Stoichiometry Lesson-2-2nd-Q PDF

Lesson 4 STOICHIOMETRY AVOGADRO’S NUMBER THE MOLE CONCEPT PERCENTAGE COMPOSITION CHEMICAL FORMULAS CHEMICAL REACTIONS AND EQUATIONS  STOICHIOMETRY "stoy-kee-ah-met-tree" derived...

Lesson 4 STOICHIOMETRY AVOGADRO’S NUMBER THE MOLE CONCEPT PERCENTAGE COMPOSITION CHEMICAL FORMULAS CHEMICAL REACTIONS AND EQUATIONS  STOICHIOMETRY "stoy-kee-ah-met-tree" derived from the Ancient Greek words stoicheion ("element") and metron ("measure") the study of the quantitative relationships or ratios between two or more substances undergoing a physical change or chemical change (chemical reaction).  LEARNING OBJECTIVES Define a mole Determine the molar mass of elements and compounds Calculate the mass of a given number of particles of an element or compound, or vice versa. Calculate the percentage composition of a compound from its formula Calculate molecular formula given the molar mass.  LEARNING OBJECTIVES Write equations for chemical reactions and balance the equations Interpret the meaning of a balanced chemical reaction in terms of the law of conservation of mass Describe the evidence that a chemical reaction has occurred Perform exercises on writing and balancing equations THE MOLE CONCEPT AVOGADRO’S NUMBER  Calculations of quantities in chemical reactions Stoichiometric calculations – calculations using balanced equations. STOICHIOMETRY  A counting unit Similar to a dozen, except instead of 12, it’s 602 billion trillion 602,000,000,000,000,000,000, 000 6.02 X 1023 (in scientific notation) This number is named in honor of Amedeo Avogadro Avogadro’s number (1776 – 1856), who studied The Mole Concept quantities of gases and discovered that no matter what the gas was, there were the same number of molecules present  In a lab, we cannot work with individual molecules. They are too small. 6.02 × 1023 atoms or molecules is an amount that brings us to lab size. It is ONE MOLE. (mol) One mole of 12C has a mass of 12.000 g.  FORMULA MASS FOR COMPOUNDS For covalent substances, the formula represents the numbers and types of atoms composing a single molecule of the substance; therefore, the formula mass may be correctly referred to as aa molecular mass. FORMULA MASS FOR COMPOUNDS CHLOROFORM (CHCl3) FORMULA MASS FOR COMPOUNDS ASPIRIN MOLECULE, C9H8O4  FORMULA MASS FOR COMPOUNDS Molecular mass - the mass of a molecule in u; Molar mass - the mass of one mole of an element or a compound, expressed in grams. Examples: Molecular Mass Molar Mass N2 28.02 u 28.02 g/mol H2O 18.02 u 18.02 g/mol C8H18 114.22 u 114.22 g/mol  MOLE-MASS RELATIONSHIP MOLE-MASS CALCULATIONS Avogadro’s Number as Conversion Factor 6.02 x 1023 particles 1 mole or 1 mole 6.02 x 1023 particles Note that a particle could be an atom OR a molecule!  MOLE-MASS RELATIONSHIP MOLE-MASS CALCULATIONS Na – 23 SAMPLE PROBLEM: O – 16.00 H – 1.01 How many moles of NaOH are present in 90 g of NaOH? *Molar Mass NaOH = 40 g/mol 90 g NaOH 1 mol = 2.25 mol NaOH 40 g  MOLE-MASS RELATIONSHIP MOLE-MASS CALCULATIONS How many moles and how many atoms are contained in 10.0 g of nickel? Molar mass Ni = 58.69 g/ mol To find Mass of Nickel to mole 10 g Ni 1 mol Ni = 0.170 mol Ni 58.69 g Ni  MOLE-MASS RELATIONSHIP MOLE-MASS CALCULATIONS How many moles and how many atoms are contained in 10.0 g of nickel? To find determine the no. of atoms 0.170 moles Ni 6.022×1023 atoms Ni = 1.02 ×1023 atoms Ni 1 mol Ni  MOLE-MASS RELATIONSHIP MOLE-MASS CALCULATIONS How many atoms of O2 are present in 78.1 g of oxygen? 78.1 g O2 1 mol O2 6.02 X 1023 molecules O2 2 atoms O 32.0 g O2 1 mol O2 1 molecule O2 = 3 x 1023 atoms O  Ca – 40.08 O – 16.00 Activity: Mass, No. of Particles, and Moles H – 1.01 P – 30.97 1. Calculate the formula mass of Hydroxyapatite, Mg – 24.31 C – 12.01 Ca10 (PO4)6 (OH)2, the enamel that forms the outer covering of our teeth. 2. Determine the mass in grams of the following substances. A. 1.25 mol Mg B. 0. 692 mol C8H16 C. 4.85 x 10-3 mol H2O2  PERCENT COMPOSITION AND CHEMICAL FORMULAS  PERCENT COMPOSITION FROM CHEMICAL FORMULA Law of Definite Proportion- any sample of a given compound always consists of the same elements in the same proportions by mass. The relative amounts of the elements in a compound are expressed as the percent composition or the percent by mass of each element in the compound.  PERCENT COMPOSITION FROM CHEMICAL FORMULA mass of element % mass of element = mass of compound x 100% Sample Problem: Sodium bicarbonate (NaHCO3), also called baking soda, is used as natural cleaner and is an active ingredient in antacids. Calculate the molar mass of each element present in one mole of NaHCO3. Solution: The atomic masses of each element with three significant figures are found to be: 23.0 for Na 1.01 for H 12.01 for C 16.0 for O Sample Problem: PERCENT COMPOSITION Sodium bicarbonate (NaHCO3), also called baking soda, is used as natural cleaner and is an active ingredient in FROM CHEMICAL FORMULA antacids. Calculate the molar mass of each element present The atomic masses in one mole of NaHCO3. Find the percentage composition of each element of each element. with three significant figures Solution: are found to be: 23.0 for Na Mass of Na = 1 mol Na x 23.0 g Na = 23.0 g Na 1.01 for H 1 mol Na 12.01 for C 16.0 for O Mass of H = 1 mol H x 1.01 g H = 1.01 g H 1 mol H Mass of C = 1 mol C x 12.0 g C = 12.01 g C 1 mol C Mass of O = 3 mol O x 16.0 g O = 48.0 g O 1 mol O  PERCENT COMPOSITION Solution: FROM CHEMICAL FORMULA Mass of Na = 1 mol Na x 23.0 g Na = 23.0 g Na The atomic masses 1 mol Na of each element with three Mass of H = 1 mol H x 1.01 g H = 1.01 g H significant figures 1 mol H are found to be: 23.0 for Na Mass of C = 1 mol C x 12.0 g C = 12.01 g C 1.01 for H 1 mol C 12.01 for C 16.0 for O Mass of O = 3 mol O x 16.0 g O = 48.0 g O 1 mol O The ,mass of one mole of NaHCO3 Molar mass NaHCO3 = 23.0 g + 1.01 g + 12.0 g + 48.0 g = 84.01 g/mol  PERCENT COMPOSITION FROM CHEMICAL FORMULA Solution: The atomic masses of each element % Mass of Na = 23.0 g Na 100 = 27.4 % with three 84.0 g NaHCO3 significant figures are found to be: % Mass of H = 1.01 g H 100 = 1.20 % 23.0 for Na 84.0 g NaHCO3 1.01 for H 12.01 for C % Mass of C = 12.0 g C 100 = 14.3 % 16.0 for O 84.0 g NaHCO3 % Mass of O = 48.0 g O 100 = 57.14 % 84.0 g NaHCO3 100 %  EMPIRICAL AND MOLECULAR FORMULAS OF A COMPOUND Empirical Formula A chemical formula that represents a simple whole number ratio of the number of moles of elements in the compound. Examples: MgO, Cu2S, CH2O, etc. Molecular Formula A formula that shows the actual number of atoms of each type in a molecule. Examples: C4H10, C6H6, C6H12O6.  EMPIRICAL AND MOLECULAR FORMULAS OF A COMPOUND Formula of Some Compounds FORMULA (NAME) CLASSIFICATION OF FORMULA CH Empirical C2H2 (Ethyne) Molecular C6H6 (Benzene) Molecular CH2O (methanol or formaldehyde) Empirical and Molecular C2H4O2 (ethanoic acid or acetic acid) Molecular C6H12O6 (glucose) Molecular  EMPIRICAL AND MOLECULAR FORMULAS OF A COMPOUND Empirical Formula A chemical formula that represents a simple whole number ratio of the number of moles of elements in the compound. Examples: MgO, Cu2S, CH2O, etc. Molecular Formula A formula that shows the actual number of atoms of each type in a molecule. Examples: C4H10, C6H6, C6H12O6.  Empirical Formula : Example: A compound containing carbon, hydrogen, and oxygen has the following composition (by mass percent): 68.12% C, 13.73% H, and 18.15% O, Determine its empirical formula. Solution: 1)Use mass percent to calculate mole and mole ratio of C:H:O Mole of C = 68.12 g x (1 mol C/12.01 g) = 5.672 mol C Mole of H = 13.73 g x (1 mol H/1.008 g) = 13.62 mol H Mole of O = 18.15 g x (1 mol O/16.00 g) = 1.134 mol O 2) Divide all moles by mole of O (smallest value) to get simple ratio: 5.672 mol C /1.134 mol O = 5; 13.62 mol H /1.134 mol O = 12, and 1.134 mol O /1.134 mol O = 1; Mole ratio: 5C:12H:1O 🡺 Empirical formula = C5H12O Empirical Formula Empiricalformula from mass of elements in a sample of compound Example: When 1.96 g of phosphorus is burned, 4.49 g of a phosphorus oxide is obtained. Calculate the empirical formula of the phosphorus oxide. Solution: 1) Calculate moles of P and O in sample and obtain a simple mole ratio; Mole of P = 1.96 g P x (1 mol/30.97 g) = 0.0633 mol P; Mole of O = (4.49 g – 1.96 g) x (1 mol/16.00 g) = 0.158 mol O; 2) Divide by mole of P (smaller value) to get a simple mole ratio: 0.0633 mol P/0.0633 = 1 mol P; 0.158 mol O/0.0633 = 2.5 mol O Mole ratio: 1 mol P to 2.5 mol O, OR 2 mol P to 5 mol O Empirical formula = P2O5 Molecular Formula Molecular formula is derived from empirical formula and molecular mass, which is obtained independently Empirical formula = CxHyOz; molecular formula = (CxHyOz)n, where n = (molecular mass /empirical formula mass) Example: A compound has an empirical formula C3H6O and its molecular formula is 116.2 u. What is the molecular formula? Solution: Empirical formula mass = (3 x 12.01 u) + (6 x 1.008 u) + 16.00 u = 58.1 u Molecular formula = (C3H6O)n; where n = (116.2 u/58.1 u) = 2 Incorrect molecular formula = (C3H6O)2; C= 3x2, H= 6x2, O= 1x2 Correct molecular formula = C6H12O2 Evidences and Types of Chemical Reactions What is a Chemical Reaction? A chemical reaction is in which the bonds are broken within reactant molecules, and new bonds are formed within product molecules in order to form a new substance EVIDENCES OF CHEMICAL REACTION 1) Evolution of light or heat.  EVIDENCES OF CHEMICAL REACTION 2) Temperature change (increase or decrease) to the surroundings. 3) Formation of a gas (bubbling or an odor) other than boiling. EVIDENCES OF CHEMICAL REACTION 4) Color change (due to the formation of a new substance). EVIDENCES OF CHEMICAL REACTION EVIDENCES OF CHEMICAL 5) Formation of a precipitate (a new solid forms) from the reaction of two aqueous REACTION solutions.  CHEMICAL EQUATION Depict the kind of reactants and products and their relative amounts in a reaction. reactants products 4 Al(s) + 3 O2(g) 2 Al2O3(s) The letters (s), (g), and (l) are the physical states of compounds. The numbers in the front are called stoichiometric coefficients.  Showing Phases in Chemical Equations H2O(s) H2O(l) H2O(g) Solid Phase – the substance is relatively rigid and has a definite volume and shape. NaCl(s) Liquid Phase – the substance has a definite volume, but is able to change shape by flowing. H2O(l) Gaseous Phase – the substance has no definite volume or shape, and it shows little response to gravity. Cl2(g) Additional Symbols Used in Chemical Equations Additional Symbols Used in Chemical Equations  TYPES OF CHEMICAL Synthesis or Decomposition Single Double Combination or Analysis Replacement Replacement TYPES OF CHEMICAL REACTION TYPES OF CHEMICAL REACTION TYPES OF CHEMICAL REACTION TYPES OF CHEMICAL REACTION TYPES OF CHEMICAL REACTION How do we know which reactions will occur and which ones will not? TYPES OF CHEMICAL REACTION TYPES OF CHEMICAL REACTION TYPES OF CHEMICAL REACTION COMBUSTION REACTION a) All involve oxygen (O2) as a reactant, combining with another substance b) All combustion reactions are are exothermic c) Complete combustion of a hydrocarbon always produces CO2 and H2O d) Incomplete combustion of a hydrocarbon will produce CO and possibly C (black carbon soot) as well Ex: CH4 + 2O2 => CO2 + 2H2O (complete combustion – blue flame) Ex: CH4 + O2 => C + 2H2O (incomplete combustion – yellow flame, soot) TRUE OR FALSE 1) C4H8 + 6O2 → 4CO2 + 4H2O COMBUSTION DOUBLE 2) HCl + NaOH → H2O + NaCl REPLACEMENT 3) 2KNO3(s) → 2KNO2(s) + O2(g) COMBINATION Balancing of Chemical Reactions  BALANCING CHEMICAL EQUATION Balanced Equation – one in which the number of atoms of each element as a reactant is equal to the number of atoms of that element as a product. What is the relationship between conservation of mass and the fact that a balanced equation will always have the same number of atoms of each element on both sides of an equation? A balanced equation assures that the law of conservation of mass- the total mass of reactants must equal the total mass of products. BALANCING CHEMICAL EQUATION  GUIDELINES FOR BALANCING EQUATIONS ❑ The subscripts of a compound are fixed; they SnO2 + 2H2 → Sn + 2H2O cannot be changed to balance an equation. ❑ The coefficients used SUBSCRIPT COEFFICIENT should be the smallest whole numbers possible. ❑ The coefficient multiplies every number in the formula. Rules for Balancing Chemical Reactions __H2 + __ O2 → __H2O Balancing is about finding the right coefficients! Rules for Balancing Chemical Reactions 1) You can change the coefficients, but 1) NEVER the subscripts! __H2 + __ O2 → __H2O Off Limits! Rules for Balancing Chemical Reactions 2) The coefficients must reduced to represent the lowest possible numbers. 4H2 + 2 O2 → 4H2O Rules for Balancing Chemical Reactions 3 ) It is OK to use fraction coefficients, but you must get rid of them in the end (multiply through by denominator). H2 + ½ O 2 → H 2 O Rules for Balancing Chemical Reactions 4) Often, it is helpful to save the following elements until the end (do other elements first): H, C, O Rules for Balancing Chemical Reactions 5) Do a final balance check for each element! 2H2 + O2 → 2H2O Practice 1) K + Br → KBr 2) HgO → Hg + O2 3) Na + H2O → NaOH + H2 4) CaO + H2O → Ca(OH)2 5) Al + HCl → AlCl3 + H2 Mass Relationships in Chemical Reactions  Converting grams to moles. Determine how many moles there are in 5.17 grams of Fe(C5H5)2. Given units match Goal 5.17 g Fe(C5H5)2 = 0.0278 moles Fe(C5H5)2 Use the molar mass to convert grams to Fe(C5H5)2 moles. 2 x 5 x 1.001 = 10.01 2 x 5 x 12.011 = 120.11 1 x 55.85 = 55.85 66  Stoichiometry (more working with ratios) Ratios are found within a chemical equation. 2HCl + 1Ba(OH)2 → 2H2O + 1 BaCl2 coefficients give MOLAR RATIOS 2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2 67  Mole – Mole Conversions When N2O5 is heated, it decomposes: 2N2O5(g) → 4NO2(g) + O2(g) a. How many moles of NO2 can be produced from 4.3 moles of N2O5? 2N2O5(g) → 4NO2(g) + O2(g) 4.3 mol ? mol Units match 4.3 mol N2O5 = 8.6 moles NO2 b. How many moles of O2 can be produced from 4.3 moles of N2O5? 2N2O5(g) → 4NO2(g) + O2(g) 4.3 mol ? mol 4.3 mol N2O5 = 2.2 mole O2 68  gram ↔ mole and gram ↔ gram conversions When N2O5 is heated, it decomposes: 2N2O5(g) → 4NO2(g) + O2(g) a. How many moles of N2O5 were used if 210g of NO2 were produced? 2N2O5(g) → 4NO2(g) + O2(g) ? moles 210g Units match 210 g NO2 = 2.28 moles N2O5 b. How many grams of N2O5 are needed to produce 75.0 grams of O2? 2N2O5(g) → 4NO2(g) + O2(g) ? grams 75.0 g 75.0 g O2 = 506 grams N2O5 69  Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? Determine the limiting reactant. 4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) 0.15 mol 0.10 mol ? moles Based on: 0.15 mol KO2 KO2 = 0.1125 mol O2 It was limited by the amount of KO2. Based on: 0.10 mol H2O = 0.150 mol O2 H2O H2O = excess (XS) reactant! What is the theoretical yield? Hint: Which is the smallest amount? The is based upon the limiting reactant? 70  Theoretical yield vs. Actual yield THEORETICAL Suppose the theoretical yield for an YIELD -the maximum amount of experiment was calculated to be pro- 19.5 grams, and the experiment was could be duct which performed, but only 12.3 gramsproduced of product were recovered.byDetermine the complete reaction of ACTUAL YIELD limiting reactant the % yield. -the amount of pro- duct formed from the actual chemical reaction and Theoretical is = 19.5 g based on limiting reactant yield usually Actual less than yield = 12.3 g experimentally recovered the theoretical yield 71  Limiting/Excess Reactant Problem with % Yield 4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) 120.0 g 47.0 one Hide g ?g Based on: 120.0 g KO2 = 40.51 g O2 KO2 72  Limiting/Excess Reactant Problem with % Yield 4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) 120.0 Hideg 47.0 g ?g Based on: 120.0 g KO2 = 40.51 g O2 KO2 Based on: 47.0 g H2O = 125.3 g O2 H2O Question if only 35.2 g of O2 were recovered, what was the percent yield? 73 SAMPLE PROBLEMS: 1. The pollutant SO2 can be removed from the emission of an industrial plant by making it react with CaCO3 and O2. 2SO2 + 2CaCO3 + O2 2CaSO4 + 2CO2 A. If 5.25 g SO2 reacts with excess of CaCO3 and O2, how many grams of CaSO4 are formed? B. If the actual yield of CaSO4 is 10.8 g what is the percent yield? 1. Aluminum reacts with iron (III) oxide, Fe2O3 to produce Iron and Aluminum oxide, Al2O3. Determine the percentage yield if 3.00 g of Fe was produced when 3.50 g Al was allowed to react with 5.00 g Fe2O3. 74 ASSIGNMENT: 1. If 145 g of N2H4 and 175 g N2O4 are allowed to react, 2 N2H4 + N2O4 3N2 + 4H2O, determine the a) limiting and excess reactant b. mass of water 2. In an experiment, 5.00 g aluminum is heated with 25.0 g sulfur to form aluminum sulfide. The equation fro the reaction is: 2 Al + 3S Al2S3 a) how many grams of aluminum sulfide will be formed? b) How many grams of excess reactant will remain? c) Determine the limiting and excess reactant 3. If 9398 g of fructose was used in the preparation of wine, what is the percent yield, if after the fermentation , 327 g of ethanol was produced? C6H12O6 2C2H6O + 2CO2 75  Hierarchy Style Slide Describe Concept: Lorem ipsum dolor amet, conse Donec sit Concept amet arcu elit. Add Text Here: Add Text Here: Lorem ipsum Lorem ipsum dolor amet, Type 1 Type 2 dolor amet, consectetur consectetur Add Text Here: Add Text Here: Lorem ipsum Lorem ipsum dolor amet, Topic 1 Topic 2 Topic 3 dolor amet, consectetur consectetur *Note:- you can add more hierarchy texts by first selecting the above shape and then go to “Design” tab and click “Add shape” by selecting at particular column of the above to “Add a Shape there”. Make sure to delete this after you have understood ☺. 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