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Chapter 3 Stoichiometry Chapter 3 Chemical Stoichiometry Stoichiometry – The study of quantities of materials consumed and produced in chemical reactions. • Since atoms are so small, we must use the average mass – we will count by weighing! Copyright © Cengage Learning. All rights reserved 2 Section 3.1 Counting by Weighing Ex:A pile of marbles weighs 394.80g. You randomly count out 10 marbles and determine their weight to be 37.60 g. A) What is the average mass of 1 marble? B) How many marbles are in the pile? Avg. Mass of 1 Marble = 37.60 g = 3.760g / marble 10 marbles 394.80 g = 105.0marbles 3.760g Return to TOC Copyright © Cengage Learning. All rights reserved 3 Section 3.3 The Mole by Weighing Counting • By observing proportions in which elements combine, 19th century chemists calculated relative atomic masses. • The modern system is based on 12C • 12C • The most accurate method for comparing masses of atoms involves the use of the mass spectrometer. is assigned a mass of exactly 12 atomic mass units (amu), and the masses of all other atoms are given relative to this. Return to TOC Copyright © Cengage Learning. All rights reserved 4 Section 3.2 Atomic Masses Counting by Weighing Schematic Diagram of a Mass Spectrometer Return to TOC Copyright © Cengage Learning. All rights reserved 5 Section 3.2 Atomic Masses Counting by Weighing Since elements occur in nature as mixtures of isotopes, chemists use the average atomic mass (aka atomic mass or atomic weight) Atomic mass can be used to “weigh out” large numbers of atoms Return to TOC Copyright © Cengage Learning. All rights reserved 6 Section 3.4 The Mole Molar Mass Avogadro’s number • The number equal to the number of carbon atoms in exactly 12 grams of pure 12C. • 1 mole = 6.022 x 1023 units of anything! • Named in honor of Avogadro (although he did not discover it!) Measuring moles Ex: 1 mole C = 6.022 x 1023 C atoms = 12.01 g C Return to TOC Copyright © Cengage Learning. All rights reserved 7 Section 3.4 Molar Mass Molar Mass is the mass in grams of one mole of a substance: Ex: Molar Mass of H = 1.008 g/mol Molar Mass of O = 16.00 g/mol (Use 4 significant figures when determining molar mass) Practice: Determine the molar mass of: 1. H2O (2 × 1.008 g) + 16.00 g = 18.02 g/mol 2. Ba(NO3)2 137.33 g + (2 × 14.01 g) + (6 × 16.00 g) = 261.35 g/mol Return to TOC Copyright © Cengage Learning. All rights reserved 8 Section 3.4 Molar Mass 1 mol = 6.022 x 1023 particles = molar mass (g) = 22.42 L of any gas at STP Return to TOC Copyright © Cengage Learning. All rights reserved 9 Section 3.4 Mole Map Molar Mass 2 2 2 2 Return to TOC Copyright © Cengage Learning. All rights reserved 10 Section 3.3 Concept Check The Mole Counting by Weighing 1. Calculate the number of iron atoms in a 4.48 mole sample of iron. (4.48 mol Fe) × (6.022×1023 Fe atoms / 1 mol Fe) = 2.70×1024 Fe atoms 2. Calculate the number of aluminum atoms in a 10.0 g sample of aluminum. (10.0 g Al) × (1 mol Al / 26.98 g Al) x (6.022×1023 Al atoms / 1 mol Al) 2.23×1023 Al atoms 3. Calculate the mass of 5.00 x 1020 atoms of cobalt. (5.00 ×1020 Co atoms) x (1 mol Co / 6.022 ×1023 atoms Co) x (58.93 g Co / 1 mol Co) = 4.89 x 10-2 g Co Return to TOC Copyright © Cengage Learning. All rights reserved 11 Section 3.3 Concept Check The Mole by Weighing Counting 4. Calculate the number of liters in 6.8g of carbon dioxide at STP (assuming ideal gas behavior). (6.8 g CO2) × (1 mol CO2 / 44.01 g CO2 ) x (22.42 L CO2 / 1 mol CO2 ) = 3.5 L CO2 5. Determine the density of carbon dioxide at STP (assuming ideal gas behavior). D = m/v *assume 1 mol = 44.01 / 22.42 1.963 g/mol CO2 6. What is the mass of the carbonate ions present in a sample of calcium carbonate containing 4.86 mol? (4.86 mol CaCO3) × (1 mol CO3 / 1 mol CaCO3) x (60.01 g CO3 / 1 mol CO3 ) = 292 g CO32Copyright © Cengage Learning. All rights reserved Return to TOC 12 Section 3.3 The Mole by Weighing Counting Done for today! Return to TOC Copyright © Cengage Learning. All rights reserved 13 Section 3.6 Percent Composition of Compounds Calculating Percent Composition • Calculating any percentage: "The part, divided by the whole, multiplied by 100" • Percentage Composition Calculate the percent of each element in the total mass of the compound mass of element in compound mass % = × 100% mass of compound Return to TOC Copyright © Cengage Learning. All rights reserved 14 Section 3.6 Percent Composition of Compounds Example: Determine the percent of iron, by mass, in iron(III) oxide, (Fe2O3): mass of element in compound mass % = × 100% mass of compound 2( 55.85 g) 111.70 g mass % Fe = = × 100% = 69.94% 2( 55.85 g)+ 3( 16.00 g) 159.70 g Return to TOC Copyright © Cengage Learning. All rights reserved 15 Ex: Calculate the percent composition by mass of magnesium carbonate. Molar mass of MgCO3 (remember, 4 SF!): 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 100.00 Check! Section 3.7 Determining the Formula of a Compound Formulas Empirical formula • Formula written as the simplest whole-number ratio Molecular formula Actual formula of the compound Return to TOC Copyright © Cengage Learning. All rights reserved 17 • Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Ex: NaCl MgCl2 CaO Al2(SO4)3 • Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2O C6H12O6 C12H22O11 Empirical: H2O CH2O C12H22O11 Section 3.7 Determining the Formula of a Compound The Poem: Percent to mass, Mass to mole Divide by small, Multiply ‘til whole! Example: The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is 146 g/mol. a. What is the empirical formula? b. What is the molecular formula? Return to TOC Copyright © Cengage Learning. All rights reserved 19 A) To determine Empirical Formula, Use the poem! Percent to mass, Mass to mole: Divide by small . . . Carbon: Hydrogen: Oxygen: Multiply ‘til whole . . . (If the answer is already whole, skip this step!) Carbon: 1.50 x 2 3 Hydrogen: 2.50 x 2 5 Oxygen: 1.00 x 2 2 Empirical formula: C3H5O2 The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. b) Determine the Molecular Formula Use the empirical formula. Divide the molecular mass by the empirical formula mass. It’s all about M/E Mass of Molecular – Given (146 g/mol) Mass of Empirical: C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g (C3H5O2) x 2 = C6H10O4 Section 3.8 Chemical Equations • Done for today . . . Return to TOC Copyright © Cengage Learning. All rights reserved 24 Warm up: A compound of B and H is 81.10% B. What is its empirical formula? Copyright © Cengage Learning. All rights reserved 25 • Percent to mass – In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H. • Mass to mole 1 mol B 81.10 g B • = 7.502 mol B 10.81 g B 18.90 g H • 1 mol H = 18.75 mol H 1.008 g H • Divide by small 7.502 mol B = 1mol B 7.502 mol • Multiply til whole 18.72 mol H = 2.499 mol H = 2.5 mol H 7.502 mol 2.5 mol H 2 = 5 mol H Answer : B2 H5 1 mol B 2 = 2 mol B Copyright © Cengage Learning. All rights reserved 26 Section 3.8 Chemical Equations • A representation of a balanced chemical equation: C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g) reactants • • products Reactants on the left, products on the right. Since atoms are neither created nor destroyed, all atoms present in the reactants must be present in the products, in the same number. Return to TOC Copyright © Cengage Learning. All rights reserved 27 Section 3.8 Chemical Equations 1C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g) • The coefficients (numbers in front) give the atomic/molecular/mole ratios Ex: 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water. (s) = solid (g) = gas (l) = liquid (aq) = aqueous: dissolved in water Return to TOC Copyright © Cengage Learning. All rights reserved 28 Section 3.9 Balancing Chemical Equations 1. Determine what reaction is occurring. What are the reactants, the products, and the physical states involved? 2. Write the unbalanced equation. 3. Balance the equation systematically, starting with the most complicated molecule(s). 4. Check - The same number of each type of atom needs to appear on both reactant and product sides. Balancing equations takes practice! Return to TOC Copyright © Cengage Learning. All rights reserved 29 Section 3.9 Balancing Chemical Equations Return to TOC Copyright © Cengage Learning. All rights reserved 30 Section 3.9 Balancing Chemical Equations Important • • • • The number of atoms of each type of element must be the same on both sides of a balanced equation. Subscripts can NOT be changed to balance an equation. A balanced equation tells us the ratio of the number of molecules/formula units which react and are produced in a chemical reaction. Coefficients can be fractions, although they are usually given as lowest integer multiples. Copyright © Cengage Learning. All rights reserved Return to TOC 31 Section 3.9 Balancing Chemical Equations Exercise Which of the following correctly balances the chemical equation given below? There may be more than one correct balanced equation. If a balanced equation is incorrect, explain what is incorrect about it. CaO + C CaC2 + CO2 I. II. III. IV. CaO2 + 3C CaC2 + CO2 2CaO + 5C 2CaC2 + CO2 CaO + (2.5)C CaC2 + (0.5)CO2 4CaO + 10C 4CaC2 + 2CO2 Return to TOC Copyright © Cengage Learning. All rights reserved 32 Section 3.9 Balancing Chemical Equations 1. NH3 + O2 NO + H2O 2. SnO2 + H2 Sn + H2O 3. Fe + H2S04 Fe2(SO4)3 + H2 4. KOH + H3PO4 + H2O 5. C2H6 + O2 H2O + CO2 6. KNO3 + H2CO3 K2CO3 + HNO3 7. B2Br6 + HNO3 + HBr K3PO4 B(NO3)3 Return to TOC Copyright Cengage Learning. All rights reserved 33 Section 3.9 Balancing Chemical Equations Concept Check Which of the following are true concerning balanced chemical equations? There may be more than one true statement. I. The number of molecules is conserved. II. The coefficients tell you how much of each substance you have. III. Atoms are neither created nor destroyed. IV. The coefficients indicate the mass ratios of the substances used. V. The sum of the coefficients on the reactant side equals the sum of the coefficients on the product side. Copyright © Cengage Learning. All rights reserved Return to TOC 34 Section 3.10 Stoichiometric Calculations: Amounts of Reactants and Products Stoichiometry is the calculation of chemical quantities from balanced equations. The four quantities involved are: • particles (atoms, ions, unit formulas or molecules) • moles • mass • volume - of gaseous reactants or products Remember: Atoms and mass are always conserved in chemical reactions. Return to TOC Copyright © Cengage Learning. All rights reserved 35 Section 3.10 Stoichiometric Calculations: Amounts of Reactants and Products 1. Begin by balancing the equation! 2. Convert the unit given to moles of that substance. 3. Use the balanced equation to set up a mole ratios. Every stoich problem has a mole-mole ratio! 4. Convert moles of given to moles of the wanted. If needed, convert moles to unit wanted. Return to TOC Copyright © Cengage Learning. All rights reserved 36 Volume 1 mol A = 22.42 L A Mass 1 mol A = (molar mass) g A Mol A (given) Molecules 1 mol A = 6.022 x 1023 particles A Get x & y from the coefficients in the balanced equation 1 mol A x x mol B = __ mol B y mol A Volume 1 mol B = 22.42 L B Mol B (wanted) Mass 1 mol B = (molar mass) g B Molecules 1 mol B = 6.022 x 1023 particles B Stoichiometry Calculations Practice 1. How many moles of ammonia are produced from 0.35 moles of nitrogen? 3H2 2NH3 N2 + 0.35 mol N2 X 2 mol NH3 1 mol N2 = 7.0 x 10-1 mol NH3 2. How many moles of aluminum are needed to react completely with 49.0 moles of chlorine? 2Al + 3Cl2 2AlCl3 49.0 mol Cl2 x 2 mol Al 3 mol Cl2 = 3.27 x 101 mol Al 3. How many moles of aluminum are needed to react completely with 49.0 g of chlorine? 2Al + 3Cl2 2AlCl3 49.0 g Cl2 X 1 mol Cl2 70.90 g Cl2 X 2 mol Al 3 mol Cl2 = 4.61 x 10-1 mol Al 4. How many grams of chlorine are needed to react completely with Al to produce 8.1 mol of aluminum chloride? 2Al 3Cl2 2AlCl3 + 8.1 mol AlCl3 X 3 mol Cl2 2 mol AlCl3 = 8.6 x 102 g Cl2 X 70.90 g Cl2 1 mol Cl2 5. How many grams of aluminum chloride are produced by the reaction of 5.4 g of Al? 2Al 5.4 g Al + 3Cl2 2AlCl3 1 mol Al X 26.98 g Al X 2 mol AlCl3 X 133.3 g AlCl3 1 mol AlCl3 2 mol Al = 2.7 x 101 g AlCl3 6. How many liters of carbon monoxide are needed to react with 4.8 g of oxygen? 2CO + O2 2CO2 4.8 g O2 1 mol O2 32.00 g O2 2 mol CO 1 mol O2 = 6.7 L CO 22.42 L CO 1 mol CO 7. What mass of ammonium is necessary to react with 2.1 x 1024 molecules of oxygen? NH4+ + 2O2 2H2O + NO2 2.1 x 1024 molec O2 1 mol O2 1 mol NH4 18.04g NH4 6.022 x 1023 2 mol O2 molec O2 = 3.1 x 101 g NH4 1 mol NH4 8. How many liters of ammonium are necessary to produce 3.0 mol of water? NH4+ + 2 O 2 2 H2 O + 3.0 mol H2O NO2 1 mol NH4 22.42 L NH4 2 mol H2O 1 mol NH4 = 3.4 x 101 L of NH4 9. How many molecules of water are produced from 222 moles of oxygen? NH4+ + 2 O2 2 H2O + NO2 222 mol O2 2 mol H2O 2 mol O2 6.022 x 1023 molec H2O 1 mol H2O = 1.34 x 1026 molec H2O 10. How many moles of nitrogen dioxide are produced from 46 moles of oxygen? NH4 + + 2 O 2 2 H2 O + 46 mol O2 NO2 1 mol NO2 2 mol O2 = 2.3 x 101 mol NO2 Section 3.11 The Concept of Limiting Reagent Done for today Return to TOC Copyright © Cengage Learning. All rights reserved 49 Copyright © Cengage Learning. All rights reserved 50 Limiting and Excess reagents The limiting reagent is consumed first and therefore “limits” the amount of product that can be formed. The excess reagent is the reactant that is left over or remaining The limiting and excess reagents are always reactants (located on the left side of the equation!) Section 3.11 The Concept of Limiting Reagent Ex: Mixture of CH4 and H2O Molecules Reacting Return to TOC Copyright © Cengage Learning. All rights reserved 52 Section 3.11 The Concept of Limiting Reagent CH4 and H2O form H2 and CO – notice that the methane is limiting, the water was in excess. Return to TOC Copyright © Cengage Learning. All rights reserved 53 Section 3.11 The Concept of Limiting Reagent Limiting Reactants Return to TOC Copyright © Cengage Learning. All rights reserved 54 How to identify a limiting problem Two amounts of reactants are given – You need to solve for the amount of product. Find which reactant you will run out of first (the limiting reagent) and use this reactant to solve the problem. Sample Problem 1 If 10.0 g of aluminum is combined with 10.0g of oxygen gas, how many grams of aluminum oxide can be made? Step One: Predict & balance the equation. 4Al+ 3O2 2Al2O3 + Step 2: Use mol/coefficient ratio • Convert reactants to moles. Aluminum : 10.0g x 1 mol = 0.371 mol Al 26.98 g Oxygen: 10.0g x 1 mol = 0.313 mol O2 32.00 g • Divide each moles by its coefficient. Aluminum: 0.371 / 4 = 0.0928 Oxygen: 0.313 / 3 = 0.104 Step 3: Identify the Limiting Reagent. The limiting reagent is the smallest number. Aluminum is the limiting reagent! Step 4: Use the limiting reagent to solve the problem. 10.0gAl x 1 mol Al x 2 mol Al203 x 101.98 g Al203 1 mol Al203 26.98 g Al 4 mol Al Answer: 1.89 x 101 g Aluminum Oxide If the problem asks how much excess reagent is left: • Use the limiting to see how much of the excess you need (do a stoich problem from limiting reagent to excess reagent) • Subtract the amount you need from the original amount given in the problem. ** Be careful of units!! Sample problem 2: If 10.0 g of aluminum is combined with 10.0g of oxygen gas, how many grams of the excess reagent remains unreacted? 4 Al + 3 O2 2 Al2O3 (Excess Reagent given) – (What you need) = Excess USE THE LIMITING REAGENT to figure out how much you need. 10.0 g Al 1 mol Al 3 mol O2 26.98 g Al 4 mol Al 32.00 g O2 1 mol O2 = 8.90 g O2 10.0g O2 - 8.90 g O2 = 1.1 g O2 remains unreacted 6.4 mol 2H2 + 3.4 mol O2 2 H2O Problem 3: How many grams of water are produced given 6.4 mol of hydrogen and 3.4 mol of oxygen? Step #1: Find the limiting & excess reagent. 6.4 mol H2 / 2= 3.2 3.4 mol O2 / 1= 3.4 H2 is the limiting reagent O2 is the excess reagent 6.4 mol 2H2 + 3.4 mol O2 2H2O Step #2 – Use the limiting reagent to solve the problem! 6.4 mol H2 X 2 mol H2O X 18.02 g H2O 2 mol H2 1 mol H2O =1.2 x 102 g H2O 6.4 mol 2H2 + 3.4 mol O2 2H2O How much of the excess reagent remains unreacted? (Excess Reagent given) – (What you need) = Excess USE THE LIMITING REAGENT to figure out how much you need. 6.4 mol H2 3.4 mol O2 - x 1 mol O2 = 3.2 mol O2 2 mol H2 3.2 mol O2 = 0.2 mol O2 remains unreacted Section 3.11 The Concept of Limiting Reagent Concept Check Which of the following reaction mixtures could produce the greatest amount of product? Each involves the reaction symbolized by the equation: 2H2 + O2 2H2O a) b) c) d) e) 2 moles of H2 and 2 moles of O2 2 moles of H2 and 3 moles of O2 2 moles of H2 and 1 mole of O2 3 moles of H2 and 1 mole of O2 Each produce the same amount of product. Return to TOC Copyright © Cengage Learning. All rights reserved 64 Section 3.11 The Concept of Limiting Reagent Percent Yield • An important indicator of the efficiency of a particular laboratory or industrial reaction. Return to TOC Copyright © Cengage Learning. All rights reserved 65 Section 3.11 The Concept of Limiting Reagent Calculating Percent Yield 1. Actual yield - what you got by actually performing the reaction 2. Theoretical yield - what stoichiometric calculation says the reaction SHOULD have produced Αctual Yield Percent Yield = Τheoretical Yield 100 Return to TOC Copyright © Cengage Learning. All rights reserved 66 Section 3.2 Atomic Masses Counting by Weighing Calculating % yield video Return to TOC Copyright © Cengage Learning. All rights reserved 67 Example 1 – Find the percent yield when 68.0 g silver reacts with oxygen to form 50.0 g silver oxide. Step 1: Determine and balance the equation: 68.0 g 4 Ag 50.0 g + O2 2 Ag2O Step 2: Calculate the theoretical yield from the reactant given. 68.0 g 4 Ag 68.0 g Ag 1 + O2 50.0 g 2 Ag2O mol Ag 2 mol Ag2O 231.8 g Ag2O 107.9 g Ag 4 mol Ag 1 mol Ag2O = 73.0 g Ag2O Theoretical yield Step 3: Plug numbers into the equation and... solve! Actual Yield (product given) Theoretical Yield X 100 = % Yield (product calculated) 50.0 g (product given) 73.0 g X 100 = % Yield (product calculated) % Yield = 68.5% Example 2 – Find the actual product made when the incomplete reaction between 16.5 g hydrogen gas and oxygen gas produces a 70.0% yield of water. Step 1: Determine and balance the equation: 16.5 g 2 H2 70.0% + O2 2 H2O Step 2: Calculate the theoretical yield from the reactant given. 16.5 g 2 H2 16.5 g H2 1 + O2 mol H2 2.016 g H2 2 H2O 2 mol H2O 2 mol H2 18.02 g H2O 1 mol H2O = 1.47 x 102 g H2O Theoretical yield Step 3: Plug numbers into the equation and... solve! Actual Yield (product given) Theoretical Yield X 100 = % Yield (product calculated) Actual Yield (product given) 1.47 x 102 g X 100 = 70.0 (product calculated) Actual Yield = 1.03 x 102 g H2O