Chapter 62 Chemical Equations C + O 2 CO 2 Substances on left –Reactants –Starting materials Substances on right –Products + and reacts to produce
Chapter 63 Reading a Chemical Reaction C(s) + O 2 (g) CO 2 (g) (s) solid (l) liquid (g) gas (aq) aqueous or in water
Chapter 64 Balancing Chemical Equations Must have the same number of each element on both sides of the equation
Chapter 65 Suggestions for Balancing Chemical Equations 1.If element occurs once on each side, balance first 2.Balance any reactants or products that exist as free elements last –CANNOT change subscripts –CANNOT add/delete products or reactants
Chapter 66 Example 6.1 Balance the following equation, which represents the chemical reaction involved when an airbag deploys: Checking, we count two Na atoms and six N atoms on each side. The equation is balanced. NaN 3 N 2 + Na Solution We can see initially that the sodium atoms are balanced, but the nitrogen atoms are not. For this sort of problem, we will use the concept of the least common multiple. There are three nitrogen atoms on the left (reactants side) and two on the right (products side). The least common multiple of 2 and 3 is 6. Therefore, we need three N 2 and two NaN 3 : We now have two sodium atoms on the left. We can get two on the right by placing the coefficient 2 in front of Na: 2 NaN 3 2 Na + 3 N 2 (balanced)
Chapter 67 Example 6.1 (cont.) Exercise 6.1A The reaction between hydrogen and nitrogen to give ammonia, called the Haber process, is typically the first step in the industrial production of fertilizer. Balance the following reaction for the Haber process: H 2 + N 2 NH 3 Exercise 6.1B Iron ores such as Fe 2 O 3 are smelted, by reaction with carbon, to produce metallic iron and carbon dioxide. Balance the following reaction: Fe 2 O 3 + CCO 2 + Fe
Chapter 68 Example 6.2 When a fuel such as methane is burned in sufficient air, the products are carbon dioxide and water. Balance the following equation for this combustion: CH 4 + O 2 CO 2 + H 2 O (not balanced) In this equation, oxygen appears in two different products and by itself in O 2 ; we leave the oxygen for last and balance the other two elements first. Carbon is already balanced, with one atom on each side of the equation. For hydrogen, the least common multiple of 2 and 4 is 4, and so we place the coefficient 2 in front of H 2 O to balance hydrogen. Now we have four hydrogen atoms on each side: CH 4 + O 2 CO 2 + 2 H 2 O (not balanced) Solution The equation is balanced. Generally, when you have a chemical species that only contains one type of element, it is easier to balance that element last. Now for the oxygen. There are four oxygen atoms on the right. If we place a 2 in front of O 2 on the left, the oxygen atoms balance. CH 4 + 2 O 2 CO 2 + 2 H 2 O (balanced)
Chapter 69 Does it takes more oxygen (per molecule) to burn butane than it does to burn methane? How much more CO 2 is produced? Butane is another common fuel. Balance the following equation describing the combustion of butane: C 4 H 10 + O 2 CO 2 + H 2 O Exercise 6.2 Example 6.2 (cont.)
Chapter 610 Avogadros Number Number of carbon-12 atoms in a 12-g sample of carbon-12 6.022 x 10 23
Chapter 611 Mole Similar to a dozen EXCEPT 1 mol has Avogadros number of items –Abbreviated mol Easier to weigh out moles instead of molecules 1 mol H 2 O has how many moles of O? H?
Chapter 612 Formula Masses Average mass of a formula unit relative to that of a carbon-12 atom –C formula mass = 12.011 –O formula mass = 15.9994
Chapter 613 Molar Mass Mass of 1 mol of a substance –Different for every compound Numerically equivalent to formula mass Units of gram/mole Can be used to convert between moles and mass
Chapter 620 Example 6.4 Calculate the molecular mass of sulfur dioxide (SO 2 ), an irritating gas formed when sulfur is burned. Solution We think about the problem in the following way. Add the atomic mass of sulfur to twice the atomic mass of oxygen. Exercise 6.4 Calculate the formula mass of (a) C 6 H 4 Cl 2 and (b) H 3 PO 4. 1 x the atomic mass of S = 1 x 32.1u = 32.1u 2 x the atomic mass of O = 2 x 16.0 u = 32.0 u Formula mass of SO 2 = 64.1 u
Chapter 621 Example 6.5 Calculate the formula mass of sodium azide, NaN 3, used in automobile airbags. Exercise 6.5 Calculate the formula mass of (a) NaHCO 3 (the main ingredient in baking soda) and (b) (NH 4 ) 2 SO 4 (a fertilizer commonly used by home gardeners). Remember, everything within the parentheses must be multiplied by 2. Solution To determine a formula mass, we add the atomic masses of the constituent elements: 1 x atomic mass of Na = 1 x 23.0 u = 23.0 u 3 x atomic mass of N = 3 x 14.0 u = 42.0 u Formula mass of NaN 3 = 65.0 u
Chapter 622 Example 6.6 How many grams of N 2 are in 0.400 mol N 2 ? Solution The molar mass of N 2 is 28.0 g/mol. Therefore: = 11.2 g N 2 28.0 g N 2 1mol N 2 ? g N 2 = 0.400 mol N 2 x Calculate the mass, in grams, of (a) 55.5 mol H 2 O and (b) 55 mol uranium. Exercise 6.6
Chapter 623 Example 6.7 Calculate the number of moles of NaN 3 in a 10.0-g sample of the solid. Solution In this case we need the molar mass of NaN 3. In Example 6.5 we calculated 65.0 u as the formula mass of NaN 3. The molar mass is then 65.0 g/mol. To convert from a mass in grams to an amount in moles, we must use the inverse of the molar mass (1 mol NaN 3 /65.0 g NaN 3 ) to get the proper cancellation of units. When we start with grams, we must have grams in the denominator of our conversion factor: ? mol NaN 3 = 10.0 g NaN 3 x= 0.154 mol NaN 3 65 g NaN 3 1 mol NaN 3 Calculate the amount, in moles, of (a) 3.71 g Fe and (b) 165 g butane, C 4 H 10. Exercise 6.7
Chapter 624 Example 6.10 Nitrogen monoxide (nitric oxide), one of the air pollutants discharged by internal combustion engines, combines with oxygen to form nitrogen dioxide, a reddish-brown gas that irritates the respiratory system and eyes. The equation for this reaction is State the molecular, molar, and mass relationships indicated by the equation. 2 NO + O 2 2 NO 2 Solution Molecular: Two molecules of NO react with one molecule of O 2 to form two molecules of NO 2. Molar: 2 mol of NO react with 1 mol of O 2 to form 2 mol of NO 2. Mass: 60.0 g of NO react with 32.0 g of O 2 to form 92.0 g of NO 2. State the molecular, molar, and mass relationships indicated by this equation. Hydrogen sulfide, a gas that smells like rotten eggs, burns in air to produce sulfur dioxide and water according to the equation Exercise 6.10 2 H 2 S + 3 O 2 2 SO 2 + 2 H 2 O
Chapter 625 Example 6.11 When 0.105 mol of propane is burned in a plentiful supply of oxygen, how many moles of oxygen is consumed? C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O Multiply the given quantity (0.105 mol C 3 H 8 ) by the factor on the right to get an answer with the desired units (moles of oxygen): 5 mol O 2 1 mol C 3 H 8 ? mol O 2 = 0.105 mol C 3 H 8 x= 0.525 mol O 2 For the combustion of propane in Example 6.11: (a) How many moles of carbon dioxide is formed when 0.529 mol of C 3 H 8 is burned? (b) How many moles of water is produced when 76.2 mol of C 3 H 8 is burned? (c) How many moles of carbon dioxide is produced when 1.010 mol of O 2 is consumed? Exercise 6.11 Solution The equation tells us that 5 mol O 2 is required to burn 1 mol C 3 H 8. We can write 1 mol C 3 H 8 5 mol O 2 where we use the symbol to mean is chemically equivalent to. From this relationship we can construct conversion factors to relate moles of oxygen to moles of propane. The possible conversion factors are 1 mol C 3 H 8 5 mol O 2 1 mol C 3 H 8 and
Chapter 626 Example 6.12 Calculate the mass of oxygen needed to react with 10.0 g of carbon in the reaction that forms carbon dioxide. Solution 1. The balanced equation is C + O 2 CO 2 2. The molar masses are 2 x 16.0 + 32.0 g/mol for O 2 and 12.0 g/mol for C. 3. We convert the mass of the given substance, carbon, to an amount in moles: ? mol C = 10.0 g C x 1 mol C 12.0 g C = 0.833 mol C 4. We use coefficients from the balanced equation to establish the stoichiometric factor that relates the amount of oxygen to that of carbon: 1 mol C 0.833 mol C x 1 mol O 2 = 0.833 mol O 2
Chapter 627 5. We convert from moles of oxygen to grams of oxygen: Example 6.12 (cont.) We can also combine the five steps into a single setup. Note that the units in the denominators of the conversion factors are chosen so that each cancels the unit in the numerator of the preceding term: (The slightly different answers are due to rounding in the intermediate steps.) Exercise 6.12A Calculate the mass of oxygen (O 2 ) needed to react with 0.334 g of nitrogen (N 2 ) in the reaction that forms nitrogen dioxide. Exercise 6.12B Calculate the mass of carbon dioxide formed by burning 775 g of each of (a) methane (CH 4 ) and (b) butane (C 4 H 10 ). Assume excess oxygen is available. 0.833 mol O 2 x 32.0 g O 2 1 mol O 2 = 26.7 g O 2
Chapter 628 Example 6.13 The decomposition of sodium azide produces nitrogen gas that is used to inflate automobile airbags. What mass of nitrogen, in grams, can be made from 60.0 g of sodium azide? Solution 1.We start by writing the balanced chemical equation, which we did in Example 6.1: 2 NaN 3 2 Na + 3 N 2 2.The molar mass of NaN 3 is 65.0 g/mol and the molar mass of N 2 is 2 x 14.0 = 28.0 g/mol. 3.We convert the mass of the given substance, sodium azide, to an amount in moles: 60.0 g NaN 3 x 1 mol NaN 3 65.0 g NaN 3 = 0.923 mol NaN 3 4.We use coefficients from the balanced equation to establish the stoichiometric factor that relates the amount of nitrogen gas to that of sodium azide: 0.923 mol NaN 3 x 3 mol N 2 2 mol NaN 3 = 138 mol N 2
Chapter 629 5. We convert from moles of nitrogen gas to grams of nitrogen gas: 1.38 mol N 2 x 28.0 g N 2 1 mol N 2 = 38.6 g N 2 Example 6.13 (cont.) As is usually the case, all of the steps outlined above can be combined into a single setup: 60.0 g NaN 3 x 1 mol NaN 3 65.0 g NaN 3 x 3 mol N 2 2 mol NaN 3 x 28.0 g N 2 1 mol N 2 = 38.8 g N 2 Notice that the units of the numerator in one stoichiometric factor are the units in the denominator of the next stoichiometric factor. In this way, the correct cancellation of units occurs and the units of the final numerator are the units of your answer. Exercise 6.13A Ammonia reacts with phosphoric acid (H 3 PO 4 ) to form ammonium phosphate [(NH 4 ) 3 PO 4 ]. How many grams of ammonia are needed to react completely with 74.8 g of phosphoric acid? Exercise 6.13B a. The decomposition of potassium chlorate (KClO 3 ) produces potassium chloride (KCl) and O 2 gas. How many grams of oxygen can be made from 2.47 g of potassium chlorate?
Chapter 630 Solutions Homogeneous mixture of two or more substances Solute – what is being dissolved Solvent – what is doing the dissolving Aqueous solutions – water is solvent
Chapter 631 Solubility Soluble: an appreciable quantity dissolves Insoluble: very little, if any, quantity dissolves Dilute solution: little solute in a lot of water Concentrated solution: lots of solute in the solvent
Chapter 632 Measurement of Solubility Molarity (M): amount of solute, in moles, per liter of solution
Chapter 634 Percent Concentration If both solute and solvent are liquids, use percent by volume
Chapter 635 Percent by mass is commonly used for commercial solutions –35.0% HCl means 35.0 g HCl for every 100.0 g of solution
Chapter 636 Note the differences between mass percent, percent by volume, and molarity –10% HCl solution is considerably different than 10 M HCl –Require different amounts of HCl
Chapter 637 Example 6.18 Calculate the molarity of a solution made by dissolving 3.50 mol of NaCl in enough water to produce 2.00 L of solution. Solution Molarity (M) = moles of solute liters of solution 3.50 mol NaCl 2.00 L solution = 1.75 M NaCl = We read 1.75 M NaCl as 1.75 molar NaCl. Exercise 6.18A Calculate the molarity of a solution that has 0.0500 mol of NH 3 in 5.75 L of solution. Exercise 6.18B Calculate the molarity of a solution made by dissolving 0.750 mol of H 3 PO 4 in enough water to produce 775 mL of solution.
Chapter 638 Example 6.19 What is the molarity of a solution in which 333 g of potassium hydrogen carbonate is dissolved in enough water to make 10.0 L of solution? Now use this value as the numerator in the defining equation for molarity. The solution volume, 10.0 L, is the denominator: molarity x 3.33 mol KHCO 3 10.0 L solution =.333 M KHCO 3 Exercise 6.19 Calculate the molarity of each of the following solutions: a. 18.0 mol of H 2 SO 4 in 2.00 L of solution b. 3.00 mol of KI in 2.39 L of solution c. 0.206 mol of HF in 752 mL of solution. (HF is used for etching glass.) Solution First, prepare a setup to convert from mass of KHCO 3 to moles of KHCO 3 : 333 g KHCO 3 x 1 mol KHCO 3 100.1 g KHCO 3 = 3.33 g KHCO 3
Chapter 639 Example 6.20 How many grams of NaCl is required to prepare 0.500 L of typical over-the-counter saline solution (about 0.15 M NaCl)? Exercise 6.20 How many grams of potassium hydroxide is required to prepare each of the following solutions? a. 2.00 L of 6.00 M KOH b. 100.0 mL of 1.00 M KOH Solution First we calculate the moles of NaCl: 15 mol NaCl 0.500 L solution x 1 L solution = 0.075 mol NaCl Then we use the molar mass to calculate the grams of NaCl: 58.4 g NaCl 0.075 mol NaCl x 1 mol NaCl = 4.4 g NaCl
Chapter 640 Example 6.21 Concentrated hydrochloric acid has a concentration of 12.0 M HCl. How many milliliters of this solution would one need to get 0.425 mol of HCl? Solution liters of HCl solution = moles of solute molarity 0.425 mol HCl 12.0 M HCl 0.425 mol HCl 12.0 mol HCl/L = 0.0354 L = We would need 0.0354 L (35.4 mL) of the solution to have 0.425 mol. Remember that molarity is moles per liter of solution, not per liter of solvent. Exercise 6.21A How many milliliters of 15.0 M aqueous ammonia (NH 3 ) solution do you need to get 0.445 mol of NH 3 ? Exercise 6.21B How many grams of HNO 3 are in 500 mL of rain that has a concentration of 2.0 x 10 –5 M HNO 3 ? =
Chapter 641 Example 6.22 Two-stroke engines use a mixture of 120 mL of oil dissolved in enough gasoline to make 4.0 L of fuel. What is the percent by volume of oil in this mixture? Solution Percent by volume = 120 mL oil 4000 mL solution x 100% = 3.0% Exercise 6.22 What is the percent by volume of a solution made by dissolving 235 mL of ethanol in enough water to make exactly 500 mL of solution (approximately the concentration of ethanol in distilled spirits)?
Chapter 642 Example 6.23 Describe how to make 775 mL of vinegar (about a 5.0% by volume solution of acetic acid in water). Solution Volume of solute = percent by volume x volume of solution 100% Lets begin by rearranging the equation for percent by volume to solve for volume of solute: Take 38.85 mL of acetic acid and add enough water to make 775 mL of solution. Exercise 6.23 Ethanol used for medicinal purposes is generally of a grade referred to as USP (an abbreviation of United States Pharmacopoeia, the official publication of standards for pharmaceutical products). USP ethanol is 95% CH 3 CH 2 OH by volume. Describe how to make 100 mL USP-grade ethanol. Substituting, we have 5.0% x 775 mL 100% = 39 mL =
Chapter 643 Example 6.24 What is the percent by mass of a solution of 25.0 g of NaCl dissolved in 475 g (475 mL) of water? Solution Percent by mass = mass of NaCl mass of solution x 100% 25.0 g NaCl 500 g solution = x 100% = 5.00% NaCl Exercise 6.24 Hydrogen peroxide from the local drugstore is a 3% by weight solution of H 2 O 2 in water. What is the percent by mass of a solution of 9.40 g of H 2 O 2 dissolved in 335 g (335 mL) of water?
Chapter 645 Avogadros hypothesis – equal volumes of gases at constant pressure and temperature have the same number of molecules
Chapter 646 Volume Relationships Law of combining volumes – when all measurements are made at same temperature and pressure, volumes of gaseous reactants and products are in small whole-number ratio
Chapter 647 Gas Laws – Kinetic Molecular Theory 1.All matter is composed of tiny discrete particles called molecules 2.Molecules in a gas are in rapid constant motion and move in straight lines 3.Molecules of a gas are tiny compared with distances between gas molecules 4.There is little attraction between molecules of a gas
Chapter 648 5.Molecules collide with each other, with energy being conserved in the collision 6.Temperature (T) is a measure of the average kinetic energy of the gas molecules
Chapter 649 Boyles Law At constant T, volume (V) of a gas varies inversely with its pressure (P) V 1/P Or PV = a
Chapter 655 Example 6.3 What volume of oxygen is required to burn 0.556 L of propane C 3 H 8 if both gases are measured at the same temperature and pressure? C 3 H 8 (g) + 5 O 2 (g)3 CO 2 (g) + 4 H 2 O(g) Solution The coefficients in the equation indicate that five volumes of O 2 (g) is required for every volume of C 3 H 8 (g). Thus, we use 5 L O 2 (g)/1 L C 3 H 8 (g) as the ratio to find the volume of oxygen required. Exercise 6.3 Using the equation in Example 6.3, calculate the volume of CO 2 (g) produced when 0.492 L of propane is burned if the two gases are compared at the same temperature and pressure. = 2.78 L O 2 (g) 5 L O 2 (g) 1LC 3 H 8 (g) ? L O 2 (g) = 0.556 L C 3 H 8 (g) x
Chapter 656 Example 6.8 Calculate the density of (a) nitrogen gas and (b) oxygen gas, both at STP. Calculate the density of He at STP. Exercise 6.8A Estimate the density of air at STP (assume 78% N 2 and 22% O 2 ) and compare this value to the value of He you calculated in part (a). Exercise 6.8B b. Similarly for oxygen gas: 32.0 g/ mol 22.4 L/ mol = 1.43 g/L Solution a. At STP, 1 mol (28.0 g) of N 2 occupies 22.4 L: 28.0 g/ mol 22.4 L/ mol = 1.25 g/L
Chapter 657 Example 6.9 The density of diethyl ether vapor at STP is 3.30 g/L. Calculate the molar mass of diethyl ether. Solution To solve, we simply multiply the density by the molar volume. The units of liters cancel and the grams per mole, the units of molar mass, remain. 30.3 g 22.4 L 1 L1 mol = 73.9 g/mol x The density of an unknown gas at STP is 2.30 g/L. Calculate its molar mass. Exercise 6.9A An unknown gaseous compound contains only hydrogen and carbon and its density is 1.34 g/mol at STP. What is the formula for the compound? Exercise 6.9B
Chapter 658 Example 6.13 (cont.) b. Phosphorus reacts with oxygen to form tetraphosphorus decoxide. The equation is P 4 + O 2 P 4 O 10 (not balanced) How many grams of tetraphosphorus decoxide can be made from 3.50 g of phosphorus?
Chapter 659 Example 6.14 A gas is enclosed in a cylinder fitted with a piston. The volume of the gas is 2.00 L at 0.524 atm. The piston is moved to increase the gas pressure to 5.15 atm. Which of the following is a reasonable value for the volume of the gas at the greater pressure? 0.20 L 0.40 L1.00 L16.0 L Solution The pressure increase is almost 10-fold. The volume should drop to about one-tenth of the initial value. We estimate a volume of 0.20 L. (The calculated value is 0.203 L.) Exercise 6.14 A gas is enclosed in a 10.2-L tank at 1208 mmHg. Which of the following is a reasonable value for the pressure when the gas is transferred to a 30.0-L tank? 25 lb/in. 2 300 mmHg400 mmHg3600 mmHg
Chapter 660 Example 6.15 A cylinder of oxygen has a volume of 2.25 L. The pressure of the gas is 1470 pounds per square inch (psi) at 20 ° C. What volume will the oxygen occupy at standard atmospheric pressure (14.7 psi) assuming no temperature change? Solution It is most helpful to first separate the initial from the final condition: Because the final pressure in Example 6.15 is less than the initial pressure, we expect the final volume to be larger than the original volume, and it is. Then use the equation V 1 P 1 = V 2 P 2 and solve for the desired volume or pressure. In this case, we solve for V 2 : V1P1V1P1 V 2 = P2P2 2.25 L x 1470 psi V 2 = 14.7 psi = 225 L
Chapter 661 Example 6.15 (cont.) Exercise 6.15A A sample of air occupies 73.3 mL at 98.7 atm and 0 ° C. What volume will the air occupy at 4.02 atm and 0 ° C? Exercise 6.15B A sample of helium occupies 535 mL at 988 mmHg and 25 ° C. If the sample is transferred to a 1.05-L flask at 25 ° C, what will be the gas pressure in the flask?
Chapter 662 Example 6.16 A balloon indoors, where the temperature is 27 ° C, has a volume of 2.00 L. What would its volume be (a) outdoors, where the temperature is –23°C and (b) in a hot car parked in the sun where the temperature is 47 ° C? (Assume no change in pressure in either case.) Solution First, convert all temperatures to the Kelvin scale: T(K) = T ( ° C) + 273 The initial temperature is (27 + 273) = 300 K, and the final temperatures are (a) ( – 23 + 273) = 250 K and (b) (47 + 273) = 320 K. a. We start by separating the initial from the final condition: Solving the equation V1V1 T1T1 V2V2 T2T2 =
Chapter 663 Example 6.16 (cont.) for V 2,we have = 1.67 L V1T2V1T2 T1T1 V2 =V2 = V2 =V2 = 2.00 L x 250K 300 K As we expected, the volume decreased because the temperature decreased. b. We have the same initial conditions as in (a), but different final conditions: Exercise 6.16A a. A sample of oxygen gas occupies a volume of 2.10 L at 25°C. What volume will this sample occupy at 150°C? (Assume no change in pressure.) In this case, since the temperature increases, the volume must also increase: V1T2V1T2 T1T1 V2 =V2 = V2 =V2 = 2.00 L x 320K 300 K = 2.13 L
Chapter 664 b. A sample of hydrogen occupies 692 L at 602 ° C. If the pressure is held constant, what volume will the gas occupy after being cooled to 23 ° C? Example 6.16 (cont.) Exercise 6.16B At what Celsius temperature will the initial volume of oxygen in Exercise 6.16A occupy 0.750 L? (Assume no change in pressure.)
Chapter 665 Example 6.17 Use the ideal gas law to calculate (a) the volume occupied by 1.00 mol of nitrogen gas at 244 K and 1.00 atm pressure, and (b) the pressure exerted by 0.500 mol of oxygen in a 15.0-L container at 303 K. Exercise 6.17A Determine (a) the pressure exerted by 0.0330 mol of oxygen in an 18.0-L container at 40 ° C, and (b) the volume occupied by 0.200 mol of nitrogen gas at 25 ° C and 0.980 atm. Exercise 6.17B Determine the volume of nitrogen gas produced from the decomposition of 130 g sodium azide (about the amount in a typical automobile air bag) at 25°C and 1 atm. Solution a. We start by solving the ideal gas equation for V: V = nRT P V = 1.00 mol 1.00 atm 0.0821 L atm mol K x x 244 K = 20.0 L b. Here we solve the ideal gas equation for P: P = nRT V P = 0.500 mol 15.0 L 0.0821 L atm mol K x 303 K = 0.829 atm x