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Quantitative Composition of Compounds

5 pancakes 2 eggs. Practice using the following mouth-washing, diet-buster recipe:. 5 pancakes 2 eggs. x. =. 20 pancakes. 8 eggs. 3 blocks cream cheese + 5 eggs + 1 cup sugar = 1 cheese cake. How many cheese cakes can we make out of 15 eggs?. 1 cake 5 eggs. 15 eggs x.

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Quantitative Composition of Compounds

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  1. 5 pancakes 2 eggs Practice using the following mouth-washing, diet-buster recipe: 5 pancakes 2 eggs x = 20 pancakes 8 eggs 3 blocks cream cheese + 5 eggs + 1 cup sugar = 1 cheese cake. How many cheese cakes can we make out of 15 eggs? 1 cake 5 eggs 15 eggs x = 3 cheese cakes How much sugar do we need for 5 cheese cakes? (5) Chapter 7 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 … except you don’t get to lick the spoon! Solve it in your head: 2 eggs makes 5 pancakes, so four times more eggs 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? makes 20 (5x4) pancakes. You can solve it using conversion factor:

  2. Suppose you want to ‘whip’ a batch of hydrogen iodide, following the balanced chemical equation: H2 + I2 2 HI How much H2 and I2 should you use to make 10 g of HI? A common mistake is that H2 and I2 react in one-to-one mass ratio so: The coefficients balancing the equation refer to number of atoms, not masses. 5 g H2 + 5 g I2 10 g HI Introducing the mole. The mole is like a dozen, but much, much more. The mole is Avogadro’s Number of items. 1 mole = 602,214,179,000,000,000,000,000 or 6.022 x 1023. 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. 1 mole of anything: donuts, pancakes, atoms, molecules, ions… is always 6.022 x 1023 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.

  3. Amadeo Avogadro 1 mole 6.022 x 1023 species 1 mole molar mass 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 6C, which, in turn, weights exactly 12 g. 12 A mole of atoms weighs the same number of grams as the atomic mass. The mole 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. H2 + I2 = 2 HI 1 molecule 1 molecule 2 molecule 2 H atoms 2 I atoms 2 x (1 atom H, 1 atom I) 12 molecules 12 molecules 24 molecules Conversion factors: 6.022 x 1023 6.022 x 1023 1.204 x 1024 molecules molecules molecules 1 mole 1 mole 2 mole 2.0158 g 253.81 g 255.8258 g or any number of molecules 1 mole of H2 weighs 2 x 1.0079 g = 2.0158 g

  4. Mole - mass - atoms Q1: How many atoms in 0.5 mole Au? conversions 6.022 x 1023 atoms Au 1 mole Au 0.5 mole Au x = 3.011 x 1023 atoms Au Q1a: How many moles in 7.12 x 1024 atoms of Cu? 1 mole Cu 6.022 x 1023 atoms Cu 7.12 x 1024 atoms Cu x = 11.8 mol Cu Q2: What is the mass of 0.5 mol Au? 196.967 g Au 1 mole Au 0.5 mole Au x = 98.4835 g Au Q3: How many atoms in 15.00 g Au? 6.022 x 1023 atoms Au 1 mole Au 1 mole Au 196.987 g Au = 4.59 x 1022 atoms Au 15.00 g Au x x

  5. 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. What is the % composition of CH2O? Percent Composition Total mass = 12.01 g + 2.016 g + 16.00 g = 30.026 g 12.01 g 30.026 g x 100 %C = %C = 40.00 % %H = 6.71 % %O = 53.29 % + Practice: What is the % composition of glucose? Check your answer: it is the same as in CH2O! 100.00 % Types of Formulas CH2O is the empirical formula for glucose, C6H12O6 Empirical Formula: the formula of a compound that expresses the smallest whole number ratio of the atoms present. Molecular Formula: the formula that states the actual number of each kind of atom found in one molecule of the compound. 1 molecule of C9H8O4 = 9 atoms of C, 8 atoms of H and 4 atoms of O. Formulas describe the relative number of atoms (or moles) of each element in a formula unit. It’s always a whole number ratio. If we can determine the relative number of moles of each element in a compound, we can determine a formula for the compound. 1 mole of C9H8O4 = 9 mol of C, 8 mol of H and 4 mol of O atoms.

  6. Dr. Ent burned 0.5 g of the sample and obtained the total of over 1 g of products. How is that possible? Oxygen from air is a reactant! 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.

  7. 2 mol H atms 1 mol H2O 1 mol C atms 1 mol CO2 1 mol H2O 18.01 g H2O 1 mol CO2 44.01 g CO2 1 mol O at. 16.00 g O 180.15 30.03 Molar mass sample Molar mass emp. formula Combustion analysis shown 0.300 g H2O and 0.733 g CO2 from 0.500 g of sample. Find the empirical and molecular formula if the molar mass of the compound is 180.15 g/mol. 1. Determine the mass in grams of each element present, if necessary. Remember, % means “out of 100”. 2. Convert grams of CO2 and H2O (or C and H) into moles of C and H atoms. x 0.3 g H2O x = 0.0333 mol H at. 3. Convert moles of C into grams of C. Do the same for H. 2. x 0.733 g CO2 x 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. = 0.0166 mol C at. 1.008 g H 1 mol H 0.0333 mol H x = 0.0336 g H 3. 5. Divide all number of moles with the smallest to obtain the subscripts of the empirical formula. 12.01 g C 1 mol C 0.0166 mol C x = 0.199 g C g O = 0.5 – (0.0336 + 0.199) = 0.267 g O 4. 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. 0.267 g O x = 0.0169 mol O at. Empirical formula Molar mass emp. form. H: 0.0333 / 0.0166 = 2 C: 0.0166 / 0.0166 = 1 O: 0.0169 / 0.0166 ~ 1 5. CH2O 30.026 6. = = 6 Note: steps 3, 4 apply only for finding formulas from combustion analysis. There is 6 CH2O units in the compound. Molecular formula: C6H12O6.

  8. 1 mol O 16.00 g O 1 mol H 1.008 g H 1 mol C 12.01 g C 180.155 30.026 Molar mass sample Molar mass emp. formula 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. 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. Thank you, Dr. Ent! 6.70 g H x 53.3 g O x 40.0 g C x = 3.33 mol O = 3.33 mol C = 6.65 mol H Skip steps 3 and 4, they apply for combustion analysis only. 1. Determine the mass in grams of each element present, if necessary. Remember, % means “out of 100”. 5. C: 3.33 / 3.33 = 1 H: 6.65 / 3.33 = 2 O: 3.33 / 3.33 = 1 Empirical formula CH2O. 2. Convert grams of CO2 and H2O (or C and H) into moles of C and H atoms. Emp. Formula mass = 30.026 3. Convert moles of C into grams of C. Do the same for H. 6. = = 6 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. Thus, there are 6 (CH2O) units. Molecular formula: C6H12O6. Practice (answer in parenthesis): 1. A compound has an empirical formula of NO2. 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? (N2O4) 5. Divide all number of moles with the smallest to obtain the subscripts of the empirical formula. 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. (NO2) 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. 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)

  9. 1 mol Fe2O3 1 mol Al2O3 1 mol Fe2O3 2 mol Al moles of desired substance moles of starting substance Chapter 9 Calculations from Chemical Equations The molar mass of an element is its atomic mass in grams. It contains 6.022 x 1023 atoms (Avogadro’s number) of the element. The molar mass of a compound is the sum of the atomic masses of all its atoms. For instance: molar mass of NaCl is 22.99 + 35.45 = 65.44 g For calculations of mole-mass-number_of_particle relationships: Remember me? Conversions go through moles. 1. Use balanced equation. D Al + Fe2O3 Al2O3 + Fe 2 2 2 mol 1 mol 1 mol 2 mol 2. The coefficient in front of a formula represents the number of moles of the reactant or product. To quantitatively convert from one quantity to another we introduce mole ratio: Mole ratio is found from the coefficients of the balanced equation. Mole ratio = Which conversion factor will be used depends on starting and desired substance A mole of a compound weighs the sum of all atoms in the compound.

  10. 2 moles NaCl 1 mole Cl2 Mole – Mole Conversions - Molecules Example 1:How many moles of NaCl result from the complete reaction of 3.4 mol of Cl2? Assume that there is more than enough Na. 2 Na(s) + Cl2(g)  2 NaCl(s) desired substance 1 mole 2 moles 3.4 moles Cl2 x = 6.8 moles NaCl starting substance The following examples refer to the equation: Ca5(PO4)3F(s) + 5H2SO4(aq) 3H3PO4(aq) + HF(aq) + 5CaSO4(s) 1 mole 5 moles 3 moles 1 mole 5 moles Example 2: Calculate the number of moles of phosphoric acid (H3PO4) formed by the reaction of 10 moles of sulfuric acid (H2SO4) on phosphate rock: 3 moles H3PO4 5 moles H2SO4 10 moles H2SO4 x = 6 moles H3PO4 Example 3: Calculate the number of moles of Ca5(PO4)3F needed to produce 6 moles of H3PO4. 1 mole Ca5(PO4)3F 3 moles H3PO4 = 2 moles Ca5(PO4)3F 6 moles H3PO4 x

  11. Mass – Mole conversion Example 4: Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4. g mole Molar mass of H3PO4 = 97.994 Ca5(PO4)3F(s) + 5H2SO4 3H3PO4 + HF + 5CaSO4 1 mole 5 moles 3 moles 1 mole 5 moles 1. Convert the starting substance into moles. 1 mole H3PO4 97.994 g H3PO4 784 g H3PO4 x = 8.00 moles H3PO4. 2. Convert moles of starting substance into moles of desired substance. 5 moles H2SO4 3 moles H3PO4 8.00 moles H3PO4 x = 13.3 moles H2SO4. 3. Convert moles of desired substance into the units specified in the problem. done. Ex. 5: Calculate the mass of phosphate rock, Ca5(PO4)3F needed to yield 200. g of HF. Molar masses: Ca5(PO4)3F = 504.31 g/mol; HF = 20.008 g/mol 1 mole HF 20.008 g HF 1 mole Ca5(PO4)3F 1 mole HF Step 1, = 10.0 moles HF x = 10.0 moles 200. g HF x Ca5(PO4)3F Step 2 504.3 g ph.r. 1 mole ph.r Step 3: = 5.00 kg Ca5(PO4)3F. 10.0 moles Ca5(PO4)3F x

  12. Step_by_step: Mass – mass conversion Ex. 6: Calculate the number of grams of H2SO4 necessary to yield 392 g of H3PO4. g mole Ca5(PO4)3F(s) + 5H2SO4 3H3PO4 + HF + 5CaSO4 Molar mass H3PO4 = 97.994 1 mole 5 moles 3 moles 1 mole 5 moles g mole 1. Convert the starting substance into moles. Molar mass H2SO4 = 98.086 1 mole H3PO4 97.994 g H3PO4 392 g H3PO4 x = 4.00 moles 5 moles H2SO4 3 moles H3PO4 2. Convert moles of starting substance into moles of desired substance. 4.00 moles H3PO4 x = 6.67 moles 3. Convert moles of desired substance into the units specified in the problem. 98.086 g 1 mole H2SO4 6.67 moles H2SO4 x = 654 g H2SO4. Combined steps: 1 mole H3PO4 97.994 g H3PO4 5 moles H2SO4 3 moles H3PO4 98.086 g 1 mole H2SO4 x x 392 g H3PO4x = 654 g H2SO4. Example 7: Find the mass of glucose that can be synthesized from 58.5 g of CO2, assuming that there is more than enough water to react with all the CO2. Molar masses are 44.01 g (CO2) and 180.16 (glucose). 1 mole CO2 44.01 g CO2 1 mole glucose 6 moles CO2 180.16 g glucose 1 mole glucose = 39.9 g glucose 58.5 g CO2x x x

  13. Conversion – General Case Mass to moles of starting compound Moles of starting compound to moles of desired compound Moles of desired comp. to units desired. Step 2 Step 3 Step 1 Mass – mass: All 3 steps Example 8: Calculate the mass of NH3 formed by the reaction of 112 grams of H2. N2 + 3H2 2NH3 grams H2 moles H2 moles NH3  grams NH3 Molar masses: H2: 2.016 g/mol; NH3: 17.034 g/mol 1 mole H2 2.016 g H2 2 moles NH3 3 moles H2 17.034 g NH3 1 mole NH3 112 g H2 x x x = 1420 g NH3 = 1.42 kg NH3. Starting compound Step 1 result Step 2 Step 3 Moles – moles: Step 2 only Example 9: Calculate the moles of NH3 formed by the reaction of 1.5 moles of H2. 2 moles NH3 3 moles H2 x = 1.00 mole NH3. 1.50 moles of H2 Starting compound Step 2 result Moles – mass: Step 2 and Step 3 only Example 10: Calculate the mass of NH3 formed by the reaction of 1.50 moles of H2. 17.034 g NH3 1 mole NH3 2 moles NH3 3 moles H2 x x 1.50 moles of H2 = 17.0 g NH3. Starting compound result Step 2 Step 3

  14. Conversion – General Case (cont’d) Mass – moles: Step 1 and Step 2 only Example 11: Calculate the moles of NH3 formed by the reaction of 150. g H2. 1 mole H2 2.016 g H2 2 moles NH3 3 moles H2 N2 + 3H2 2NH3 x x = 49.6 g NH3. 150. g H2 Starting compound Step 1 Step 2 result Mass – particles: All 3 steps Example 12: Calculate the # molecules of NH3 formed by the reaction of 150. g H2. 1 mole H2 2.016 g H2 2 moles NH3 3 moles H2 6.022 x 1023 molecules NH3 1 mole NH3 = 2.23 x 1025 molecules NH3. 112 g H2 x x x Starting compound Step 1 Step 3 result Step 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. One bicycle needs 1 frame, 1 seat and 2 wheels, therefore not more than 3 bicycles can be made. The number of seats is the limiting part (reactant); one frame and two wheels are parts inexcess; 3 bicycles is the yield.

  15. Limiting Reactant and yield Calculations (cont’d) Example 13: How many moles of Fe3O4 can be obtained by reacting 16.8 g Fe with 10.0 g H2O? Which substance is the limiting reactant? Which substance is in excess? How much of the reactant in excess remains unreacted? • Strategy: • Write and balance equation. • Calculate the number of moles of product for each reactant; • The reactant that gives the least moles of (the same!) product is the limiting reactant. • 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. • Find the yieldfrom the limiting reactant. Balanced equation: 3 Fe (s) + 4 H2O (g) Fe3O4 (s) + 4 H2 (g) Yield 1 mol Fe 55.85 g Fe 1 mol Fe3O4 3 mol Fe yield x x from Fe: 16.8 g Fe = 0.100 mol Fe3O4. limiting reactant Least moles Fe3O4? 1 mol H2O 18.02 g H2O 1 mol Fe3O4 4 mol H2O x x From H2O: = 0.139 mol Fe3O4. 10.0 g H2O Reacted H2O 18.02 g H2O 1 mol H2O 1 mol Fe 55.85 g Fe 4 mol H2O 3 mol Fe x x = 7.01 g H2O. x 16.8 g Fe Excess: 10.0 g – 7.01 g= 2.99 g H2O. Answer: Yield is 0.100 mol Fe3O4, Fe is the limiting reactant, 2.99 g H2O is in excess.

  16. Percent Yield Calculations done so far assumed that the reaction gives maximum (100%) yield. Many reactions (especially organic) do not give the 100% yield, due to: side reactions, reversible reactions, product losses due to human factor. Theoretical yield: Amount calculated from the chemical equation. Strategy: Find limiting reactant. Calculate theoretical yield. Calculate percent yield. Actual yield: Amount obtained experimentally. Actual yield Theor. yield Percent yield: x 100 % Example 14: If 65.0 g CCl4 was prepared by reacting 100. g CS2 and 100. g of Cl2, calculate the percent yield. CS2 + 3 Cl2 CCl4 + S2Cl2 Molar masses: CS2: 76.15; Cl2: 70.90; CCl4: 153.81 g/mol 1 mol CS2 76.15 g CS2 1 mol CCl4 1 mol CS2 100. g CS2 x x = 1.31 mol CCl4. 1 mol Cl2 70.90 g Cl2 1 mol CCl4 3 mol CS2 100. g Cl2 x x = 0.470 mol CCl4. Limiting reactant 153.81 g CCl4 1 mol CCl4 0.470 mol CCl4 x = 72.3 g CCl4. Theoretical yield 65.0 g CCl4 Actual yield 65.0 g CCl4 72.3 g CCl4 HW, Chp. 7: 1, 5, 15, 26, 33 Chp. 9: 3, 7, 13, 15, 23, 29 x 100 % = 89.9 % Percent yield

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