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Chapter 11

Chapter 11. The Mole. Calculation Relationships. Particles (atoms/molecules/ formula unit). Mass. Molar mass / Formula mass or atomic mass. Avagadro ’ s number 6.022 x 10 23 particles/mol. Moles. Molar volume = 22.4 L/mol. Volume of gases at STP. Vocabulary. Atomic Mass (A) -.

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Chapter 11

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  1. Chapter 11 The Mole

  2. Calculation Relationships Particles (atoms/molecules/ formula unit) Mass Molar mass / Formula mass or atomic mass Avagadro’s number 6.022 x 1023 particles/mol Moles Molar volume = 22.4 L/mol Volume of gases at STP

  3. Vocabulary Atomic Mass (A) - The mass of an atom expressed relative to the mass assigned to carbon-12 Formula Mass - The sum of the atomic masses of all the elements in a compound Mole (mol) - The number of atoms of an element relative to the number of atoms in exactly 12g of carbon-12 Avagadro’s - Number (N) The number of particles (atoms, molecules or formula units) in a mol (6.022 x 1023 particles/mol) The mass in grams (g) of one mole of a substance Molar Mass (M) -

  4. The Mole Mole (mol) - The number of atoms of an element relative to the number of atoms in exactly 12g of carbon-12 Avagadro’s - Number (N) The number of particles (atoms, molecules or formula units) in a mol (6.022 x 1023 particles/mol) So what’s the relationship? It’s a relationship between numbers Example: Describe the following?

  5. So… The Mole 12 Eggs = 1 Dozen Eggs 12 Donuts = 1 Dozen Donuts 12 Baseballs = 1 Dozen Baseballs The Relationship? 12 Eggs 12 Donuts 12 Baseballs 1 Dozen Donuts 1 Dozen Eggs 1 Dozen Baseballs So how does this relate to moles?

  6. The Mole 12 Eggs = 1 Dozen Eggs As… 6.022 x 1023 C atoms = 1 mole C atoms 12 Eggs 6.022 x 1023 C atoms Like… 1 Dozen Eggs 1 mole C atoms

  7. Calculating Particles Atoms - Individual elements (ex. Na, Zn, Si) Molecules - Diatomic elements or molecular substances (ex. O, N, H, Group 7A or CH4, NH3) Formula Units - ionic substances (ex. NaCl, KBr, LiF) ** All three (3) are calculated using Avagadro’s Number ** 1 mol = 6.022 x 1023 atoms (single element) 6.022 x 1023 molecules (covalent molecules) 6.022 x 1023 formula units (ionic compounds)

  8. Rules to follow when solving word Problems! Write down WHAT you know. Given in the problem. Write down WHAT you want to know. What the problem asks for. Determine HOW to get there. Conversion factors you will use. Practice solving word Problems using the following conversion factor

  9. Example A piece of foil contains 0.36 mol Al. Calculate the number of atoms in the piece of foil. 6.022 x 1023 atoms Al 1 mol Al = 2.1679 x 1023 atoms Al = 2.2 x 1023 atoms Al x 0.36 mol Al - - - - - - - - - - Example 13 mol of sodium carbonate (Na2CO3) was used to neutralize an acid before throwing it away. How many formula (fu) units were used. 6.022 x 1023 fu Na2CO3 1 mol Na2CO3 = 7.8286 x 1024 fu Na2CO3 = 7.8 x 1024 fu Na2CO3 x - - - - 13 mol Na2CO3 - - - - - - - Example It takes 50 mols of propane (C3H8) to cook a steak on a gas barbeque. How many molecules are burned. 6.022 x 1023 molecules C3H8 1 mol C3H8 = 3.011 x 1025 molecues C3H8 = 3 x 1025 molecues C3H8 x - - - - - - 50 mol C3H8 - - - - - -

  10. Calculating Formula Mass H H 1.00794 g ---- ----- 1.00794 g O ----- 15.999 g Water (H2O) Formula Mass = Sum of atomic mass for each element Formula Mass = 1.008 + 1.008 + 15.999 = 18.015g

  11. Example Calculate the formula mass for ammonia (NH3) Step 1 Write down the different elements and their atomic mass N 14.007g H 1.008g Step 2 Count how many of each element is present and multiply the quantity by the atomic mass N 14.007g x 1 H 1.008g x 3 = 14.007 g = 3.024 g Step 3 Add all the totals together Formula mass = 14.007 + 3.024 = 17.031 g

  12. Example Calculate the formula mass of acetic acid (HC2H3O2) Step 2 H 1.008 g x 4 = 4.032 g C 12.011 g x 2 = 24.022 g O 15.999 g x 2 = 31.998 g Step 1 H 1.008 g C 12.011 g O 15.999 g Step 3 Formula mass = 4.032 + 24.022 + 31.998 = 60.052 g Step 2 H 1.008 g x 4 C 12.011 g x 2 O 15.999 g x 2 Example Calculate the formula mass of carbonic acid (H2CO3) Step 1 H 1.008 g C 12.011 g O 15.999 g Step 2 H 1.008 g x 2 = 2.016 g C 12.011 g x 1 = 12.011 g O 15.999 g x 3 = 47.997 g Step 2 H 1.008 g x 2 C 12.011 g x 1 O 15.999 g x 3 Step 3 Formula mass = 2.016 + 12.011 + 47.997 = 62.024 g

  13. Calculating Moles 1 mol of element = atomic mass of element 1 mol of compound = formula or molar mass of compound Example 1 mol of Fe = 55.847 g 1 mol of Na = 22.990 g 1 mol of carbonic acid (H2CO3) = 62.024 g 1 mol of sodium chloride (NaCl) = 58.443 g

  14. Example How many mols of oxygen are in 52 g of oxygen 1 mol O2 x - - - 52 g O2 = 1.625 mol O2 = 1.6 mol O2 31.998 g O2 - - - Example How many mols of calcium are in 37 g of calcium 1 mol Ca x = 0.92 mol Ca = 0.9232 mol Ca - - - - 37 g Ca 40.08 g Ca - - - - Example How many mols of sodium chloride are in 100 g NaCl 1 mol NaCl x = 2 mol NaCl = 1.711 mol NaCl - - - - - 100 g NaCl 58.443 g NaCl - - - - -

  15. Example How many grams of hydrogen are in 1.67 mol of hydrogen 2.016 g H2 x 1.67 mol H2 - - - - - = 3.367 g H2 = 3.37 g H2 - - - - 1 mol H2 Example How many grams of carbon are in 0.85 mol of carbon 12.011 g C x = 10.209 g C = 10. g C 0.85 mol C - - - - - - - - 1 mol C Example How many grams of ammonia are in 2.3 x 105 mol ammonia 17.031 g NH3 x - - - - - - = 3.917 x 106 g NH3 = 3.9 x 106 g NH3 2.3 x 105 mol NH3 1 mol NH3 - - - - -

  16. Calculating Volume 1 mol of gas @ STP = 22.4 L STP = Standard Temperature and Pressure T = 25oC = 298 K (Room Temperature) P = 1 atm Example A chemical reaction produces 0.82 mol of oxygen gas. What volume of O2 was produced 22.4 L O2 = 18.368 L O2 = 18 L O2 x - - - - - 0.82 mol O2 1 mol O2 - - - - -

  17. Example How many liters of oxygen gas are in 3.250 mol of oxygen gas 22.4 L O2 = 72.800 L O2 x = 72.80 L O2 - - - - - 3.250 mol O2 1 mol O2 - - - - - Example An average size room has a volume of around 4000L of air at STP. How many moles of air are in a average size room. 1 mol air x = 178.571 mol air = 200 mol air - - - 4000 L air 22.4 L air - - - Example How many liters of nitrogen gas are in 100 g N2 22.4 L N2 1 mol N2 - - - - - = 79.960 L N2 = 80 L N2 x x 100 g N2 - - - 28.014 g N2 - - - - 1 mol N2 - - - - -

  18. Multistep Conversions X Mass (g) Particles Moles (mol) Example You need 250g of table sugar, or sucrose (C12H22O11) to bake a cake. How many sucrose molecules will be in the cake? - - - - - - - - 1 mol C12H22O11 6.022 x 1023 molecules C12H22O11 • - - - - - - - - - - 250 g C12H22O11 x x - - - - - - - - 1 mol C12H22O11 - - - - - - - - - - 342.30254 g C12H22O11 = 4.4 x 1023 molecules C12H22O11 = 4.398 x 1023 molecules C12H22O11

  19. Example An extra strength aspirin tablet contains 0.500g of aspirin (C9H8O4). How many molecules of aspirin are in one tablet - - - - - - - 1 mol C9H8O4 6.022 x 1023 molecules C9H8O4 • - - - - - - - - 0.500 g C9H8O4 x x - - - - - - - 1 mol C9H8O4 - - - - - - - - 180.16171 g C9H8O4 = 1.67 x 1021 molecules C9H8O4 = 1.671 x 1021 molecules C9H8O4 Example In a small firework, 3.467 x 1022 atoms Sr are used to make the firework burn red. How many grams of strontium are in the firework. - - - - - 1 mol Sr 87.620 g Sr 3.467 x 1022 atoms Sr x = 5.044 g Sr x - - - - • - - - - - - - 1 mol Sr - - - - - - - 6.022 x 1023 atoms Sr

  20. Vocabulary Tells you the percentage of the mass made up by each element in the compound Percent Composition - Gives the simplest whole number ratio of the atoms of the element Empirical Formula - Gives actual number of atoms of each element in the molecular compound Molecular Formula -

  21. Calculating Percent Composition % of element = mass of element x 100 formula mass compound Example What is the percent composition of hydrogen and oxygen in water? Chemical Formula = H2O Formula mass = 2(1.00794g) + 15.999g = 18.01488g % of H = 2.01588g x 100 18.01488g = 11.2% % of O = 100 – 11.2 = 88.8%

  22. Example Find the percent composition of a compound that contains 2.30g Na, 1.60g O and 0.100g H . Total mass = 2.30g + 1.60g + 0.100g = 4.00g % of Na = 2.30g x 100 4.00g = 57.5 % % of O = 1.60g x 100 4.00g = 40.0 % % of H = 0.100g x 100 4.00g = 2.5 %

  23. Example A sample of an unknown compound with a mass of 0.562g has the following percent compositions: 13.0% C, 2.2% H and 84.8% F. What is the mass of each element? Formula mass = 0.562 % of C = mass C x 100 0.562g Mass C = 0.0731g = 13.0 % Mass H = 0.0124g % of H = mass H x 100 0.562g = 2.2 % Mass F = 0.477g % of F = mass F x 100 0.562g = 84.8 %

  24. Calculating Empirical Formula Example A compound was analyzed and found to contain 13.5g Ca, 10.8g O, and 0.675g H. What is the empirical formula for the compound? Step 1 Convert the mass of each element into moles and box the smallest mole value 1 mol Ca - - - - 13.5 g Ca x = 0.337 mol Ca - - - - 40.08 g Ca 1 mol O - - - 10.8 g O x = 0.675 mol O - - - 15.9994 g O 1 mol H - - - 0.675 g H x = 0.670 mol H - - - 1.00797 g H

  25. Step 2 Divide each mole value by the smallest mole value and round answers to the nearest whole number 0.337 mol Ca = 1 mol Ca 0.337 0.675 mol O = 2 mol O 0.337 0.670 mol H = 1.98 = 2 mol H 0.337 Step 3 Express mole ratio of each element and empirical formula using mole values Mole ratio 1 Ca : 2 O : 2 H Empirical Formula CaO2H2 = Ca(OH)2

  26. Step 4 Check answer by comparing initial percent composition to final percent composition . Initial: Total mass = 13.5g + 10.8g + 0.675g = 24.975g % of Ca = 13.5g x 100 24.975g = 54.1 % % of O = 10.8g x 100 24.975g = 43.2 % = 2.7 % % of H = 0.675g x 100 24.975g Final: Formula mass = 40.08g + 2(15.999) + 2(1.00794g) = 74.094g % of Ca = 40.08g x 100 74.094g = 54.1 % % of O = 31.998g x 100 74.094g = 43.2 % = 2.7 % % of H = 2.0159g x 100 74.094g

  27. Example Determine the empirical formula for a compound containing 2.128g Cl and 1.203 g Ca 1 mol Cl - - - - 2.128 g Cl x Step 1 = 0.060 mol Cl - - - - 35.453 g Cl 1 mol Ca - - - 1.203 g Ca x = 0.030 mol Ca - - - 40.078 g Ca Step 2 0.060 mol Cl = 2 mol Cl 0.030 0.030 mol Ca = 1 mol Ca 0.030 Step 3 Mole ratio 1 Ca : 2 Cl Empirical Formula CaCl2

  28. Step 4 Initial: Formula mass = 2.128g + 1.203g = 3.331g % of Cl = 2.128g x 100 3.331g = 63.9 % % of Ca = 1.203g x 100 3.331g = 36.1 % Final: Formula mass = 2(35.453) + 40.078g = 110.986g % of Cl = 70.906g x 100 110.986g = 63.9 % % of Ca = 40.078g x 100 110.986g = 36.1 %

  29. Calculating Molecular Formula Example -carotene, compound in carrots, can be broken down to form vitamin A. The empirical formula for -carotene is C5H7. The molar mass of -carotene is 536 g/mol. What is the molecular formula for -carotene Step 1 Calculate the empirical formula Empirical formula = C5H7 Step 2 Calculate empirical formula mass and divide that value into the molar mass. Empirical Formula Mass = formula mass of empirical formula Empirical Formula Mass = 5(12.01125) + 7(1.00797) = 67.111g molar mass 536g 67.111g = 7.987 = 8 = empirical mass

  30. Step 3 Multiply the empirical formula by the ratio of molar mass to empirical formula Empirical Formula C5H7 = C5 x 8 H7 x 8 = C40H56 Molecular Formula C40H56

  31. Example Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol. 1 mol N Step 1 - - - - 4.90 g N x = 0.350 mol N - - - - 14.007 g N 1 mol O - - - 11.2 g O x = 0.700 mol O - - - 15.999 g O 0.350 mol N 0.700 mol O = 1 mol N = 2 mol O 0.350 0.350 Mole ratio 1 N : 2 O Empirical Formula NO2

  32. Step 2 Empirical Formula Mass = 14.007 + 2(15.999) = 46.005g molar mass 92.0g 46.005g = = 2 empirical mass Step 3 Empirical Formula NO2 = N1 x 2 O2 x 2 = N2O4 Molecular Formula N2O4

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