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Chapter 7 Quantitative Composition of Compounds Making new chemicals is much like following a recipe from a cook book... 1 cup of flour + 2 eggs + ½ tsp baking powder → 5 pancakes … except you don’t get to lick the spoon! 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? You can solve it using conversion factor that translates eggs into pancakes: 5 pancakes x 8 eggs 2 eggs = 20 This is the quantity you need 5 pancakes 2 eggs Solve it in your head: 2 eggs makes 5 pancakes, so four times more eggs makes 20 (5x4) pancakes. This is the quantity you want to cancel Practice using the following mouthwashing, diet-buster recipe: pancakes 3 blocks cream cheese + 5 eggs + 1 cup sugar = 1 cheese cake. How many eggs should you use with 9 blocks of cheese? 5 eggs 9 blocks cheese x 3 blocks cheese= 15 eggs How much sugar do we need for 5 cheese cakes? (5) 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. 5 g H2 + 5 g I2 10 g HI The coefficients in a balanced equation refer to number of atoms or molecules, not their masses. Counting atoms is impossible. However, since all atoms of one type have the same mass, they could be measured by weighing. The mass of just one atom is too small to be measured on a balance. Remember: 1 amu = 1.6 x 10-24 g. Introducing the mole. The mole is like a dozen, but 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. 1 mole of anything: donuts, pancakes, is always 6.022 x 1023 atoms, molecules, ions… 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. 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 126C, which, in turn, weights exactly 12 g. A mole of atoms weighs the same number of grams as the atomic mass. One mole of H atoms weighs 1.008 g. One mole of C atoms weighs 12.01 g. Atomic number 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 1 molecule 2 H atoms 1 molecule 2 I atoms 12 molecules 12 molecules 6.022 x 1023 molecules 1 mole 2.016 g 6.022 x 1023 molecules 1 mole 253.8 g = Amadeo Avogadro Atomic mass The mole 1 H 1.008 1 mol of naturally occurring H atoms has a mass of 1.008 g 2 HI 2 molecule 2 x (1 atom H, 1 atom I) 24 molecules 1.204 x 1024 molecules 2 mole 255.9 g or any number of 1 mole of H2 weighs 2 x 1.008 g = 2.016 g molecules Conversion factors: 1 mole 6.022 x 1023 species 1 mole molar mass Mole - mass - atoms conversions Q1: How many atoms in 0.500 mol Au? 6.022 x 1023 atoms Au 0.500 mole Au x 1 mole Au = 3.011 x 1023 atoms Au Q1a: How many moles in 7.12 x 1024 atoms of Cu? 1 mole Cu 7.12 x 1024 atoms Cu x 6.022 x 1023 atoms Cu = 11.8 mol Cu Q2: What is the mass of 0.500 mol Au? 197.0 g Au 0.500 mole Au x 1 mole Au = 98.5g Au Q3: # molecules in 15.00 g H2O? 2.016 Molar mass H2O = 16.00 18.02 g 15.00 g H2O x 6.022 x 1023 molec. H2O 1 mole H2O x 1 mole H2O 18.02 g H2O = 5.013 x 1023 molec. H2O Percent Composition 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. Practice: What is the % composition of glucose, C6H12O6? Check: identical to CH2O! Types of Formulas What is the % composition of CH2O? Total mass = 12.01 g + 2.016 g + 16.00 g = 30.03 g 12.01 g %C = 30.026 g x 100 %C = 40.00 % Rounding to 30.03 g produces errors: %H = 6.71 % percentages add up to 99.98 % ! + %O = 53.29 % 100.00 % CH2O is the empirical formula for glucose, C6H12O6 Empirical Formula: the formula of a Molecular Formula: the formula that states compound that expresses the the actual number of each kind of atom found smallest whole number ratio of the in one molecule of the compound. atoms present. 1 molecule of aspirin, C9H8O4 = 9 atoms of Formulas describe the relative C, 8 atoms of H and 4 atoms of O. number of atoms (or moles) of each element in a formula unit. It’s always 1 mole of C H O = 9 mol of C atoms, 8 mol 9 8 4 a whole number ratio. of H atoms and 4 mol of O atoms. If we determine the relative number of moles of each element in a compound, we can find the formula of that compound. Empirical formulas of organic compounds (consisting of C,H, O only) can be found by combustion analysis. 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. 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! Calculating Empirical and Molecular Formula Example 1. Percent composition of a compound is found to be 31.9% K, 28.9% Cl, and some O. Find the empirical formula. If the molar mass of the compound is 122.55 g mol-1, find the molecular formula. 1. Determine the mass in grams of each element present, if necessary. Remember, % means “out of 100”. 2. Convert grams of each element into moles. 3. Divide all number of moles with the smallest to obtain the subscripts of the empirical formula. 4. Divide the molar mass of the compound by the molar mass of the empirical formula to get number of empirical formula units in the molecular formula (n). Multiply all subscripts in empirical formula by n. 1. Assume 100.0 g of the compound. 39.2 31.9 g K 28.9 g Cl O??g O g O = 100.0 g – ( 31.9 g + 28.9 g) = 100.0 g – 60.8 g = 39.2 g 2. 1 mol K = 0.816 mol K 39.10 g K 1 mol Cl 28.9 g Cl x = 0.815 mol Cl 35.45 g Cl 1 mol O 39.2 g O x = 2.45 mol O 16.00 g O 31.9 g K x 3. K: 0.816 / 0.815 = 1.00 Cl: 0.815 / 0.815 = 1.00 O: 2.45 / 0.815 = 3.01 Empirical formula: K Cl O3 4. Molar mass of empirical formula: 39.1 g + 35.45 g + (3 x 16.00 g) = 122.55 Molar mass compound 122.55 n = Molar mass emp. formula = 122.55 = 1 There is 1 unit of KClO3 in molecular formula. (KClO3) x 1 = KClO3 Example 2. Find the empirical and molecular formulas if the % composition is 40.0% C, 6.71% H, 53.3% O, and the molar mass of the compound is 180.16 g/mol. 40.0 g C, 1. Assume that you have 100.00 g sample; 6.71 g H, the mass of each element is equal to the % 53.3 g O. composition. 1 mol C 2. 40.0 g C x 12.01 g C = 3.33 mol C 1 mol H 1 mol O 6.70 g H x 53.3 g O x 1.008 g H 16.00 g O = 6.66 mol H 1. Determine the mass in grams of each element present, if necessary. Remember, % means “out of 100”. 2. Convert grams of each element into moles of atoms of that element. 3. Divide all number of moles with the smallest to obtain the subscripts of the empirical formula. 4. Divide the molar mass of the compound by the molar mass of the empirical formula to get number of empirical formula units in the molecular formula (n). Multiply all subscripts in the empirical formula by n. 3. C: 3.33 / 3.33 = 1 H: 6.66 / 3.33 = 2 O: 3.33 / 3.33 = 1 = 3.33 mol O Empirical formula CH2O. Emp. Formula mass = 30.03 g/mol Molar mass compound 180.2 = = 6.00 Molar mass emp. formula 30.03 Thus, there are 6 (CH2O) units. Molecular formula: C6H12O6. 4. 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) 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) Chapter 9 Calculations from Chemical Equations contains 6.022 x 1023 atoms of particles. is expressed in g/mol. Molar mass: of an element is its atomic mass in grams. of a compound is the sum of the atomic masses of all atoms. Example: molar mass of NaCl is 22.99 + 35.45 = 65.44 g/mol For calculations of mole – mass – number_of particles relationship: Conversions go through moles. 1. Use balanced equation. ∆ 2 Al + Fe2O3 Al2O3 + 2 Fe 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. 1 mol Fe2O3 2 mol Al 1 mol Fe2O3 1 mol Al2O3 Mole ratio is found from the coefficients of the balanced equation: moles of desired substance Mole ratio = moles of starting substance Which conversion factor will be used depends on starting and desired substance. Mole – Mole Conversions 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 NaCl = 6.8 moles NaCl 3.4 moles Cl2 x 1 mole Cl2 2 moles These are exact numbers ! starting substance The following examples refer to the equation: Ca5(PO4)3F(s) + 5 H2SO4(aq)→ 3 H3PO4(aq) + HF(aq) + 5 CaSO4(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: 10 moles H2SO4 x 3 moles H3PO4 5 moles H2SO4 = 6 moles H3PO4 Example 3: Calculate the number of moles of Ca5(PO4)3F needed to produce 6 moles of H3PO4. 6 moles H3PO4 x 1 mole Ca5(PO4)3F 3 moles H3PO4 = 2 moles Ca5(PO4)3F Mass – Mole conversion 3.024 Example 4: Calculate the 30.97 number of moles of H2SO4 + 64.00 necessary to yield 784 g of 97.99 H3PO4. g 97.99 Molar mass of H3PO4 = mole Ca5(PO4)3F(s) + 5H2SO4 → 3H3PO4 + HF + 5CaSO4 1 mole 5 moles 3 moles 1 mole 5 moles 1. Convert the starting 1 mole H3PO4 substance into moles. 784 g H3PO4 x = 8.00 moles H3PO4. 97.99 g H PO 3 4 2. Convert moles of starting substance into 5 moles H2SO4 8.00 moles H3PO4 x moles of desired = 13.3 moles H2SO4. 3 moles H3PO4 substance. 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 = 504.3 g/mol; HF = 20.01 g/mol Step 1, 200. g HF x 1 mole HF x 1 mole Ca5(PO4)3F = 10.0 moles Ca5(PO4)3F 20.01 g HF 1 mole HF Step 2 504.3 g ph.r. = 5.04 kg Ca (PO ) F. Step 3: 5 4 3 10.0 moles Ca5(PO4)3F x 1 mole ph.r. Step_by_step: Mass – mass conversion Ex. 6: Calculate the number of grams of H2SO4 necessary to yield 392 g of H3PO4. Ca5(PO4)3F(s) + 5H2SO4 → 3H3PO4 + HF + 5CaSO4 1 mole 5 moles 3 moles 1 mole 5 moles Molar mass H3PO4 = 97.99 g mole g Molar mass H2SO4 = 98.09 mole 1. Convert the starting substance into moles. 1 mole H3PO4 392 g H3PO4 x = 4.00 moles 97.99 g H3PO4 5 moles H2SO4 2. Convert moles of starting substance 4.00 moles H3PO4 x 3 moles H3PO4 into moles of desired substance. = 6.67 moles 3. Convert moles of desired substance into the units specified in the problem. 6.67 moles H2SO4 x 392 g H3PO4 x 98.09 g = 654 g H2SO4. 1 mole H2SO4 Combined steps: 1 mole H3PO4 x 5 moles H2SO4 x 98.09 g = 654 g H2SO4. 97.99 g H3PO4 3 moles H3PO4 1 mole 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 180.2 (glucose). sunlight 6 CO 2(g) + 6 H 2O(l) → 6 O 2(g) + C6 H12 O 6(aq) 58.5 g CO2 x 1 mole CO2 1 mole glucose 180.2 g glucose x x 44.01 g CO2 6 moles CO2 1 mole glucose = 39.9 g glucose Conversion – General Case Mass to moles of starting compound Step 1 Moles of starting compound to moles of desired compound Step 2 Moles of desired comp. to units desired. Step 3 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: 17.03 g/mol 1 mole H2 2 moles NH3 17.03 g NH3 x x = 631 g NH3 2.016 g H2 3 moles H2 1 mole NH3 Starting Step 1 compound 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 NH 1.50 moles of H2 x 3 moles H 3 = 1.00 mole NH3. 2 112 g H2 x 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. 17.03 g NH3 2 moles NH 1.50 moles of H2 x 3 moles H 3 x 1 mole NH = 17.0 g NH3. 3 2 Starting result Step 2 compound Step 3 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 H 2 moles NH N2 + 3H2 → 2NH3 150. g H2 x 2.016 g H2 x 3 moles H 3 = 49.6 mol NH3. 2 2 Starting Step 1 Step 2 result compound Mass – particles: All 3 steps Example 12: Calculate the # molecules of NH3 formed by the reaction of 150. g H2. 1 mole H2 2 moles NH3 6.022 x 1023 molecules NH3 = 2.99 x 1025 150. g H2 x x x 2.016 g H2 3 moles H2 1 mole NH3 molecules NH3. Starting Step 1 Step 3 result compound 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 in excess; 3 bicycles is the yield. 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 many grams of the reactant in excess remains unreacted? Strategy: 1. Write and balance equation. 2. Calculate # moles of a product that can be obtained from 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. 1. Balanced equation: 3 Fe (s) + 4 H2O (g) Fe3O4 (s) + 4 H2 (g) 2. # moles of Fe3O4: 1 mol Fe3O4 1 mol Fe x = 0.100 mol Fe3O4. from Fe: 16.8 g Fe x 55.85 g Fe 3 mol Fe Least moles Fe3O4? 1 mol Fe3O4 1 mol H2O from H2O: 10.0 g H2O x 18.02 g H O x 4 mol H O = 0.139 mol Fe3O4. 2 2 3. Limiting reactant: Fe 18.02 g H2O 1 mol Fe x 4 mol H2O x = 7.23 g H2O. x 16.8 g Fe 4. Reacted 1 mol H2O 55.85 g Fe 3 mol Fe H2O: Yield is 0.100 mol Fe3O4; Excess H2O: 10.0 g – 7.23 g = 2.77 g H2O. Answer: Limiting reactant Fe; 5. Yield: 0.1 mol Fe3O4. Excess H2O is 2.77 g H2O. 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. Actual yield: Amount obtained experimentally. Actual yield Percent yield: x 100 % Theor. yield Strategy: Find limiting reactant. Calculate theoretical yield. Calculate percent yield. Example 14: If 65.0 g CCl4 was prepared by CS2 + 3 Cl2 CCl4 + S2Cl2 reacting 100. g CS2 and 100. g of Cl2, calculate the percent yield. Molar masses: CS2: 76.15; Cl2: 70.90; CCl4: 153.8 g/mol 1 mol CS2 1 mol CCl4 100. g CS2 x x = 1.31 mol CCl4. 76.15 g CS2 1 mol CS2 1 mol Cl2 1 mol CCl4 x = 0.470 mol CCl4. 70.90 g Cl2 3 mol Cl2 Limiting reactant 153.8 g CCl4 0.470 mol CCl4 x = 72.3 g CCl4. Theoretical yield 1 mol CCl4 100. g Cl2 x 65.0 g CCl4 Actual yield 65.0 g CCl4 x 100 % = 89.9 % Percent yield 72.3 g CCl4 HW, Chapter 7 (p.139): 1 (a-e):Determine the molar masses of these compounds: KBr, Na2SO4, Pb(NO3)2; C2H5OH, HC2H3O2 5. Calculate the number of grams In each of the following: 0.550 mol Au; 15.8 mol H2O; 12.5 mol Cl2; 3.15 mol NH4NO3. 15. 1 molecule of tetraphosphorus decoxide contains: how many moles? How many grams? How many P atoms? How many O atoms? How many total atoms? 26. A 7.52 g sample of ajoene (garlic odor) was found to contain 3.09 g S, 0.453 g H, 0.513 g O and the rest, C. Calculate the percent composition. 39. Ethanedioic acid, a compound that is present in many vegetables, has a molar mass of 90.04 g/mol and a composition of 26.7% C, 2.2% H and 71.1% O. What is the molecular formula? HW, Chapter 9 (p.183): 3. Calculate the number of grams in these quantities: 2.55 mol Fe(OH)3; 125 kg CaCO3; 10.5 mol NH3; 72 millimol HCl; 500.0 mL of liquid Br2 (d=3.119 g/mL) 7(a-c). Balance the equation for the synthesis of sucrose: CO2 + H2O C12H22O11 + O2 and set up the mole ratio of: CO2 to H2O; H2O to C12H22O11; O2 to CO2. 13. Carbonates react with acids to form salt, water and carbon dioxide gas. When 50.0 g of CaCO3 are reacted with sufficient HCl, how many grams of CaCl2 will be produced? (Balance eq. first!) 15. In a blast furnace iron(III) oxide reacts with carbon to produce molten iron and CO: Fe2O3 + 3C 2 Fe + 3 CO How many kilograms of iron would be formed from 125 kg of Fe2O3? 23. Determine limiting reactant and the one in excess. KOH + HNO3 KNO3 + H2O 16.0 g 12.0 g 2 NaOH + H2SO4 Na2SO4 + H2O 10.0 g 10.0 g 29. Elemental silicon can be produced by the reduction of SiO2 (sand) with carbon: SiO2 + 2 C Si + 2 CO When 35.0 kg of SiO2 react with 25.3 kg of C, and 14.4 kg of Si is recovered, what is the percent yield of the reaction?