 1 mole of S atoms or 1 mole (NH 4 ) 2 S 1 mole (NH 4 ) 2 S formula units mole S
13 13 Sample calculation: How many H atoms are in 35.6 g of (NH 4 ) 2 S? (NH 4 ) 2 S formula mass = 68.1 amu 1 mole (NH 4 ) 2 S = 68.1 g 1 mole (NH 4 ) 2 S = moles H atoms 1 mole H atoms = 6.02 x H atoms Strategy: mass (NH 4 ) 2 S moles (NH 4 ) 2 S moles H atoms H 35.6 g (NH 4 ) 2 S x 1 mol (NH 4 ) 2 S x 8 mol H x 6.02 x H 68.1 g (NH 4 ) 2 S 1 mol (NH 4 ) 2 S 1 mol H = x 10 H atoms
14 14 Fig 6.7 Transitions allowed in solving chemical-formula bases problems: Drill problem: How many g of (NH 4 ) 2 S are required to obtain moles of NH 4 +? (NH 4 ) 2 S = 68.1 g/mol (molar mass) moles NH 4 + mole (NH 4 ) 2 S g (NH 4 ) 2 S 0.50 moles NH 4 + x 1 mole (NH 4 ) 2 S + 2 moles NH 4 x 68.1 g (NH 4 ) 2 S 1 mol (NH 4 ) 2 S = (NH 4 ) 2 S
15 15 6. Empirical & Molecular Formula (supplemental material & Lab Exp 8) C 6 H 12 O 6 Molecular Formula CH 2 O Empirical Formula The Empirical Formula (E.F.) is the simplest ratio of atoms in a compound. acid C 2 H 4 O 2 CH 2 O Molecular Formula Empirical Formula Both compounds have the same composition: 40.0 % C, 6.7 % H, 53.3 % O
16 16 If we are given the experimentally determined composition of a compound, we can calculate the E.F. Sample calculation. Composition: 40.0 % C, 6.7 % H, 53.3 % O Step 1: assume a 100 g sample and convert to For C: 40.0 g C x 1 mole C = 3.33 moles C 12.0 g C For H: 6.7 g H x 1 mole H = 6.7 moles H 1.0 g H For O: 53.3 g O x 1 mole O = 3.33 moles O 16.0 g O Step 2: divide each number of moles by the smallest of the numbers to obtain mole ratios = E.F. For C: 3.33/3.33 = 1.0 For H: 6.7/3.33 = 2.0 For O: 3.33/3.33 = 1.0 Empirical Formula =
17 17 Drill problem: Composition of Borazole = 40.28%B, 52.20%N, 7.52%H Molar mass = 80.5 amu Calculated the molecular formula B: 40.28g x 1 mole = moles = g N: 52.20g x 1 mole = moles = H: 7.52g x 1 mole = 7.45 moles = g E.F.= E.F. mass: (2x1.01) = Molar mass = amu = 3 E.F. mass amu Molecular Formula = 3 x (BNH 2 ) = B 3 N 3 H 6 Borazole
18 18 Mole ratios must be within 0.1 of a whole number. If they are not, each result must be multiplied by the same multiplication factor until every value is of a whole number. Example for a hypothetical set of mole ratios obtained from % composition: for C 1.98 for H for O The result for C is not of a whole number; therefore, each result must be multiplied by an integer until all of the values are. For the above example, the multiplication factor that works is 4: x 4 = for C 1.98 x 4 = 7.92 for H x 4 = for O Empirical Formula =
19 7. Chemical Equations & Stoichiometric Calculations (Ch 6.6, 6.7, 6.8) Summary of submicroscopic and macroscopic levels of a chemical equation: 2 Na(s) + 2 H 2 O 2 NaOH(aq) + H 2 (g) 2 atoms + 2 molecules 2 formula units + 1 molecule 2 moles + 2 moles 2 moles + 1 mole 19 2x23=46g + 2x18=36g 2x40=80g + 1x2=2g reactants products Law of Conservation of Mass
20 20 The coefficients in a balanced equation give the numerical relationships among formula units consumed or produced in a chemical reaction. Keys to the calculations = N H 2 2 NH 3 1 mole N 2 3 moles H 2 2 moles NH 3 3 moles H 2 1 mole N 2 1 mole N 2 1 mole N 2 3 moles H 2 2 moles NH 3 2 moles NH 3 2 moles NH 3 3 mole H 2
21 21 Figure 6.9 In solving stoichiometric calculations, only the following transitions are allowed: Sample calculation: How many g O 2 are needed to convert 45 g glucose (C 6 H 12 O 6 ) into CO 2 and H 2 O? First write a balanced equation and calculated the molecular masses of glucose and oxygen: M.W. C 6 H 12 O O 2 6 CO amu + 32 amu Then set up the calculations according to Figure H 2 O 45 g glu x 1 mol glu x mol O 2 x 32 g O g glu 1 mol glu 1 mol O 2 = O 2
22 22 Drill Problem: Figure 6.10 The chemical equation for the deployment of airbags is 2 NaN 3 (s) 2 Na(s) + 3 N 2 (g) How many g NaN 3 would have to decompose on order to generate 253 million molecules of N 2? F.W. NaN 3 = 65.0 amu Avogadro s # = 6.02 x molecules N 2 moles N 2 moles NaN 3 g NaN x10 8 molecules N 2 x 1 mole N 2 = x moles N x molecules N x moles N 2 x moles NaN 3 x 65.0 g NaN 3 moles N 2 mol NaN 3 = x g NaN 3
23 23 8. Heat of Reaction (Ch 9.5 & supplemental material) When the chemical energy stored in reactants is greater than that stored in the products, energy is released by the reaction, and it is termed exothermic. The change in energy is called the enthalpy change and is represented by H; the value is negative for an exothermic reaction. Example: combustion of propane is exothermic. The heat released can be represented with energy on the product side: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l) kcal However, the reaction is also represented by placing the enthalpy change to the right of the equation: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l) H =
24 24 An endothermic reaction is a chemical reaction in which a continuous input of energy is needed for the reaction to occur. Energy is a reactant. Example: photosynthesis is endothermic H = positive 6 CO 2 (g) + 6 H 2 O(g) kcal C 6 H 12 O 6 (aq) + 6 O 2 (g) 6 CO 2 (g) + 6 H 2 O(g) C 6 H 12 O 6 (aq) + 6 O 2 (g) H =
25 25 Sample Problem: How much energy is produced when 0.50 g of butane, C 4 H 10, is burned in a butane lighter? C 4 H 10 = 58.1 g/mole 2 C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O(l) kcal The equation shows that 1365 kcal of heat are produced when 2 moles of butane undergo combustion. 0.50g C 4 H 10 x 1 mol C 4 H 10 x 1365 kcal = kcal 58.1 g C 4 H 10 moles C 4 H 10 Is this reaction exothermic or endothermic? What is the sign for H?
26 26 9. Percent Yield (supplemental material) The theoretical yield in a reaction is the amount of product that could be obtained if a given reactant reacted completely. In most cases the actual yield is smaller that the theoretical yield because of side reactions, incomplete reaction, or other experimental limitations. The discrepancy between the theoretical yield and the actual yield is reported as the percent yield, which is calculated as shown: % yield = actual yield x 100 theoretical yield The theoretical yield is calculated from the given amount of the specified reactant. The actual yield is identified in the problem.
27 27 Fe 2 O C 2 Fe + 3 CO 1. Calculate the theoretical yield: g Fe 2 O 3 mol Fe 2 O 3 mol Fe g Fe Pure iron can be produced from iron oxide in a blast furnace. Sample Problem. When 884 g of Fe 2 O 3 was reduced with excess carbon, 507 g of Fe were obtained. What was the percent yield? 884g Fe 2 O 3 x 1 mol Fe 2 O 3 x 2 moles Fe x 55.85g Fe = g Fe 159.7g Fe 2 O 3 1 mol Fe 2 O 3 1 mol Fe 2. Calculate the % yield: % yield = actual yield x 100 = 507 g x 100 = % theoretical yield g
28 Limiting Reagent (supplemental material) Two batteries are required for these flashlights to work. So, if you have 10 flashlights and 17 batteries, how many working flashlights do you have? The batteries are the LIMITING REAGENT. The flashlights are IN EXCESS.
29 29 Reaction A: Stoichiometric amounts of reactants are used. NaOH + HCl NaCl + H 2 O 1 mol + 1 mol 1 mol + 1 mol Reaction B: One reactant is limiting. NaOH + HCl NaCl + H 2 O 0.6 mol mol mol + mol Which is the limiting reagent? Which reagent is in excess? What are the amounts of each product? How much of each compound is present at the end of the reaction? NaOH = HCl = NaCl = H 2 O = Please take note that you must calculate the theoretical yield of any reaction only from the limiting reagent!
30 30 Sample Problem: If you have 25 moles of N 2 reacting with 45 moles of H 2 according to the following reaction, which is the limiting reagent? N 2 (g) + 3 H 2 (g) 2 NH 3 (g) moles of A x mole ratio = moles of B available needed 25 moles N 2 x 3 moles H 2 = moles H 2 1 mole N 2 For 25 moles N 2 we would need moles H 2 but only 45 moles H 2 are available. H 2 = limiting reagent You can do the same analysis starting with H 2 45 moles H 2 x 1 moles N 2 = moles N 2 moles H 2 We have more N 2 than we need. N 2 =
31 31 Here is another way to finding the limiting reagent. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 25 moles 45 moles The limiting reagent has the lowest mole-to-coefficient ratio: 25 moles N 2 = moles H 2 = 15 1 mole N 2 3 moles H 2
32 32 Demo: Identify the limiting reagent HC 2 H 3 O 2 (aq) + NaHCO 3 (s) H 2 O(l) + CO 2 (g) + NaC 2 H 3 O 2 (aq) A B 0.19 mol mol in Rxn I 0.19 mol mol in Rxn II 0.19 mol mol in Rxn III 0.19 mol mol in Rxn IV Reagent A and B are mixed and CO 2 evolution is measured. Which is the limiting reagent? Rxn I Rxn II Rxn III Rxn IV B is limiting B is limiting stoichiometric amounts (balanced) A is limiting
33 33 The following drill problem summarizes the important chemical calculations you have learned in this chapter: When aqueous solutions of CaCl 2 and AgNO 3 are mixed, a white precipitate of AgCl forms. If 3.33 g CaCl 2 are reacted with 8.50 g AgNO 3 and 5.63 g AgCl are obtained, what is the % yield? CaCl 2 (aq) + 2 AgNO 3 (aq) 2 AgCl(s) + Ca(NO 3 ) 2 (aq) Calculated formula masses Convert g to moles Check if you have a limiting reagent Calculate theoretical yield Calculate % yield
34 34 CaCl 2 (aq) + 2 AgNO 3 (aq) 2 AgCl(s) + Ca(NO 3 ) 2 (aq) 3.33 g 8.50 g 5.63 g amu amu amu 3.33 g CaCl 2 x 1 mol = mol CaCl g 8.50 g AgNO 3 x 1 mol = mol AgNO g Mole-to-coefficient ratios: CaCl 2 = AgNO 3 = Theoretical yield: mol AgNO 3 x 2 mol AgCl x g AgCl = g AgCl 2 mol AgNO 3 1 mol AgCl % Yield: actual yield x 100 = 5.63 g x 100 = % theoretical yield g
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
Chapter Three Calculations with Chemical Formulas and Equations Mass and Moles of a Substance Chemistry requires a method for determining the numbers of molecules in a given mass of a substance. This allows
The Mole Concept Ron Robertson r2 c:\files\courses\1110-20\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
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
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
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
Chapter 3 Calculation with Chemical Formulas and Equations Practical Applications of Chemistry Determining chemical formula of a substance Predicting the amount of substances consumed during a reaction
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.
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
Chapter 3: Stoichiometry Goal is to understand and become proficient at working with: 1. Chemical equations (Balancing REVIEW) 2. Some simple patterns of reactivity 3. Formula weights (REVIEW) 4. Avogadro's
Chapter 4: The Mole Atomic mass provides a means to count atoms by measuring the mass of a sample The periodic table on the inside cover of the text gives atomic masses of the elements The mass of an atom
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
Chapter 3 Insert picture from First page of chapter Stoichiometry: Ratios of Combination Copyright McGraw-Hill 2009 1 3.1 Molecular and Formula Masses Molecular mass - (molecular weight) The mass in amu
CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS FOLLOW UP PROBLEMS 3.1A Plan: The mass of carbon must be changed from mg to g. The molar mass of carbon can then be used to determine the number of moles.
Chemistry I: Using Chemical Formulas Formula Mass The sum of the average atomic masses of all elements in the compound. Units are amu. Molar Mass - The mass in grams of 1 mole of a substance. Substance
Chapter 3 Chemical Reactions and Reaction Stoichiometry 許富銀 ( Hsu Fu-Yin) 1 Stoichiometry The study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry.
Chapter 3: Stoichiometry Goal is to understand and become proficient at working with: 1. Avogadro's Number, molar mass and converting between mass and moles (REVIEW). 2. empirical formulas from analysis.
Chapter 3: Stoichiometry Mole - Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical Equations 3.4 Calculating the Amounts
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
PERIODIC TABLE OF ELEMENTS 4/23/14 Chapter 7: Chemical Reactions 1 CHAPTER 7: CHEMICAL REACTIONS 7.1 Describing Reactions 7.2 Types of Reactions 7.3 Energy Changes in Reactions 7.4 Reaction Rates 7.5 Equilibrium
Chapter 3: Molecular analysis Read: BLB 3.3 3.5 H W : BLB 3:21a, c, e, f, 25, 29, 37,49, 51, 53 Supplemental 3:1 8 CHEMICAL FORMULA Formula that gives the TOTAL number of elements in a molecule or formula
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
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 10-24 g Atomic weight: Average mass of all isotopes of a given
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
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
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
Chemistry 100 Bettelheim, Brown, Campbell & Farrell Ninth Edition Introduction to General, Organic and Biochemistry Chapter 4 Chemical Reactions Chemical Reactions In a chemical reaction, one set of chemical
Chapter 4 Chemical Equations & Stoichiometry Chemical reactions are best described using equations which tells us what compounds we started with (reactants), what we did to them (reaction conditions) and
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
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
Measuring Atomic Masses Mass Spectrometer used to isolate isotopes of an element and determine their mass. 1 An element sample is heated to vaporize it and the gaseous atoms are zapped with an electron
Chemical Equations & Stoichiometry Chapter Goals Balance equations for simple chemical reactions. Perform stoichiometry calculations using balanced chemical equations. Understand the meaning of the term
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
Chapter 7: Stoichiometry - Mass Relations in Chemical Reactions How do we balance chemical equations? How can we used balanced chemical equations to relate the quantities of substances consumed and produced
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
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
Chemical Reactions Chapter 6 Chemical Arithmetic Balancing Equations The mole Gram - mole conversions Mole - mole relationships in chemical equations Mass relationships in chemical equations Per cent yeilds
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.
Chapter 3 Stoichiometry of Formulas and Equations Chapter 3 Outline: Mole - Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing
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
The Mole, Avogadro s Number, and Molar Mass Example: How many atoms are present in 2.0 kg of silver? (1 amu = 1.6605402x10-24 g) Example: How many molecules are present in 10. mg of smelling salts, (NH
Formula Conventions (1 of 24) Superscripts used to show the charges on ions Mg 2+ the 2 means a 2+ charge (lost 2 electrons) Subscripts used to show numbers of atoms in a formula unit H 2 SO 4 two H s,
CHEMICAL REACTIONS A chemical reaction is a rearrangement of atoms in which some of the original bonds are broken and new bonds are formed to give different chemical structures. In a chemical reaction,
Chapter 3. Stoichiometry: Mole-Mass Relationships in Chemical Reactions Concept 1. The meaning and usefulness of the mole The mole (or mol) represents a certain number of objects. SI def.: the amount of
Stoichiometry Review There are 20 problems in this review set. Answers, including problem set-up, can be found in the second half of this document. 1. N 2 (g) + 3H 2 (g) --------> 2NH 3 (g) a. nitrogen
Chapter 5 Chemical Quantities and Reactions 5.1 The Mole Collection Terms A collection term states a specific number of items. 1 dozen donuts = 12 donuts 1 ream of paper = 500 sheets 1 case = 24 cans 1
1 1. Classify the following reaction? Mg(CN) 2 (s) + H 2 SO 4 (aq) MgSO 4 (aq) + 2 HCN(g) a) Single-Replacement Reaction b) Double-Replacement Reaction c) Redox Reaction d) Combination Reaction e) Neutralization
Chapter 11 Chemical Calculations For the past several weeks we have been working on our qualitative understanding of first atoms, then molecules, and finally chemical reactions. In this chapter we enter
CHAPTER 3 Calculations with Chemical Formulas and Equations MOLECULAR WEIGHT (M. W.) Sum of the Atomic Weights of all atoms in a MOLECULE of a substance. FORMULA WEIGHT (F. W.) Sum of the atomic Weights
Lecture 5, The Mole What is a mole? Moles Atomic mass unit and the mole amu definition: 12 C = 12 amu. The atomic mass unit is defined this way. 1 amu = 1.6605 x 10-24 g How many 12 C atoms weigh 12 g?
What coefficients mean: 2 Na + Cl 2 2NaCl 2 Na 1 Cl 2 2NaCl 4 Na 2 Na + Cl 2 4Cl 2 6 moles Na 2NaCl 10 atoms Na ONLY WORKS FOR MOLES, MOLECULES, ATOMS 1. How many moles of H 2 and O 2 must react to form
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
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
MOLES AND CALCULATIONS USING THE MOLE CONCEPT INTRODUCTORY TERMS A. What is an amu? 1.66 x 10-24 g B. We need a conversion to the macroscopic world. 1. How many hydrogen atoms are in 1.00 g of hydrogen?
Stoichiometry Stoichiometry (greek): Stoicheion element, metry to measure Balanced Chemical equation: Skills Tells Ex. formula writing balancing equations substances involved in the chemical rxn relationship
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)
score /10 pts. Name Class Date 7.1 Stoichiometry and Percent Yield Mole Ratios An example: The combustion of propane is used to heat many rural homes in winter. Balance the equation below for the combustion
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
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
Chapter 9 Chemical Composition (Moles) Which weighs more, 1 atom of He or 1 atom of O? Units of mass: Pound Kilogram Atomic mass unit (AMU) There are others! 1 amu = 1.66 x 10-24 grams = mass of a proton
Introductory Chemistry, 3 rd Edition Nivaldo Tro Quantities in Car an octane and oxygen molecules and carbon dioxide and water Chemical Reactions Roy Kennedy Massachusetts Bay Community College Wellesley
1. Balance the following equation. What is the sum of the coefficients of the reactants and products? 1 Fe 2 O 3 (s) + _3 C(s) 2 Fe(s) + _3 CO(g) a) 5 b) 6 c) 7 d) 8 e) 9 2. Which of the following equations
Chapter 3 Mass Relationships in Chemical Reactions 國防醫學院生化學科王明芳老師 2011-9-20 1 Balancing Chemical Equations A balanced chemical equation shows that the law of conservation of mass is adhered to. In a balanced
PERCENTAGE COMPOSITION STOICHOMETRY UNIT 3 1. A sample of magnesium weighing 2.246 g burns in oxygen to form 3.724 g of magnesium oxide. What are the percentages of magnesium and oxygen in magnesium oxide?
Lecture 5 Outline 5.1 Stoichiometry,, the mole etc. 5.2 Chemical Equations 5.3 Molarity 5.4 Limiting reagents and yields 5.5 Reaction enthalpies and Gibbs free energy 5.6 Catalyst Lecture 5 Stoichiometry
1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) A) 1 B) 2 C) 4 D) 5 E) Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) Al (s) + H 2 O (l)? Al(OH)
Quantities of Reactants and Products CHAPTER 3 Chemical Reactions Stoichiometry Application of The Law of Conservation of Matter Chemical book-keeping Chemical Equations Chemical equations: Describe proportions
CHEM1001 2014-J-2 June 2014 22/01(a) What is the molarity of the solution formed when 0.50 g of aluminium fluoride is dissolved in 800.0 ml of water? 2 The molar mass of AlF 3 is: molar mass = (26.98 (Al)
MASS RELATIONS IN CHEMISTRY; STOICHIOMETRY [MH5; Ch. 3] Each element has its own unique mass. The mass of each element is found on the Periodic Table under the chemical symbol for the element (it is usually
Stoichiometry: Calculations with Chemical Equations Objectives Use chemical equations to predict amount of product from given reactants Determine percentage yield Determine limiting reactant Working with
Ch. 6 Chemical Composition and Stoichiometry The Mole Concept [6.2, 6.3] Conversions between g mol atoms [6.3, 6.4, 6.5] Mass Percent [6.6, 6.7] Empirical and Molecular Formula [6.8, 6.9] Bring your calculators!
AP Chemistry Unit #3 Chapter 3 Zumdahl Stoichiometry C6H12O6 + 6 O2 6 CO2 + 6 H2O Students should be able to: Calculate the atomic weight (average atomic mass) of an element from the relative abundances
CHEMICAL REACTIONS A chemical reaction is a rearrangement of atoms in which some of the original bonds are broken and new bonds are formed to give different chemical structures. In a chemical reaction,
Chapter 7 Part II: Chemical Formulas and Equations Mr. Chumbley Chemistry 1-2 SECTION 3: USING CHEMICAL FORMULAS Molecules and Formula Unit We have not yet discussed the different ways in which chemical
Chapter 3 (Hill/Petrucci/McCreary/Perry Stoichiometry: Chemical Calculations This chapter deals with quantitative relationships in compounds and between compounds in chemical reactions. These quantitative
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.
Chapter 6 Notes Chemical Composition Section 6.1: Counting By Weighing We can weigh a large number of the objects and find the average mass. Once we know the average mass we can equate that to any number
Chapter 3 Calculations with Chemical Formulas and Equations Concept Check 3.1 You have 1.5 moles of tricycles. a. How many moles of seats do you have? b. How many moles of tires do you have? c. How could
MASS RELATIONSHIPS IN CHEMICAL REACTIONS 1. The mole, Avogadro s number and molar mass of an element. Molecular mass (molecular weight) 3. Percent composition of compounds 4. Empirical and Molecular formulas