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The Mole II

The Mole II. Mathematical Relationships with Chemical Equations… Stoichiometry. Suppose…. You have to make a number of BLTs (bacon, lettuce, tomato sandwiches).

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The Mole II

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  1. The Mole II Mathematical Relationships with Chemical Equations…Stoichiometry

  2. Suppose… • You have to make a number of BLTs (bacon, lettuce, tomato sandwiches). • Each sandwich requires 2 slices of bread, 4 slices of bacon, 2 slices of tomato, 1 leaf of lettuce, and 1 tbsp of mayonnaise. • How many sandwiches can you make with 24 slices of bacon? 14 pieces of bread? 7 slices of tomato?

  3. molecules 1 molecule C3H8 + 5 molecules O2 3 molecules CO2 + 4 molecules H2O amount (mol) 1 mol C3H8 + 5 mol O2 3 mol CO2 + 4 mol H2O mass (amu) 44.09 amu C3H8 + 160.00 amu O2 132.03 amu CO2 + 72.06 amu H2O mass (g) 44.09 g C3H8 + 160.00 g O2 132.03 g CO2 + 72.06 g H2O total mass (g) 204.09 g 204.09 g Information Contained in a Balanced Equation Viewed in Terms of Reactants C3H8(g) + 5O2(g) Products 3CO2(g) + 4H2O(g)

  4. Stoichiometry • The relation between the quantities of substances that take part in a chemical reaction or form a compound • “"Stoichiometry" is derived from the Greek words στοιχείον (stoikheion, meaning element) and μέτρον (metron, meaning measure.” (http://en.wikipedia.org/wiki/Stoichiometry)

  5. Info in a Chemical Equation 2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g) 2 mol C8H18= 25 molO2 = 16 molCO2 = 18 molH2O • So…

  6. Summary of the mass-mole-number relationships in a chemical reaction. MASS(g) of compound A MASS(g) of compound B MM (g/mol) of compound A MM (g/mol) of compound B molar ratio from balanced equation AMOUNT(mol) of compound A AMOUNT(mol) of compound B Avogadro’s number (molecules/mol) Avogadro’s number (molecules/mol) MOLECULES (or formula units) of compound A MOLECULES (or formula units) of compound B

  7. Questions to Ask When Solving a Problem • Are grams mentioned? If yes you will need to use a molar mass conversion. • Are you comparing two different substances? If yes you will need a conversion using the coefficients in an equation. • Are particles (atoms or molecules) involved? If yes you will need a conversion with Avogadro’s number. You may need any number of these conversions depending on the problem.

  8. Calculating Amounts of Reactants and Products PROBLEM: • In a lifetime, the average American uses 1750 lb(794 g) of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores, such as chalcocite, or copper(I) sulfide, by a multistep process. After an initial grinding, the first step is to “roast” the ore (heat it strongly with oxygen gas) to form powdered copper(I) oxide and gaseous sulfur dioxide. • How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? • (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? • (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?

  9. 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) 3mol O2 (a) 10.0mol Cu2S 2mol Cu2S 2mol SO2 64.07g SO2 1 mol SO2 2mol Cu2S 10.0mol Cu2S (b) 103g Cu2O 1 mol Cu2O 1kg Cu2O 143.10g Cu2O 3mol O2 32.00g O2 1 kg O2 (c) 2.86kg Cu2O 2mol Cu2O 1 mol O2 103g O2 Calculating Amounts of Reactants and Products SOLUTION: (a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? = 15.0mol O2 (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? = 641g SO2 (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? = 0.959kg O2

  10. Now suppose… • You have to make BLTs (bacon, lettuce, tomato sandwiches). • Each sandwich requires 2 slices of bread, 4 slices of bacon, 2 slices of tomato, 1 leaf of lettuce, and 1 tbsp of mayonnaise. • How many sandwiches can you make with 24 slices of bacon and 14 pieces of bread?

  11. Limiting Reactants • In most chemical changes you do not have the exact amounts of all reactants. • In this case, one reactant will run out first. This reactant will limit the amount of product and so is called the limiting reactant or limiting reagent. • The reactant(s) left over is called the excess reactant.

  12. Calculating Amounts of Reactant and Product in a Limiting-Reactant Problem PROBLEM: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine (N2H4) and dinitrogen tetraoxide(N2O4), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x102 g of N2H4 and 2.00x102 g of N2O4 are mixed?

  13. N2H4(l) + N2O4(l) N2(g) + H2O(l) 1 mol N2H4 3 mol N2 32.05g N2H4 2mol N2H4 28.02g N2 1 mol N2 1 mol N2O4 92.02g N2O4 3 mol N2 1mol N2O4 Calculating Amounts of Reactant and Product in a Limiting-Reactant Problem SOLUTION: 2 3 4 1.00x102g N2H4 = 4.68mol N2 2.00x102g N2O4 = 6.51mol N2 N2H4 is the limiting reactant because it produces less product, N2, than does N2O4. 4.68mol N2 = 131g N2

  14. Excess Reactant • How much reactant was left in the previous problem? • 6.51 mol N2 possible but only 4.68 mol made 6.51 - 4.68 = 1.83 mol N2 not made

  15. How many grams of aluminum carbonate are formed when 20.0 grams of sodium carbonate react with 25.0 grams of aluminum nitrate? How much of which reactant is left over?

  16. Percent Yield • Reactions do not always go to completion for many reasons. Percent yield is an expression of how much of the expected product actually formed. • The theoretical amount generally comes from the stoichiometry of the reaction.

  17. SiO2(s) + 3C(s) SiC(s) + 2CO(g) 1 mol SiO2 60.09 g SiO2 40.10 g SiC 1 mol SiC 51.4 g 66.73 g Calculating Percent Yield PROBLEM: Silicon carbide (SiC) is an important ceramic material that is made by allowing sand (silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When 100.0 g of sand is processed, 51.4 g of SiC is recovered. What is the percent yield of SiC from this process? SOLUTION: 100.0 g SiO2 = 1.664 mol SiO2 mol SiO2 = mol SiC = 1.664 1.664 mol SiC x 100 =77.0% = 66.73 g

  18. Calculating Amounts of Reactants and Products for a Reaction in Solution PROBLEM: Specialized cells in the stomach release HCl to aid digestion. If they release too much, the excess can be neutralized with antacids. A common antacid contains magnesium hydroxide, which reacts with the acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10M HCl to simulate the acid concentration in the stomach. How many liters of “stomach acid” react with a tablet containing 0.10g of magnesium hydroxide?

  19. Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) 1 mol Mg(OH)2 58.33g Mg(OH)2 2 mol HCl 1 mol Mg(OH)2 1L 0.10mol HCl Calculating Amounts of Reactants and Products for a Reaction in Solution SOLUTION: = 3.4x10-3 mol HCl 0.10g Mg(OH)2 3.4x10-3 mol HCl = 3.4x10-2 L HCl

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