Chapter 7: Stoichiometry  Mass Relations in Chemical Reactions


 Lynne Montgomery
 2 years ago
 Views:
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 highpressure 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
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 informationChapter 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 informationAtomic Masses. Chapter 3. Stoichiometry. Chemical Stoichiometry. Mass and Moles of a Substance. Average Atomic Mass
Atomic Masses Chapter 3 Stoichiometry 1 atomic mass unit (amu) = 1/12 of the mass of a 12 C atom so one 12 C atom has a mass of 12 amu (exact number). From mass spectrometry: 13 C/ 12 C = 1.0836129 amu
More informationChemical 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 bookkeeping Chemical Equations Chemical equations: Describe proportions
More informationCa 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 informationChapter 3! Stoichiometry: Calculations with Chemical Formulas and Equations. Stoichiometry
Chapter 3! : Calculations with Chemical Formulas and Equations Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2 H 2 O (g) Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2
More informationChapter 3. Stoichiometry: Ratios of Combination. Insert picture from First page of chapter. Copyright McGrawHill 2009 1
Chapter 3 Insert picture from First page of chapter Stoichiometry: Ratios of Combination Copyright McGrawHill 2009 1 3.1 Molecular and Formula Masses Molecular mass  (molecular weight) The mass in amu
More informationThe Mole Concept. The Mole. Masses of molecules
The Mole Concept Ron Robertson r2 c:\files\courses\111020\2010 final slides for web\mole concept.docx The Mole The mole is a unit of measurement equal to 6.022 x 10 23 things (to 4 sf) just like there
More informationChemical Equations & Stoichiometry
Chemical Equations & Stoichiometry Chapter Goals Balance equations for simple chemical reactions. Perform stoichiometry calculations using balanced chemical equations. Understand the meaning of the term
More informationChemistry B11 Chapter 4 Chemical reactions
Chemistry B11 Chapter 4 Chemical reactions Chemical reactions are classified into five groups: A + B AB Synthesis reactions (Combination) H + O H O AB A + B Decomposition reactions (Analysis) NaCl Na +Cl
More informationPart One: Mass and Moles of Substance. Molecular Mass = sum of the Atomic Masses in a molecule
CHAPTER THREE: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS Part One: Mass and Moles of Substance A. Molecular Mass and Formula Mass. (Section 3.1) 1. Just as we can talk about mass of one atom of
More informationChapter 6 Chemical Calculations
Chapter 6 Chemical Calculations 1 Submicroscopic Macroscopic 2 Chapter Outline 1. Formula Masses (Ch 6.1) 2. Percent Composition (supplemental material) 3. The Mole & Avogadro s Number (Ch 6.2) 4. Molar
More informationFormulas, Equations and Moles
Chapter 3 Formulas, Equations and Moles Interpreting Chemical Equations You can interpret a balanced chemical equation in many ways. On a microscopic level, two molecules of H 2 react with one molecule
More informationChem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses
Chem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses B. Calculations of moles C. Calculations of number of atoms from moles/molar masses 1. Avagadro
More informationChem 31 Fall 2002. Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations. Writing and Balancing Chemical Equations
Chem 31 Fall 2002 Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Writing and Balancing Chemical Equations 1. Write Equation in Words you cannot write an equation unless you
More informationMoles 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 informationMass 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 informationCalculating Atoms, Ions, or Molecules Using Moles
TEKS REVIEW 8B Calculating Atoms, Ions, or Molecules Using Moles TEKS 8B READINESS Use the mole concept to calculate the number of atoms, ions, or molecules in a sample TEKS_TXT of material. Vocabulary
More informationIB Chemistry 1 Mole. One atom of C12 has a mass of 12 amu. One mole of C12 has a mass of 12 g. Grams we can use more easily.
The Mole Atomic mass units and atoms are not convenient units to work with. The concept of the mole was invented. This was the number of atoms of carbon12 that were needed to make 12 g of carbon. 1 mole
More information1. How many hydrogen atoms are in 1.00 g of hydrogen?
MOLES AND CALCULATIONS USING THE MOLE CONCEPT INTRODUCTORY TERMS A. What is an amu? 1.66 x 1024 g B. We need a conversion to the macroscopic world. 1. How many hydrogen atoms are in 1.00 g of hydrogen?
More informationStoichiometry. What is the atomic mass for carbon? For zinc?
Stoichiometry Atomic Mass (atomic weight) Atoms are so small, it is difficult to discuss how much they weigh in grams We use atomic mass units an atomic mass unit (AMU) is one twelfth the mass of the catbon12
More informationChapter Three: STOICHIOMETRY
p70 Chapter Three: STOICHIOMETRY Contents p76 Stoichiometry  The study of quantities of materials consumed and produced in chemical reactions. p70 31 Counting by Weighing 32 Atomic Masses p78 Mass Mass
More informationChem 115 POGIL Worksheet  Week 4 Moles & Stoichiometry
Chem 115 POGIL Worksheet  Week 4 Moles & Stoichiometry Why? Chemists are concerned with mass relationships in chemical reactions, usually run on a macroscopic scale (grams, kilograms, etc.). To deal with
More informationThe 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 informationMolecular 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 informationCHEMICAL 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 information2 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 information10.3 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 MoleMass and MoleVolume Relationships 103 Percent Composition and Chemical Formulas 1 CHEMISTRY & YOU What does the percent composition
More informationChapter 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 informationChapter 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 informationChapter 1 The Atomic Nature of Matter
Chapter 1 The Atomic Nature of Matter 6. Substances that cannot be decomposed into two or more simpler substances by chemical means are called a. pure substances. b. compounds. c. molecules. d. elements.
More informationCalculations 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 informationChemical 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 informationStoichiometry. Lecture Examples Answer Key
Stoichiometry Lecture Examples Answer Key Ex. 1 Balance the following chemical equations: 3 NaBr + 1 H 3 PO 4 3 HBr + 1 Na 3 PO 4 2 C 3 H 5 N 3 O 9 6 CO 2 + 3 N 2 + 5 H 2 O + 9 O 2 2 Ca(OH) 2 + 2 SO 2
More informationMoles. Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations
Moles Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations Micro World atoms & molecules Macro World grams Atomic mass is the mass of an
More informationChemical 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 informationBalance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O
Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O Ans: 8 KClO 3 + C 12 H 22 O 11 8 KCl + 12 CO 2 + 11 H 2 O 3.2 Chemical Symbols at Different levels Chemical symbols represent
More informationSample Problem: STOICHIOMETRY and percent yield calculations. How much H 2 O will be formed if 454 g of. decomposes? NH 4 NO 3 N 2 O + 2 H 2 O
STOICHIOMETRY and percent yield calculations 1 Steps for solving Stoichiometric Problems 2 Step 1 Write the balanced equation for the reaction. Step 2 Identify your known and unknown quantities. Step 3
More informationName Date Class STOICHIOMETRY. SECTION 12.1 THE ARITHMETIC OF EQUATIONS (pages 353 358)
Name Date Class 1 STOICHIOMETRY SECTION 1.1 THE ARITHMETIC OF EQUATIONS (pages 353 358) This section explains how to calculate the amount of reactants required or product formed in a nonchemical process.
More informationMoles. Moles. Moles. Moles. Balancing Eqns. Balancing. Balancing Eqns. Symbols Yields or Produces. Like a recipe:
Like a recipe: Balancing Eqns Reactants Products 2H 2 (g) + O 2 (g) 2H 2 O(l) coefficients subscripts Balancing Eqns Balancing Symbols (s) (l) (aq) (g) or Yields or Produces solid liquid (pure liquid)
More informationIntroductory 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 informationChapter 5, Calculations and the Chemical Equation
1. How many iron atoms are present in one mole of iron? Ans. 6.02 1023 atoms 2. How many grams of sulfur are found in 0.150 mol of sulfur? [Use atomic weight: S, 32.06 amu] Ans. 4.81 g 3. How many moles
More information1. 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 informationChemical Calculations: The Mole Concept and Chemical Formulas. AW Atomic weight (mass of the atom of an element) was determined by relative weights.
1 Introduction to Chemistry Atomic Weights (Definitions) Chemical Calculations: The Mole Concept and Chemical Formulas AW Atomic weight (mass of the atom of an element) was determined by relative weights.
More informationLecture Notes Chemistry E1. Chapter 3
Lecture Notes Chemistry E1 Chapter 3 http://inserbia.info/news/wpcontent/uploads/2013/05/tamiflu.jpg http://nutsforhealthcare.files.wordpress.com/2013/01/tamiflumoa.jpg The Mole A mole is a certain
More informationCalculations and Chemical Equations. Example: Hydrogen atomic weight = 1.008 amu Carbon atomic weight = 12.001 amu
Calculations and Chemical Equations Atomic mass: Mass of an atom of an element, expressed in atomic mass units Atomic mass unit (amu): 1.661 x 1024 g Atomic weight: Average mass of all isotopes of a given
More informationStoichiometry Exploring a StudentFriendly Method of Problem Solving
Stoichiometry Exploring a StudentFriendly Method of Problem Solving Stoichiometry comes in two forms: composition and reaction. If the relationship in question is between the quantities of each element
More informationChapter 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 informationProblem 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 informationMASS 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 information1. What is the molecular formula of a compound with the empirical formula PO and a grammolecular mass of 284 grams?
Name: Tuesday, May 20, 2008 1. What is the molecular formula of a compound with the empirical formula PO and a grammolecular mass of 284 grams? 2 5 1. P2O 5 3. P10O4 2. P5O 2 4. P4O10 2. Which substance
More informationIB Chemistry. DP Chemistry Review
DP Chemistry Review Topic 1: Quantitative chemistry 1.1 The mole concept and Avogadro s constant Assessment statement Apply the mole concept to substances. Determine the number of particles and the amount
More informationSample Exercise 3.1 Interpreting and Balancing Chemical Equations
Sample Exercise 3.1 Interpreting and Balancing Chemical Equations The following diagram represents a chemical reaction in which the red spheres are oxygen atoms and the blue spheres are nitrogen atoms.
More informationK 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 equilibriumconstant expression for any reaction
More informationChemistry 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 informationChapter 3 Mass Relationships in Chemical Reactions
Chapter 3 Mass Relationships in Chemical Reactions Student: 1. An atom of bromine has a mass about four times greater than that of an atom of neon. Which choice makes the correct comparison of the relative
More informationName Class Date. Section: Calculating Quantities in Reactions. Complete each statement below by writing the correct term or phrase.
Skills Worksheet Concept Review Section: Calculating Quantities in Reactions Complete each statement below by writing the correct term or phrase. 1. All stoichiometric calculations involving equations
More informationTHE MOLE / COUNTING IN CHEMISTRY
1 THE MOLE / COUNTING IN CHEMISTRY ***A mole is 6.0 x 10 items.*** 1 mole = 6.0 x 10 items 1 mole = 60, 00, 000, 000, 000, 000, 000, 000 items Analogy #1 1 dozen = 1 items 18 eggs = 1.5 dz.  to convert
More informationstoichiometry = the numerical relationships between chemical amounts in a reaction.
1 REACTIONS AND YIELD ANSWERS stoichiometry = the numerical relationships between chemical amounts in a reaction. 2C 8 H 18 (l) + 25O 2 16CO 2 (g) + 18H 2 O(g) From the equation, 16 moles of CO 2 (a greenhouse
More informationChapter 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 information0.786 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 informationHonors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1. A chemical equation. (C4.4)
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 informationIntroduction to Chemistry
1 Copyright ç 1996 Richard Hochstim. All rights reserved. Terms of use. Introduction to Chemistry In Chemistry the word weight is commonly used in place of the more proper term mass. 1.1 Atoms, Ions, and
More informationPractice 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 informationCalculations 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 informationChapter 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 informationHow much does a single atom weigh? Different elements weigh different amounts related to what makes them unique.
How much does a single atom weigh? Different elements weigh different amounts related to what makes them unique. What units do we use to define the weight of an atom? amu units of atomic weight. (atomic
More informationChapter 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 informationChemical Calculations: Formula Masses, Moles, and Chemical Equations
Chemical Calculations: Formula Masses, Moles, and Chemical Equations Atomic Mass & Formula Mass Recall from Chapter Three that the average mass of an atom of a given element can be found on the periodic
More informationStoichiometry. Unit Outline
3 Stoichiometry Unit Outline 3.1 The Mole and Molar Mass 3.2 Stoichiometry and Compound Formulas 3.3 Stoichiometry and Chemical Reactions 3.4 Stoichiometry and Limiting Reactants 3.5 Chemical Analysis
More informationCalculation of Molar Masses. Molar Mass. Solutions. Solutions
Molar Mass Molar mass = Mass in grams of one mole of any element, numerically equal to its atomic weight Molar mass of molecules can be determined from the chemical formula and molar masses of elements
More informationF321 MOLES. Example If 1 atom has a mass of 1.241 x 1023 g 1 mole of atoms will have a mass of 1.241 x 1023 g x 6.02 x 10 23 = 7.
Moles 1 MOLES The mole the standard unit of amount of a substance (mol) the number of particles in a mole is known as Avogadro s constant (N A ) Avogadro s constant has a value of 6.02 x 10 23 mol 1.
More informationChem 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 informationOther Stoich Calculations A. mole mass (mass mole) calculations. GIVEN mol A x CE mol B. PT g A CE mol A MOLE MASS :
Chem. I Notes Ch. 12, part 2 Using Moles NOTE: Vocabulary terms are in boldface and underlined. Supporting details are in italics. 1 MOLE = 6.02 x 10 23 representative particles (representative particles
More informationReactions. 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 informationChapter 4. Chemical Composition. Chapter 4 Topics H 2 S. 4.1 Mole Quantities. The Mole Scale. Molar Mass The Mass of 1 Mole
Chapter 4 Chemical Composition Chapter 4 Topics 1. Mole Quantities 2. Moles, Masses, and Particles 3. Determining Empirical Formulas 4. Chemical Composition of Solutions Copyright The McGrawHill Companies,
More informationCHEM 110: CHAPTER 3: STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS
1 CHEM 110: CHAPTER 3: STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS The Chemical Equation A chemical equation concisely shows the initial (reactants) and final (products) results of
More informationCHAPTER 12 GASES AND THEIR BEHAVIOR
Chapter 12 Gases and Their Behavior Page 1 CHAPTER 12 GASES AND THEIR BEHAVIOR 121. 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 122. Nitrogen
More informationFormulae, stoichiometry and the mole concept
3 Formulae, stoichiometry and the mole concept Content 3.1 Symbols, Formulae and Chemical equations 3.2 Concept of Relative Mass 3.3 Mole Concept and Stoichiometry Learning Outcomes Candidates should be
More informationAtomic mass is the mass of an atom in atomic mass units (amu)
Micro World atoms & molecules Laboratory scale measurements Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12 C weighs 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00
More informationUnit 2: Quantities in Chemistry
Mass, Moles, & Molar Mass Relative quantities of isotopes in a natural occurring element (%) E.g. Carbon has 2 isotopes C12 and C13. Of Carbon s two isotopes, there is 98.9% C12 and 11.1% C13. Find
More informationMOLE 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 informationConcept 1. The meaning and usefulness of the mole. The mole (or mol) represents a certain number of objects.
Chapter 3. Stoichiometry: MoleMass 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 informationThe 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 = 1.6605402x1024 g) Example: How many molecules are present in 10. mg of smelling salts, (NH
More informationAS1 MOLES. oxygen molecules have the formula O 2 the relative mass will be 2 x 16 = 32 so the molar mass will be 32g mol 1
Moles 1 MOLES The mole the standard unit of amount of a substance the number of particles in a mole is known as Avogadro s constant (L) Avogadro s constant has a value of 6.023 x 10 23 mol 1. Example
More informationUnit 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 informationSTOICHIOMETRY OF COMBUSTION
STOICHIOMETRY OF COMBUSTION FUNDAMENTALS: moles and kilomoles Atomic unit mass: 1/12 126 C ~ 1.66 1027 kg Atoms and molecules mass is defined in atomic unit mass: which is defined in relation to the 1/12
More informationSample Exercise 15.1 Writing EquilibriumConstant Expressions
Sample Exercise 15.1 Writing EquilibriumConstant 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 informationMOLAR 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 informationEquilibrium 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 informationChemistry 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 informationChemical 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 informationCHAPTER 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 informationChemistry  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 informationProblem 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 informationCONSERVATION OF MASS During a chemical reaction, matter is neither created nor destroyed.  i. e. the number of atoms of each element remains constant
1 CHEMICAL REACTINS Example: Hydrogen + xygen Water H + H + +  Note there is not enough hydrogen to react with oxygen  It is necessary to balance equation. reactants products + H + H (balanced equation)
More informationChapter 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 informationThe 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 informationChemical Reactions 2 The Chemical Equation
Chemical Reactions 2 The Chemical Equation INFORMATION Chemical equations are symbolic devices used to represent actual chemical reactions. The left side of the equation, called the reactants, is separated
More informationLiquid phase. Balance equation Moles A Stoic. coefficient. Aqueous phase
STOICHIOMETRY Objective The purpose of this exercise is to give you some practice on some Stoichiometry calculations. Discussion The molecular mass of a compound is the sum of the atomic masses of all
More information