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  1. Stoichiometry Chemistry

  2. Introduction • Billions of pounds of chemicals are produced each year across the world. • These chemicals help manufacture: • Medicines • Computer chips and electronic instruments • Fertilizers and pesticides • Glass • Paper, plastics, synthetic fibers…….

  3. Introduction • In order to determine the cost of producing such essential items we use each and every day, chemists and chemical engineers perform calculations based on balanced chemical equations • Which chemical do you think is used the most industrially? • Sulfuric acid • 165 million mega tons (1000 kg) • Value of $8,000,000,000 annually… • yes 8 billion dollars…

  4. Uses of H2SO4 • Production of fertilizers,. • Used in the manufacture of chemicals such as hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs. • It is used in petroleum refining to wash impurities out of gasoline and other refinery products. • Sulfuric acid is used in processing metals. In cleaning iron and steel before plating them with tin or zinc. • Rayon is made with sulfuric acid. • It serves as the electrolyte in the lead-acid storage battery commonly used in motor vehicles (acid for this use, containing about 33% H2SO4).

  5. Stoichiometry? Big Deal…. • What are some items you use everyday? • Clothes? • Soap and shampoo? • CD’s and iPods? • Medicine? • Company’s that make these and every other substance do not want the cost to make these items to be higher than the cost they’re sold at. • Profit ring a bell? • Companies carry out chemical rxns economically to keep prices down (we hope at least)

  6. Why did I have to learn how to balance an equation? • Chemists use balanced equations as a basis to calculate how much reactant is needed or product is formed in a reaction. • If you know the quantity of 1 piece, you can calculate the quantity of anything else in the reaction • We use grams and mols • Sometimes L, tons, molecules are used

  7. What IS stoichiometry? • The calculation of quantities in chemical reactions • We will use a version of the mole road map for this • Equations must be balanced • Allows the chemist (you) to keep track of the amounts of R and P using ratios of moles or particles

  8. Still not convinced?

  9. Still not convinced? Air bags must inflate in less than a second. The effectiveness of air bags is based on the rapid conversion of a small amount of sodium azide into a large volume of gas. The gas fills up an air bag, preventing the driver and passengers from incurring any life threatening injuries. 2NaN3(s) 2 Na(s) + 3 N2(g) Sensors trigger an igniter Fills up the airbag

  10. Interpreting a Chemical Equation • What information can you get from a balanced chemical equation? • Nitrogen monoxide is present in car exhaust. UV light catalyzes the reaction of nitrogen monoxide and O2 to produce nitrogen dioxide smog. • 2 NO(g) + O2 (g)  2 NO2(g) UV

  11. Interpreting a Chemical Equation UV • 2 NO(g) + O2 (g)  2 NO2(g) • We see that 2 molecules of NO react w/ 1 molecule of O2 to produce 2 molecules of NO2. • These coefficients indicate relative numbers of reactant and product molecules… it’s a ratio! • The coefficients indicate a ratio of moles, or the MOLE RATIO, of reactants and products in every balanced chemical equation.

  12. Verifying the Conservation of Mass Law • “Mass is neither created nor destroyed during a chemical reaction.” • The combined masses of the reactants must equal the combined masses of the products. • NO + O2 NO2 • 2 NO + O2  2 NO2 • Which one obeys the Law of Conservation of Mass? • THIS IS WHY YOU BALANCE AN EQUATION

  13. Mole-Mole Relationships • Remember, the coefficients in a chemical equation indicate the mole ratio of the reactants and products • Consider the synthesis reaction of nitrogen and oxygen to give nitrogen monoxide: N2(g) + O2(g)  2 NO(g) • We see that 1 mol of nitrogen reacts with 1 mol of oxygen to produce 2 mol of nitrogen monoxide

  14. Mole-Mole N2(g) + O2(g)  2 NO(g) We can write several mole ratios for this equation 1 mol N2 : 1 mol O2 1 mol N2 : 2 mol NO 1 mol O2 : 2 mol NO Each can become a unit factor to convert between units

  15. Mole-Mole Relationships • N2(g) + O2(g)  2 NO(g) • How many moles of oxygen react with 2.25 mol of nitrogen? • To cancel units, we will use a mole ratio as our unit factor: • 1 mol O2 / 1 mol N2

  16. Mole-Mole Relationships N2(g) + O2(g)  2 NO(g) Calculate moles of nitrogen monoxide produced by the reaction if 2.25 mol of N2 are used. Plan: Use the mole ratio 2 mol NO/1 mol N2 as the unit factor.

  17. Mole- Mole Relationships • Whenever we have a balanced chemical equation, we can always convert from moles of one substance to moles of anothersubstanceusing a mole ratio as a unit factor.

  18. Mole-Mole Relationships • Carbon monoxide is produced in a furnace by passing oxygen gas over hot coal. The balanced equation is: • 2 C(s) + O2(g)  2 CO(g) • How many moles of oxygen react with 2.50 mol of carbon? • Mole ratio is 2 mol C / 1 mol O2

  19. Mole-Mole relationships 2 C(s) + O2(g)  2 CO(g) How many moles of CO are produced from 2.50 mol of carbon? Mole ratio is 2 mol C / 2 mol CO

  20. Stoichiometry • We will apply mole ratios in order to relate quantities of reactants and products. • Stoichiometry is a term used to refer to the relationship between quantities in a chemical reaction according to a balanced chemical equation. • We can determine mass-mass relationships, mass-volume relationships (w/ a gas), and volume-volume relationships (w/ 2 gases)

  21. MASS A  MASS B PROBLEMS • An unknown mass of substance is calculated from a given mass of reactant or product. • After balancing the equation, proceed as follows: • Convert the given mass to moles using the molar mass of the substance as a unit factor • Convert the moles of the given to moles of the unknown using the mole ratio (coefficients) • Convert the moles of unknown to grams using the molar mass of the unknown as a unit factor.

  22. 1) GIVEN MASS  MOLES 2) GIVEN MOLES  UNKNOWN MOLES 3) UNKNOWN MOLES  UNKNOWN MASS • Calculate the mass of tin(IV)chloride produced by the reaction of 1.25 g of metallic tin with yellow chlorine gas. Sn(s) + 2 Cl2(g)  SnCl4(s) • First verify the equation is balanced, then calculate the moles of tin. The molar mass of Sn is 118.71 g/mol. • Next, find the moles of SnCl4 using the mole ratio (1 mol Sn = 1 mol SnCl4) • Last, calculate the mass of product using the molar mass of SnCl4 (260.51 g/mol)

  23. Do you have to complete these in 3 separate steps? • Once you get comfortable with the calculations, this is what it looks like as one step: 1.25 g Sn x x x = 2.74 g SnCl4 • Until then, complete them in three separate steps: • Given mass  moles (using molar mass) • Given moles  unknown moles (using mole ratio) • Unknown moles  unknown mass (using molar mass) 1 mol Sn 118.71 g Sn 1 mol SnCl4 1 mol Sn 260.51 g SnCl4 1 mol SnCl4

  24. Mass-Mass Calculate the mass of iron, Fe produced from 122.5 g of aluminium, Al. Fe2O3 + Al  Fe + Al2O3 Molar mass Mole ratio Molar mass g Fe g Al

  25. Mass Mass Concept Map Mass of Given Mass of unknown convert to moles using molar mass of given convert to mass using molar mass of unknown Moles of given Moles of Unknown use molar ratio

  26. Mass-Volume Problems • An unknown volume of gas is calculated from a given mass of reactant or product. • After balancing the equation, proceed as follows: • Convert the given mass to moles using the molar mass as a unit factor. • Convert the moles of the given to moles of the unknown using the mole ratio (coefficients) • Convert the moles of unknown to liters using the molar volume(at STP it’s 22.4 L/mol) as a unit factor. • Of course this can be reversed; we can find the mass of an unknown substance from a given volume of gas.

  27. 1) GIVEN MASS  MOLE2) GIVEN MOLES  UNKNOWN MOLES 3) UNKNOWN MOLES  VOLUME • 0.165 g of aluminum metal reacts with dilute hydrochloric acid. What is the volume of hydrogen gas produced at STP? 2 Al(s) + 6 HCl(aq)  2 AlCl3 + 3 H2(g) • First, calculate the moles of aluminum (26.98 g/mol) • Next, use the mole ratio to find the moles of H2. (2 mol Al = 3 mol H2)

  28. 1) GIVEN MASS  MOLE 2) GIVEN MOLES  UNKNOWN MOLES 3) UNKNOWN MOLES  VOLUME Last, multiply by the molar volume, 22.4 L/mol, to get the volume of H2 gas produced. 1 step:

  29. Mass-Volume Practice • Baking soda can be used as a fire extinguisher. When heated, it decomposes to carbon dioxide gas which can smother a fire. If a sample of NaHCO3 (84.01 g/mol) produces 0.500 L of CO2 at STP, what is the mass of the sample? 2 NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g) • Verify the equation is balanced, find the moles of CO2 by using 22.4 L/mol.

  30. Mass-Volume Concept Map Mass of given/unknown Volume of Given/ unknown gas Convert to moles using molar mass of given Convert to volume using molar volume of a gas Moles of given/unknown Moles of Given/Unknown Use molar ratio

  31. Volume-Volume Problems • We can convert from a given volume of gas to an unknown volume of gas in a single conversion using the mole ratio. • Sulfuric acid is produced by the conversion of sulfur dioxide to sulfur trioxide using heat and a platinum catalyst. The sulfur trioxide is then passed through water to produce sulfuric acid. 2 SO2(g) + O2(g) 2 SO3(g) • Calculate the liters of sulfur trioxide produced from 37.5 L of SO2. • From the balanced equation we see that 2 volumes of SO2 = 2 volumes of SO3. Pt/

  32. Sulfuric acid continued • Calculate how much oxygen gas needs to react with 37.5 L SO2 in order to produce that amount of sulfur trioxide. 2 SO2(g) + O2(g) 2 SO3(g) • Summary, for volume-volume problems, you only need to use the volume-ratio.

  33. Summary

  34. Percent Yield When the product from a reaction is less than expected, a percentage is calculated Theoretical yield The maximum amount of product that could be formed from given amounts of reactions This is calculated by you on paper! Actual yield The amount of product that actually forms in laboratory from an experiment Percent Yield- ratio of the two

  35. Limiting Reactants • Recipe: • ¼ cup butter • 1 10.5 oz bag of mini-marshmallows • 5 cup crispy rice cereal • Given that we have 1 box of crispy rice cereal (approx. 10 c. cereal), 1 stick of butter (1/2 c. butter per stick) and 1 bag (10.5 oz) of marshmallows, how many recipes can we make? • ONE

  36. Limiting Reactants • What did we run out of first?? • Marshmallows limited how many rice-crispy treats we could make. • It’s called the limiting reactant. • If we go and get more butter, can we make more recipes? • If we go and get more cereal, can we make more recipes? • If we go and get more marshmallows, can we make more recipes? NO NO YES!

  37. Limiting Reactants • In problems where we are given the amounts of two reactants, we determine the limiting reactant by: • Calculate the mass of product that can be produced from the first reactant. • Calculate the mass of product that can be produced from the second reactant. • Repeat with each reactant… hope that it’s not a big reaction… • State the limiting reactant (the one that produces the least amount of product) and the corresponding mass of product formed.

  38. Limiting Reactants • Lets think of this as a balanced equation. Assume the recipe looks like this: 3 Sugar + Butter + 8 Cereal  Rice Crispy Treats • Using the above equation, if I had 3 moles of sugar, 2 moles of butter, and 14 moles of cereal, how many recipes could I make?

  39. What is the limiting reactant? • In other words, what do we run out of first? • SUGAR!!! • We can only make one recipe w/ that amount of sugar. It limits the number of recipes we can make.

  40. Limiting Reactants 3 C6H12O6 + CH3(CH2)14COOH + 8(C6H12O6)12  Rice Crispy Treats • Using the above equation, if I had 986 g of sugar (C6H12O6), 155 g of butter (CH3(CH2)14COOH), and 12,000 g of cereal ((C6H12O6)12), how many recipes (moles) could I make? • Sugar • Butter • Cereal

  41. What is the limiting reactant? BUTTER!

  42. Last question… • You determined the limiting reactant in the previous problem to be butter. Calculate how many grams of sugar are needed in order to completely react with that amount of butter (155 g butter). • Why? Butter is leaving excess cereal and sugar which is wasted money ($$CHA-CHING!) • Start with the grams of your limiting reactant (butter) and do a mass-mass calculation to determine grams of sugar. • We only need 327 g of sugar to “react” with the butter to make these treats.

  43. Ok…one more question How many grams of sugar would be left over? We have 986 g of sugar. We only need 327 g sugar. 986 g – 327 g = 659 g sugar is left over! Manufacturers need to do these calculations all the time to be sure they’re not wasting anything.