1 / 81

Unit 6 Chapter 12 Chemical Quantities or

"Our Friend the Mole". Unit 6 Chapter 12 Chemical Quantities or. Stoichiometry stoichiometry —using balanced chemical equations to obtain info. 12.1 Counting Particles of Matter Particles of matter are too small and numerous to count SI unit of chemical quantity =

shona
Download Presentation

Unit 6 Chapter 12 Chemical Quantities or

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. "Our Friend the Mole" Unit 6 Chapter 12 Chemical Quantities or

  2. Stoichiometry stoichiometry—using balanced chemical equations to obtain info

  3. 12.1 Counting Particles of Matter Particles of matter are too small and numerous to count • SI unit of chemical quantity = the mole (abbreviated mol) • The mole is a counting unit

  4. How do you measure how much? You can measure mass = grams or volume = liters or you can count pieces = MOLES

  5. Moles 1 mole = 6.02 x 1023 representative particles • 602 000 000 000 000 000 000 000 Treat it like a very large dozen • 6.02 x 1023 is called Avogadro's number

  6. Representative particles = The smallest pieces of a substance representativeparticles = • ATOMS • IONS • MOLECULES • FORMULA UNITS

  7. Representative particles For a molecular (covalent) compounds it is a molecule compound with all nonmetals CO, BF3 , Cl2 1 mole = 6.02 x 1023 molecules

  8. Representative particles For an element it is an atom • Unless it is diatomic one symbol, no charge: Br,Cs 1 mole = 6.02 x 1023 atoms

  9. Representative particles For an ionic compound it is a formula unit compound with metal and nonmetal - KI, Na2SO4 1 mole = 6.02 x 1023 formula units

  10. Representative particles For ions it is 1 mol ions one symbol with charge (monatomic) or more than one symbol with charge (polyatomic: Na+ , N3- , (C2H3O2) – 1 mole ions = 6.02 x 1023 ions

  11. Molar Mass = The mass of 1 mole of an element in grams. • We can make conversion factors from these. • Example - We can write this as 12.01 g C = 1 mol • To change grams of a compound to moles of a compound. • Or moles to grams

  12. Molar Masses • the atomic masses on the periodic table • have a unit of amu (atomic mass unit) GAM = gram atomic mass = the atomic mass (listed on the periodic table) written in grams • 1 atom Xe = 131.30 u • GAM of Xe = 131.301 g

  13. molar mass—the mass, in g, of 1 mole of a substance Add up the gram atomic mass of all elements in compound to calculate molar mass of a substance

  14. Find the molar mass of methane, CH4. CH4 = 1(12.0) + 4(1.0) = 12.0 + 4.0 = 16.0 g Find the molar mass of calcium hydroxide, Ca(OH)2. Ca(OH)2 = 1(40.1) + 2(16.0) + 2(1.0) = 74.1 g

  15. 1. What is the molar mass of 1 mole of phosphorus? 2. What is the molar mass of a molecule of H2? 3. What is the molar mass of a formula unit of MgCl2? • 30.97 g • 2.016 g • 94 g

  16. Molar Volume: volume-to-mole and mole-to-volume conversions At STP, all gases occupy the same amount of space: MOLAR VOLUME of any gas at STP: 22.4 L = 1 mol

  17. dimensional analysis = using the units (dimensions) to solve problems steps for success: 1) identify unknown (read carefully) 2) identify known (read carefully) “Play checkers” with the units, moving them diagonally, canceling when appropriate. All units should cancel except those of the desired answer. 3) plan solution 4) calculate 5) check (sig.figs., units, and math)

  18. CONVERSION FACTOR SUMMARY: 6.02 x 10^23 representative particles 1 MOLE & 1 MOLE 6.02 x 10^23representativeparticles

  19. CONVERSION FACTOR SUMMARY: MOLAR MASS (g) & 1 MOLE 1 MOLE MOLAR MASS (g) for a gas at STP: 22.4 L & 1 MOLE 1 MOLE 22.4 L

  20. Calculation question How many molecules of CO2 are the in 4.56 moles of CO2 ? 4.56 moles CO2 X 6.02 x 1023 molecules CO2 1 1 mole CO2 = 27.5 x 1023 molecules CO2

  21. For example • How many moles is 5.69 g of NaOH? 5.69 g NaOH x 1 mole= 0.142 mol NaOH 1 40.0 gNaOH need to change grams to moles for NaOH • 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g • 1 mole NaOH = 40.00 g

  22. Examples • How much would 2.34 moles of carbon weigh? 2.34 mol C X 12.01 g C 1 1mol C = 28.10 g C

  23. Calculation question How many moles of salt is 5.87 x 1022 formula units? 5.87 X 1022 formula units NaCl X 1 mole NaCl_ 1 6.02 x 1023f.u. NaCl = 0.0975 moles NaCl

  24. Examples • How many moles of magnesium in 4.61 g of Mg? 4.61 g Mg X _1 mol Mg 1 24.3 g Mg = 0.1897 mol Mg

  25. example What is the volume, in L, of 0.495 mol of NO2 gas at STP? 0.495 mol NO2 x 22.4 L NO2 = 11.1 L NO2 1 1 mol NO2

  26. How many moles are found in 84 L of neon gas at STP? 84 L Ne x 1 mol Ne = 3.8 mol Ne 1 22.4 L Ne

  27. There are many types of mole problems: 1 step: r.p. mol & mol r.p. mass mol & mol mass 2 step: mass r.p. & r.p. mass mass volume & volume mass r.p. volume & volume r.p.

  28. Moles of Compounds • 1 mole of a compound contains as many moles of each element as are indicated by the subscripts in the formula for the compound. • Example: 1 mole of ammonia (NH3) has 1 mole of nitrogen atoms and 3 moles of hydrogen atoms.

  29. Percent composition • composition of a compound is the percent by mass of each element in the compound. Find the mass of each element divide by the total mass of compound.

  30. Determine the percent composition of calcium chloride (CaCl2). • Step 1: determine the molar masses of each element in compound and the compound Ca = 40.078 g Cl2 = 2 x 35.453 g = + 70.906 g CaCl2=110.984 g

  31. Step 2: divide the molar mass of each element by the molar mass of the compound and times by 100 ****make sure they add up to 100

  32. Empirical Formula The lowest whole number ratio of elements in a compound. CH2 H2O

  33. You can determine the empirical formula from percent composition and mole ratios *** percent means “parts per hundred” assume you have 100 g of the compound Step 1: change the percent sign to g (grams) Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N

  34. Step 2:calculate the number of moles for each element by converting grams to moles using 1 mole = molar mass 38.67 g C x 1mol C = 3.220 mole C 1 12.01 g C 16.22 g H x 1mol H = 16.1 mole H 1 1.01 g H 45.11 g N x 1mol N = 3.220 mole N 1 14.01 g N

  35. Step 3: divide each number of moles by the smallest number of moles 3.220 mole C = 1 mol C 3.220 16.1 mole H = 5 mol H 3.220 3.220 mole N = 1 mol N 3.220 So empirical formula is CH5N

  36. Determine the empirical formula of a compound with percent composition of 38.4 % Mn, 16.8 % C, 44.7 % O 38.4 g Mn X 1mol Mn = 0.699 mole Mn 1 55 g Mn 16.8 g C X 1mol C = 1.339 mole C 1 12 g C 45.7 g O X 1mol O = 2.798 mole O 1 16 g O

  37. 0.699 mole Mn = 1 mol Mn 0.699 1.339 mole C = 2 mol C 0.699 2.798 mole O = 4 mol O 0.699 So formula is MnC2O4

  38. For many compounds, the empirical formula is not the true formula A molecular formula tells the exact number of atoms of each element in a molecule

  39. Example: acetic acid molecular formula = C2H4O2 empirical formula = CH2O The molecular formula for a compound is always a whole-number multiple of the empirical formula.

  40. To determine molecular formula you need to know the molar mass of the compound Divide the actual molar mass by the molar mass of the empirical formula. Multiply the empirical formula by this number. molar mass of compound molar mass of empirical formula

  41. Example: A compound has an empirical formula of ClCH2 and a molar mass of 98.96 g/mol. What is its molecular formula? molar mass of ClCH2 = 49.0 g/mol 98.96 = 2.01 49.0 2 X (ClCH2) Empirical formula = ClCH2 Molecular formula = Cl2C2H4

  42. Example • A compound has an empirical formula of CH2O and a molar mass of 180.0g/mol. What is its molecular formula?

  43. Example • Ibuprofen is 75.69 % C, 8.80 % H, 15.51 % O, and has a molar mass of about 207 g/mol. What is its molecular formula?

  44. Chapter 12.2 “Using the Mole" Unit 6 Chemical Quantities Continued

  45. remember our friend the mole? Recall that the mole should be treated like a dozen 1 dozen = 12 pieces & 1 mole = 602 000 000 000 000 000 000 000 pieces (or 6.02 X 1023 r.p)

  46. remember our friend the mole? You still need to know how and when to use:  Avogadro’s number (6.02 x 1023), representative particles  Molar mass, grams in one mole,  Molar volume of a gas at STP (22.4 L)

  47. Conversion factor review 1 mole = 6.02 x 1023 r.p 1 mole = molar mass (g) 1 mole = 22.4 L (gas @STP)

  48. So what about this silly mole? Our useful friend the mole allows us to do calculations called stoichiometry—using balanced chemical equations to obtain info ex: 2C + O2 2CO

  49. Mole - Mole (MOL – MOL) Conversions new conversion factor – # of mol A = # of mol B # = coefficients in balanced equation

  50. I.Mole - Mole (MOL – MOL) Conversions the most important, most basic stoich calculation B. uses the coefficients of a balanced equation to compare the amounts of reactants and products C. coefficients are mole ratios

More Related