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The Mole SI base unit for amount of substance

The Mole SI base unit for amount of substance. Use mole and molar mass to make conversions among moles, mass, and number of representative particles. Relating mass to number of atoms. Mole Amount of substance containing as many particles as atoms in exactly 12g of carbon-12

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The Mole SI base unit for amount of substance

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  1. The MoleSI base unit for amount of substance Use mole and molar mass to make conversions among moles, mass, and number of representative particles

  2. Relating mass to number of atoms • Mole • Amount of substance containing as many particles as atoms in exactly 12g of carbon-12 • Avogadro’s Number (6.022 x 1023) • # particles in exactly one mole of pure substance • At same temperature/pressure, equal volumes of all gases contain same # of molecules • Molar mass • Mass of one mole of pure substance

  3. Mole: quantity of substance whose mass in grams is same as its atomic number • Iron has atomic weight of 55.845 g, so one mole of iron weighs 55.845 grams

  4. Calculate the number of particles in: • 2.50 moles of Neon, Ne 2.50 mol Ne 6.02 x 1023 atoms Ne = 1.5 x 1024 atoms Ne 1 mole Ne • 0.050 moles of iron, Fe 0.050 mol Fe 6.02 x 1023 atoms Fe = 3.01 x 1022 atoms Fe 1 mole Fe

  5. Calculate the number of molecules in each of the following amounts: • 4.27 mol of tungsten(VI) oxide, WO3 • 0.00300 mol of stronium nitrate, Sr(NO3)2 • 72.5 mol of toluene, C6H5CH3 • 5.11 x 10-7 mol of Vitamin E, C29H50O2 • 1500 mol of hydrazine, N2H4 • 0.989 mol nitrobenzene, C6H5NO2

  6. Calculate the number of moles in: • 9.03 X 1023 atoms of Cu 9.03 X 1023 atoms of Cu 1 mole Cu = 1.5 moles Cu 6.02 x 1023 atoms Cu • 3.76 X 1025 molecules of SO2 3.76 X 1025 atoms of SO2 1 mole SO2 = 62.46 moles SO2 6.02 x 1023 atoms SO2 • 8.6 X 1018 electrons Fe 8.6 X 1018 electrons Fe 1 mole Fe = 1.43 x 10-23 moles Fe 6.02 x 1023 atoms Fe

  7. Homework: Read 11.1, pp. 309-312 Q pg. 312, #9 Q pp. 346-347, #75, 76, 89 b/c, 90 a/b, 94, 96

  8. Asbestos, a known cancer-causing agent, has a typical formula, Ca3Mg5(Si4O11)2(OH)2. How many atoms of each element are given in the formula? What is the total number of atoms represented in the formula of cobalt(II) chloride hexahydrate? CoCl2 . 6H2O How many atoms of each kind are represented in the following formulas? Na3PO4 Ca(H2PO4)2  Fe3(AsO4)2  MgSO4.7H2O Calculate molecular masses of: SO2 P4O10 UF6 NH3  Atomic Weights & Masses

  9. Determine molecular mass: methane, CH4  potassium perchlorate, KClO4 phosphorus trichloride, PCl3 sulfuric acid, H2SO4 silicon dioxide, SiO2 nitrogen(IV) oxide, NO2 nitrogen(V) oxide, N2O5 glucose, C6H12O6 What is the molecular weight of each of these common chemicals compounds? sodium bicarbonate, NaHCO3  laughing gas, N2O Potassium permanganate, KMnO4 limestone, CaCO3 Epsom salts, MgSO4.7H2O ozone, O3

  10. Mass  Mole/Mole  Mass • 0.039 g palladium = ? moles 0.039 g palladium 1 mole Pd = 3.66 x 10-4 mole Pd 106.42 g Pd • 1.002 mol Cr = ? grams 1.002 mol Cr 51.996 g Cr = 52.1 g Cr 1 mole Cr

  11. Calculate the number of moles in each of the following masses: • 8200 g of iron • .0073 kg of tantalum • 0.00655 g of antimony • 5.64 kg of barium • 3.37 x 10-6 g of molybdenum • 25 g of helium, He (6.25 moles) • 12.5 g of methane, CH4 (0.78 moles) • 0.364 g of iodine, I2 (1.43 x 10-3 moles) • 40.0 g of sodium, Na (1.74 moles)

  12. Calculate the mass of: • 2.00 moles of water, H2O (36.04 g) • 4.38 moles of chlorine, Cl2 (310.54 g) • 0.025 moles of ammonia, NH3 (0.43 g) • 1.8 moles of oxygen, O2 (57.60 g) • 50 mol of aluminum • 4.08 x 10-8 mol of neon • 7 mol of titanium • 0.0086 mol of xenon • 3.29 x 104 mol of lithium

  13. Mass to atoms • Calculate # molecules of carbon dioxide formed if 300 g of iron was produced. Fe2O3 + 3CO  2Fe + 3CO2 300 g Fe 1 mol Fe 3 mol CO 6.02 x 1023 molecules 55.8 g Fe 2 mol Fe 1 mol CO

  14. Calculate the number of molecules (atoms in last 5) in: • 12.5 g of nitrogen, N2 (2.69 X 1023 molecules) • 0.76 g of ammonia, NH3 (2.68 X 1022 molecules) • 0.60 g of hydrogen, H2(1.79 X 1023 molecules) • 0.0082 g of gold • 812 g of molybdenum • 2.00 x 102 mg of americium • 10.09 kg of neon • 0.705 mg of bismuth

  15. Calculate the mass of each of the following: • 8.22 x 1023 atoms of rubidium • 9.96 x 1026 atoms of tellurium • 2.94 x 1017 atoms of hafnium

  16. Determining number of moles of ions from a formula • Use conversion factor by breaking the molecule into its ions. • How many moles of Cl ions are there in 2.50 mole ZnCl2? • 2.50 mol ZnCl2 2 molCl = 5.00 molCl 1 mol ZnCl2

  17. Practice Problem 31 • A sample of silver chromate (Ag2CrO4) has a mass of 25.8 g. • molar mass = 331.8 g/mol • 25.8 g Ag2CrO4 1 mol Ag2CrO4 = 0.0778 mol Ag2CrO4 331.8 g Ag2CrO4 • 0.0778 mol Ag2CrO4 6.02 x 1023formula units = 4.68 x 1022formula units Ag2CrO4 1 mole Ag2CrO4 • How many Ag+ ions are present? • 4.68 x 1022formula units Ag2CrO4 2 Ag+ ions = 9.36 x 1022 Ag+ ions 1 formula units Ag2CrO4 • How many CrO4- are present? • 4.68 x 1022formula units Ag2CrO4 1 Cl- ion = 4.68 x 1022Cl- ions 1 formula units Ag2CrO4 • What is the mass in grams of one formula unit of silver chromate? • 331.8 g Ag2CrO4 1 mol = 5.51 x 10-22formula units Ag2CrO4 6.02 x 1023formula units

  18. Homework: Read 11.2-11.3, pp. 313-327 Q pg. 319, #19 Q pg. 323, #33, 34 (answers in the back-I want you to look them over and make sure you understand them) Q pg. 347, #98 a/b, 99 c/d, 100a, 101c, 103 a/b, 104, 111b, 112d, 121, 131

  19. Homework: Read 11.3, pp. 320-327 Q. pp. 347-348, #

  20. What is the mass percentage composition of • CaO? (Ca = 71.47%; O = 28.53%) • MgCl2 (Mg = 25.53%; Cl = 74.47%) • Na2SO4 (Na = 32.37%; S = 22.58%; O = 45.05%) • Fe2O3 (Fe = 69.94%; O = 30.06%) • C7H5N3O6 (C = 37.01%; H = 2.22%; N = 18.50%; O = 42.27%) • AlBr3.6H2O (Al = 7.20%; Br = 63.95%; H = 3.23%; O = 25.62%)

  21. Determine the percent composition of Ca3(PO4)2 (Ca = 38.76%; P = 19.97%; O = 41.27% ) • Calculate the percentage of nitrogen in the two important nitrogen fertilizers, ammonia, NH3 and urea, CO(NH2)2 (NH3: N = 82.22%/CO(NH2)2:  N = 46.65%)) • A forensic scientist analyzes a sample for sodium arsenate, Na2AsO4(s), a source of arsenic.  Calculate the percentage composition of sodium arsenate.(Na = 24.78%; As = 40.52%; O = 34.61%)

  22. Calculate the percentage composition of the following compounds • lithium bromide, LiBr • anthracence, C14 H10 • ammonium nitrate, NH4NO3 • nitrous acid, HNO2 • nitric acid • Ammonia • mercury (II) sulfate • antimnoy (V) flouride

  23. Calculating empirical formula using the following rhyme: Percent to mass Mass to mole Divide by small Multiply 'til whole

  24. Here's an example of how it works. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? • (1) Percent to mass: Percent to mass: the assumption of 100 grams is purely for convenience sake. This means that the percentages transfer directly into grams. If you assumed that 36.7 grams were present, you would have to multiply 36.7 by each percentage. Assuming 100 grams makes it much easier. • Assume 100 g of the substance, then 72.2 g magnesium and 27.8 g nitrogen. • (2) Mass to moles: • for Mg: 72.2 g Mg x (1 mol Mg/24.3 g Mg) = 2.97 mol Mgfor N: 27.8 g N x (1 mol N/14.0 g N) = 1.99 mol N • (3) Divide by small: Divide by small: make sure you divide ALL answers from #2 by the smallest value. • for Mg: 2.97 mol / l.99 mol = 1.49for N: 1.99 mol / l.99 mol = 1.00 • (4) Multiply 'til whole: Multiply 'til whole: multiply ALL values from #3 by the same factor. This factor is selected so as to produce ALL whole numbers as answers. Often this factor is chosen by trial-and-error. • for Mg: 2 x 1.49 = 2.98 (i.e., 3)for N: 2 x 1.00 = 2.00 • and the formula of the compound is Mg3N2.

  25. Here's the example problem: A compound is analyzed and found to contain 68.54% carbon, 8.63% hydrogen, and 22.83% oxygen. The molecular weight of this compound is known to be approximately 140 g/mol. What is the empirical formula? What is the molecular formula? • 1) Percent to mass • carbon: 68.54 gramshydrogen: 8.63 gramsoxygen: 22.83 grams • (2) Mass to moles • carbon: 68.54 / 12.011 = 5.71 molhydrogen: 8.63 / 1.008 = 8.56 moloxygen: 22.83 / 16.00 = 1.43 mol • (3) Divide by small • carbon: 5.71 ÷ 1.43 = 3.99hydrogen: 8.56 ÷ 1.43 = 5.99oxygen: 1.43 ÷ 1.43 = 1.00 • (4) Multiply 'til whole • The empirical formula of the compound is C4H6O.

  26. Problem-A compound contains 48.38% carbon, 8.12% hydrogen, and the rest is oxygen. • 1) Percent to mass. • carbon: 48.38 gramshydrogen: 8.12 gramsoxygen: 43.50 grams • Note that the oxygen percentage came from 100% - (48.38 + 8.12) = 43.5% • (2) Mass to moles. • carbon: 48.38 / 12.011 = 4.028 molhydrogen: 8.12 / 1.008 = 8.056 moloxygen: 43.50 / 16.00 = 2.719 mol • (3) Divide by small: • carbon: 4.028 ÷ 2.719 = 1.48hydrogen: 8.056 ÷ 2.719 = 2.96oxygen: 2.719 ÷ 2.719 = 1.00 • Note that the carbon value (4.028) is half the hydrogen value (8.056). That means there is one carbon for every two hydrogens in the answer. • (4) Multiply 'til whole: • carbon: 1.48 x 2 = 3 hydrogen: 2.96 x 2 = 6 oxygen: 1 x 2 = 2 • The empirical formula is C3H6O2

  27. Problem-don’t round off too early-trial and error • Step One: percent to mass. • Carbon: 48.38 gHydrogen: 8.12 gOxygen: 100 minus (48.38 + 8.12) = 43.50 g • Step Two: mass to moles. • Carbon: 48.38 g ÷ 12.011 g/mol = 4.03 molHydrogen: 8.12 g ÷ 1.008 g/mol = 8.04 molOxygen: 43.50 g ÷ 15.999 g/mol = 2.72 mol • Here is where the problem with rounding comes into play. Should the 2.72 be rounded off or not? If you do, then the answer is C4H8O3. Is it right or wrong? Let us continue. • Step Three: divide by small. (much less error using 2.72 than the other two) • Carbon: 4.03 mol ÷ 2.72 mol = 1.48Hydrogen: 8.04 mol ÷ 2.72 mol = 2.96Oxygen: 2.72 mol ÷ 2.72 mol = 1 • Step Four: multiply 'til whole. • Doubling each value gives C = 3, H = 5, O = 2, so the empirical formula is C3H5O2.

  28. Calculating Molecular Formula • Empirical formula  molecular formula determined if molar mass of compound is known • Calculate mass of empirical formula and divide molar mass of compound by mass of empirical formula to find ratio between molecular formula and the formula • Multiply all subscripts by ratio to find molecular formula

  29. NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O.  Calculate the empirical formula of NutraSweet and find the molecular formula.  (The molar mass of NutraSweet is 294.30 g/mol) • Start with the number of grams of each element, given in the problem. • If percentages are given, assume that the total mass is 100 grams so that the mass of each element = the percent given.

  30. Empirical Formula Practice Problems • A compound is found to have (by mass) 48.38% carbon, 8.12% hydrogen and the rest oxygen. What is its empirical formula? • A compound is found to have 46.67% nitrogen, 6.70% hydrogen, 19.98% carbon and 26.65% oxygen. What is its empirical formula? • A compound is known to have an empirical formula of CH and a molar mass of 78.11 g/mol. What is its molecular formula? • Another compound, also with an empirical formula if CH is found to have a molar mass of 26.04 g/mol. What is its molecular formula? • A compound is found to have 1.121 g nitrogen, 0.161 g hydrogen, 0.480 g carbon and 0.640 g oxygen. What is its empirical formula? (Note that masses are given, NOT percentages.) • A compound containing only carbon, hydrogen and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. What is its empirical formula?

  31. The molecular formulas of some substances are as follows.  Write their empirical formulas. (a) Acetylene, C2H2 (used in oxyacetylene torches) (b) Glucose, C6H12O6 (chief sugar in blood) (c) Octane, C8H18 (component of gasoline) (d) Ammonium nitrate, NH4NO3 (component of fertilizer) • (a)  CH  (b)  CH2O  (c)  C4H9    (d)  NH4NO3

  32. A 8.48 g sample of an unknown organic compound containing only carbon, hydrogen and oxygen was burned completely to produce 12.42 g of carbon dioxide and 5.08 g of water. Find the percentage of carbon, hydrogen and oxygen in the compound. • A 11.61 g sample of an unknown organic compound containing only carbon, hydrogen and oxygen was burned completely to produce 17.40 g of carbon dioxide and 4.74g of water. Find the percentage of carbon, hydrogen and oxygen in the compound. • A 15.96 g sample of an unknown organic compound containing only carbon, hydrogen and oxygen was burned completely to produce 44.01 g of carbon dioxide and 9.03 g of water. Find the percentage of carbon, hydrogen and oxygen in the compound

  33. Determine the Molar mass of the following compounds • P2O3 (109.05 g/mol) • BaSO4 (233.29 g/mol) • Mg(C2H3O2)2 (142.39 g/mol) • lithium carbonate (73.89 g/mol) • Calculate the Percent Composition of oxygen in the following compounds • SO3 (60.00%) • CH3COOH (53.29%) • Ca(NO3)2 (58.50%) • Ammonium Sulfate (48.43%) • Calculate the empirical formula for the following compounds from percent composition • 0.0130 mol C, 0.0390 mol H, 0.0065 mol (C2H6O ) • 11.66 g iron, 5.01 g oxygen (Fe2O3 ) • 40.0 percent C, 6.7 percent H, and 53.3 percent O by mass (CH2O )

  34. Calculate the empirical formula for the following • 15.8% carbon and 84.2% sulfur (CS2 ) • 43.6% phosphorus and 56.4% oxygen (P2O5) • 28.7% K, 1.5% H, 22.8% P and 47.0% O (KH2PO4) • Calculate the molecular formula for the following • empirical formula CH, molar mass = 78 g/mol (C6H6) • empirical formula NO2, molar mass = 92.02 g/mol (N2O4) • caffeine, 49.5% C, 5.15% H, 28.9% N, 16.5% O by mass, molar mass = 195 g. (C8H10N4O2) • Calculate the empirical formula for each compound. • 7.8% carbon and 92.2% chloride (CCl4) • 10% C, 0.8% H, and 89.1%Cl (CHCl3) • 30.4% N and 69.6%O (NO2) • 22.6% N and 77.4% O (NO3) • 40% C, 6.7% H, and 53.3% O (CH2O-empirical formula for sugar)

  35. Homework: Read 11.4, pp. 328-337 Q pg. 341, #67, 69 Q pp. 348-349, #136c, 137, 138, 143, 150

  36. Formula of a hydrate We already learned how to name them

  37. Calculate the percentage of water in each of the following hydrates: • stronium chloride hexahydrate, SrCl2 • 6H2O • zinc sulfate heptahydrate, ZnSO4 • 7H2O • calcium flourophosphatedihydrate, CaFPO3 • 2H2O • beryllium nitrate trihydrate, Be(NO3)2 • 3H2O

  38. Determining the formula for a hydrate • A 344 gram sample of hydrated calcium sulfate and the sample is heated to evaporate the water. The dry sample of calcium sulfate has a mass of 272 grams. What is the mole ratio between the calcium sulfate, CaSO4 and water, H2O? What is the formula of the hydrate? • Step 1: Calculate the difference between the hydrated sample and the dry sample. • 344 grams (hydrate) - 272 grams (dry sample) = 72.0 grams of water • Step 2: Convert mass to moles for each sample. • 1.9970 mol CaSO4 • 3.9964 mol H2O • Dividing both mole values by 2 gives a ratio of 1 to 2. The formula for the hydrate is written as CaSO4 * 2H2O

  39. An empty crucible has a mass of 12.770 grams. The crucible and hydrate have a mass of 13.454 grams. After heating, the crucible and anhydrous salt have a mass of 13.010 grams. What is the formula of this hydrate of • MgSO4. ?H2O? • Mass of hydrate = 13.454 - 12.770 = .684 grams • Mass of anhydrous salt = 13.010 - 12.770 = .240 grams • Mass of water = 13.454 - 13.010 = .444 grams • Moles of anhydrous salt = .240 grams MgSO4x 1 mol MgSO4= .00199 moles MgSO4 • 1 120.367 g MgSO4 • Moles of water = .444 grams H2O x 1 mol H2O = .0246 moles H2O • 1 18.0148 g H2O •  Ratio of moles of water to moles of anhydrous salt = .0246/.00199 = 12 • Therefore the formula is MgSO4 .12H2O

  40. Homework: Read 11.5, pp. 338-344 Q pg. 341, #69 Q pg. 349, #151, 152, 154 Test practice, pg. 351, all questions Use link for quiz and submit as before. http://www.glencoe.com/qe/science.php?qi=978

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