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S t o i c h i o m e t r y

S t o i c h i o m e t r y. S t o i c h i o m e t r y. … is the mathematics of chemical equations. Stoichiometry involves calculations of the amounts of reactants and products in a reaction. Coefficients show Mole Ratios. Mole ratios :

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S t o i c h i o m e t r y

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  1. S t o i c h i o m e t r y

  2. S t o i c h i o m e t r y … is the mathematics of chemical equations. Stoichiometry involves calculations of the amounts of reactants and products in a reaction.

  3. Coefficients show Mole Ratios • Mole ratios: • how many moles of products are produced with a given number of moles of reactants. • Here, the mole ratio is 1:1:2 (1H2 : 1Cl2 : 2 HCl) • This means two moles of HCl will be produced when one mole of H2 and one mole of Cl2 react.

  4. Mole ratios in water decomposition 2H2O  2H2 + O2

  5. What can we do with a chemical equation? • The mole concept allows us to use the quantitative information from a balanced equation on a larger, more practical scale. Consider the following balanced equation: • The coefficients tell us that two molecules of H2 react with each molecule of O2 to form two molecules of H2O. It follows that the relative numbers of moles are identical to the relative numbers of molecules:

  6. Ratio of eggs to cookies Proportional Relationships 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. I have 5 eggs. How many cookies can I make? Conversion Factor 5 dozen 5 eggs 2 eggs 150 cookies = 12.5 dozen cookies Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  7. List of conversion units • Counting 1 mole = 6.02 x1023 molecules • Mass 1 mole = molar mass in grams • Gas volume 1 mole = 22.4 Liters (at STP)

  8. Avogadro’s Number • Recall that one mole of a substance is equal to 6.022 x1023 units, atoms or molecules. For a good time call 602-1023 !! Avogadro –chemist, mathematician… ladies man.

  9. Converting mol to # of particles: • To convert between moles and number of particles, use Avogadro’s number as a conversion factor: • Example: How many atoms are in 12.0 moles of carbon? 12.0 mol C x 6.022 x1023 atoms = 1 mol C 7.23 x1024 atoms C

  10. Stoichiometry Problem Mapping • The mol-mol ratio can be thought of as a “bridge” between different units, where the two different units would appear on either side. The most common units in stoichiometry problems are typically grams, number of particles, and volume (in liters).

  11. Mole map

  12. Mol to mol conversions • These stoichiometric relations (equation coefficients) can be used to convert between quantities of reactants and products in a chemical reaction: 2H2 + O2 -> 2H2O • Ex: The number of moles of H2O produced from 1.57 mol of O2 can be calculated as follows: hey, mol-mol problems are easy!

  13. Mol-mol PracticeProblems 1) Use the equation below to determine how many moles of aluminum are needed to create 1.38 moles of aluminum oxide: 2) According to the equation below, how many moles of oxygen are needed to combust 0.75 moles of C2H2?

  14. Volume-Volume Conversions Volume (L) of reactant gas moles of reactant gas moles of product gas Volume (L) of product gas • Similarly to mass-mass conversions, we can use the molar volume of any gas at STP (standard temperature and pressure) to convert from reactant gas volume to product volume. • Stepwise map for conversion: • Note: Standard temperature and pressure are 25˚C and 1 atmosphere.

  15. Vol-vol Example Volume (L) of reactant gas moles of reactant gas moles of product gas Volume (L) of product gas • How many liters of oxygen is formed from the decomposition of 4.8 liters of water vapor at STP? 2H2O(g) → 2H2(g) + O2(g) • ANSWER: • ? L O2 = 4.8 L H2O x 1 mol H2O x 1 mol O2 x 22.4 L O2 22.4 L H2O 2 mol H2O 1 mol O2 = 2.4 L O2

  16. Vol-vol Practice Problem Use the equation below to determine how many liters of oxygen are needed to combust 35 L of C2H2?

  17. Mass-mass Example Mass (g) of reactant moles of reactant moles of product Mass (g) of product • How many grams of oxygen are formed from the decomposition of 6.50 g of water? 2H2O(g) → 2H2(g) + O2(g) • ANSWER: • ? g O2 = 6.50 g H2O x 1 mol H2O x 1 mol O2 x 32.00 g O2 18.02 g H2O 2 mol H2O 1 mol O2 = 9.61 g O2

  18. Mass-mass Practice Problems 1) Use the equation below to determine how many grams of aluminum are needed to create 2.8 grams of aluminum oxide: 2) According to the equation below, how many grams of oxygen are needed to combust 3.00 grams of C2H2?

  19. Mixed unit example • How many grams of oxygen are formed from the decom- position of 5.2 liters of water at STP? 2H2O(g) → 2H2(g) + O2(g) • ANSWER: • 5.2 L H2O x 1 mol H2O x 1 mol O2 x 32.00 g O2 22.4 L H2O 2 mol H2O 1 mol O2 = 3.7 g O2

  20. Stoichiometry Practice • http://library.thinkquest.org/C0110275/case8.htm (simple problems) • http://www2.hn.psu.edu/faculty/dmencer/stoichiometry/stoich.htm (limiting reactants)

  21. Unlimited Reactant… • We can use the coefficients of a balanced reaction to determine the amount of product –or “yield” -that can be formed. • Here’s a problem: If you are given 6.0 grams of C2H2 and you were given an unlimited supply of oxygen, how much CO2 could you make? • First, balance the equation: 2C2H2(g) + 5O2(g) ---> 4CO2(g) + 2 H2O(l) • Then we need to know how many moles of C2H2 we start with. Convert grams to moles using the molar mass of C2H2 (26.04 g/mol).

  22. …Unlimited Reactant • Mass to moles conversion: 6.0 grams C2H2 x 1 mol/26.04 g/mol C2H2 = 0.23 moles C2H2. • In the balanced equation, two moles of C2H2 reacts with 5 moles of O2, so … • 0.23 moles C2H2 x 4 mol CO2/2 mols C2H2 = 0.46 mol CO2 • To get the number of grams of CO2 created you need to convert it using molar mass of CO2 • 0.46 mol CO2 x 44.02 g CO2/mol CO2 = 20. g CO2

  23. Limiting vs. excess reactant

  24. Identify the limiting reactant

  25. Concept of limiting reactant • What was limiting? • What is in excess? cookies sandwiches AND oranges

  26. Limiting Reactant… • Determining the limiting reagent is necessary to find out how much product is actually going to be formed, because one reactant could run out before the other one. • Here’s a problem: If you are given 8.0 grams of C2H2 and 7.0 grams of oxygen, how much CO2 could you make? Which reactant is limiting? • First, balance the equation: 2C2H2(g) + 5O2(g) ---> 4CO2(g) + 2 H2O(l) • Then we need to know how many moles of each reactant we have. Convert from grams to moles using calculated molar masses. • 8.0 grams C2H2 x 1 mol/26.04 g/mol C2H2 = 0.31 moles C2H2. • 7.0 grams O2 x 1 mol/32.00 g/mol O2 = 0.22 moles O2.

  27. …Limiting Reactant • Next, use the molar ratios (given by the coefficients) to convert both to moles of CO2. • 0.31 moles C2H2 x 4 mol CO2/2 mols C2H2 = 0.38 mol CO2 • 0.22 moles O2 x 4 mol CO2/5 mols O2 = 0.18 mol CO2 • Compare the amounts. Which yields less product? This is the limiting reactant. • Finally, convert the lowest amount from moles of CO2 back to grams of CO2. This is your product yield. • 0.18 mol CO2 x 44.02 g CO2/mol CO2 = 7.9 g CO2

  28. Percent Yield • Tells how much product was formed and recovered, compared to the maximum possible amount (100%) • Formula: % Yield = experimental product mass x 100% theoretical product mass

  29. The End

  30. Molar Volume at STP LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS NUMBER OF PARTICLES MOLES Molar Mass (g/mol) 6.02  1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  31. LAB: Stoichiometry and Recovery of Cu Caution: NO2 gas is harmful. Conc. HNO3 and KOH must be handled carefully. Uses of copper compounds: Essential to life in trace amounts, used in electroplating, as a colorant, in fireworks, as a preservative (rot control, fungicide, algal herbicide).

  32. Mass to mass conversions • Using the conversion map below…. • We can convert from grams of a reactant to grams of a product, and vice versa. • The next two examples illustrate this type of conversion.

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