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Stoichiometry

Stoichiometry. Chapter 3. Cookery and Chemistry. Chefs have recipes, chemists have recipes. Recipes in chemistry can be seen on chemical equation . Instead of using cups and teaspoons , chemists use moles .

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Stoichiometry

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  1. Stoichiometry Chapter 3

  2. Cookery and Chemistry • Chefs have recipes, chemists have recipes. • Recipes in chemistry can be seen on chemical equation. • Instead of using cups and teaspoons, chemists use moles. • Instead of eggs, butter, sugar, etc. Chemists use chemical compounds as ingredients.

  3. How to make • chocolate chips cookies? • Ingredients: • a cup of butter • a half cup of sugar • a cup of brown sugar • a teaspoon of vanilla • 2 pieces of eggs • 2,5 cups of flour • a teaspoon of baking soda • a teaspoon of salt • 2 cups of chocolate chips • How to make a soap? • Ingredients: • 75 grams of texapon • 30 grams of coconut oil • 30 grams of glycerin • 50% of Salt solution • 50% of Citric acid • and so on….

  4. The Relation between cookies and chemistry: STOICHIOMETRY • Reaction equation tells us about what you need to react (reactant) to get a product. (like the cookies recipe) • STOICHIOMETRY is derived from Greek languages: stoicheion (element) and metron (measure) • Usage: STOICHIOMETRY is used to measure the amount of substances involved in chemical reactions.

  5. Example: CH4 + 2O2 CO2 + 2H20 • This reaction tells us that by mixing 1 mole of methane with 2 moles of oxygen we will get 1 mole of carbon dioxide and 2 moles of water. • If we want to get 10 moles of water, how many moles of methane and oxygen is needed? How many grams of CO2 is produced?

  6. What is a Mole? • The unit of measurement which is used to count the number of atoms, molecules, or particles. • 1 mole of any substance = NA = 6.02 x 1023 atoms, molecules, or particles. • e.g. 1 mole of silver = 6.02 x 1023 atoms of silver

  7. Particles in a Mole Amadeo Avogadro(1776 – 1856) 1 mole = 602213673600000000000000 or 6.022 x 1023 Amedeo Avogadro(1766-1856) never knew his own number; it was named in his honor by a French scientist in 1909. its value was first estimated by Josef Loschmidt, an Austrian chemistry teacher, in 1895. ? quadrillions thousands trillions billions millions There is Avogadro's number of particles in a mole of any substance.

  8. Analogy

  9. Molar Mass (MM) • When we measure one mole of a substance on a balance, it called “molar mass” and the unit is g/mol (gram per mole).

  10. Molar Mass Examples 12.01 g/mol 26.98 g/mol 65.39 g/mol • carbon • aluminum • zinc

  11. Molar Mass Examples • water • sodium chloride • H2O • 2(1.01) + 16.00 = 18.02 g/mol • NaCl • 22.99 + 35.45 = 58.44 g/mol

  12. Molar Mass Examples • sodium bicarbonate • sucrose • NaHCO3 • 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol • C12H22O11 • 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol

  13. Mole Conversion ٪NA X MM xNA ٪MM X 22,4L 22,4L٪ MM = Molar Mass NA = 6.02 x 1023 STP = Standard Temperature Pressure

  14. Stoichiometry has 5 basic steps • Write and balance the equation • Write down all given information • Convert everything into moles • Use mole ratio to solve the problem • Convert everything into the required unit (Mass, particles, volume)

  15. What is a mole ratio? • Mole ratio is based on the coefficient of the balanced chemical equation. • e.g. CH4 + 2O2 CO2 + 2H2O • Remember : “The ratio of coefficient = the ratio of mole” • The mole ratio = 1 : 2 : 1 : 2

  16. Example of stoichiometric problem H2 + O2 H2O (not balance) Question: • If 3 moles of oxygen are completely react with hydrogen, how many grams off water produced? • If 72 grams of water are produced, how many moles of oxygen are needed?

  17. Example of stoichiometric problem N2 + H2 NH3 (not balance) Question: • How many molecules of ammonia are produced when 2 grams of nitrogen is reacted with hydrogen ? • How many grams of oxygen are needed to produce 10 grams of ammonia?

  18. Stoichiometry in Real Life : Ethane gas (C2H6) is burnt at STP by following a reaction below: C2H6 + O2 CO2 + H2O (not balance) Question: • How many liters of oxygen are needed to burn 12 moles of C2H6? • How many liters of CO2 are produced if we burn 14 moles of O2?

  19. Water from a Camel Camels store the fat tristearin (C57H110O6) in the hump. As well as being a source of energy, the fat is a source of water, because when it is used the reaction takes place. 2 C57H110O6(s) + 163 O2(g)  114 CO2(g) + 110 H2O(l) What mass of water can be made from 1.0 kg of fat?

  20. B2H6 + O2 B2O3 + H2O Rocket Fuel The compound diborane (B2H6) was at one time considered for use as a rocket fuel. How many grams of liquid oxygen would a rocket have to carry to burn 10 kg of diborane completely? (The products are B2O3 and H2O). B2H6 + O2 B2O3 + H2O Chemical equation Balanced chemical equation 3 3 10 kg x g

  21. Lithium Hydroxide ScrubberModified by Apollo 13 Mission Astronaut John L. Swigert holds the jury-rigged lithium hydroxide scrubber used to remove excess carbon dioxide from the damaged Apollo 13 spacecraft.

  22. Percentage composition Formula: n = number of element Percentage composition tell you the percent of mass of the element which made up the compound.

  23. Example Calculate the percentage composition of C and N in urea, CO(NH2)2. Answer: Molar mass of urea = Ar C + Ar O + (2 x Ar N) + (4 x Ar H) = 12 + 16 + 28 + 4 = 60 % C = (1 x 12) x 100% = 20% 60 % N = (2 x 14) x 100% = 46.67% 60

  24. Exercise 1. Calculate the percentage composition of nitrogen in : • (NH4)2SO4 • NH4NO3 2. Determine the mass of nitrogen in: - 100 grams of Ca(NO3)2 - 200 grams of (C2H5)2NH

  25. Exercise 3. Calculate the percentage composition of carbon and oxygen in : • C6H12O6 • CH3OCH3 4. Chlorophyll contains 4.8% of magnesium. Assume that each molecules of chlorophyll contain 1 atom Mg. determine the relative molecular mass (Mr) of chlorophyll. (Ar Mg=24)

  26. Limiting Reactants • Available Ingredients • 4 slices of bread • 1 jar of peanut butter • 1/2 jar of jelly • Limiting Reactant • bread • Excess Reactants • peanut butter and jelly

  27. Limiting Reactants • Limiting Reactant • used up in a reaction • determines the amount of product • Excess Reactant • added to ensure that the other reactant is completely used up • cheaper & easier to recycle

  28. 6 green used up 6 red left over Limiting Reagents 3.9

  29. Method 1 • Pick A Product • Try ALL the reactants • The lowest answer will be the correct answer • The reactant that gives the lowest answer will be the limiting reactant

  30. Limiting Reactant: Method 1 • 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl2 2 AlCl3 • Start with Al: • Now Cl2: 10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3 27.0 g Al 2 mol Al 1 mol AlCl3 = 49.4g AlCl3 35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3 71.0 g Cl2 3 mol Cl2 1 mol AlCl3 = 43.9g AlCl3

  31. Method 2 • Convert one of the reactants to the other REACTANT • See if there is enough reactant “A” to use up the other reactants • If there is less than the GIVEN amount, it is the limiting reactant • Then, you can find the desired species

  32. Do You Understand Limiting Reagents? 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Fe2O3 needed g Fe2O3 needed OR g Fe2O3 mol Fe2O3 mol Al needed g Al needed 1 mol Fe2O3 160. g Fe2O3 1 mol Al = x x x 27.0 g Al 2 mol Al 1 mol Fe2O3 Start with 124 g Al need 367 g Fe2O3 In one process, 124 g of Al are reacted with 601 g of Fe2O3 Calculate the mass of Al2O3 formed. 367 g Fe2O3 124 g Al Have more Fe2O3 (601 g) so Al is limiting reagent 3.9

  33. 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Al2O3 g Al2O3 1 mol Al x 27.0 g Al 1 mol Al2O3 102. g Al2O3 = x x 2 mol Al 1 mol Al2O3 Use limiting reagent (Al) to calculate amount of product that can be formed. 234 g Al2O3 124 g Al 3.9

  34. Types of Formulas • Empirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. Ionic formula are always empirical formula • Molecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

  35. Empirical Formula Molecular formula the true or actual ratio of the atoms in a compound C6H12O6 the simplest whole number ratio of the atoms in a compound Example CH2O

  36. Learning Check Timberlake LecturePLUS 1. What is the empirical formula for C4H8? A) C2H4 B) CH2 C) CH 2. What is the empirical formula for C8H14? A) C4H7 B) C6H12 C) C8H14

  37. Calculating Empirical Just find the lowest whole number ratio and It is not just the ratio of atoms, it is also the ratio of moles of atoms

  38. In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen In one molecule of CO2 there is 1 atom of C and 2 atoms of O

  39. Learning Check 1. Find the empirical formula of a compound that contains 42 g of nitrogen and 9 g of hydrogen. (Ar N = 14 Ar H = 1)

  40. Learning Check 2. Find the empirical formula of a compound containing 20 g of calcium, 6 g of carbon and 24 g of oxygen. (Ar Ca = 20 Ar C = 12 Ar O = 16)

  41. Convert the grams to mol for each element Write the number of mol as a subscript in a chemical formula Divide each number by the lowest number. Multiply the result to get rid of any fractions.

  42. The Answer • 1. –Convert the grams to mol for each element N = mass = 42 g = 3 mol H = mass = 9 g = 9 mol Molar mass 14 g/mol 1 g/mol Molar mass

  43. Write the number of mol as a subscript in a chemical formula • 3 mol of N - 9 mol of H N3H9 - Divide each number by the lowest number. NH3

  44. 4. Calculate the empirical formula of a compound composed of 37 % C, 16 % H, and 47 % N. (Ar C = 12 Ar H = 1 Ar N = 14) Pretend that you have a 100 gram sample of the compound. 1. 2. change the % to grams 3. Convert the grams to mol for each element 4. Write the number of mol as a subscript in a chemical formula 5. Divide each number by the lowest number. 6. Multiply the result to get rid of any fractions.

  45. Example • Calculate the empirical formula of a compound composed of 37 % C, 16 % H, and 47 %N. • Assume 100 g so • 37 g C = 3.1 mol C 12 g/mol • 16 g H = 16 mol H 1 g H • 47 g N = 3.4 mole N 14 g N

  46. 3.1 mol C • 16 mol H • 3.4 mol N • C3.1H16N3.4 If we divide all of these by the smallest one It will give us the empirical formula

  47. Example • The ratio is 3.1 mol C = 1 mol C 3.1 mol C 1 mol C • The ratio is 16 mol H = 5 mol H 3.1 mol C 1 mol C • The ratio is 3.4 mol N = 1 mol C 3.1 mol C 1 mol C • C1H5N1 isthe empirical formula or CH5N

  48. Empirical Molecular

  49. 6. Caffeine has a molar mass of 194 g. what is its molecular formula?

  50. Find x if 194 g 97 g = 2 C4H5N2O1 2 X C8H10N4O2

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