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Chemical Composition Chemical Quantities

Chemical Composition Chemical Quantities. Chapters 9&10 Pg. 221-293. Chemical Composition. Atomic Masses - Because the mass of one atom is so small- on the order of 10 -23 gram- a special unit – the atomic mass unit - is used to described the mass of an atom.

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Chemical Composition Chemical Quantities

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  1. Chemical CompositionChemical Quantities Chapters 9&10 Pg. 221-293

  2. Chemical Composition • Atomic Masses- • Because the mass of one atom is so small- on the order of 10-23 gram- a special unit – the atomic mass unit- is used to described the mass of an atom. • A system has been devised for expressing the masses of atoms on a relative scale- that is, a scale based on experimental comparison with a standard. The scale for expressing atomic masses is based on the mass of carbon- 12. 1 carbon 12 atom weighs 12 amu. • Masses on the periodic table are weighted averages of all the naturally occurring isotopes for an element.

  3. Atomic Mass = the mass of an atom expressed relative to the mass of carbon –12 • Na = • K = • Cl = • Notice that we will be rounding to the 1/10th place for all calculations. Formula Mass / Molecular Mass = the sum of the atomic masses for all the atoms in a compound. H2O H: 2 atoms X 1.0 amu = 2.0 amu O: 1 atom X 16.0 amu = 16.0 amu = 18.0 amu HC2H3O2 = ?

  4. The Mole- Three Faces • Avogadro’s Number One mole refers to Avogadro’s number ___________________ • 1 mole of silver = • 1 mole of CO2 = Just like a dozen represents 12 of anything, and a gross represents 144 of anything.

  5. 2. Molar Mass One mole of particles has a special mass associated with it: the formula mass or molecular mass of the substance expressed in grams.

  6. Question: Find the molar mass of the following substances: • Hydrogen gas • Carbon Monoxide • Glucose (C6H12O6)

  7. 3. Molar Volume of an Ideal Gas Suppose the mole of particles happens to be a gas. In this case, the term mole has a special volume associated with it. If the mole of gas particles behaves ideally, the gas will occupy a volume of approximately 22.4 L at STP. • 1 mole of He gas occupies 22.4 L • 1 mole of hydrogen gas (H2) occupies 22.4 L • 1 mole of carbon dioxide gas (CO2) occupies 22.4 L

  8. Mole Conversions • Because the mole measures mass, number of particles, and volume of a gas, it is the central unit in converting the amount of a substance from one type of measurement to another. • These conversions will require use of the Factor Label Method we learned at the start of the year.

  9. FLM • How many apples are there in 3.75 dozen? 3.75 dozen X 12 apples = 45 apples 1 dozen Unit you want Unit you are trying to get rid of

  10. Mass and Moles • If you know the mass of a given amount of substance, you can calculate the number of moles of the substance. • Your fact you will use in this conversion is the molar mass of the substance.

  11. Example: Suppose a laboratory procedure calls for 0.65 moles of NaCl to be added to a solution. • Question: Suppose you produced 45.0 g of magnesium sulfate in the lab. How many moles of magnesium sulfate does this represent?

  12. Particles and Moles • This conversion requires only one fact that does not change from atom to atom or compound to compound: Avagadro’s Number = 6.02 X 1023 “stuff” = 1 mole of “stuff” Stuff could be atoms, molecules, particles, electron, protons, ions, ….anything!

  13. Example: Determine the number of molecules of water in 2.55 moles of water. • Question: How many moles of C6H12O6 contain 7.9 X 1024 molecules?

  14. Multi- Step Conversions • Using a mole map and the FLM, link your conversion facts together and solve. • Question: You need 250.g of table sugar or sucrose (C12H22O11) to bake a cake. How many sucrose molecules will be in the cake? Note: There is no direct conversion between grams and molecules. You must first convert grams to moles, and then moles to molecules.

  15. Question: A student fills a 1.0 L flask with carbon dioxide at standard temperature and pressure. How many molecules of gas are in the flask?

  16. Percent Composition by Mass Percent = part per hundred • Percentages are calculated by dividing the quantity under consideration by the total quantity involved and multiplying the result by 100. For example, if 7 persons in 20 wear glasses, then the percentage of persons wearing glasses is

  17. Similarly, we can calculate the percentage composition of an element in a compound by dividing the mass of the element by the formula mass of the substance and multiplying the result by 100. • Question: Calculate the percentage of oxygen by mass in CuSO4.

  18. Empirical Formula from Percent Composition Empirical Formula = • Example: The empirical formula for N2O4 is NO2, and the empirical formula for C6H6 is CH, but the empirical formula for H2O is still H2O. • Since we calculate percent composition from chemical formulas, we are able to reverse the process and determine formulas from percent composition. However, when we use this process, we are able to determine only the empirical formula.

  19. Question: Calculate the empirical formula of a substance that contains 50.00 % sulfur and 50.00% oxygen by mass. • Solution: Assume we have 100 grams of the substance. If so, then based on the percentages, we would have 50.00g of S and 50.00g of O. If we divide the given masses by the atomic masses of the two elements, we transform the mass into moles.

  20. As strange as the previous formula looks, it is accurate because it shows the relative number of moles of each element. Our task is to make the formula look like a formula, that is, to convert it to one with whole numbers. Usually this is accomplished by dividing each element by the smallest number of moles.

  21. Determining Molecular Formula • The empirical formula for a compound indicates the simplest ratio of the atoms in the compound. However, it does not tell the actual numbers of atoms in each molecule of the compound. • Molecular Formula =

  22. Question: The empirical formula for glucose is CH2O. Its empirical formula mass is 30.0 g / mol. Experiments show that the molar mass of glucose is 180 g / mol. What is the molecular formula of glucose?

  23. Mole Problems Involving Chemical Equations Consider the following equation: N2 + 3H2 2NH3 How do we interpret the coefficients? • Molecules OR • Moles of molecules 1 mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3

  24. We can construct a mole diagram for solving mole problems involving equations just as we did for single substances. Note that the diagram represents simply the joining of two single mole maps. It does not matter whether A and B are both reactants, products, or one of each of a given chemical reaction. The important point is that (moles of ) A and (moles of ) B are connected by the coefficients the two substances have in the BALANCED chemical reaction.

  25. Gas volume of A at STP Gas volume of B at STP Mass of A Mass of B Moles of A Moles of B Number of Particles of A Number of Particles of B

  26. All equation problems are solved using the steps listed below: • Examine the chemical equation and make sure that it is balanced. • Refer to the preceding diagram and prepare a solution map as follows: Quantity of A  moles of A  moles of B  Quantity of B given in problem from rxn from rxn calculate 3. Set up the problem, using FLM and the solution map 4. Perform the calculations on the numbers and units.

  27. Mole Problems • Question: In the equation N2 + H2 NH3, how many moles of N2 are needed to produce 5.0 moles of NH3? Tip- BALANCE EQUATION FIRST!

  28. Mass – Mass Problems • Question: In the equation CH4 + O2 CO2 + H2O how many grams of CO2 are formed when 8.0 grams of CH4 reacts? Tip- BALANCE EQUATION FIRST

  29. Mass-Volume Problems • Question: In the equation CO (g) + O2 (g)  CO2 (g) how many liters of CO2 (g) at STP is produced by the reaction of 64.0 grams of O2 (g)? Tip- BALANCE EQUATION FIRST

  30. Volume – Volume Problems • Question: In the equation NH3 (g) + O2 (g)  NO (g) + H2O (l) how many liters of NH3 (g) at STP are needed to react with 200. liters of O2 (g) at STP? Tip- BALANCE EQUATION FIRST

  31. Problems Involving Numbers of Particles • Question: In the equation C2H6 + O2 CO2 + H2O How many molecules of C2H6 are needed to produce 27 grams of H2O? Tip- BALANCE EQUATION FIRST

  32. Limiting Reactants and Percent Yield Consider the recipe for tollhouse cookies. The number of batches of cookies you can bake depends on the amount of each of the starting materials you have.

  33. Ingredients: • 2 1/4 cups all-purpose flour • 1 teaspoon baking soda • 1 teaspoon salt • 1 cup (2 sticks) butter or margarine, softened • 3/4 cup granulated sugar • 3/4 cup packed brown sugar • 1 teaspoon vanilla extract • 2 large eggs • 2 cups (12-ounce package) NESTLÉ TOLL HOUSE Semi-Sweet Chocolate Morsels • 1 cup chopped nuts Original Nestlé Toll House Chocolate Chip Cookies

  34. If you are in short supply of any of the ingredients, then the recipe must be modified in order to make the cookies. For example, say you only had 1 1/4 cups of semi sweet chocolate chips. • How would that affect the rest of your recipe? • How many cups of flour would you wind up using?

  35. When the quantities of reactants are available in the exact ratio described by the balanced equation, the chemists say that the reactants are in stoichiometric proportions. When this is the case, all the reactants will take part in the reaction and there will be no reactants left over one the reaction is complete. More often than not, one of the reactants is available in limited quantities. The reactants are then said to be in non-stoichiometric proportions.

  36. Limiting Reactant • The reactant that limit the amount of product formed in a chemical reaction. It is completely used up in the chemical reaction.

  37. To determine the limiting reactant in an equation, you must complete two separate mass- mass problems. • First, write the balanced equation for the reaction. • Then, calculate the mass of one of the reactants needed based on the other reactant. • Compare the calculated needed value to what you actually are given. • If the calculated needed value is more than what is given, the substance is the limiting reactant. If not, the other reactant will by default be the limiting reactant.

  38. Question: Identify the limiting reactant when 10.0 g of water reacts with 4.5 g of sodium to produce sodium hydroxide and hydrogen gas.

  39. Question: Identify the limiting reactant when 12.5 L H2S at STP is bubbled through a solution containing 24.0 g of KOH to from K2S and H2O.

  40. Once the limiting reactant is identified, it is easy to calculate the theoretical mass or volume of the products. Remember, the limiting reactant will be used up completely in the chemical reaction. The other reactant will be in excess meaning there will be reactant left over.

  41. Question: If 3.5 g of Zn and 3.5 g of S are mixed together and heated, what mass of ZnS will be produced?

  42. Question: What mass of barium nitride (Ba3N2) is produced from the reaction between 22.6 g barium and 4.2 g nitrogen gas?

  43. Percent Yield • Expected yield- the amount of a product that should be produced based on calculations. • Actual yield- the amount of product that is really obtained from a chemical reaction. Why would the expected yield be different than the actual yield? • Loss of product in the chemical reaction • Error in measuring the starting materials • Side reactions take place between the reactants leading to a different product than was expected. • Reaction did not occur because of temperature, pressure, etc. was not ideal for the reaction.

  44. Percent yield = percent of the expected yield that was obtained. Percent yield = Actual yield X 100 Expected yield • Question: Determine the percent yield for the reaction between 2.80 g Al(NO3)3 and excess NaOH if 0.966 g Al(OH)3 is recovered.

  45. Question: Determine the percent yield for the reaction between 15.0 g N2 and 15.0 g H2 if 10.5 g NH3 is produced.

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