Quantitative Composition of Compounds

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: 8 eggs x 5 pancakes 2 eggs = 20 pancakes This is the quantity you need 5 pancakes 2 eggs Solve it in your head: 2 eggs makes5 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: 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 = 15 eggs 3 blocks cheese 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: 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. 5 g H 2 + 5 g I 2 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 Numberof items. 1 mole = 602,214,179,000,000,000,000,000 or 6.022 x 10 23. 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.

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 12 atoms 6 C, which, in turn, weights exactly12 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 massrefers to: the sum of protons and neutrons in a single atom, weighted average mass of all isotopes of an element and alsoto the number of grams in one mole of atoms. H 2 + I 2 = 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 6.022 x 10 23 6.022 x 10 23 1.204 x 10 24 molecules molecules molecules 1 mole 1 mole 2 mole 2.016 g 253.8 g 255.9 g or any number of 1 mole of H 2 weighs 2 x 1.008 g = 2.016 g molecules Amadeo Avogadro Atomic number Atomic mass 1 H 1.008 Conversion factors: 1 mole 6.022 x 10 23 species 1 mole molar mass The mole 1 mol of naturally occurring H atoms has a mass of 1.008 g

Mole -mass -atoms conversions Q1: How many atoms in 0.500 mol Au? 0.500 mole Au x 6.022 x 1023 atoms Au 1 mole Au = 3.011 x 10 23 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 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 H 2 O? 2.016 Molar mass H 2 O = 16.00 18.02 g 15.00 g H 2 O x 1 mole H 2 O 18.02 g H 2 O x 6.022 x10 23 molec. H 2 O 1 mole H 2 O = 5.013 x10 23 molec. H 2 O

Percent Composition What is the % composition of CH 2 O? Percent composition is % massthat 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, C 6 H 12 O 6? Check: identical to CH 2 O! Types of Formulas Empirical Formula: the formula of a compound that expresses the smallest whole number ratioof the atoms present. 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 determine the relative number of moles of each element in a compound, we can find the formula of that compound. Total mass = 12.01 g + 2.016 g + 16.00 g %C = 12.01 g = 30.03 g 30.026 g x100 %C = 40.00 % %H = 6.71 % Rounding to 30.03 g produces errors: percentages add up to 99.98 %! + %O = 53.29 % 100.00 % CH 2 O is the empirical formula for glucose, C 6 H 12 O 6 Molecular Formula: the formula that states the actualnumber of each kind of atom found in one molecule of the compound. 1 moleculeof aspirin, C 9 H 8 O 4 = 9 atomsof C, 8 atoms of H and 4 atoms of O. 1 moleof C 9 H 8 O 4 = 9 molof C atoms, 8 mol of H atoms and 4 mol of O atoms.

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 unitsin the molecular formula (n). Multiply all subscripts in empirical formula by n. 1. 2. 3. Assume 100.0 g of the compound. 31.9 g K 28.9 g Cl 39.2 O?? g O g O = 100.0 g ( 31.9 g + 28.9 g) = 100.0 g 60.8 g = 39.2 g 31.9 g K x 1 mol K 39.10 g K 28.9 g Cl x 39.2 g O x 1 mol O 16.00 g O = 0.816 mol K 1 mol Cl = 0.815 mol Cl 35.45 g Cl K: 0.816 / 0.815 = 1.00 Cl: 0.815 / 0.815 = 1.00 O: 2.45 / 0.815 = 3.01 = 2.45 mol O Empirical formula: K Cl O 3 4. Molar mass of empirical formula: 39.1 g + 35.45 g + (3 x16.00 g) = 122.55 Molar mass compound 122.55 n= Molar mass emp. formula = 122.55 = 1 There is 1 unit of KClO 3 in molecular formula. (KClO 3 ) x1 = KClO 3

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. 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. 1. Assume that you have 100.00 g sample; the mass of each element is equal to the % composition. 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 3. C: 3.33 / 3.33 = 1 H: 6.66 / 3.33 = 2 O: 3.33 / 3.33 = 1 = 3.33 mol C 53.3 g O x 1 mol O 16.00 g O = 6.66 mol H = 3.33 mol O 4. Molar mass compound Molar mass emp. formula Thus, there are 6 (CH 2 O) units. Practice(answer in parenthesis): Empirical formula CH 2 O. 40.0 g C, 6.71 g H, 53.3 g O. Emp. Formula mass = 30.03 g/mol 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 ) 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 ) = 180.2 30.03 = 6.00 Molecular formula: C 6 H 12 O 6.

Chapter 9 Calculations from Chemical Equations contains 6.022 x 10 23 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 NaClis 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. 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe 2 mol 1 mol 1 mol 2 mol 2. The coefficientin 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 Fe 2 O 3 2 mol Al 1 mol Fe 2 O 3 1 mol Al 2 O 3 Mole ratio = Mole ratio is found from the coefficients of the balanced equation: moles of desiredsubstance 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 Cl 2? Assume that there is more than enough Na. desired substance 2 moles NaCl 3.4 moles Cl 2 x = 6.8 moles NaCl 1 mole Cl 2 starting substance The following examples refer to the equation: 2 Na(s) + Cl 2 (g) 2 NaCl(s) 1 mole 2 moles Ca 5 (PO 4 ) 3 F(s) + 5 H 2 SO 4 (aq) 3 H 3 PO 4 (aq) + HF(aq) + 5 CaSO 4 (s) 1 mole 5 moles 3 moles 1 mole 5 moles Example 2: Calculate the number of molesof phosphoric acid (H 3 PO 4 ) formed by the reaction of 10 moles of sulfuric acid (H 2 SO 4 ) on phosphate rock: 3 moles H 3 PO 10 moles H 4 2 SO 4 x 5 moles H 2 SO = 6 moles H 3 PO 4 4 These are exact numbers! Example 3: Calculate the number of molesof Ca 5 (PO 4 ) 3 F needed to produce 6 moles of H 3 PO 4. 1 mole Ca 5 (PO 4 ) 3 F 6 moles H 3 PO 4 x = 2 moles Ca 3 moles H 3 PO 5 (PO 4 ) 3 F 4

Mass Mole conversion Example 4: Calculate the number of molesof H 2 SO 4 necessary to yield 784 g of H 3 PO 4. Molar mass of H 3 PO 4 = 97.99 3.024 30.97 64.00 97.99 g mole 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. 784 g H 3 PO 4 x 1 mole H 3PO 4 = 8.00 moles H 2. Convert moles of 97.99 g H 3 PO 3 PO 4. 4 starting substance into 5 moles H moles of desired 8.00 moles H 3 PO 4 x 2 SO 4 = 13.3 moles H 3 moles H substance. 3 PO 2 SO 4. 4 3. Convert moles of desired substance into the units specified in the problem. done. Ex. 5: Calculate the massof phosphate rock, Ca 5 (PO 4 ) 3 F needed to yield 200. g of HF. Molar masses: Ca 5 (PO 4 ) 3 F = 504.3 g/mol; HF = 20.01 g/mol Step 1, 1 mole HF 1 mole Ca 200. g HF x 5 (PO 4 ) 3 F x = 10.0 moles Ca Step 2 20.01 g HF 1 mole HF 5 (PO 4 ) 3 F Step 3: 504.3 g ph.r. 10.0 moles Ca5 (PO 4 ) 3 F x = 5.04 kg Ca 5 (PO 4 ) 3 F. 1 mole ph.r.

Mass mass conversion Step_by_step: Ex. 6: Calculate the number of gramsof 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. Molar mass H 3 PO 4 = 97.99 392 g H 3 PO 4 x 1 mole H 3PO 4 = 4.00 moles 97.99 g H 3 PO 4 2. Convert moles of starting substance 5 moles H 4.00 moles H 3 PO 4 x 2 SO 4 into moles of desired substance. 3 moles H 3 PO 4 = 6.67 moles 3. Convert moles of desired substance into the units specified in the problem. 6.67 moles H 2 SO 4 x 98.09 g 1 mole H 2 SO 4 = 654 g H 2 SO 4. g mole Molar mass H 2 SO 4 = 98.09 g mole 392 g H 3 PO 4 x 1 mole H 3PO 4 x 5 moles H 2 SO 4 x 98.09 g = 654 g H 97.99 g H 3 PO 4 3 moles H 3 PO 4 1 mole H 2 SO 2 SO 4. 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 180.2 (glucose). 58.5 g CO 2 x sunlight 6 CO2(g) + 6H2O(l) 6O2(g) + C6H12O6(aq) 1 mole CO 2 1 mole glucose 180.2 g glucose x x 44.01 g CO 2 6 moles CO 2 1 mole glucose Combined steps: = 39.9 g glucose

Mass to moles of starting compound Conversion General Case Moles of starting compound to moles of desired compound Step 1 Step 2 Step 3 Moles of desired comp. to units desired. Mass mass: All 3 steps 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 Molar masses: H 2 : 2.016 g/mol; NH 3 : 17.03 g/mol 112 g H 2 x 1 mole H 2 2 moles NH 3 17.03 g NH x x 3 = 631 g NH 2.016 g H 2 3 moles H 2 1 mole NH 3 Starting 3 compound Step 1 Step 2 Step 3 result Moles moles: Step 2 only Example 9: Calculate the moles of NH 3 formed by the reaction of 1.5 moles of H 2. 2 moles NH 1.50 moles of H 3 2 x = 1.00 mole NH 3 moles H 3. Starting 2 compound Step 2 result Moles mass: Step 2 and Step 3 only Example 10: Calculate the mass of NH 3 formed by the reaction of 1.50 moles of H 2. 1.50 moles of H 2 Starting compound x 2 moles NH 3 3 moles H 2 Step 2 x 17.03 g NH 3 1 mole NH 3 Step 3 = 17.0 g NH 3. result

Conversion General Case (cont d) Mass moles: Step 1 and Step 2 only Example 11: Calculate the moles of NH 3 formed by the reaction of 150. g H 2. 1 mole H 2 moles NH 150. g H x 3 2 x 2 = 49.6 mol NH 3 moles H 3. 2.016 g H 2 2 Step 1 Step 2 result Starting compound N 2 + 3H 2 2NH 3 Mass particles: All 3 steps Example 12: Calculate the # molecules of NH 3 formed by the reaction of 150. g H 2. 150. g H 2 x 1 mole H 2 2.016 g H 2 Step 1 Starting compound x 2 moles NH 3 3 moles H 2 Step 2 x 6.022 x 10 23 molecules NH 3 1 mole NH 3 Step 3 = 2.99 x 10 25 molecules NH 3. result 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.

Limiting Reactant and yield Calculations (cont d) 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 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 leastmoles 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: 3Fe (s) + 4H 2 O (g) Fe 3 O 4 (s) + 4 H 2 (g) 2. # moles of Fe 3 O 4 : 16.8 g Fe 1 mol Fe from Fe: x 55.85 g Fe x 1 mol Fe 3O 4 = 0.100 mol Fe 3 mol Fe 3 O 4. Least moles Fe 3 O 4? 10.0 g H 2 O x 1 mol H 2O 18.02 g H 2 O x 1 mol Fe 3O 4 from H 4 mol H 2 O = 0.139 mol Fe 2 O: 3O 4. 3. Limiting reactant: Fe 16.8 g Fe x 1 mol Fe 4. Reacted 55.85 g Fe x 4 mol H 2O x 18.02 g H 2O 3 mol Fe 1 mol H 2 O = 7.23 g H 2O. H 2 O: Excess H 2 O: 10.0 g 7.23 g= 2.77 g H 2 O. Yieldis 0.100 mol Fe 3 O 4 ; Answer: Limiting reactant Fe; 5.Yield: 0.1 mol Fe 3 O 4. ExcessH 2 O is 2.77 g H 2 O.

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: Theor. yield x100 % Example 14: If 65.0 g CCl 4 was prepared by CS reacting 100. g CS 2 and 100. g of Cl 2, calculate 2 + 3 Cl 2 CCl 4 + S 2 Cl 2 the percent yield. Molar masses: CS 2 : 76.15; Cl 2 : 70.90; CCl 4 : 153.8 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 x 1 mol CCl 4 = 0.470 mol CCl 70.90 g Cl 2 3 mol Cl 4. 2 Limiting reactant 0.470 mol CCl 4 x 153.8 g CCl 4 1 mol CCl 4 = 72.3 g CCl 4. Theoretical yield Strategy: Find limiting reactant. Calculate theoretical yield. Calculate percent yield. 65.0 g CCl 4 Actual yield 65.0 g CCl 4 72.3 g CCl 4 x 100 % = 89.9 % Percent yield

HW, Chapter 7 (p.142): 1(a-e):Determine the molar masses of these compounds: KBr, Na 2 SO 4, Pb(NO 3 ) 2 ; C 2 H 5 OH, HC 2 H 3 O 2 5. Calculate the number of grams In each of the following: 0.550 mol Au; 15.8 mol H 2 O; 12.5 mol Cl 2 ; 3.15 mol NH 4 NO 3. 19. 1 molecule of tetraphosphorus decoxide contains: how many moles? How many grams? How many P atoms? How many O atoms? How many total atoms? 32. 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. 47. 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.191): 3. Calculate the number of grams in these quantities: 2.55 mol Fe(OH) 3 ; 125 kg CaCO 3 ; 10.5 mol NH 3 ; 72 millimol HCl; 500.0 ml of liquid Br 2 (d=3.119 g/ml) 7(a-c). Balance the equation for the synthesis of sucrose: CO 2 + H 2 O C 12 H 22 O 11 + O 2 and set up the mole ratio of: CO 2 to H 2 O; H 2 O to C 12 H 22 O 11 ; O 2 to CO 2. 13. Carbonates react with acids to form salt, water and carbon dioxide gas. When 50.0 g of CaCO 3 are reacted with sufficient HCl, how many grams of CaCl 2 will be produced? (Balance eq. first!) 15. In a blast furnace iron(iii) oxide reacts with carbon to produce molten iron and CO: Fe 2 O 3 + 3C 2 Fe + 3 CO How many kilograms of iron would be formed from 125 kg of Fe 2 O 3? 23. Determine limiting reactant and the one in excess. KOH + HNO 3 KNO 3 + H 2 O 16.0 g 12.0 g 2 NaOH + H 2 SO 4 Na 2 SO 4 + H 2 O 10.0 g 10.0 g