# CHAPTER 10 - PowerPoint PPT Presentation

1 / 81
CHAPTER 10

## CHAPTER 10

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript

1. CHAPTER 10 CHEMICAL QUANTITIES

2. What is a Mole? • In chemistry you will do calculations using a measurement called a mole. • The mole, the SI unit that measures the amount of substances, is a unit just like the dozen. • The mole can be related to: • the number of particles(ion,atoms, etc.) • the mass (in grams) • volume of an element (STP conditions)

3. The Number of Particles in a Mole • Because atoms, molecules and ions are very small, the number of individual particles in an object is very large. Instead of counting these particles individually, you can count them using a term that represents a specific number of particles. • Here’s the number: 602 214 199 000 000 000 000 000. • A mole is the Avogadro’s number of particles…. rather like a dozen is twelve, the mole is just that, an Avogadro’s number of particles (The number above was named for Amadeo Avogadro, an Italian chemist who worked on gases in the nineteenth century.) • Representive particles- species present in the substance (usually atoms, molecules, or formula units)

4. Examples: • How many moles of a magnesium is 1.25 X 1023 atoms of magnesium? The answer: 0.21 moles • How many moles is 2.80 X 1024 atoms of silicon? The answer: 4.65 moles • When given # of particles, divide by 6.02 x 1023

5. Mass of a Mole of a Compound • Gram atomic mass- the atomic mass of an element expressed in grams (use the Periodic Table) • The gram atomic mass of any element contains 1 mole of atoms of that element. • 1 atom Carbon = 12 amu…atomic mass units • 1 mole Carbon = 12 grams (6.02 x 1023 atoms) Examples: The formula of hydrogen is H2. What is the gram atomic mass? Note: The subscript = the # of atoms Pg. 179

6. Finding Formula Mass • Element # of Atoms x Atomic Mass = Total Mass • Examples: Find gram formula mass for: • Aluminum Chloride • Lithium Sulfide • Calcium Hydroxide

7. Relating Moles and Grams • Moles = grams given Formula Mass • Use Periodic table to determine the formula mass of substance

8. The Molar Mass of a Substance Molar Mass- the mass in grams of a mole of any substance Mol = gr Fm Examples: • How many grams are in 9.45 mol of dinitrogen trioxide (N2O3)? • Find the number of moles are in 92.2 g of iron (III) oxide (Fe2O3) • Find the number of grams for 2.5 moles of Calcium Bromide. 1. 718.2 gr. 2. .578 moles 3. 500gr pg. 182

9. The Volume of a Mole of Gas • The volume of gas changes when temperature or pressure changes. • Because of the variation gas is measured in Standard Temperature and Pressure(STP) • Standard temperature = 0o C (273 0Kelvin)(32 0F) • Standard pressure is 101.3 kPa (KiloPascals) • Molar Volume of gas= 22.4 L and is measured at STP • 1 mole of any gas at STP will occupy 22.4 liters of volume Calculate the number of liters if you have16 grams of Sulfur Dioxide@STP

10. % Composition with Formulas • Use the atomic masses on Periodic Table • Find the % composition, by mass, for the formulas of : • H2O Na2SO4

11. Calculating the Percent Composition of a Compound • The percent composition of a compound has as many percent values as there are different elements in a compound grams of element A grams of compound Example: An 8.20-g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. What is the percent composition of this compound? pg. 305 • Percent Composition- percent by mass of each element of a compound % mass of element A= X 100

12. Using Percent as a Conversion Factor • You can use percent composition to calculate the number of grams of an element contained in a specific amount of a compound. Example: calculate the mass of a carbon in 82.0 g of propane (C3H8)

13. Calculating Empirical Formulas • Empirical Formula- gives the lowest whole-number ratio of atoms (of the element) in a compound. • Empirical formula for H2O2 is HO • Empirical formula for CO2 is CO2 Examples-Find the Empirical Formula if given: --6 gr of Magnesium and 4 gr of Oxygen --79.8 gr of Carbon and 20.2 gr of Hydrogen --67.6 gr.Mercury (Hg), 10.8 gr. Sulfur, 21.6 gr. Oxygen pg. 309

14. Molecular Formulas • Molecular Compounds- gives the actual ratio of atoms in a compound. • This can be calculated once you have found the Empirical formula. • Example: CO2 Same molecular & Empirical • C6H12O6 = molecular CH2O = empirical Example: Calculate the molecular formula of the compound whose molar mass is 60.0 g and empirical formula is CH4N

15. Calculating Molecular Formulas • Steps to Solve: • Find the Formula Mass of the Empirical Formula Divide total grams given Formula Mass of Empirical Multiply this answer by the Empirical Formula, this will equal the Molecular Formula

16. STOICHIOMETRY USING THE REACTION EQUATION LIKE A RECIPE

17. USING EQUATIONS • Nearly everything we use is manufactured from chemicals. • Soaps, shampoos, conditioners, cd’s, cosmetics, medications, and clothes. • For a manufacturer to make a profit the cost of making any of these items can’t be more than the money paid for them.

18. USING EQUATIONS • Chemical processes carried out in industry must be economical, this is where balanced equationshelp. • Equations are a chemist’s recipe. • Eqs tell chemists what amounts of reactants to mix and what amounts of products to expect.

19. USING EQUATIONS • When you know the quantity of one substance in a reaction, you can calculate the quantity of any other substance consumed or created in the reaction. • Quantity meaning the amount of a substance in grams, liters, molecules, or moles.

20. USING EQUATIONS • The calculation of quantities in chemical reactions is called stoichiometry. • When you bake cookies you probably use a recipe. • A cookie recipe tells you the amounts of ingredients to mix together to make a certain number of cookies.

21. If you need a larger number of cookies than the yield of the recipe, the amounts of ingredientscan be doubled or tripled. • In a way, a cookie recipe provides the same kind of information that a balanced chemical equ. provides • Ingredients are the reactants • Cookies are the products.

22. Imagine you are in charge of manufacturing for Rugged Rider Bicycle Company. • The business plan for Rugged Rider requires production of 128 custom-made bikes each day. • One of your responsibilities is to be sure that there are enough parts available at the start of each day.

23. Assume that the major components of the bike are the frame(F), the seat (S), the wheels (W), the handlebars(H), and the pedals (P). • The finished bike has a “formula” of FSW2HP2. • The balanced equation for the production of 1 bike is. F +S+2W+H+2P FSW2HP2

24. + + + + H P F S 2W + + + + FSW HP 2 2

25. Now in a 5 day workweek, Rugged Riders is scheduled to make 640 bikes. How many wheels should be in the plant on Monday morning to make these bikes? • What do we know? • Number of bikes = 640 bikes • 1 FSW2HP2=2W (balanced eqn) • What is unknown? • # of wheels = ? wheels

26. = 1280 wheels • The connection between wheels and bikes is 2 wheels per bike. We can use this information as a conversion factor to do the calculation. 2 W 640 FSW2HP2 1 FSW2HP2

27. We can derive the same kind of information from a normal chemical reaction equation. • For instance the synthesis reaction of ammonia: N2(g) + 3H2(g)  2NH3(g) • What kinds of information can we glean from this equation? • Well for starters…

28. PARTICLES SUMMARY • 1 molecule of N2 reacts with 3 molecules of H2 to produce 2 molecules of ammonia. • N2 and H2 will always react to form ammonia in this 1:3:2 ratio of molecules. • So if you started with 10 molecules of N2 it would take 30 molecules of H2 and would produce 20molecules of NH3

29. PARTICLES SUMMARY • It isn’t possible to count such small numbers of molecules and allow them to react. • You could react Avogadro’s number of N2 molecules and make them react with 3 times Avogadro’s number of H2 molecules forming 2 times Avogadro’s number of NH3 molecules.

30. 1X 3X 2X

31. MOLES SUMMARY • We have recently learned that Avogadro’s number of particles is the same as a mole of a substance. • On the basis of the particle interpretation we just discussed, the equation also tells you the number of moles of reactants and products.

32. MOLES SUMMARY • 1 mole of N2 molecules reacts with 3 moles of H2 molecules to make 2 moles of NH3 molecules. • The coefficients of the balanced chemical equation indicate the numbers of moles of reactants and products in a chemical reaction.

33. MOLES SUMMARY • This is the most important information that a reaction equation provides. • Using this information, you can calculate the amounts of reactants and products.

34. MASS SUMMARY • A balanced chemical equation must also obey the law of conservation of mass. • Mass can be neither created nor destroyed in ordinary chemical or physical processes. • Remember that mass is related to the number of atoms in a compound through the mole.

35. MASS SUMMARY • The mass of 1 mol of N2 molecules is 28 g; the mass of 3 mols of H2 molecules is 6 g for a total mass of reactnts of 34 g. • The mass of 2 moles of NH3 molecules is 2 * 17g or 34 g. • As you can see the reactants mass is equal to the mass of the products.

36. VOLUME SUMMARY • Remember that 1 mole of any gas at STP occupies 22.4 L of space. • It follows that 22.4 L of N2 reacts with 67.2 L of H2 to form 44.8 L of Ammonia gas.

37. 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L

38. USING EQUATIONS • You can see how much information is stored in a simple balanced reaction eqn • We can combine this information with our knowledge of mole conversions to perform important common stoichiometric calculations.

39. MOLE – MOLE CALCULATIONS • A balanced rxn eqn is essential for all calculations involving amounts of reactants and products. • If you know the number of moles of 1 substance, the balanced eqn allows you to calc. the number of moles of all other substances in a rxn eqn.

40. MOLE – MOLE CALCULATIONS • Let’s go back to our synthesis of ammonia rxn. N2(g) + 3H2(g)  2NH3(g) • The MOST important interpretation of this rxn is that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

41. MOLE – MOLE CALCULATIONS • These connections of the coefficients allows us to set up conversion factors called mole ratios. • The mole ratios are used to calculate the connections in moles of compounds in our reaction equation. • We can start calculating…

42. MOLE – MOLE CALCULATIONS • Sample Mole – Mole problem: How many moles of ammonia are produced when .60 moles of N2 are reacted with H2? N2(g) + 3H2(g)  2NH3(g) Given: .60 moles of N2 Uknown: ____ moles of NH3

43. MOLE – MOLE CALCULATIONS • According to the reaction equation, for every 1 mole of N2 reacted we form 2 mols of NH3. • To determine the number of moles of NH3, the given quantity of N2 is multiplied by the mole ratio from the rxn eqn in such a way that the units of “mol N2” cancel

44. = 1.2 mol NH3 MOLE – MOLE CALCULATIONS • Solve for the unknown: N2(g) + 3H2(g)  2NH3(g) 2 mol NH3 .6 mol N2 1 mol N2

45. MOLE – MOLE EXAMPLE #2 • This equation shows the formation of aluminum oxide. 4Al(s) + 3O2(g)  2Al2O3(s) • How many moles of aluminum are needed to form 3.7 mol Al2O3? Given: 3.7 moles of Al2O3 Uknown: ____ moles of Al

46. MOLE – MOLE EXAMPLE 2 • This equation shows the formation of aluminum oxide. 4Al(s)+ 3O2(g)  2Al2O3(s) • How many moles of aluminum are needed to form 3.7 mol Al2O3? Given: 3.7 moles of Al2O3 Uknown: ____ moles of Al

47. Coefficients in the balanced equation Coefficients in the balanced equation = 7.4 mol Al MOLE – MOLE CALCULATIONS • Solve for the unknown: 4Al(s) + 3O2(g)  2Al2O3(s) 4 mol Al 3.7 mol Al2O3 2 mol Al2O3

48. MASS – MASS CALCULATIONS • No lab balance measures moles directly, instead the mass of a substance is usually measured in grams. • From the mass of a reactant or product, the mass of any other reactant or product in a given chemical equation can be calculated.

49. MASS – MASS CALCULATIONS • The mole – mole connection is still vital to do these calcs. • If the given sample is measured in grams, the mass can be converted to moles by using the molar mass. • Then the mole ratio from the balanced equation can be used to calculate the number of moles of the unknown.

50. MASS – MASS CALCULATIONS • If it is mass of the unknown that needs to be determined, the number of moles of the unknown can be multiplied by the molar mass of the desired compound. • As in mole-mole calcs, the unknow can be either a reactant or a product.