Moles and Stoichiometry

Moles and Stoichiometry I. Moles A. Definition. 1. A mole (mol) = 6.022 x 10 23 units. 6.022 x 10 23 = Avogadro's constant, with a dimension of particles mol 1 or mol 1 b. A mass in grams equal to the formula mass of a substance contains 6.02 x 10 23 formula units. c. The practical unit of formula mass or atomic mass = grams/mol. 2. The following should be familiar to you. Type of Formula Composition of 10.0 g of substance Formula substance mass in moles _ in no. of units Na element 23.0 g/mol 10.0 /23.0= (0.435)(6.02 x10 23 ) = 0.435 mol 2.62 x 10 23 atoms H 2 molecular 2.0 g/mol 5.0 moles H 2 3.01 x 10 24 molecules of H 2 10.0 moles H 6.02 x 10 24 atoms of H C 2 H 4 molecular 28.0 g/mol 0.357 mol C 2 H 4 2.15 x 10 23 molecules C 2 H 4 0.714 mol C 4.30 x 10 23 atoms C 1.428 mol H 8.60 x 10 23 atoms H Al 2 O 3 ionic 102 g/mol 0.098 mol Al 2 O 3 5.90 x 10 22 formula units 0.196 mol Al 3+ 1.18 x 10 23 Al 3+ ions 0.294 mol O 2-1.77 x 10 23 O 2- ions 1

3. Additional Example. Consider a 48.60 g sample of the compound C 4 H 10 O 2 a. How many moles of C 4 H 10 O 2 are present? MM of C 4 H 10 O 2 = 4(12.0) + 10(1.0) + 2(16.0) = 90.0g/mol 48.60g Mol C 4 H 10 O 2 = = 0.540 mol 90.0g/mol b. How many moles and how many grams of C are present in the sample? The formula shows 4 C atoms for every molecule there are 4 moles of C for every mole of C 4 H 10 O 2. Moles C = 0.540 mole C 4 H 10 O 2 x 4 mol C 1 mol C 4 H 10 O 2 = 2.16 mol C grams C = (2.16 mol C)(12.0 g/mol C) = 25.92 g or 48.0 g C grams C =48.60g 90.0 g compound = 25.92g C 25.92 g moles C = = 2.16 mol C 12.0 g/mol c. How many O atoms are present in the sample? 2 mol O Mol O = 0.540 mole C 4 H 10 O 2 x = 1.08 mol O 1 mol C 4 H 1 0 O 2 Atoms O = 1.08 mol O x 6.02x10 23 atoms/mol = 6.50x10 23 d. How many atoms of H and how many grams of H are present in the sample? 10 mol H Moles of H = 0.540 mole C 4 H 10 O 2 x = 5.40 mol H 1 mol C 4 H 10 O 2 grams H = 5.40 molx1.0g/mol = 5.40g Note: mol H could also be obtained by using the moles of C: Mol H = 2.16 mol Cx 10mol H = 5.40 mol 4 mol C 2

B. Empirical formulas from analysis data. 1. A compound was analyzed and found to contain 29.1% Na, 40.6% S, and 30.3% O by mass. Calculate the empirical formula. moles of Na = 29.1g / 23.0 g/mol= 1.265 mol.----------> 1.000----------> 2 moles of S = 40.6g / 32.1g/mol = 1.265 mol -----------> 1.000 ---------> 2 moles of O = 30.3g / 16.0g/mol = 1.894 mol ------------>1.500--------- > 3 formula : Na 2 S 2 O 3 Sodium thiosulfate 2. A 0.401 g sample of a compound was analyzed and found to contain 0.320 g of C and 0.081 g of H. In another experiment the molar mass was estimated to be equal to 30. Calculate the empirical and molecular formulas of the compound. moles C = 0.320g / 12.0g/mol = 0.0267 -------------> 1.00 ---------> 1 moles H = 0.081g / 1.0 g/mol = 0.081---------------> 3.04 ----------> 3 empirical formula = CH 3 empirical formula mass (EFM) = 12.0 + 3(1.0) = 15 30 15 = 2 = empirical formula units / molecule C 2H 6 = molecular formula. 3. Analysis by combustion. A 1.000 g sample of a compound containing only C, H, and O was burned to give 2.197 g of CO 2 and 1.199 g of H 2 O. Calculate the empirical formula. + O 2 (C x H y O z )---------------> CO 2 + H 2 O 1.000 g 2.197 g 1.199 g all the C goes to form CO 2 mol C = mol CO 2 = 2.197 g 44.0 g/mol = 0.04992 mol grams C in the sample = (.04922 mol )(12.0 g/mol) = 0.600 g all the H ends up in H 2 O 2(1.199 g) mol H = 2xmol H 2 O = = 0.1331 mol 18.0 g/mol grams H in sample = (0.1331 mol)(1.0 g/mol) = 0.133 g total mass of sample = 1.000 g = g of C + g of H + g of O therefore g of O = 1.000 g - 0.600 g - 0.133 g = 0.266 g 0.266 g mol O = = 0.0166 mol 16.0 g/mol mol C =.04992--------------> 3.00 3

mol H = 0.1331 -------------> 8.01 mol O = 0.0166 -------------> 1.00 empirical formula = C 3 H 8 O II. Stoichiometry: Calculations from balanced chemical equations. A. Information in a balanced chemical equation: 4Al(s) + 3O2(g) ------> 2Al2O3(s) 1. In terms of atoms and molecules. 4 Al atoms + 3 O 2 molecules -----> 2 Al 2 O 3 formula units. 2. In terms of atomic mass units. (4x27) = 108 u of Al + (3x32) = 96 u of O 2 -----> (2x102) = 204 u of Al 2 O 3 Note the Law of Conservation of mass (mass of products = mass of reactants) 3. In terms of moles. 4 moles of Al + 3 moles of O 2 -----> 2 moles of Al 2 O 3 4. In terms of grams. 108 g of Al + 96 g of O 2 -----> 204 g Al 2 O 3 5. In terms of any mass unit. 108 lb. of Al + 96 lb. of O 2 ----> 204 lb. of Al 2 O 3 108 slugs of Al + 96 slugs of O 2 ----> 204 slugs of Al 2 O 3 B. Calculations from balanced equations. 1. General. 108 tons of Al + 96 tons of O 2 ----> 204 tons of Al 2 O 3 a. A balanced chemical equation gives stiochiometeric information directly in terms of moles of reactants and products. Example, in the above equation, each mole of Al that reacts requires 3 4 mole of O 2 and produces 1 2 mole of Al 2O 3. b. In the laboratory, we do not obtain information directly in terms of moles. The normal laboratory measurements involve obtaining 1) the masses of pure solids and some pure liquids. 2) the volumes of solutions of known concentration. 3) the volumes of gases at known temperatures and pressures. 4) the volume of liquids of known densities. c. In stoichiometric calculations information is given, and asked for, in terms of these laboratory measurements. Therefore, the sequence to be followed is: 4

1) CONVERT LABORATORY INFORMATION TO MOLES 2) USE THE BALANCED CHEMICAL EQUATION TO GET INFORMATION IN TERMS OF MOLES. 3) RECONVERT MOLES BACK INTO LABORATORY UNITS. 2. Conversions to moles-these relationships should be familiar to you. a. Direct mass measurements of pure solids and liquids. Moles = b. Pure liquids of known densities. Mass in grams formula mass Mass = (density)x(volume) The densities are usually in g/ml and the volumes are in ml Mass in grams Moles = formula mass c. Volumes of solutions of known Molarity: Molarity (M) = Moles of Solute Liters of Solution = mmol Solute ml of Solution moles of solute = (M)(no. of liters of solution) mmoles of solute = (M)(no. of ml of solution) d. Volumes of gases: Use the Ideal gas Law (see Chapter 5.1-5.6) 1) for a pure gases PV = nrt = g MM RT. Where V = volume of the gas in L, P = pressure of the gas, MM= molar mass n = number of moles of the gas, T = temperature in K (= C + 273.15), R = gas constant. The value of R depends on the dimensions of pressure. R = 62.4 Torr L mol K = 0.08206 atm L mol K = 3.184 Pa m3 mol K 2) For mixtures of gases use partial pressures and Dalton s Law. P x V = n x RT for a component x of a gaseous mixture (P x = Partial pressure of x) P tot = P x 3) This is used when collecting a gas over water. A standard laboratory technique for collecting a gaseous product from a reaction is to allow the gas to displace water in a 5

container. The gas collected is saturated with water vapor. Therefore the total pressure in the container is due to a mixture of the gas (X) and H 2 O, so that P Tot = P X + P H 2 O P H2 O is the vapor pressure of water and is a function of temperature only, this can be looked up in standard tables and subtracted from P tot (usually atmospheric pressure) to obtain the partial pressure of the gas. C. Stoichiometric calculations. 1. Suppose that 80.0 g of Al was reacted according to the equation 4Al(s) + 3O 2 (g) ----> 2Al 2 O 3 (s) a. How many moles of O 2 was consumed? 80.0 g moles Al = 27.0 g/mol = 2.96 mol of Al moles O 2 = 2.96 mol Al x 3 mol O 2 4 mol Al = 2.22 mol O 2 b. What volume of O 2 measured at 27 C and 750 Torr pressure would be used? V = n Torr L O 2 RT (2.22 mole)(62.4 P = mol K )(300 K) 750 Torr = 55.3 L c. How many grams of Al 2 O 3 would be formed? moles Al 2 O 3 = 2.96 mol Al x ( 2 mol Al 2O 3 4 mol Al ) = 1.48 mol Al 2 O 3 g Al 2 O 3 = (1.48 mol)(102 g/mol) = 151 g Al 2 O 3 6

2. Consider the reaction: 4KO 2 (s) + 2H 2 O(l) ----> 4KOH(aq) + 3O 2 (g). Suppose that 7.50 g of KO 2 was placed in 50.0 ml of water. a. What volume of O 2 would be collected H 2 O at 27 C and 757 Torr atmospheric pressure? 7.50 g moles KO 2 = 71.1 g/mol = 0.105 mol KO 2 moles of O 2 produced = (0.105 mol KO 2 ) x ( 3 mol O 2 4 mol KO 2 ) = 7.875x10-2 = n O2 Since the gas is collected over water, the vapor pressure of water must be subtracted from the atmospheres pressure to get the partial pressure of O 2, therefore P O2 = Patm -P H2 O, where P H2 O = vapor pressure of water. At 27 C, P H2 O = 27 Torr. P O2 = 757 Torr - 27 Torr = 730 Torr V = n o2rt P O2 = (7.875x10-2 )(62.4)(300) 730 = 2.02 L b. How many grams of KOH are formed? moles of KOH formed = 0.105 mol KO 2 x ( g KOH formed = 0.105 mol (56.1 g/mol) = 5.89 g c. How many ml of water remains? 4 mol KOH 4 mol KO 2 ) = 0.105 mol KOH moles H 2 O required = 0.105 mol KO 2 x ( 2 mol H 2O 4 mol KO 2 ) = 0.0525 mol H 2 O g H 2 O =( 0.0525 mol) x (18.0 g/mol) = 0.945 g H 2 O. This is equivalent to 1 ml of H 2 O (the density of H 2 O = 1.0 g/ml). 49.0 ml of H 2 O will remain. d. What is the final concentration of KOH? M KOH = 0.105 mol 49.0x10-3 L = 2.14M 7

3. Limiting reagents. a. It is not necessary, and many times not desirable, to mix stoichiometric amounts of reactants. Under these conditions, the reagent used up first (the limiting reagent) determines the amount of product; the other reactant is the excess reactant. In such cases the first step is to decide which reactant is limiting. b. Consider the reaction 3Mg + N 2 ----> Mg 3 N 2 Suppose that 2.50 g Mg and 1.00 g N 2 are mixed and the reaction takes place. 1) What reactant will be in excess and how many grams of this reactant remains unreacted? 2.50 g moles Mg = 24.3 g/mol = 0.1029 mol moles N 1.00 g 2 = 28.0 g/mol = 0.0357 mol The stoichiometric Mg molar ratio = 3 N 1, the experimental molar ratio is 2 0.1029 0.0357 = 2.88 which is less than 3 N 2 is in excess and Mg is the limiting reagent and will determine the amount of product. moles of N 2 required = 0.1029 mol Mg x ( 1 mol N 2 3 mol Mg moles N 2 left = 0.0357 mol - 0.0343 mol = 0.0014 mol g in excess = (0.0014 mol) x (28.0 g/mol) = 3.92x10-2 g. b. How many grams of Mg 3 N 2 will be formed. 4. Yields. ) = 0.0343 mol moles Mg 3 N 2 = 0.1029 mol Mg x ( 1 mol Mg 3N 2 3 mol Mg ) = 3.43x10-2 mol grams of Mg 3 N 2 = (3.43x10-2 mol) x (100.9 g/mol) = 3.46 g a. The above calculations give the maximum amount of product that could be obtained in a reaction. In practice, this is never obtained, we do not get 100% of what we expect. b. The percent yield, or yield, is: experimental amount of product Yield = x100 theoretical amount of product The theoretical amount of product is that calculated in the above examples. c. There are a number of reasons for low yields. 1) Poor technique. 2) An unfavorable equilibrium. In such cases the yield can be improved by having one of the reactants in excess. 3) Competing reaction or consecutive reactions. 8

d. When new reactions are reported in the literature, they are usually described under conditions which maximize the yield. 9

Problems 1. On analysis a 2.75 g sample of a compound was found to contain 1.55 g of phosphorus and 1.20 g of sulfur. Calculate the empirical formula of this compound. (P 4 S 3 ) 2. A sample of a compound contains 3.96 g of carbon, 0.66 g of hydrogen, and 3.52 g of oxygen. In another experiment it was found that a 2.05 g sample of the gaseous compound occupied a volume of 584 ml at 125 C and 400 Torr pressure. a. What is the empirical formula of this compound? b. What is the molar mass of this compound? c. What is the molecular formula of this compound? (C 3 H 6 O 2 ; 218; C 9 H 18 O 6 ) 3. A sample of a compound containing only nitrogen and sulfur was burned in oxygen completely converting the nitrogen to N 2 O 3 and the sulfur to SO 2. The oxides formed were trapped and weighed giving 0.198 g of N 2 O 3 and 0.512 g of SO 2. What is the empirical formula of the compound? (N 2 O 3 ) 4. Calculate the empirical formulas from the following percent compositions. a. 34.3% Na; 17.9% C; 47.8% O b. 39.3% C; 8.2% H; 52.5% O c. 72.3% Fe; 27.7% O d. 69.9% Fe; 30.1% O e. 77.7% Fe; 22.3% O f. 9.7% Al; 38.4% Cl; 51.9% O (Answer: a.naco 2 b. C 2 H 5 O 2 c. Fe 3 O 4 d. Fe 2 O 3 e. FeO f. AlCl 3 O 9 ) 5. A 5.00 g sample of a compound containing C, H, and N was burned in oxygen to give 9.778 g of CO2 and 7.000 g of H2O. Calculate the empirical formula of the compound. (C 2 H 7 N) 6. A 3.500 g sample of a mixture of NaCl and KCl was dissolved in water and the chloride precipitated as AgCl. If the mass of the AgCl precipitate is 7.522 g, calculate the mass percent NaCl in the sample. (42.9% NaCl) 7. Calcium hydroxide reacts with H 3 PO 4 to give Ca 3 (PO 4 ) 2. Suppose the 0.850 g of Calcium hydroxide was placed in 200 ml of a 0.045M H 3 PO 4 solution. What reactant would be in excess and how many grams of Ca 3 (PO 4 ) 2 would be formed. (H 3 PO 4, 1.19 g Ca 3 (PO 4 ) 2 ) 8. What volume of H 2 could be collected over water at 27 C and 774 Torr atmospheric pressure by the reaction of 0.256 g of Na with excess water to give H 2 and NaOH? (139 ml) 10

9. A hydrocarbon was analyzed and found to contain 84.1% C and 15.9% H by mass. In an experiment it was found that a 0.488 g sample of the gaseous compound occupied a volume of 215 ml at 50 C and 400 Torr. Calculate the molecular formula of the compound. (C 8 H 18 ) 10. Aluminum sulfide reacts with oxygen to give aluminum sulfate. How much oxygen would be consumed when 6.60 g of aluminum sulfide reacts? (8.45 g) 11. Propene, C 3 H 6, burns in oxygen to form carbon dioxide and water. a) Write the balanced equation for this reaction. b) Explain what the equation states in a quantitative way. c) Per mole of propene, how many moles of oxygen would be required? d) Suppose 0.84 g of propene is burned. How many moles of Propene is present? How many moles of oxygen would be required for the complete combustion? How many moles of water and carbon dioxide would be formed? Calculate the grams of oxygen and the grams of carbon dioxide and water formed. e) Show that the masses in (d) are in accordance with the law of conservation of mass. (d. 0.02 mol C 3 H 6 ; O 2, 0.09 mol, 2.88g; CO 2, 0.06 mol, 2.64 g; H 2 O, 0.06 mol, 1.08g) 12. Calcium carbonate decomposes on heating to give calcium oxide and CO 2 (g). What volume of CO 2, measured at 100 C and 757 torr, would be generated by the decomposition of 5.0 g of calcium carbonate? (1.54 L) 13. Naphthalene, C 10 H 8, reacts with O 2 to give CO and H 2 O. What volume of CO could be collected over water at 24 C and 750 torr total pressure by the complete reaction of 1.50 g of naphthalene? (Vapor pressure of water at 24 C = 22 torr.) (2.98 L) 14. a. A compound containing only carbon and hydrogen when reacted with O 2 produced 1.62 g of H 2 O and 2.40 liter of CO 2 gas when measured at 27 C and 700 torr pressure. Assuming that all of the carbon in the compound was converted to CO 2 and all the hydrogen was converted to H 2 O, calculate the empiricalformula of the compound. (CH 2 ) b. In another experiment, a 0.30 g sample of this gaseous compound was found to occupy a volume of 137 ml at 27 C and 730 torr pressure. Calculate the molar mass of the compound. (56.1 g/mol) c) what is the molecular formula of the compound? (C 4 H 8 ) 11

d) Write the balanced equation for the reaction of this compound with O 2. (C 4 H 8 + 6 O 2 ----> 4 CO 2 + 4 H 2 O) 15. Aluminum reacts with HCl to give AlCl 3 and H 2 (g). What volume of H 2 would be collected over water at 28 C and 748 torr pressure when 5.0 g of Al is placed in 250 ml of a 2.0 M HCl solution? (Vapor pressure of water at 28 C = 28 m torr.) (0.500 mol HCl (LIMITING REAGENT); 0.185 mol Al; 13.0 L of H 2 ) 16. Zinc sulfide reacts with O 2 (g) to give zinc (II) oxide and SO 2 (g). What volume of O 2 (g), measured at 25 C and 740 torr pressure, would be required to react with 0.25 g of zinc sulfide? What volume of SO 2 (g), measured under the same conditions, would be produced in this reaction? (volume of O 2 = 96.7 ml volume of SO 2 = 64.5 ml) 17. Phosphorus burns in O 2 to give P 2 O 5. Suppose that 10.0 g of P is ignited in a 30.0 L container of O 2 at a temperature of 100 C and a pressure of 400 torr. a. What reactant is in excess, and how many moles of that reactant will be left unreacted? (O 2 in excess, 0.113 mol left) b. How many grams of P 2 O 5 will be formed? (22.9 g) 18. CO 2 can be removed from a gaseous mixture by reacting it with Na 2 O (s) to give Na 2 CO 3 (s). A mixture of CO 2 (g) and an inert gas in a 5.0 liter container originally exerted a pressure of 500 torr at 25 C. After the gas mixture was exposed to Na 2 O, the pressure in the container decreased to 200 torr. a) What was the partial pressure of CO 2 in the gas mixture? (300 Torr) b) How many grams of CO 2 was in the gas mixture? (3.55 g) c) How many grams of Na 2 CO 3 was formed? (8.55 g) 19. Consider the following reaction: Al + H 2 SO 4 -----> Al 2 (SO 4 ) 3 + H 2 a. Balance the equation. b. For a 8.1 g samples of Al, 1) Calculate the moles of Al present. 12

2) Calculate the moles and grams of H 2 SO 4 required for complete reaction. 3) Calculate the moles and grams of Al 2 (SO 4 ) 3 and of H 2 that would be formed. 4) Show that mass is conserved in this reaction. 20. For the following reaction: HCl + O 2 -----> H 2 O + Cl 2 a. Balance the equation. b. For a 0.16 g sample of O 2, 1) Calculate the moles of O 2 present 2) Calculate the moles and grams of HCl required for complete reaction. 3) Calculate the moles and grams of H 2 O and of Cl 2 formed. 4) Show that mass is conserved in this reaction. 21. Butane, C 4 H 10, reacts with O 2 to give CO 2 and H 2 O. a. Write the balanced equation for the reaction. b. Suppose 17.4 g of C 4 H 10 and 64.0 g of O 2 are mixed and the reaction allowed to take place. 1) Which reactant will be in excess? How many moles of this reactant will be left? How may grams? 2) How many grams of CO 2 is formed? 3) How many grams of H 2 O is formed? 22. Zinc sulfide, ZnS, reacts with O 2 to give ZnO and SO 2. a. Write the balanced equation for the reaction. b. Suppose 20.0 g of ZnS and 15.0 g of O 2 are mixed and the reaction allowed to take place. 1) What reactant will be in excess? How many moles of this reactant will be left? How many grams will be left? 2) How many grams of SO 2 will be formed? c. SO 2 reacts with CaO according to the equation: CaO + SO 2 -----> CaSO 3 13

How many grams of ZnS would be required to produce enough SO 2 to react with 8.4 g of CaO? 23. Aluminum carbonate reacts with HBr according to the equation: Al 2 (CO 3 ) 3 (s) + 6 HBr(aq) 2 AlBr 3 (aq) + 3 H 2 O(l) + 3 CO 2 (g) Suppose that 5.84 g of Al 2 (CO 3 ) 3 reacts., a. how many grams of AlBr 3 (FM = 267) is formed? b. what volume of CO 2 (g) would be collected over H 2 O at 24 C and 754.4 torr total pressure? pressure of H 2 O at 24 C = 22.4 torr) c. how many ml of a 3.0 M HBr solution reaction? 24. Suppose that iron reacted with H 2 S to give Fe 2 S 3 and H 2 according to the equation: 2 Fe(s) + 3 H 2 S(aq) -----> Fe 2 S 3 (s) + 3 H 2 (g) (The vapor would be needed in this When 8.40 g of Fe reacts, a. what volume of H 2 (g) could be collected over water at 25 C and 758.8 Torr atmospheric pressure from the reaction? (vapor pressure of water at 25 C, P H2 O = 23.8Torr ) b. how many Liters of a 0.20 M H 2 S solution would be required in the c. How many grams of Fe 2 S 3 would be produced in the reaction? Answers 19-24 19. b. 1) 0.30; 2)0.45 mol, 44.1 g; 3) Al 2 (SO 4 ) 3, 0.15 mol, 51.3 g; H 2, 0.45 mol, 0.90 g 20. b. 1) 5.0x10 3 mol; 2) 0.02 mol, 0.73 g; 3) Cl 2, 0.01 mol, 0.71 g; H 2 O, 0.01 mol, 0.18 g 21. b. 1) O 2 in excess, 0.05 mol or 1.6 g remain; 2) 52.8 g; 3) 27.0 g 22. b. 1) O 2 in excess, 0.16 mol or 5.2 g remain; 2) 13.1 g; c. 14.6 g 23. a. 13.33 g; b. 1.9 L; c. 50 ml 24. a. 5.71 L; b. 1.13 ml; c. 15.6 reaction? 14

15

16