Chapter 7: Stoichiometry - Mass Relations in Chemical Reactions

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Chapter 7: Stoichiometry - Mass Relations in Chemical Reactions"

Transcription

1 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 in chemical reactions? How can we determine a compound s elemental composition and chemical formula? Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cyle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield 1

2 Law of Conservation of Mass The law of conservation of mass states that the sum of the masses of the reactants of a chemical equation is equal to the sum of the masses of the products. C(s) + O 2 (g) CO 2 (g) 12 g + 32 g = 44 g 1 mole 1 mole 1 mole Law of Conservation of Mass 2 C(s) + O 2 (g) 2 CO(g) 2 12 = 24 g + 32 g = 2 28 = 56 g 2 mole 1 mole 2 mole 2

3 Chemical Equations 2 C(s) + O 2 (g) 2 CO(g) The 2 s are called stoichiometric coefficients The symbol means reaction proceeds in this direction a symbol means the reaction is at equilibrium (s) = solid phase (l) = liquid phase (g) = gas phase (aq) = aqueous phase = heat Stoichiometric Coefficients: the ratios between reactants and/or products mol O 2 2 mol N 2 O 5 2 mol N 2 O 5 4 mol NO 2 3

4 Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cyle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield Guidelines for Balancing Chemical Equations 1. Write an epression using correct chemical formulas for the reactants and products, separated by an arrow ( ). Include symbols indicating physical states. 2. For each element, add up the numbers of atoms on each side. Check whether the epression is already balanced. If so, you re done! 3. Otherwise - if present - choose an element that appears in only one reactant and product to balance first. Insert the appropriate coefficient(s) to balance this element. 4. Choose the element that appears in the net fewest total reactants and products and balance it. Repeat the process for additional elements if necessary. 4

5 Eample 1: balance the reaction that occurs between nitrogen dioide and water to form nitric acid and nitrogen monoide 1. Write an epression using correct chemical formulas for the reactants and products, separated by an arrow ( ). Include symbols indicating physical states. NO 2 (g) + H 2 O(l) HNO 3 (aq) + NO(aq) 2. For each element, add up the numbers of atoms on each side. Check whether the epression is already balanced. If so, you re done! N = 1 N = 2 O = 3 O = 4 H = 2 H = 1 NO 2 (g) + H 2 O(l) HNO 3 (aq) + NO(aq) 3. Otherwise - if present - choose an element that appears in only one reactant and product to balance first. Insert the appropriate coefficient(s) to balance this element. = H, so try a 2 in front of HNO 3 NO 2 (g) + H 2 O(l) 2 HNO 3 (aq) + NO(aq) N = 1 N = 3 O = 3 O = 7 H = 2 H = 2 5

6 NO 2 (g) + H 2 O(l) 2 HNO 3 (aq) + NO(aq) 4. Choose the element that appears in the net fewest total reactants and products and balance it. Repeat the process for additional elements if necessary. = N, try a 3 in front of NO 2 3 NO 2 (g) + H 2 O(l) 2 HNO 3 (aq) + NO(aq) N = 3 N = 3 O = 7 O = 7 H = 2 H = 2 Done! Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cyle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield 6

7 Combustion Reactions The reaction of an organic compound with oygen to produce CO 2 + H 2 O, for eample, balance - CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O(g) TIP FOR ALL COMBUSTION REACTIONS: Since oygen appears by itself, balance the other elements first, and then O 2 CH 4 (g) + O 2 (g) CO 2 (g) + 2 H 2 O(g) CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) Done! Combustion Reactions Eample: C 2 H 6 (ethane) is combusted in oygen C 2 H 6 (g) + O 2 (g) CO 2 (g) + H 2 O(g) C 2 H 6 (g) + O 2 (g) 2 CO 2 (g) + H 2 O(g) C 2 H 6 (g) + O 2 (g) 2 CO 2 (g) + 3 H 2 O(g) C 2 H 6 (g) + 7/2 O 2 (g) 2 CO 2 (g) + 3 H 2 O(g) 2 C 2 H 6 (g) + 7 O 2 (g) 4 CO 2 (g) + 6 H 2 O(g) 7

8 Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cycle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield Stoichiometric Calculations and the Carbon Cycle Photosynthesis: 6 CO 2 (g) + 6 H 2 O(g) C 6 H 12 O 6 (aq) + 6 O 2 (g) 8

9 Mauna Loa Observatory CO 2 Concentrations since CO 2 emissions over the last 800,000 years - 9

10 Amounts of Reactants & Products = STOICHIOMETRY MM ratio MM 1. Write the balanced chemical equation 2. Convert the mass of the reactant into moles 3. Use coefficients in the balanced equation to calculate the number of moles of product (stoichiometric ratio) 4. Convert moles of products into grams (or other desired quantities) Stoichiometry Eample, p. 282 If the combustion of fossil fuels adds kilograms of carbon to the atmosphere each year as CO 2, what is the mass of added carbon dioide? Step 1: Write the balanced chemical equation C(s) + O 2 (g) CO 2 (g) Step 2: Convert quantities of known substances (C) into moles kg C 1000 g C kg 1 mol C 12.0 g C = mol C 10

11 Stoichiometry Eample, p. 282 Step 3: Use coefficients in balanced equation to calculate the number of moles of CO 2 (stoichiometric ratio) mol C 1 mol CO 2 = mol C 14 mol CO 2 Step 4: Convert moles of CO 2 into grams mol CO g CO 2 = mol CO 16 g CO 2 2 Stoichiometry Eample, p. 283 C 7 H 6 O 3 MW = C 4 H 6 O 3 MW = C 9 H 8 O 4 MW = Suppose we wish to prepare 1.00 kg of acetylsalicylic acid. How many grams of salicylic acid and how many grams of acetic anhydride are needed? 11

12 Stoichiometry Eample, p. 283 Step 1: Write the balanced chemical equation SA + AA ASA + Ac? g? g 1.00 kg Step 2: Convert quantities of known substances (ASA) into moles 1.00 kg ASA 1000 g ASA kg ASA 1 mol ASA g ASA = mol ASA Stoichiometry Eample, p. 283 SA + AA ASA + Ac Step 3: Use coefficients in balanced equation to calculate the number of moles of SA and AA (stoichiometric ratio) mol SA = mol ASA 1 mol SA 1 mol ASA = mol SA mol AA = mol ASA 1 mol AA 1 mol ASA = mol AA 12

13 Stoichiometry Eample, p. 283 Step 4: Convert moles of SA and AA into grams g SA = mol SA g SA 1 mol SA = 767 g SA g AA = mol AA g AA 1 mol AA = 567 g AA Sample Eercise 7.3 Calculating the Mass of a Product from the Mass of a Reactant. Each year, power plants in the U.S. consume about kg of natural gas (CH 4 ). How many kg of CO 2 (MW = 44.01) are released into atmosphere from these power plants. Given that natural gas is mainly CH 4 (MW = 16.04), base the calculation on its combustion reaction Step 1: Write the balanced chemical equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) kg? g 13

14 Sample Eercise 7.3 Calculating the Mass of a Product from the Mass of a Reactant. Step 2: Convert quantities of known substances (C) into moles kg CH 1000 g CH 1 mol CH = mol kg CH g CH 4 CH 4 Step 3: Use coefficients in balanced equation to calculate the number of moles of CO 2 (stoichiometric ratio) = mol CO mol CH 2 4 = mol CH 12 mol CO 2 4 Step 4: Convert moles of CO 2 into grams g CO mol CO 2 2 = g CO 2 1 mol CO 2 Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cycle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield 14

15 Percent Composition: the composition of a compound in terms of the percentage by mass of each element in the compound n molar mass of element molar mass of compound 100% n is the number of moles of the element in 1 mole of the compound 3 (1.008 g) %H = 100% = 3.086% g 1 (30.97 g) %P = 100% = 31.60% g 4 (16.00 g) %O = 100% = 65.31% g H 3 PO 4 MM = g/mol 3.086% % % = 99.96% = 100% Sample Eercise 7.4 Calculating Percent Composition from a Chemical Formula The mineral forsterite: Mg 2 SiO 4, MW = %Mg = %Si = %O = 2 (24.31 g Mg) g 1 (28.09 g Si) g 4 (16.00 g O) g 100% = 34.55% 100% = 19.96% 100% = 45.48% 34.55% % % = 99.99% = 100% 15

16 Empirical Formula from % Composition A formula showing the smallest whole number ratio of elements in a compound, e.g. Benzene:» Empirical = CH» Molecular = C 6 H 6 Glucose» Empirical = CH 2 O» Molecular = C 6 H 12 O 6 Empirical Formula from % Composition 1. Assume 100 g 2. Convert to moles 3. Divide by fewest number of moles 4. Convert the mole ratio from step 3 into small whole numbers if necessary 16

17 Sample Eercise 7.6 A sample of the carbonate mineral dolomite is 21.73% Ca, 13.18% Mg, 13.03% C, and the rest is oygen. What is its empirical formula? % O = = % 1. Assume 100 g 2. Convert to moles Ca = g Mg = g 1 mol Ca g 1 mol Mg g = mol Ca = mol Mg C = g 1 mol C g = mol C O = g 1 mol O g = mol O Sample Eercise Divide by fewest number of moles Ca = = 1.0 Mg = C = = 1.0 = 2.0 Empirical formula = CaMgC 2 O 6 O = = Convert the mole ratio from step 3 into small whole numbers if necessary NOT REQUIRED HERE 17

18 Eample Illustrating Step 4 Vanillin is a common flavoring agent. It has a molar mass of 152 g/mol and is %C and 5.30 %H; the rest is oygen. What are the empirical and molecular formulas? 1. Assume 100 g 2. Convert to moles % O = = % C = g 1 mol C g = mol C H = 5.30 g 1 mol H g = mol H O = g 1 mol O g = mol O Eample Illustrating Step 4 3. Divide by fewest number of moles C = H = O = = 2.67 = 2.67 = 1.0 Note that 2.67 = So multiply by 3 which = 8 4. Convert the mole ratio from step 3 into small whole numbers if necessary C = H = O = = = = 3 Empirical formula = C 8 H 8 O 3 18

19 Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cycle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield Empirical and Molecular Formulas Compared The molecular formula can be determined from the empirical formula if the molecular weight of the compound is known. empirical = CH 2 O formula weight (FW) = 30 g/mol molecular = C 2 H 4 O 2 Molecular weight (MW) = 60 g/mol Note that molecular formula = empirical 2 glycolaldehyde C 2 H 4 O 2 = (CH 2 O) 2 = C 2 H 4 O 2 Or more generally molecular = (empirical) n 19

20 Empirical and Molecular Formulas Compared molecular = (empirical) n it follows that - MW = (FW) n and therefore - n = MW FW e.g. glucose assume we know the MW = Empirical = CH 2 O FW = 30.0 n = = 6 So molecular = (CH 2 O) 6 = C 6 H 12 O 6 Molecular Mass and Mass Spectrometry Acetylene C 2 H 2 Benzene C 6 H 6 20

21 Sample Eercise 7.7 Using Percent Composition and Molecular weight to Derive a Molecular Fomula Pheromones are chemical substances secreted by members of a species to stimulate a response in other individuals of the same species. The percent composition of eicosene, a compound similar to the Japanese beetle mating pheromone, is 85.63% C and 14.37% H. Its molecular mass, as determined by mass spectrometry, is 280 amu. What is the molecular formula of eicosene? 1. Assume 100 g 2. Convert to moles C = g 1 mol C g = mol C H = g 1 mol H g = mol H Sample Eercise 7.7 Using Percent Composition and Molecular weight to Derive a Molecular Fomula 3. Divide by fewest number of moles C = H = = 1.00 = 2 Empirical formula = CH 2 FW = Convert the mole ratio from step 3 into small whole numbers if necessary NOT NEEDED n = MW FW = = 20 So molecular = (CH 2 ) 20 = C 20 H 40 21

22 Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cycle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield Eperimental Determination of Empirical Formulas: Combustion Analysis 22

23 Eperimental Determination of Empirical Formulas: Combustion Analysis C H y O z + O 2 (g) CO 2 (g) + y/2 H 2 O(g) mass sample Calculation Outline: ecess Some of the oygen comes from the sample, some from the ecess O 2 g CO 2 mol CO 2 mol C g C g H 2 O mol H 2 O mol H g H g of O = g of sample g of C - g of H Sample Eercise 7.8: Deriving an Empirical Formula from Combustion Analysis Data Combustion of grams of an organic compound known to contain only C, H and O produces g CO 2 and g H 2 O. What is the empirical formula of the compound? mol C = g CO 2 1 mol CO g CO 2 1 mol C mol CO 2 = mol C g C = mol C g C 1 mol C = g C mol H = g H 2 O 1 mol H 2 O 18.0 g H 2 O 2 mol H mol H 2 O = mol H g H = = mol H g H 1 mol H = g H 23

24 Sample Eercise 7.8: Deriving an Empirical Formula from Combustion Analysis Data g of O = g of sample g of C - g of H g of O = g g C g H g of O = g and so the moles of O = g = mol O 1 mol O g O Sample Eercise 7.8: Deriving an Empirical Formula from Combustion Analysis Data Now divide by the smallest number of moles - C = mol C = 3 H = mol H = 4 Empirical formula = C 3 H 4 O O = mol O = 1 24

25 Chapter Outline 7.1 Chemical reactions and the Conservation of Mass 7.2 Balancing Chemical Equations 7.3 Combustion Reactions 7.4 Stoichiometric Calculations and the Carbon Cycle 7.5 Percent Composition and Empirical Formulas 7.6 Empirical and Molecular Formulas Compared 7.7 Combustion Analysis 7.8 Limiting Reactants and Percent Yield Limiting Reagents Limiting Reagents - a reactant that is consumed completely in a chemical reaction before the other reactant(s) run out. The amount of product formed depends on the amount of the limiting reagent available. Each sandwich consists of 2 slices bread, 1 slice of cheese, and 1 slice of salami 8 slices bread 4 slices cheese 3 slices salami 25

26 Limiting Reagents 2 Br + 1 Ch + 1 Sal 1 sandwich Given: Here s the logic compare the given number of moles to the stoichiometric ratio Given ratio: 8 Br = 2.7 tip: use a ratio > 1 3 Sal Stoichiometric ratio: 2 Br = Sal Br in ecess, so need more salami! = LR Strategy for determining which reactant is the Limiting Reagent: aa + bb C 1. Convert grams of each into moles 2. Calculate the stoichiometric ratio that s the largest, e.g. a/b or b/a 3. Calculate the given mole ratio in the same way 4. Compare to identify the LR if moles A moles B given > a b stoichiometric then A is in ecess and B is the limiting reagent 26

27 Sample Eercise 7.9: Identifying the Limiting Reagent in a Reaction Miture The flame in an acetylene torch reaches temperatures as high as 3500 o C as a result of the combustion of a miture of acetylene (C 2 H 2 ) and pure oygen. If these two gases flow from high-pressure tanks at the rates of 52.0 g C 2 H 2 and 188 g O 2 per minute, which reactant is the limiting reagent, or is the miture stoichiometric? Given ratio: mol C 2 H 2 2 C 2 H O 2 4 CO H 2 O 52.0 g 188 g 1 mol C = 52.0 g C 2 H 2 H 2 2 = 2.00 mol C 26.0 g C 2 H 2 H 2 2 tip: use a ratio > mol O mol C H 1 mol O 2 2 mol O 2 = 188 g O g O = 5.88 mol O 2 2 = 2.94 Sample Eercise 7.9: Identifying the Limiting Reagent in a Reaction Miture 2 C 2 H O 2 4 CO H 2 O is moles O 2 moles C 2 H 2 given > 5 mol O 2 2 mol C 2 H 2 stoichiometric??? Given ratio: Stoichiometric ratio: 5.88 mol O mol C 2 H 2 = mol O 2 2 mol C 2 H 2 = 2.5 therefore O 2 is in ecess and so C 2 H 2 is the LR 27

28 Percent Yield Theoretical Yield is the maimum amount of product formed from given quantities of reactants (check if a limiting reagent is present). Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield 100 Sample Eercise 7.10 The industrial process for making the ammonia used in fertilizer, eplosives, and many other products is based on the reaction between nitrogen and hydrogen at high temperature and pressure. If 18.2 kg of NH 3 (MW = 17.03) is produced by a reaction miture that initially contains 6.00 kg of H 2 (MW = 2.016) and an ecess of N 2, what is the percent yield of the reaction? Actual Yield N 2 (g) + 3 H 2 (g) 2 NH 3 (g) % Yield = 100 Ecess (no LR) 6.00 kg Actual yield = 18.2 kg Theoretical Yield Calculate the theoretical yield of NH 3 based on H 2, and then calculate the % yield g g NH 3 = 6.00 kg H 2 kg mol H g H 2 2 mol NH g NH 3 3 mol H 2 mol NH 3 = g NH 3 = kg NH 3 = 33.8 kg NH 3 28

29 Sample Eercise 7.10 % Yield = Actual Yield Theoretical Yield 100 Actual Yield = 18.2 kg Theoretical Yield = kg % Yield = 18.2 kg kg 100 = 53.8 % 29

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

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

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

Chemical Reactions. Chemical Equations. Mole as Conversion Factor: To convert between number of particles and an equivalent number of moles:

Chemical Reactions. Chemical Equations. Mole as Conversion Factor: To convert between number of particles and an equivalent number of moles: 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

More information

Ca 3 N 2 (s) + 6H 2 O(l) H 2NH 3 (g) + 3Ca(OH) 2 (s) mole ratio 1 : 6 : 2 : 3 molar mass (g/mole)

Ca 3 N 2 (s) + 6H 2 O(l) H 2NH 3 (g) + 3Ca(OH) 2 (s) mole ratio 1 : 6 : 2 : 3 molar mass (g/mole) 1. STOICHIOMETRY INVOLVING ONLY PURE SUBSTANCES For all chemical reactions, the balanced chemical equation gives the mole ratios of reactants and products. If we are dealing with pure chemicals, the molar

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

Chapter 3. Stoichiometry: Ratios of Combination. Insert picture from First page of chapter. Copyright McGraw-Hill 2009 1

Chapter 3. Stoichiometry: Ratios of Combination. Insert picture from First page of chapter. Copyright McGraw-Hill 2009 1 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

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

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

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

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

Chapter 6 Chemical Calculations

Chapter 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 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

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

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

Moles and Chemical Reactions. Moles and Chemical Reactions. Molar mass = 2 x 12.011 + 6 x 1.008 + 1 x15.999 = 46.069 g/mol

Moles and Chemical Reactions. Moles and Chemical Reactions. Molar mass = 2 x 12.011 + 6 x 1.008 + 1 x15.999 = 46.069 g/mol 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

Mass and Moles of a Substance

Mass and Moles of a Substance 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

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

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

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. 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

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

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

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

Molecular Formula: Example

Molecular Formula: Example Molecular Formula: Example A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? 1 CHAPTER 3 Chemical

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

2 Stoichiometry: Chemical Arithmetic Formula Conventions (1 of 24) 2 Stoichiometry: Chemical Arithmetic Stoichiometry Terms (2 of 24)

2 Stoichiometry: Chemical Arithmetic Formula Conventions (1 of 24) 2 Stoichiometry: Chemical Arithmetic Stoichiometry Terms (2 of 24) 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,

More information

10.3 Percent Composition and Chemical Formulas. Chapter 10 Chemical Quantities Percent Composition and Chemical Formulas

10.3 Percent Composition and Chemical Formulas. Chapter 10 Chemical Quantities Percent Composition and Chemical Formulas Chapter 10 Chemical Quantities 101 The Mole: A Measurement of Matter 102 Mole-Mass and Mole-Volume Relationships 103 Percent Composition and Chemical Formulas 1 CHEMISTRY & YOU What does the percent composition

More information

Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations CHE11 Chapter Chapter Stoichiometry: Calculations with Chemical Formulas and Equations 1. When the following equation is balanced, the coefficients are. NH (g) + O (g) NO (g) + H O (g) (a). 1, 1, 1, 1

More information

Chapter 1 The Atomic Nature of Matter: Selected Answersc for Practice Exam.

Chapter 1 The Atomic Nature of Matter: Selected Answersc for Practice Exam. Chapter 1 The Atomic Nature of Matter: Selected Answersc for Practice Exam. MULTIPLE CHOICE 50. 5.80 g of dioxane (C 4 H 8 O 2 ) is how many moles of dioxane? 0.0658 mol 0.0707 mol 0.0725 mol d. 0.0804

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

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

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

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

Chemical calculations

Chemical calculations Chemical calculations Stoichiometry refers to the quantities of material which react according to a balanced chemical equation. Compounds are formed when atoms combine in fixed proportions. E.g. 2Mg +

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

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

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

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

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

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

1. P 2 O 5 2. P 5 O 2 3. P 10 O 4 4. P 4 O 10

1. P 2 O 5 2. P 5 O 2 3. P 10 O 4 4. P 4 O 10 Teacher: Mr. gerraputa Print Close Name: 1. A chemical formula is an expression used to represent 1. mixtures, only 3. compounds, only 2. elements, only 4. compounds and elements 2. What is the total number

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

Lecture Notes Chemistry E-1. Chapter 3

Lecture Notes Chemistry E-1. Chapter 3 Lecture Notes Chemistry E-1 Chapter 3 http://inserbia.info/news/wp-content/uploads/2013/05/tamiflu.jpg http://nutsforhealthcare.files.wordpress.com/2013/01/tamiflu-moa.jpg The Mole A mole is a certain

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

Stoichiometry Exploring a Student-Friendly Method of Problem Solving

Stoichiometry Exploring a Student-Friendly Method of Problem Solving Stoichiometry Exploring a Student-Friendly 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 information

Chapter 10 Chemical Calculations and Chemical Equations. An Introduction to Chemistry by Mark Bishop

Chapter 10 Chemical Calculations and Chemical Equations. An Introduction to Chemistry by Mark Bishop Chapter 10 Chemical Calculations and Chemical Equations An Introduction to Chemistry by Mark Bishop Chapter Map Making Phosphoric Acid Furnace Process for making H 3 PO 4 to be used to make fertilizers,

More information

Problem Solving. Stoichiometry of Gases

Problem Solving. Stoichiometry of Gases Skills Worksheet Problem Solving Stoichiometry of Gases Now that you have worked with relationships among moles, mass, and volumes of gases, you can easily put these to work in stoichiometry calculations.

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

1. What is the molecular formula of a compound with the empirical formula PO and a gram-molecular mass of 284 grams?

1. What is the molecular formula of a compound with the empirical formula PO and a gram-molecular 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 gram-molecular mass of 284 grams? 2 5 1. P2O 5 3. P10O4 2. P5O 2 4. P4O10 2. Which substance

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

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

K c = [C]c [D] d [A] a [B] b. k f [NO 2 ] = k r [N 2 O 4 ] = K eq = The Concept of Equilibrium. Chapter 15 Chemical Equilibrium

K c = [C]c [D] d [A] a [B] b. k f [NO 2 ] = k r [N 2 O 4 ] = K eq = The Concept of Equilibrium. Chapter 15 Chemical Equilibrium Chapter 15 Chemical Equilibrium Learning goals and key skills: Understand what is meant by chemical equilibrium and how it relates to reaction rates Write the equilibrium-constant expression for any reaction

More information

Chemistry Stoichiometry Lesson 8 Lesson Plan David V. Fansler

Chemistry Stoichiometry Lesson 8 Lesson Plan David V. Fansler Chemistry Stoichiometry Lesson 8 Lesson Plan David V. Fansler The Arithmetic of Equations Objectives: Interpret balanced chemical equations in terms of interacting moles, representative particles, masses,

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

Name Class Date. Section: Calculating Quantities in Reactions. Complete each statement below by writing the correct term or phrase.

Name 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 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

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

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

0.786 mol carbon dioxide to grams g lithium carbonate to mol

0.786 mol carbon dioxide to grams g lithium carbonate to mol 1 2 Convert: 2.54 x 10 22 atoms of Cr to mol 4.32 mol NaCl to grams 0.786 mol carbon dioxide to grams 2.67 g lithium carbonate to mol 1.000 atom of C 12 to grams 3 Convert: 2.54 x 10 22 atoms of Cr to

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

Introduction to Chemistry

Introduction 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 information

Practice questions for Chapter 8

Practice questions for Chapter 8 Practice questions for Chapter 8 2) How many atoms of nickel equal a mass of 58.69 g? (Refer to the Periodic Table.) A) 1 B) 28 C) 58.69 D) 59 E) 6.02 x 1023 Answer: E Section: 8.1 Avogadro's Number 6)

More information

Calculations with Chemical Reactions

Calculations with Chemical Reactions Calculations with Chemical Reactions Calculations with chemical reactions require some background knowledge in basic chemistry concepts. Please, see the definitions from chemistry listed below: Atomic

More information

Chapter 3 Calculation with Chemical Formulas and Equations

Chapter 3 Calculation with Chemical Formulas and Equations 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

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

Chapter 3 Molecules, Moles, and Chemical Equations. Chapter Objectives. Warning!! Chapter Objectives. Chapter Objectives

Chapter 3 Molecules, Moles, and Chemical Equations. Chapter Objectives. Warning!! Chapter Objectives. Chapter Objectives Larry Brown Tom Holme www.cengage.com/chemistry/brown Chapter 3 Molecules, Moles, and Chemical Equations Jacqueline Bennett SUNY Oneonta 2 Warning!! These slides contains visual aids for learning BUT they

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

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

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

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

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

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

Reactions. Balancing Chemical Equations uses Law of conservation of mass: matter cannot be lost in any chemical reaction

Reactions. Balancing Chemical Equations uses Law of conservation of mass: matter cannot be lost in any chemical reaction Reactions Chapter 8 Combustion Decomposition Combination Chapter 9 Aqueous Reactions Exchange reactions (Metathesis) Formation of a precipitate Formation of a gas Formation of a week or nonelectrolyte

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

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

CHAPTER 12 GASES AND THEIR BEHAVIOR

CHAPTER 12 GASES AND THEIR BEHAVIOR Chapter 12 Gases and Their Behavior Page 1 CHAPTER 12 GASES AND THEIR BEHAVIOR 12-1. Which of the following represents the largest gas pressure? (a) 1.0 atm (b) 1.0 mm Hg (c) 1.0 Pa (d) 1.0 KPa 12-2. Nitrogen

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

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

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

MOLE CONVERSION PROBLEMS. 2. How many moles are present in 34 grams of Cu(OH) 2? [0.35 moles]

MOLE CONVERSION PROBLEMS. 2. How many moles are present in 34 grams of Cu(OH) 2? [0.35 moles] MOLE CONVERSION PROBLEMS 1. What is the molar mass of MgO? [40.31 g/mol] 2. How many moles are present in 34 grams of Cu(OH) 2? [0.35 moles] 3. How many moles are present in 2.5 x 10 23 molecules of CH

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

The Mole, Avogadro s Number, and Molar Mass

The Mole, Avogadro s Number, and Molar Mass 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

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

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 OF COMBUSTION

STOICHIOMETRY OF COMBUSTION STOICHIOMETRY OF COMBUSTION FUNDAMENTALS: moles and kilomoles Atomic unit mass: 1/12 126 C ~ 1.66 10-27 kg Atoms and molecules mass is defined in atomic unit mass: which is defined in relation to the 1/12

More information

Sample Exercise 15.1 Writing Equilibrium-Constant Expressions

Sample Exercise 15.1 Writing Equilibrium-Constant Expressions Sample Exercise 15.1 Writing Equilibrium-Constant Expressions Write the equilibrium expression for K c for the following reactions: Solution Analyze: We are given three equations and are asked to write

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

Equilibrium Notes Ch 14:

Equilibrium Notes Ch 14: Equilibrium Notes Ch 14: Homework: E q u i l i b r i u m P a g e 1 Read Chapter 14 Work out sample/practice exercises in the sections, Bonus Chapter 14: 23, 27, 29, 31, 39, 41, 45, 51, 57, 63, 77, 83,

More information

Chemistry I: Using Chemical Formulas. Formula Mass The sum of the average atomic masses of all elements in the compound. Units are amu.

Chemistry I: Using Chemical Formulas. Formula Mass The sum of the average atomic masses of all elements in the compound. Units are amu. 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

More information

Chemical Quantities and Aqueous Reactions

Chemical Quantities and Aqueous Reactions 4 Chemical Quantities and Aqueous Reactions I feel sorry for people who don t understand anything about chemistry. They are missing an important source of happiness. Linus Pauling (1901 1994) 4.1 Climate

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

Chemistry - A Quantitative Science

Chemistry - A Quantitative Science 56 Ba 137.33 1 H 1.0079 2 He 4.0026 5 B 10.811 6 C 12.011 7 N 14.007 8 O 15.999 9 F 18.998 10 Ne 20.180 13 Al 26.982 14 Si 28.086 15 P 30.974 16 S 32.066 17 Cl 35.453 18 Ar 39.948 3 Li 6.941 4 Be 9.0122

More information

Problem Solving. Percentage Yield

Problem Solving. Percentage Yield Skills Worksheet Problem Solving Percentage Yield Although we can write perfectly balanced equations to represent perfect reactions, the reactions themselves are often not perfect. A reaction does not

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 8: Quantities in Chemical Reactions

Chapter 8: Quantities in Chemical Reactions Ch 8 Page 1 Chapter 8: Quantities in Chemical Reactions Stoichiometry: the numerical relationship between chemical quantities in a balanced chemical equation. Ex. 4NH 3 + 5O 2 4NO + 6H 2 O The reaction

More information

The Mole Concept. A. Atomic Masses and Avogadro s Hypothesis

The Mole Concept. A. Atomic Masses and Avogadro s Hypothesis The Mole Concept A. Atomic Masses and Avogadro s Hypothesis 1. We have learned that compounds are made up of two or more different elements and that elements are composed of atoms. Therefore, compounds

More information

Chemical Reactions 2 The Chemical Equation

Chemical 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 information

Liquid phase. Balance equation Moles A Stoic. coefficient. Aqueous phase

Liquid 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 information