Chemistry 142A, Autumn 2011 Worksheet #3 Answer Key Ch. 3, start of Ch. 4

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1 Chemistry 142A, Autumn 2011 Worksheet #3 Answer Key Ch. 3, start of Ch. 4 1) There are several important compounds that contain only nitrogen and oxygen. Calculate the mass percent of nitrogen in each of the following. (Do (b) only.) a) NO, a gas formed by the reaction of N 2 and O 2 in internal combustion engines. b) NO 2, a brown gas mainly responsible for the brownish color of photochemical smog. c) N 2 O 4, a colorless liquid used as a fuel in space shuttles. d) N 2 O, a colorless gas used as an anesthetic by dentists (known as laughing gas). 2) Give the empirical formula for each of these compounds. Do(b) only. 1

2 a) Blue = Nitrogen, Red = Oxygen. We have 2N and 4O. Reducing to the simplest whole-number ratio, we get 2O for each 1N. The empirical formula is therefore NO 2. b) Black = Carbon, White = Hydrogen. We have 3C and 6H. Reducing to the simplest whole-number ratio, we get 2H for each 1C. The empirical formula is therefore CH 2. c) Yellow = Phosphorus, Red = Oxygen. We have 4P and 10O. Reducing to the simplest whole-number ratio, we get 5O for each 2P. The empirical formula is therefore P 2 O 5. d) Black = Carbon, Red = O, White = H. We have 6C, 6O, and 12H. Reducing to the simplest whole-number ratio, we get 2H for every 1C or 1O. The empirical formula is therefore CH 2 O. 3) Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticizers. It contains only C, H, and O. Combustion of mg terephthalic acid produces mg CO 2 and 6.45 mg H 2 O. If mol of terephthalic acid has a mass of 41.5 g, determine the molecular formula for terphthalic acid. We are told that every molecule of terephthalic acid contains C, H, and O. We are also told that when mg of this compound is combusted, the products are mg of CO 2 and 6.45 mg of H 2 O. Finally, we are told that mol of terephthalic acid has a mass of 41.5 g. The first step in tackling this problem is to realize that if we can deduce what happens during combustion, we can use the known values of the combustion products to figure out the make-up of the starting materials. Combustion occurs when a compound reacts with gaseous oxygen (this is what happens when something burns). For the purposes of this problem, we should assume that the terephthalic acid is combusted completely, and that the only combustion products are the ones mentioned. We should also assume that the acid is combusted in excess oxygen. Terephthalic Acid + O 2 CO 2 + H 2 O From inspection of this reaction, it should be clear that the only source of carbon and hydrogen is the acid. Because we know the masses of CO 2 and H 2 O produced, and because we can figure out the molar mass of these two compounds, we can figure out both the mass and number of moles of C and H that were combusted. First, obtain atomic masses of C, H, and O from the Periodic Table. Carbon: g/mol Hydrogen: g/mol Oxygen: g/mol Using these values we can calculate the molar masses of CO 2 and H 2 O. CO 2 : g/mol H 2 O: g/mol Using these values, we can determine the mass and number of moles of C and H combusted. 2

3 C: H: Based on this, we know that in the original mg sample of terephthalic acid, mg of the sample is carbon and mg of the sample is hydrogen. That leaves 7.63 mg that has to be oxygen. Now that we know how many moles of C, H, and O were in the original sample, we can determine the empirical formula for terephthalic acid. To do this, we divide the number of moles of each element by the number of moles of the smallest quantity. NOTE: There is a typo above: Should be 7.16x10-4 mol H on top left, not g H! But the H/O ratio remains very close to Based on these results, we know that there are two moles of carbon for every mole of oxygen, and 1.5 moles of hydrogen for every mole of oxygen. Multiplying all factors by 2 in order to get whole number ratios, we can determine that the empirical formula is C 4 H 3 O 2. We are not finished, however, because the empirical formula may not equal the molecular formula. If this were, indeed, the molecular formula, we can determine via the periodic table that the molar mass of this compound would be g/mol. Using the values given to us in the problem statement relating the mass of a certain number of moles of this compound, we can directly calculate the actual molar mass of terephthalic acid. This is approximately twice as much as the value of the molar mass if the empirical formula were also the molecular formula. We therefore know that the molecular formula of terephthalic acid is C 8 H 6 O 4. 3

4 4). Give the balanced equation for each of the following. Use the lowest possible coefficients. Do (a) only. a) The combustion of ethanol (C 2 H 5 OH) forms carbon dioxide and water vapor. A combustion reaction refers to a reaction of a substance with oxygen gas. C 2 H 5 OH + O 2 CO 2 + H 2 O Balance C first: C 2 H 5 OH + O 2 2CO 2 + H 2 O Balance H next: C 2 H 5 OH + O 2 2CO 2 + 3H 2 O Balance O next: C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 O b) Aqueous solutions of lead(ii) nitrate and sodium phosphate are mixed, resulting in the precipitate formation of lead(ii) phosphate with aqueous sodium nitrate as the other product. Based on knowledge of ionic charges, we know that Lead(II) will have a +2 charge, Sodium will have a +1 charge, Nitrate will have a -1 charge, and Phosphate will have a -3 charge. Using these charges, we can determine the molecular formula for each compound involved in the reaction. Lead(II) Nitrate = Pb(NO 3 ) 2 Sodium Phosphate = Na 3 PO 4 Lead(II) Phosphate = Pb 3 (PO 4 ) 2 Sodium Phosphate = NaNO 3 The unbalanced reaction is therefore: Pb(NO 3 ) 2(aq) + Na 3 PO 4(aq) Pb 3 (PO 4 ) 2(aq) + NaNO 3(aq) Balance Pb first: 3Pb(NO 3 ) 2(aq) + Na 3 PO 4(aq) Pb 3 (PO 4 ) 2(aq) + NaNO 3(aq) Balance Nitrate next: 3Pb(NO 3 ) 2(aq) + Na 3 PO 4(aq) Pb 3 (PO 4 ) 2(aq) + 6NaNO 3(aq) Balance Na next: 3Pb(NO 3 ) 2(aq) + 2Na 3 PO 4(aq) Pb 3 (PO 4 ) 2(aq) + 6NaNO 3(aq) And Phosphate appears to be balanced. Make sure answers are in alphabetical order for WebAssign. 2Na 3 PO 4(aq) + 3Pb(NO 3 ) 2(aq) 6NaNO 3(aq) + Pb 3 (PO 4 ) 2(aq) 5). Balance the following equations. (Do (c) only.) a) 16Cr + 3S 8 8Cr 2 S 3 b) 2NaHCO 3 Na 2 CO 3 +CO 2 + H 2 O c) 2KClO 3 2KCl + 3O 2 d) 2Eu + 6HF 2EuF 3 + 3H 2 e) 2C 6 H O 2 12CO 2 + 6H 2 O 4

5 6). Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. What masses of iron(iii) oxide and aluminum must be used to produce 15.0 g of iron? What is the maximum mass of aluminum oxide that could be produced. Fe 2 O 3 + 2Al 2Fe + Al 2 O 3 Begin by determining the molar mass of each compound involved in the reaction. Using atomic masses from the periodic table, we will find the following: Fe 2 O 3 : g/mol Al: g/mol Fe: g/mol Al 2 O 3 : g/mol We are given a target product of 15.0 g of iron. Using this, determine the minimum masses of reactants used and the maximum mass of the other product in the reaction. Fe 2 O 3 : Al: Al 2 O 3 : 7). DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction. 2 C 6 H 5 Cl + C 2 HOCl 3 C 14 H 9 Cl 5 + H 2 O In a government lab, 1142 g of chlorobenzene is reacted with 485 g of chloral. a) What mass of DDT is formed? b) Which reactant is limiting? Which is in excess? c) What mass of the excess reactant is left over? d) If the actual yield of DDT is g, what is the percent yield? Begin by determining the molar masses of each compound by using atomic masses found on the periodic table. C 6 H 5 Cl: g/mol C 2 HOCl 3 : g/mol C 14 H 9 Cl 5 : g/mol We must figure out which reactant is limiting. C 6 H 5 Cl: C 2 HOCl 3 : 5

6 Chloral is therefore the limiting reactant, while Chlorobenzene is in excess. Assuming all of the Chloral will react, while some Chlorobenzene will be left over, we can determine the excess amount of Chlorobenzene as well as the mass of DDT formed. Finally, we are asked to calculate the % yield if only g of DDT is formed g / 1170 g * 100% = 17.1% yield a) 1170 g DDT b) Chloral is limiting, while Chlorobenzene is in excess. c) 401 g Chlorobenzene will be left over. d) 17.1% yield 8). The empirical formula of styrene is CH; the molar mass of styrene is g/mol. What is the molecular formula of styrene? If the molecular formula of styrene were the same as the empirical formula (CH), then the molar mass would be g/mol. The ratio of this value and the given molar mass is: g/mol / g/mol = ~ 8 Therefore, the molecular formula must have subscripts that are all larger than CH by a factor or 8: C 8 H 8. 9). What is the molarity of the following solution? (Do (a) only.) a mol KBr in 500. ml H 2 O 500. ml * 1 L/1000mL = L mol/0.500 L = M b. 40. grams of HF in 10.0 L H 2 O 40. g /( g/mol HF) = 2.0 mol 2.0 mol / 10.0 L = 0.20 M 10). What is the total concentration of all ions in a 0.3 M solution of Na 2 SO 4 (aq)? Na 2 SO 4 (aq) = 2 Na + (aq) + SO 4 2- (aq) 2 Na + * 0.3 M = 0.6 M Na +, 1 SO 4 2- * 0.3 M = 0.3 M SO 4 2- Sum these to get TOTAL concentration of all ions: 0.6 M M = 0.9 M 11). How much of a 2.0 M solution would you dilute to make ml of a 0.50 M solution? moles before = moles after so, M 1 V 1 = M 2 V M * V 1 = ml* 0.50 M V 1 = (100.0 ml * 0.50 M)/2.0 M V 1 = 25. ml (2 sig figs) 6

7 12). Consider the balanced reaction below. Ba(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2 NaNO 3 (aq) You add 1.0 L of 1.0 M barium nitrate solution to 1.0 L of 0.5 M sodium sulfate solution. How many moles of the precipitate will you make? 1.0 M * 1.0 L = 1.0 mol Ba(NO 3 ) mol Ba(NO 3 ) 2 / 1 = 1.0 eq. Ba(NO 3 ) M * 1.0 L = 0.5 mol Na 2 SO 4 (0.5 mol Na 2 SO 4 is really 1. mol Na + and 0.5 mol SO 4 2- ) 0.5 mol Na 2 SO 4 / 1 = 0.5 eq. Na 2 SO 4, which is less, so this is the limiting reagent! 0.5 mol Na 2 SO 4 * (1 mol BaSO 4 /1 mol Na 2 SO 4 ) = 0.5 mol BaSO 4 (s) 7

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