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Objective #3

Objective #3. Objective: Understand molecular formulas and balancing equations. Before: Introduction to molecular formulas During: Discuss molecular formulas and balancing equations After: Review molecular formulas and balancing equations. Finding the Molecular Formula.

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Objective #3

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  1. Objective #3 • Objective: Understand molecular formulas and balancing equations. • Before: Introduction to molecular formulas • During: Discuss molecular formulas and balancing equations • After: Review molecular formulas and balancing equations

  2. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

  3. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 2. Divide the molecular mass by the mass given by the emipirical formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

  4. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3. Multiply the empirical formula by this number to get the molecular formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g (C3H5O2) x 2 = C6H10O4

  5. Review: Chemical Equations C2H5OH + 3O2®2CO2 + 3H2O Chemical change involves a reorganization of the atoms in one or more substances. reactants products When the equation is balanced it has quantitative significance: 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxideand 3 moles of water

  6. Balancing Equations In order to do stoichiometric problems a balance chemical equation is needed

  7. Balancing Equations • Ca3(PO4)2 + H2SO4 CaSO4 + H3PO4 • Reactants: Ca2+ – 3, PO43- - 2, H+ – 2, SO42+ - 1 • Products: Ca2+ - 1, SO42- - 1, H+ - 3, PO43- - 1

  8. Balancing Equations ____C3H8(g) + _____ O2(g) ----> _____CO2(g) + _____ H2O(g) ____B4H10(g) + _____ O2(g) ----> ___ B2O3(g) + _____ H2O(g)

  9. Solving a Stoichiometry Problem • Balance the equation. • Convert masses to moles. • Determine which reactant is limiting. • Use moles of limiting reactant and mole ratios to find moles of desired product. • Convert from moles to grams.

  10. Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed. 1. Identify reactants and products and write the balanced equation. 4 Al + 3 O2 2 Al2O3 a. Every reaction needs a yield sign! b. What are the reactants? c. What are the products? d. What are the balanced coefficients?

  11. Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O2 2Al2O3 6.50 g Al 1 mol Al 2 mol Al2O3 101.96 g Al2O3 = ? g Al2O3 4 mol Al 1 mol Al2O3 26.98 g Al 6.50 x 2 x 101.96÷ 26.98 ÷ 4 = 12.3 g Al2O3

  12. Objective #4 • Objective: Understand limiting reactant(s) in a chemical reaction. • Before: Introduction to limiting reactants • During: Discuss limiting reactants and chemical equations • After: Review limiting reactants and chemical equations

  13. Limiting Reactant The limiting reactantis the reactant that is consumed first,limiting the amounts of products formed.

  14. LIMITING REACTANT The reactant that gives the least number of product moles “limits” the reaction. To understand this concept, let’s suppose you were an elf working for Santa Claus and your job was to make candy canes. You take one red stick and one white stick then twist them around to make one candy cane. The ratio of red to white is 1:1. Time is ticking and you find that you have 24 red stick and 17 white sticks left. What is the maximum number of candy canes you can make? The answer is 17 candy canes!The white sticks “limit” the amount of product you could make. In chemistry we do not use sticks but *MOLES* to determine which starting material will limit the maximum amount of product that can be produced in a chemical reaction.

  15. LIMITING REACTANT a) Sodium metal reacts with oxygen to produce sodium oxide. If 5.00 g of sodium reacted with 5.00 grams of oxygen, how many grams of product is formed? 4Na (s)+ O2(g)2 Na2O(s) Start with what is given, calculate the amount of product that can be theoretically made but do it twice (once for each reactant): 5.00g Na(1 mole Na)(2 mole Na2O)( 62 g Na2O) = 6.74 g of Na2O 23 g Na 4 mole Na 1 mol Na2O 5.00g O2(1 mole O2)(2 mole Na2O)( 62 g Na2O) = 19.38 g of Na2O 32 g O2 1 mole O2 1 mol Na2O Notice you can not have two different masses produced for the same product in one reaction vessel! So in this case, Na (sodium) “limits” how much sodium oxide is produced. The correct answer is 6.74 g of sodium oxide. Wrong answer

  16. LIMITING REACTANT b) How much oxygen was used in this reaction and how much of each reactant was leftover (in excess)? 4Na (s)+ O2(g)2 Na2O(s) There are two methods used to answer this question. The Law of Conservation of mass and Stoichiometry. The amount of O2 used to make 6.74 g of Na2O is calculated by: 5.00g Na(1 mole Na)(1 mole O2)( 32 g O2) = 1.74 g of O2 was used 23 g Na 4 mole Na 1 mol O2 Or use the Law of Conservation of mass: Mass of product (6.74 g) – mass of limiting reactant (5.00 g) = mass of other reactant, in this case oxygen (1.74 g). The amount of oxygen (O2) leftover can be calculated by subtracting the starting mass of oxygen from the used mass. 5.00g – 1.74 g = 3.26 g The amount of sodium (Na) leftover at the end of the reaction is “0.00 g” (zero), since it was the limiting reactant and was completely consumed in the reaction.

  17. PRACTICE PROBLEM #19 1. How much AgCl product will be produced if 100.00 g of BaCl2 reacted with excess AgNO3? 2. How many moles of carbon dioxide could be produced from 220.0 g of C2H2 and 545.0 g of O2? 3. How many grams of CO2 can be produced by the reaction of 35.5 grams of C2H2 and 45.9 grams of O2? 4. In the reaction between CH4 and O2, if 25.0 g of CO2 are produced, what is the minimum amount of each reactant needed? 5. Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag. When 10.0 g of copper was reacted with 60.0 g of silver nitrate solution, 30.0 g of silver was obtained. What is the percent yield of silver obtained? 137.57 g 13.63 mol 50.5 g 9.09 g of CH4 & 36.4 g of O2 88.3%

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