1. Quantitative Composition of Compounds Making new chemicals is much like following a recipe from a cook book... 1 cup flour + 2 eggs + ½ tsp baking powder  5 pancakes What if you want to make more (or less)? Suppose you have plenty of flour and baking powder, but only 8 eggs. How many pancakes can you make? Solve it in your head: 2 eggs makes 5 pancakes, so four times more eggs You can solve it using conversion factor: 5 pancakes 2 eggs makes 20 (5x4) pancakes. 8 eggs 5 pancakes 2 eggs x=20 pancakes Chapter 7 … except you don’t get to lick the spoon! 3 blocks cream cheese + 5 eggs + 1 cup sugar = 1 cheese cake. How many cheese cakes can we make out of 15 eggs? How much sugar do we need for 5 cheese cakes? (5) Practice using the following mouth- washing, diet-buster recipe: 15 eggs x 1 cake 5 eggs = 3 cheese cakes

2. Suppose you want to ‘whip’ a batch of hydrogen iodide, following the balanced chemical equation: H 2 + I 2  2 HI How much H 2 and I 2 should you use to make 10 g of HI? A common mistake is that H 2 and I 2 react in one-to-one mass ratio so: 5 g H 2 + 5 g I 2  10 g HI The coefficients balancing the equation refer to number of atoms, not masses. Introducing the mole. The mole is like a dozen, but much, much more. 1 mole of anything: donuts, pancakes, atoms, molecules, ions… is always 6.022 x 10 23 of that thing. 1 mole of soft drink cans is enough to cover the surface of the earth to a depth of over 200 miles. If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole. The mole is Avogadro’s Number of items. 1 mole = 602,214,179,000,000,000,000,000 or 6.022 x 10 23. We need the mole because the mass of an atom is too small to be measured on a balance. Remember: 1 amu = 1.6 x 10 -24 g.

3. 6.022 x 10 23 6.022 x 10 23 1.204 x 10 24 moleculesmoleculesmolecules H 2 +I 2 =2 HI The mole translates between the number of atoms (or molecules, ions) and grams of atoms (molecules, ions). It is defined as the mass of Avogadro’s number of atoms 6 C, which, in turn, weights exactly 12 g. 12 A mole of atoms weighs the same number of grams as the atomic mass. One mole of H atoms weighs 1.0079 g. One mole of C atoms weighs 12.011 g. Atomic mass refers to: the sum of protons and neutrons in a single atom, weighted average mass of all isotopes of an element and also to the number of grams in one mole of atoms. 1 molecule1 molecule2 molecule 2 H atoms2 I atoms2 x (1 atom H, 1 atom I) Conversion factors: 1 mole 6.022 x 10 23 species 1 mole molar mass 1 mole of H 2 weighs 2 x 1.0079 g = 2.0158 g Amadeo Avogadro The mole or any number of molecules 12 molecules12 molecules24 molecules 1 mole 1 mole 2 mole 2.0158 g 253.81 g 255.8258 g

4. Mole - mass - atomsQ1: How many atoms in 0.5 mole Au? 0.5 mole Au x 6.022 x 10 23 atoms Au 1 mole Au = 3.011 x 10 23 atoms Au Q2: What is the mass of 0.5 mol Au? 0.5 mole Au x 196.967 g Au 1 mole Au = 98.4835 g Au Q3: How many atoms in 15.00 g Au? 15.00 g Au x 1 mole Au 196.987 g Au 6.022 x 10 23 atoms Au 1 mole Au x = 4.59 x 10 22 atoms Au Q1a: How many moles in 7.12 x 10 24 atoms of Cu? 7.12 x 10 24 atoms Cu x 1 mole Cu 6.022 x 10 23 atoms Cu = 11.8 mol Cu conversions

5. Percent Composition What is the % composition of CH 2 O? Total mass = 12.01 g + 2.016 g + 16.00 g Percent composition is % mass that each element in a molecule contributes to the total molar mass of the compound. Assume that you have one mole of the compound. %C = 12.01 g 30.026 g x 100 %H = 6.71 % %O = 53.29 % 100.00 % + Practice: What is the % composition of glucose? Check your answer: it is the same as in CH 2 O! Empirical Formula: the formula of a compound that expresses the smallest whole number ratio of the atoms present. Types of Formulas Molecular Formula: the formula that states the actual number of each kind of atom found in one molecule of the compound. CH 2 O is the empirical formula for glucose, C 6 H 12 O 6 Formulas describe the relative number of atoms (or moles) of each element in a formula unit. It’s always a whole number ratio. 1 molecule of C 9 H 8 O 4 = 9 atoms of C, 8 atoms of H and 4 atoms of O. 1 mole of C 9 H 8 O 4 = 9 mol of C, 8 mol of H and 4 mol of O atoms. If we can determine the relative number of moles of each element in a compound, we can determine a formula for the compound. = 30.026 g %C = 40.00 %

6. Dr. Ent burned 0.5 g of the sample and obtained the total of over 1 g of products. How is that possible? From the mass of the products (water and carbon dioxide) we determine the number of moles of C, H, and O, and from them obtain the empirical formula of the compound. Oxygen from air is a reactant!

7. 1. Determine the mass in grams of each element present, if necessary. Remember, % means “out of 100”. 2. Convert grams of CO 2 and H 2 O (or C and H) into moles of C and H atoms. 3. Convert moles of C into grams of C. Do the same for H. 6. Divide the molar mass of the compound by the molar mass of the empirical formula. To find the molecular formula, multiply all subscripts in the empirical formula by this product. Combustion analysis shown 0.300 g H 2 O and 0.733 g CO 2 from 0.500 g of sample. Find the empirical and molecular formula if the molar mass of the compound is 180.15 g/mol. 2. 0.3 g H 2 O x 1 mol H 2 O 18.01 g H 2 O x 2 mol H atms 1 mol H 2 O = 0.0333 mol H at. 0.733 g CO 2 x 1 mol CO 2 44.01 g CO 2 x 1 mol C atms 1 mol CO 2 = 0.0166 mol C at. 3.3. 0.0333 mol H x 1.008 g H 1 mol H = 0.0336 g H 0.0166 mol C x 12.01 g C 1 mol C = 0.199 g C g O = 0.5 – (0.0336 + 0.199) = 0.267 g O 4. Add masses for C and H and subtract the sum from the mass of the sample to obtain mass of O. Convert the mass into moles of O. 4.4. 0.267 g O x 1 mol O at. 16.00 g O = 0.0169 mol O at. 5. Divide all number of moles with the smallest to obtain the subscripts of the empirical formula. 5. H: 0.0333 / 0.0166 = 2 C: 0.0166 / 0.0166 = 1 O: 0.0169 / 0.0166 ~ 1 CH 2 O Empirical formula 6. Molar mass emp. form. 30.026 Molar mass sample Molar mass emp. formula = 180.15 30.03 = 6 Molecular formula: C 6 H 12 O 6. Note: steps 3, 4 apply only for finding formulas from combustion analysis. There is 6 CH 2 O units in the compound.

8. 3. Percent composition of a compound is found to be 43.2% K, 39.1% Cl, and some O. Find the empirical formula. If the molar mass of the compound is 90.550 g mol -1, find the molecular formula. (KClO) Find the empirical and molecular formulas if the % composition is 40.0% C, 6.70% H, 53.3% O, and the molar mass of the compound is 180.155 g/mol. 2. A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34 g O. Determine an empirical formula for this substance. (NO 2 ) 1. A compound has an empirical formula of NO 2. The colorless liquid used in rocket engines has a molar mass of 92.0 g mole -1. What is the molecular formula of this substance? (N 2 O 4 ) 1. Determine the mass in grams of each element present, if necessary. Remember, % means “out of 100”. 2. Convert grams of CO 2 and H 2 O (or C and H) into moles of C and H atoms. 3. Convert moles of C into grams of C. Do the same for H. 6. Divide the molar mass of the compound by the molar mass of the empirical formula. To find the molecular formula, multiply all subscripts in the empirical formula by this product. 4. Add masses for C and H and subtract the sum from the mass of the sample to obtain mass of O. Convert the mass into moles of O. 5. Divide all number of moles with the smallest to obtain the subscripts of the empirical formula. 1. Assume that you have 100.00 g sample; the mass of each element is equal to the % composition. 40.0 g C, 6.70 g H, 53.3 g O. 2. 40.0 g C x 1 mol C 12.01 g C 6.70 g H x 1 mol H 1.008 g H 53.3 g O x 1 mol O 16.00 g O = 3.33 mol C= 6.65 mol H = 3.33 mol O Skip steps 3 and 4, they apply for combustion analysis only. 5.C: 3.33 / 3.33 = 1 H: 6.65 / 3.33 = 2 O: 3.33 / 3.33 = 1 Empirical formula CH 2 O. Emp. Formula mass = 30.026 6. Molar mass sample Molar mass emp. formula = 180.155 30.026 = 6 Molecular formula: C 6 H 12 O 6. Practice (answer in parenthesis): Thank you, Dr. Ent! Thus, there are 6 (CH 2 O) units.

9. Chapter 9 Calculations from Chemical Equations Remember me? The molar mass of an element is its atomic mass in grams. It contains 6.022 x 10 23 atoms (Avogadro’s number) of the element. The molar mass of a compound is the sum of the atomic masses of all its atoms. Conversions go through moles. For calculations of mole-mass-number_of_particle relationships: 1. Use balanced equation. 2. The coefficient in front of a formula represents the number of moles of the reactant or product. Al + Fe 2 O 3  Al 2 O 3 + Fe  2 2 2 mol 1 mol 1 mol 2 mol To quantitatively convert from one quantity to another we introduce mole ratio: Mole ratio is found from the coefficients of the balanced equation. 1 mol Fe 2 O 3 2 mol Al 1 mol Fe 2 O 3 1 mol Al 2 O 3 Mole ratio = moles of desired substance moles of starting substance A mole of a compound weighs the sum of all atoms in the compound. For instance: molar mass of NaCl is 22.99 + 35.45 = 65.44 g Which conversion factor will be used depends on starting and desired substance

10. Mole – Mole Conversions - Molecules Example 1:How many moles of NaCl result from the complete reaction of 3.4 mol of Cl 2 ? Assume that there is more than enough Na. 2 Na(s) + Cl 2 (g)  2 NaCl(s) 1 mole2 moles 3.4 moles Cl 2 x 2 moles NaCl 1 mole Cl 2 = 6.8 moles NaCl Ca 5 (PO 4 ) 3 F(s) + 5H 2 SO 4 (aq)  3H 3 PO 4 (aq) + HF(aq) + 5CaSO 4 (s) Example 2: Calculate the number of moles of phosphoric acid (H 3 PO 4 ) formed by the reaction of 10 moles of sulfuric acid (H 2 SO 4 ) on phosphate rock: 1 mole 5 moles 3 moles 1 mole 5 moles 3 moles H 3 PO 4 5 moles H 2 SO 4 10 moles H 2 SO 4 x = 6 moles H 3 PO 4 Example 3: Calculate the number of moles of Ca 5 (PO 4 ) 3 F needed to produce 6 moles of H 3 PO 4. 1 mole Ca 5 (PO 4 ) 3 F 3 moles H 3 PO 4 6 moles H 3 PO 4 x = 2 moles Ca 5 (PO 4 ) 3 F desired substance starting substance The following examples refer to the equation:

11. Example 4: Calculate the number of moles of H 2 SO 4 necessary to yield 784 g of H 3 PO 4. Ca 5 (PO 4 ) 3 F(s) + 5H 2 SO 4  3H 3 PO 4 + HF + 5CaSO 4 1 mole 5 moles 3 moles 1 mole 5 moles 1. Convert the starting substance into moles. 2. Convert moles of starting substance into moles of desired substance. 3. Convert moles of desired substance into the units specified in the problem. Molar mass of H 3 PO 4 = 97.994 784 g H 3 PO 4 x 1 mole H 3 PO 4 97.994 g H 3 PO 4 = 8.00 moles H 3 PO 4. g mole 8.00 moles H 3 PO 4 x 5 moles H 2 SO 4 3 moles H 3 PO 4 = 13.3 moles H 2 SO 4. done. Mass – Mole conversion Ex. 5: Calculate the mass of phosphate rock, Ca 5 (PO 4 ) 3 F needed to yield 200. g of HF. Molar masses: Ca 5 (PO 4 ) 3 F = 504.31 g/mol; HF = 20.008 g/mol Step 1, 200. g HF x 1 mole HF 20.008 g HF = 10.0 moles HF x Step 2 1 mole Ca 5 (PO 4 ) 3 F 1 mole HF = 10.0 moles Ca 5 (PO 4 ) 3 F Step 3: 10.0 moles Ca 5 (PO 4 ) 3 F x 504.3 g ph.r. 1 mole ph.r = 5.00 kg Ca 5 (PO 4 ) 3 F.

12. Mass – mass conversion Ex. 6: Calculate the number of grams of H 2 SO 4 necessary to yield 392 g of H 3 PO 4. Ca 5 (PO 4 ) 3 F(s) + 5H 2 SO 4  3H 3 PO 4 + HF + 5CaSO 4 1 mole 5 moles 3 moles 1 mole 5 moles 1. Convert the starting substance into moles. 2. Convert moles of starting substance into moles of desired substance. 3. Convert moles of desired substance into the units specified in the problem. Molar mass H 3 PO 4 = 97.994 392 g H 3 PO 4 x 1 mole H 3 PO 4 97.994 g H 3 PO 4 = 4.00 moles g mole 4.00 moles H 3 PO 4 x 5 moles H 2 SO 4 3 moles H 3 PO 4 = 6.67 moles Molar mass H 2 SO 4 = 98.086 g mole 6.67 moles H 2 SO 4 x 98.086 g 1 mole H 2 SO 4 = 654 g H 2 SO 4. Example 7: Find the mass of glucose that can be synthesized from 58.5 g of CO 2, assuming that there is more than enough water to react with all the CO 2. Molar masses are 44.01 g (CO 2 ) and 180.16 (glucose). 58.5 g CO 2 x 1 mole CO 2 44.01 g CO 2 x 1 mole glucose 6 moles CO 2 180.16 g glucose 1 mole glucose x = 39.9 g glucose 392 g H 3 PO 4 x 1 mole H 3 PO 4 97.994 g H 3 PO 4 5 moles H 2 SO 4 3 moles H 3 PO 4 98.086 g 1 mole H 2 SO 4 = 654 g H 2 SO 4. Step_by_step: Combined steps: xx

13. Conversion – General Case Example 8: Calculate the mass of NH 3 formed by the reaction of 112 grams of H 2. N 2 + 3H 2  2NH 3 grams H 2  moles H 2  moles NH 3  grams NH 3 Mass to moles of starting compound Moles of starting compound to moles of desired compound Moles of desired comp. to units desired. Step 1 Step 2Step 3 Moles – moles: Step 2 only Moles – mass: Step 2 and Step 3 only Mass – mass: All 3 steps Molar masses: H 2 : 2.016 g/mol; NH 3 : 17.034 g/mol 112 g H 2 x 1 mole H 2 2.016 g H 2 Step 1 2 moles NH 3 3 moles H 2 x Step 2 17.034 g NH 3 1 mole NH 3 x Step 3 = 1420 g NH 3 = 1.42 kg NH 3. Example 9: Calculate the moles of NH 3 formed by the reaction of 1.5 moles of H 2. 2 moles NH 3 3 moles H 2 x Step 2 = 17.0 g NH 3. 1.50 moles of H 2 Starting compound Example 10: Calculate the mass of NH 3 formed by the reaction of 1.50 moles of H 2. Starting compound 2 moles NH 3 3 moles H 2 x Step 2 = 1.00 mole NH 3. 1.50 moles of H 2 Starting compound 17.034 g NH 3 1 mole NH 3 x Step 3 result

14. Mass – moles: Step 1 and Step 2 only 2 moles NH 3 3 moles H 2 x Step 2 = 49.6 g NH 3. 150. g H 2 Example 11: Calculate the moles of NH 3 formed by the reaction of 150. g H 2. Starting compound x Step 1 result Conversion – General Case (cont’d) N 2 + 3H 2  2NH 3 1 mole H 2 2.016 g H 2 Mass – particles: All 3 steps 112 g H 2 x 1 mole H 2 2.016 g H 2 Step 1 2 moles NH 3 3 moles H 2 x Step 2 6.022 x 10 23 molecules NH 3 1 mole NH 3 x Step 3 = 2.23 x 10 25 molecules NH 3. Starting compound result Example 12: Calculate the # molecules of NH 3 formed by the reaction of 150. g H 2. Limiting Reactant and Yield Calculations The amount of the product(s) depends on the reactant that is used up during the reaction, i.e. limiting reactant. The number of seats is the limiting part (reactant); one frame and two wheels are parts in excess; 3 bicycles is the yield. One bicycle needs 1 frame, 1 seat and 2 wheels, therefore not more than 3 bicycles can be made.

15. Example 13: How many moles of Fe 3 O 4 can be obtained by reacting 16.8 g Fe with 10.0 g H 2 O? Which substance is the limiting reactant? Which substance is in excess? How much of the reactant in excess remains unreacted? Limiting Reactant and yield Calculations (cont’d) Balanced equation: 3 Fe (s) + 4 H 2 O (g) Fe 3 O 4 (s) + 4 H 2 (g) 16.8 g Fe x 1 mol Fe 55.85 g Fe x 1 mol Fe 3 O 4 3 mol Fe = 0.100 mol Fe 3 O 4. Strategy: 1.Write and balance equation. 2.Calculate the number of moles of product for each reactant; 3.The reactant that gives the least moles of (the same!) product is the limiting reactant. 4.Find the amount of reactant in excess needed to react with the limiting reactant. Subtract this amount from the starting quantity to obtain the amount in excess. 5.Find the yield from the limiting reactant. from Fe: 10.0 g H 2 O x 1 mol H 2 O 18.02 g H 2 O x 1 mol Fe 3 O 4 4 mol H 2 O = 0.139 mol Fe 3 O 4.From H 2 O: limiting reactant yield Yield is 0.100 mol Fe 3 O 4, Fe is the limiting reactant, 2.99 g H 2 O is in excess.Answer: 16.8 g Fe x 1 mol Fe 55.85 g Fe x 4 mol H 2 O 3 mol Fe x 18.02 g H 2 O 1 mol H 2 O = 7.01 g H 2 O. Reacted H 2 O 10.0 g – 7.01 g = 2.99 g H 2 O.Excess: Least moles Fe 3 O 4 ? Yield

16. Many reactions (especially organic) do not give the 100% yield, due to: side reactions, reversible reactions, product losses due to human factor. Percent Yield Calculations done so far assumed that the reaction gives maximum (100%) yield. Theoretical yield: Amount calculated from the chemical equation. Actual yield: Amount obtained experimentally. Percent yield: Actual yield Theor. yield x 100 % Example 14: If 65.0 g CCl 4 was prepared by reacting 100. g CS 2 and 100. g of Cl 2, calculate the percent yield. CS 2 + 3 Cl 2 CCl 4 + S 2 Cl 2 Molar masses: CS 2 : 76.15; Cl 2 : 70.90; CCl 4 : 153.81 g/mol 100. g CS 2 x 1 mol CS 2 76.15 g CS 2 x 1 mol CCl 4 1 mol CS 2 = 1.31 mol CCl 4. 100. g Cl 2 x 1 mol Cl 2 70.90 g Cl 2 x 1 mol CCl 4 3 mol CS 2 = 0.470 mol CCl 4. Theoretical yield Limiting reactant 0.470 mol CCl 4 x 153.81 g CCl 4 1 mol CCl 4 = 72.3 g CCl 4. 65.0 g CCl 4 72.3 g CCl 4 x 100 % = 89.9 % Percent yield Strategy: Find limiting reactant. Calculate theoretical yield. Calculate percent yield. 65.0 g CCl 4 Actual yield HW, Chp. 7: 1, 5, 15, 26, 33 Chp. 9: 3, 7, 13, 15, 23, 29