Chapter 6 Chemical Calculations

Size: px
Start display at page:

Download "Chapter 6 Chemical Calculations"

Transcription

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

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 information

Chapter 3. Chemical Reactions and Reaction Stoichiometry. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT

Chapter 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 information

The Mole Concept. The Mole. Masses of molecules

The Mole Concept. The Mole. Masses of molecules 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

More information

Moles. 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 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 information

Formulas, Equations and Moles

Formulas, 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

Part One: Mass and Moles of Substance. Molecular Mass = sum of the Atomic Masses in a molecule

Part 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 information

Sample Exercise 3.1 Interpreting and Balancing Chemical Equations

Sample 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

Chemistry B11 Chapter 4 Chemical reactions

Chemistry 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 information

Chapter 3! Stoichiometry: Calculations with Chemical Formulas and Equations. Stoichiometry

Chapter 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 information

Balance 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 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 information

Calculating Atoms, Ions, or Molecules Using Moles

Calculating 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 information

CHEMICAL FORMULA COEFFICIENTS AND SUBSCRIPTS. Chapter 3: Molecular analysis 3O 2 2O 3

CHEMICAL FORMULA COEFFICIENTS AND SUBSCRIPTS. Chapter 3: Molecular analysis 3O 2 2O 3 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

More information

Calculations and Chemical Equations. Example: Hydrogen atomic weight = 1.008 amu Carbon atomic weight = 12.001 amu

Calculations 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 10-24 g Atomic weight: Average mass of all isotopes of a given

More information

How 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. 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 information

Atomic Masses. Chapter 3. Stoichiometry. Chemical Stoichiometry. Mass and Moles of a Substance. Average Atomic Mass

Atomic 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 information

Sample 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

Sample 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 information

Chem 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 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 information

How To Calculate Mass In Chemical Reactions

How 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 information

Calculation of Molar Masses. Molar Mass. Solutions. Solutions

Calculation 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 information

Other Stoich Calculations A. mole mass (mass mole) calculations. GIVEN mol A x CE mol B. PT g A CE mol A MOLE MASS :

Other 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 information

Chapter 5, Calculations and the Chemical Equation

Chapter 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 information

Concept 1. The meaning and usefulness of the mole. The mole (or mol) represents a certain number of objects.

Concept 1. The meaning and usefulness of the mole. The mole (or mol) represents a certain number of objects. 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

More information

IB Chemistry. DP Chemistry Review

IB 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 information

Chem 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 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 information

Chemical Equations & Stoichiometry

Chemical 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 information

CHEMICAL REACTIONS. Chemistry 51 Chapter 6

CHEMICAL REACTIONS. Chemistry 51 Chapter 6 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,

More information

Chem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry

Chem 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 information

Stoichiometry Review

Stoichiometry Review 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

More information

Chemistry 65 Chapter 6 THE MOLE CONCEPT

Chemistry 65 Chapter 6 THE MOLE CONCEPT THE MOLE CONCEPT Chemists find it more convenient to use mass relationships in the laboratory, while chemical reactions depend on the number of atoms present. In order to relate the mass and number of

More information

Lecture 5, The Mole. What is a mole?

Lecture 5, The Mole. What is a mole? 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?

More information

CHAPTER 3 Calculations with Chemical Formulas and Equations. atoms in a FORMULA UNIT

CHAPTER 3 Calculations with Chemical Formulas and Equations. atoms in a FORMULA UNIT 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

More information

1. How many hydrogen atoms are in 1.00 g of hydrogen?

1. 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 10-24 g B. We need a conversion to the macroscopic world. 1. How many hydrogen atoms are in 1.00 g of hydrogen?

More information

Stoichiometry. Unit Outline

Stoichiometry. 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 information

Lecture Topics Atomic weight, Mole, Molecular Mass, Derivation of Formulas, Percent Composition

Lecture 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 information

Chemical Calculations: The Mole Concept and Chemical Formulas. AW Atomic weight (mass of the atom of an element) was determined by relative weights.

Chemical 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 information

stoichiometry = the numerical relationships between chemical amounts in a reaction.

stoichiometry = 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 information

Moles. Moles. Moles. Moles. Balancing Eqns. Balancing. Balancing Eqns. Symbols Yields or Produces. Like a recipe:

Moles. 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 information

Chapter 5 Chemical Quantities and Reactions. Collection Terms. 5.1 The Mole. A Mole of a Compound. A Mole of Atoms.

Chapter 5 Chemical Quantities and Reactions. Collection Terms. 5.1 The Mole. A Mole of a Compound. A Mole of Atoms. 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

More information

Introductory Chemistry, 3 rd Edition Nivaldo Tro. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, Maqqwertd ygoijpk[l

Introductory Chemistry, 3 rd Edition Nivaldo Tro. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, Maqqwertd ygoijpk[l 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

More information

1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) 3 (s) + H 2 (g)

1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) 3 (s) + H 2 (g) 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)

More information

Chemical Calculations: Formula Masses, Moles, and Chemical Equations

Chemical 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 information

The Mole. 6.022 x 10 23

The 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 information

THE MOLE / COUNTING IN CHEMISTRY

THE 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 information

The Mole and Molar Mass

The 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 information

W1 WORKSHOP ON STOICHIOMETRY

W1 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 information

MASS RELATIONSHIPS IN CHEMICAL REACTIONS

MASS RELATIONSHIPS IN CHEMICAL REACTIONS 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

More information

Calculations with Chemical Formulas and Equations

Calculations with Chemical Formulas and Equations 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

More information

Ch. 6 Chemical Composition and Stoichiometry

Ch. 6 Chemical Composition and Stoichiometry 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!

More information

Chapter 3 Stoichiometry

Chapter 3 Stoichiometry Chapter 3 Stoichiometry 3-1 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 information

Chapter 3 Mass Relationships in Chemical Reactions

Chapter 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 information

CONSERVATION OF MASS During a chemical reaction, matter is neither created nor destroyed. - i. e. the number of atoms of each element remains constant

CONSERVATION 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 information

Chapter Three: STOICHIOMETRY

Chapter Three: STOICHIOMETRY p70 Chapter Three: STOICHIOMETRY Contents p76 Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions. p70 3-1 Counting by Weighing 3-2 Atomic Masses p78 Mass Mass

More information

We know from the information given that we have an equal mass of each compound, but no real numbers to plug in and find moles. So what can we do?

We know from the information given that we have an equal mass of each compound, but no real numbers to plug in and find moles. So what can we do? How do we figure this out? We know that: 1) the number of oxygen atoms can be found by using Avogadro s number, if we know the moles of oxygen atoms; 2) the number of moles of oxygen atoms can be found

More information

Chapter 1 The Atomic Nature of Matter

Chapter 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 information

Chapter 6 Notes. Chemical Composition

Chapter 6 Notes. Chemical Composition 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

More information

Chapter 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 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 McGraw-Hill Companies,

More information

IB Chemistry 1 Mole. One atom of C-12 has a mass of 12 amu. One mole of C-12 has a mass of 12 g. Grams we can use more easily.

IB Chemistry 1 Mole. One atom of C-12 has a mass of 12 amu. One mole of C-12 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 carbon-12 that were needed to make 12 g of carbon. 1 mole

More information

Chem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry Answers

Chem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry Answers Key Questions & Exercises Chem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry Answers 1. The atomic weight of carbon is 12.0107 u, so a mole of carbon has a mass of 12.0107 g. Why doesn t a mole of

More information

The 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. 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 information

Unit 10A Stoichiometry Notes

Unit 10A Stoichiometry Notes Unit 10A Stoichiometry Notes Stoichiometry 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 information

Stoichiometry. What is the atomic mass for carbon? For zinc?

Stoichiometry. 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 catbon-12

More information

Study Guide For Chapter 7

Study 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 information

Appendix D. Reaction Stoichiometry D.1 INTRODUCTION

Appendix 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 information

Ch. 10 The Mole I. Molar Conversions

Ch. 10 The Mole I. Molar Conversions Ch. 10 The Mole I. Molar Conversions I II III IV A. What is the Mole? A counting number (like a dozen) Avogadro s number (N A ) 1 mole = 6.022 10 23 representative particles B. Mole/Particle Conversions

More information

Answers and Solutions to Text Problems

Answers and Solutions to Text Problems Chapter 7 Answers and Solutions 7 Answers and Solutions to Text Problems 7.1 A mole is the amount of a substance that contains 6.02 x 10 23 items. For example, one mole of water contains 6.02 10 23 molecules

More information

CHEM 110: CHAPTER 3: STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS

CHEM 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 information

2. The percent yield is the maximum amount of product that can be produced from the given amount of limiting reactant.

2. 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 information

Element of same atomic number, but different atomic mass o Example: Hydrogen

Element 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 carbon-12, 6p + + 6n 0 = atomic mass of 12.0 o For chlorine-35, 17p + + 18n 0 = atomic mass of 35.0 atomic mass

More information

MOLECULAR MASS AND FORMULA MASS

MOLECULAR MASS AND FORMULA MASS 1 MOLECULAR MASS AND FORMULA MASS Molecular mass = sum of the atomic weights of all atoms in the molecule. Formula mass = sum of the atomic weights of all atoms in the formula unit. 2 MOLECULAR MASS AND

More information

Atomic mass is the mass of an atom in atomic mass units (amu)

Atomic 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 information

STOICHIOMETRY UNIT 1 LEARNING OUTCOMES. At the end of this unit students will be expected to:

STOICHIOMETRY UNIT 1 LEARNING OUTCOMES. At the end of this unit students will be expected to: STOICHIOMETRY LEARNING OUTCOMES At the end of this unit students will be expected to: UNIT 1 THE MOLE AND MOLAR MASS define molar mass and perform mole-mass inter-conversions for pure substances explain

More information

UNIT (4) CALCULATIONS AND CHEMICAL REACTIONS

UNIT (4) CALCULATIONS AND CHEMICAL REACTIONS UNIT (4) CALCULATIONS AND CHEMICAL REACTIONS 4.1 Formula Masses Recall that the decimal number written under the symbol of the element in the periodic table is the atomic mass of the element. 1 7 8 12

More information

Matter. Atomic weight, Molecular weight and Mole

Matter. 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 information

Unit 2: Quantities in Chemistry

Unit 2: Quantities in Chemistry Mass, Moles, & Molar Mass Relative quantities of isotopes in a natural occurring element (%) E.g. Carbon has 2 isotopes C-12 and C-13. Of Carbon s two isotopes, there is 98.9% C-12 and 11.1% C-13. Find

More information

Chapter 1: Moles and equations. Learning outcomes. you should be able to:

Chapter 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 information

The Mole Concept and Atoms

The Mole Concept and Atoms Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 4 24 September 2013 Calculations and the Chemical Equation The Mole Concept and Atoms Atoms are exceedingly

More information

Name Date Class STOICHIOMETRY. SECTION 12.1 THE ARITHMETIC OF EQUATIONS (pages 353 358)

Name 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 information

Stoichiometry. Web Resources Chem Team Chem Team Stoichiometry. Section 1: Definitions Define the following terms. Average Atomic mass - Molecule -

Stoichiometry. Web Resources Chem Team Chem Team Stoichiometry. Section 1: Definitions Define the following terms. Average Atomic mass - Molecule - Web Resources Chem Team Chem Team Section 1: Definitions Define the following terms Average Atomic mass - Molecule - Molecular mass - Moles - Avagadro's Number - Conservation of matter - Percent composition

More information

Moles Lab mole. 1 mole = 6.02 x 1023. This is also known as Avagadro's number Demo amu amu amu

Moles 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 information

3.3 Moles, 3.4 Molar Mass, and 3.5 Percent Composition

3.3 Moles, 3.4 Molar Mass, and 3.5 Percent Composition 3.3 Moles, 3.4 Molar Mass, and 3.5 Percent Composition 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 Copyright

More information

Chapter 3. Mass Relationships in Chemical Reactions

Chapter 3. Mass Relationships in Chemical Reactions Chapter 3 Mass Relationships in Chemical Reactions This chapter uses the concepts of conservation of mass to assist the student in gaining an understanding of chemical changes. Upon completion of Chapter

More information

Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1. A chemical equation. (C-4.4)

Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1. A chemical equation. (C-4.4) Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1 1. 2. 3. 4. 5. 6. Question What is a symbolic representation of a chemical reaction? What 3 things (values) is a mole of a chemical

More information

SCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGraw-Hill Ryerson, 2001

SCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, Chemistry 11, McGraw-Hill Ryerson, 2001 SCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGraw-Hill Ryerson, 2001 1. A small pin contains 0.0178 mol of iron. How many atoms of iron are in the pin? 2. A sample

More information

Chemical formulae are used as shorthand to indicate how many atoms of one element combine with another element to form a compound.

Chemical formulae are used as shorthand to indicate how many atoms of one element combine with another element to form a compound. 29 Chemical Formulae Chemical formulae are used as shorthand to indicate how many atoms of one element combine with another element to form a compound. C 2 H 6, 2 atoms of carbon combine with 6 atoms of

More information

Stoichiometry. Lecture Examples Answer Key

Stoichiometry. 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 information

CHEM 120 Online: Chapter 6 Sample problems Date: 2. Which of the following compounds has the largest formula mass? A) H2O B) NH3 C) CO D) BeH2

CHEM 120 Online: Chapter 6 Sample problems Date: 2. Which of the following compounds has the largest formula mass? A) H2O B) NH3 C) CO D) BeH2 CHEM 120 Online: Chapter 6 Sample problems Date: 1. To determine the formula mass of a compound you should A) add up the atomic masses of all the atoms present. B) add up the atomic masses of all the atoms

More information

Formulae, stoichiometry and the mole concept

Formulae, 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 information

Unit 3 Notepack Chapter 7 Chemical Quantities Qualifier for Test

Unit 3 Notepack Chapter 7 Chemical Quantities Qualifier for Test Unit 3 Notepack Chapter 7 Chemical Quantities Qualifier for Test NAME Section 7.1 The Mole: A Measurement of Matter A. What is a mole? 1. Chemistry is a quantitative science. What does this term mean?

More information

Chemical Composition. Introductory Chemistry: A Foundation FOURTH EDITION. Atomic Masses. Atomic Masses. Atomic Masses. Chapter 8

Chemical 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 information

CHAPTER 8: CHEMICAL COMPOSITION

CHAPTER 8: CHEMICAL COMPOSITION CHAPTER 8: CHEMICAL COMPOSITION Active Learning: 1-4, 6-8, 12, 18-25; End-of-Chapter Problems: 3-4, 9-82, 84-85, 87-92, 94-104, 107-109, 111, 113, 119, 125-126 8.2 ATOMIC MASSES: COUNTING ATOMS BY WEIGHING

More information

CHEMICAL REACTIONS AND REACTING MASSES AND VOLUMES

CHEMICAL REACTIONS AND REACTING MASSES AND VOLUMES CHEMICAL REACTIONS AND REACTING MASSES AND VOLUMES The meaning of stoichiometric coefficients: 2 H 2 (g) + O 2 (g) 2 H 2 O(l) number of reacting particles 2 molecules of hydrogen react with 1 molecule

More information

CHEM 101/105 Numbers and mass / Counting and weighing Lect-03

CHEM 101/105 Numbers and mass / Counting and weighing Lect-03 CHEM 101/105 Numbers and mass / Counting and weighing Lect-03 Interpretation of Elemental Chemical Symbols, Chemical Formulas, and Chemical Equations Interpretation of an element's chemical symbol depends

More information

AS1 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

AS1 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 information

Chapter 5. Chemical Reactions and Equations. Introduction. Chapter 5 Topics. 5.1 What is a Chemical Reaction

Chapter 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 McGraw-Hill Companies,

More information

Stoichiometry of Formulas and Equations

Stoichiometry of Formulas and Equations sil07204_ch03_69-107 8/22/05 15:16 Page 69 CHAPTER THREE Stoichiometry of Formulas and Equations Key Principles The mole (mol) is the standard unit for amount of substance and consists of Avogadro s number

More information

Unit 6 The Mole Concept

Unit 6 The Mole Concept Chemistry Form 3 Page 62 Ms. R. Buttigieg Unit 6 The Mole Concept See Chemistry for You Chapter 28 pg. 352-363 See GCSE Chemistry Chapter 5 pg. 70-79 6.1 Relative atomic mass. The relative atomic mass

More information

EXPERIMENT 12: Empirical Formula of a Compound

EXPERIMENT 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 information

F321 MOLES. Example If 1 atom has a mass of 1.241 x 10-23 g 1 mole of atoms will have a mass of 1.241 x 10-23 g x 6.02 x 10 23 = 7.

F321 MOLES. Example If 1 atom has a mass of 1.241 x 10-23 g 1 mole of atoms will have a mass of 1.241 x 10-23 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 information

Chemical Equations and Chemical Reactions. Chapter 8.1

Chemical Equations and Chemical Reactions. Chapter 8.1 Chemical Equations and Chemical Reactions Chapter 8.1 Objectives List observations that suggest that a chemical reaction has taken place List the requirements for a correctly written chemical equation.

More information

MOLAR MASS AND MOLECULAR WEIGHT Themolar mass of a molecule is the sum of the atomic weights of all atoms in the molecule. Molar Mass.

MOLAR MASS AND MOLECULAR WEIGHT Themolar mass of a molecule is the sum of the atomic weights of all atoms in the molecule. Molar Mass. Counting Atoms Mg burns in air (O 2 ) to produce white magnesium oxide, MgO. How can we figure out how much oxide is produced from a given mass of Mg? PROBLEM: If If 0.200 g of Mg is is burned, how much

More information