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Joseph Gauthier, Homework 1 Due: Sep 6 2007, midnight Inst: Vandenbout This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. This homework has problems that should review what you have already learned in high school or at least previous to our course. You will find most of this information in the first 4 chapters of your textbook. 001 (part 1 of 1) 10 points What is the symbol for the element silver? 1. Au 2. Ag correct 3. Su 4. Si Explanation: The symbol for silver is Ag. 002 (part 1 of 1) 10 points A pure substance composed of two or more elements chemically combined is 1. a solution. 2. an atom. 3. a compound. correct 4. a mixture. Explanation: A compound is a substance composed of two or more elements chemically combined. 003 (part 1 of 1) 10 points A factor of 1.0 10-6 corresponds to the prefix 1. milli. 2. nano. 3. mega. 4. micro. correct Explanation: Micro refers to 1.0 10-6 . 004 (part 1 of 1) 10 points What is 0.00875 in scientific notation? 1. 8.8 10-3 2. 8.75 103 3. 8.75 10-3 correct 4. 0.0875 101 1 Explanation: 0.00875 = 8.75 10-3 (move decimal 3 places to right) 005 (part 1 of 1) 10 points What is the proper solution to the following equation? Answer with correct significant figures. (23.1 + 5.61 + 1.008) 7.6134 8.431 = 1. 1908 2. 1907.566 3. 1907.5661 4. 2000 5. 1907 6. 1900 7. 1907.6 8. 1910 correct 9. 1907.56 Explanation: For the addition, the position of the greatest magnitude is the tenths 23.1 so the solution to this part will have 3 significant figures. 29.7 7.6134 8.431 = 1907.6 Gauthier, Joseph Homework 1 Due: Sep 6 2007, midnight Inst: Vandenbout The smallest number of sigfigs is 3, so the answer is rounded to 1910. 006 (part 1 of 1) 10 points Find the volume of the Hope Diamond given the density of diamond (carbon) is 3.51 g/cm3 and the weight is 44 karats. (1 karat = 0.200 g) 1. 772.00 cm3 2. 31 cm3 3. 63 cm3 4. 2.5 cm3 correct Explanation: density = 3.51 g/cm3 mdiamond = 44 karat = 8.8 g V = m = 44 karat 0.200 g 1 karat 3. 10 protons, 14 electrons 4. 12 protons, 14 electrons 5. 14 protons, 12 electrons 6. 12 protons, 12 electrons 7. 10 protons, 10 electrons 8. 14 protons, 14 electrons 9. 14 protons, 10 electrons 2 Explanation: Mg has atomic number 12; it has 12 protons and 12 electrons; Mg2+ has only 10 electrons. 009 (part 1 of 1) 10 points Which element does NOT exist as a diatomic gas at room temperature and atmospheric pressure? 1. N m 8.8 g = = 2.5 cm3 3 density 3.51 g/cm 007 (part 1 of 1) 10 points What is the formula of aluminum carbonate? 1. Al3 (CO3 )2 2. Al2 (CO3 )3 correct 3. Al2 (CO3 )5 4. AlCO3 5. Al2 (CO2 )3 Explanation: Aluminum forms Al3+ ions and the carbonate ion is CO2- , giving a formula of 3 Al2 (CO3 )3 . 008 (part 1 of 1) 10 points How many protons and electrons are present in the Mg2+ ion? 1. 12 protons, 10 electrons correct 2. 10 protons, 12 electrons 2. Ar correct 3. H 4. Cl 5. O Explanation: The diatomic gases at room temperature and atmospheric pressure are H2 , N2 , O2 , F2 , and Cl2 . 010 (part 1 of 1) 10 points Which of the following would be expected to have chemical properties similar to those of sulfur? 1. F 2. Ar 3. Si Gauthier, Joseph Homework 1 Due: Sep 6 2007, midnight Inst: Vandenbout 4. Mg 5. P 6. Se correct Explanation: 011 (part 1 of 1) 10 points How many neutrons are there in one atom of 48 22 Ti? 1. 22 2. 70 3. 48 4. 26 correct 5. None of these Explanation: The standard nuclide symbol takes the form A Z E, where E is the element symbol, A is the mass number (number of n + number of p+ ) and Z is the atomic number (number of p+ ). The atomic number for 48 Ti is 22, and its 22 mass number is 48. 48 = p+ + n = 22 + n n = 48 - 22 = 26 neutrons 012 (part 1 of 1) 10 points Which of the following is FALSE? 1. The mass of neutrons and electrons are about the same. correct 2. The mass of protons and neutrons are about the same. 3. The mass of a hydrogen atom is about the same as a proton. 4. The mass of a proton is much greater than the mass of an electron. Explanation: A proton weighs 1 amu, as does a neutron. 1 amu. Hydrogen An electron weighs 2000 3 has one proton, one electron, and usually no neutrons, so H weighs about 1 amu. 013 (part 1 of 1) 10 points Which has more molecules: a mole of CO2 or a mole of H2 ? 1. a mole of CO2 2. They both have the same number of molecules. correct 3. It depends on the pressure and temperature. 4. a mole of H2 Explanation: 014 (part 1 of 1) 10 points Which of the following would contain the largest number of atoms? 1. All contain the same number of atoms. 2. 1.00 gram of beryllium (Be) correct 3. 1.00 gram of uranium (U) 4. 1.00 gram of iron (Fe) 5. 1.00 gram of sulfur (S) Explanation: Be has the smallest atomic weight and U has the largest atomic weight. The one with the smallest atomic weight would have the most moles per 1.0 gram. It would therefore also have the most atoms per 1.0 gram. The one with the largest atomic weight would have the fewest moles per 1.0 gram. It would therefore also have the fewest atoms per 1.0 gram: ? Be atoms = 1.0 g Be 1 mol Be 9.0 g Be 6.02 1023 Be atoms 1 mol Be = 6.7 1022 Be atoms Gauthier, Joseph Homework 1 Due: Sep 6 2007, midnight Inst: Vandenbout 015 (part 1 of 1) 10 points What is the formula weight of lithium phosphate? 1. 101.9 2. 108.9 3. 112.7 4. 115.8 correct Explanation: Lithium ions have a charge of +1; phosphate ions have a charge of -3. The formula for lithium phosphate is Li3 PO4 . The formula weight of a compound is based on the atomic weights of the elements in the compound. Each mole of Li3 PO4 contains 3 mol of Li, 1 mol of P, and 4 mol of O. We know the atomic masses of each of these elements from the periodic table. Using these atomic masses we calculate the grams each of of these elements in one mole of Li3 PO4 : 6.94 g Li ? g from Li = 3 mol Li 1 mol Li = 20.82 g Li 30.97 g P ? g from P = 1 mol P 1 mol P = 30.97 g P 16 g O ? g from O = 4 mol O 1 mol O = 64 g O To get the formula mass of Li3 PO4 we add the masses of the component parts: ? g/mol Li3 PO4 = 20.82 g + 30.97 g + 64 g 115.79 g Li3 PO4 = mol Li3 PO4 016 (part 1 of 1) 10 points A compound is 69.9% iron and 30.1% oxygen by mass. How many moles of iron could by obtained from 5.41 grams of the compound? 1. 211 moles 2. 0.0969 moles 3. 0.0292 moles 4. 0.865 moles 5. 0.0677 moles correct 6. 3.78 moles Explanation: X g Fe 69.9 g Fe = 100 g compound 5.41 g compound (69.9 g Fe) (5.41 g compound) (100 g compound) = 3.7815 g Fe 4 X g Fe = ? mol Fe = 3.7815 g Fe = 0.06770 mol Fe 1 mol Fe 55.85 g Fe 017 (part 1 of 1) 10 points Determine the percentage by weight of Na in Na2 SO4 . 1. 11.2638 % 2. 32.3704 % correct 3. 22.5743 % 4. 45.0554 % 5. 16.1852 % Explanation: part 100% whole In this case, the part would be the mass of Na in a certain mass of Na2 SO4 (the whole). The simplest amount of Na2 SO4 to use is the mass of one mole of Na2 SO4 . Percent = Gauthier, Joseph Homework 1 Due: Sep 6 2007, midnight Inst: Vandenbout Each mole of Na2 SO4 contains 2 moles of Na, 1 mole of S, and 4 moles of O. We know the atomic masses of each of these elements from the periodic table. Using these atomic masses we calculate the grams of each element in one mole of Na2 SO4 : 22.9898 g Na 1 mol Na 3.2 g For 3.1 mol O: 3.1 mol O 5 ? grams from Na = 2 moles Na = 45.9795 , ? grams from S = 1 mole S 32.065 g S 1 mol S = 32.065 , and 15.9994 g O ? grams from O = 4 moles O 1 mol O = 63.9976 . To find the mass of one mole of Na2 SO4 we add the masses of the component parts: molar mass Na2 SO4 = 45.9795 + 32.065 + 63.9976 142.042 g Na2 SO4 = , mol Na2 SO4 so the percentage of Na in Na2 SO4 is 45.9795 g Na 100% 142.042 g Na2 SO4 = 32.3704 % . 16 g = 49.6 g 1 mol O For 4.23 1023 molec O2 : 4.23 1023 O2 molec 1 mol O2 mol 6.022 1023 O2 molec 32 g = 22.5 g 1 mol O2 3.1 moles of O atoms has the greatest mass. 019 (part 1 of 1) 10 points Calculate the number of carbon atoms in 6.50 grams of ethane (C2 H6 ). 1. 1.30 1023 carbon atoms 2. 3.01 1023 carbon atoms 3. 2.61 1023 carbon atoms correct 4. 2.78 1024 carbon atoms 5. 1.17 1024 carbon atoms Explanation: m = 6.50 g To solve this problem, we will need both the molecular mass of ethane (30.0 g/mol) and Avogadro's number (6.022 1023 ). First find the number of moles of ethane using the molecular mass: ? mol C2 H6 = 6.50 g C2 H6 1 mol C2 H6 30.0 g C2 H6 = 0.2167 mol C2 H6 . Percent Na = 018 (part 1 of 1) 10 points Which one has the greatest mass? 1. 4.23 1023 O2 molec 2. 3.2 10-3 kg of O2 molec 3. 3.1 moles of O atoms correct 4. 3.2 g of O atoms Explanation: Convert all to g or kg and compare: For 3.2 10-3 kg O2 : 1000 g = 3.2 g 3.2 10-3 kg O2 1 kg For 3.2 g O: Now find the number of ethane molecules using Avogadro's number: nC2 H6 = 0.2167 mol C2 H6 6.022 1023 molec 1 mol C2 H6 = 1.305 1023 molec C2 H6 . You know from the formula C2 H6 that there are 2 carbon atoms per 1 ethane molecule so Gauthier, Joseph Homework 1 Due: Sep 6 2007, midnight Inst: Vandenbout that nC = 1.305 1023 molec C2 H6 2 atoms C 1 molec ethane the coefficient of H2 O is 1. 1. 2. 2. correct 3. 5. 4. 4. 6 = 2.61 1023 atoms C. 020 (part 1 of 1) 10 points Saccharin, a sweeting agent, has mass compostion 45.89% C, 2.75% H, 7.65% N, 26.20% O, and 17.50% S. What is the empirical formula of saccharin? 1. C5 H5 NO2 S 2. CH4 NO3 S 3. C7 H5 NO3 S correct 4. C6 HNO3 S 5. C6 H3 NO2 S Explanation: A balanced equation has the same number of each kind of atom on both sides of the equation. We find the number of each kind of atom using equation coefficients and composition stoichiometry. For example, we find there are 4 H atoms on the product side: 2H =4H 1 H2 O ? H atoms = 2 H2 O The balanced equation is C + 4 HNO3 CO2 + 4 NO2 + 2 H2 O, 6. C7 H5 NOS Explanation: For 100 g of saccharin, 45.89 g moles of C = = 3.821 mol 12.01 g/mol 2.75 g moles of H = = 2.728 mol 1.0079 g/mol 7.65 g moles of N = = 0.546 mol 14.01 g/mol 26.20 g moles of O = = 1.638 mol 16.0 g/mol 17.50 g moles of S = = 0.5459 mol 32.06 g/mol Dividing by 0.546 mol gives a ratio of 7.0 C : 5.0 H : 1.0 N : 3.0 O : 1.0 S. The empirical formula is C7 H5 NO3 S. 021 (part 1 of 1) 10 points Using the smallest possible integer coefficients to balance the equation C + HNO3 CO2 + NO2 + H2 O , and the coefficient of H2 O is 2. 022 (part 1 of 1) 10 points Iron reacts with Cl2 according to the equation 2 Fe(s) + 3 Cl2 (g) 2 FeCl3 (s) . How many moles of Cl2 is needed to react with 4.4 mol Fe? 1. 3.0 2. 4.4 3. 3.3 4. 6.6 correct Explanation: nFe = 4.4 mol The balanced equation for the reaction indicates that 3 mol Cl2 are needed for every 2 mol Fe reacted. Using this mole ratio we find Gauthier, Joseph Homework 1 Due: Sep 6 2007, midnight Inst: Vandenbout the moles Cl2 needed to completely react 4.4 mol Fe: 3 mol Cl2 2 mol Fe 7 ? mol Cl2 = 4.4 mol Fe = 6.6 mol Cl2 023 (part 1 of 1) 10 points For the reaction ? C6 H6 + ? O2 ? CO2 + ? H2 O 27.7 grams of C6 H6 are allowed to react with 97.02 grams of O2 . How much CO2 will be produced by this reaction? Correct answer: 93.64 grams. Explanation: mC6 H6 = 27.7 g mO2 = 97.02 g The balanced equation is 2 C6 H6 + 15 O2 12 CO2 + 6 H2 O . FW of C6 H6 is 78.1118 g/mol, giving 0.35462 mol C6 H6 . FW of O2 is 31.9988 g/mol, giving 3.03199 mol O2 . 15 = 2.65965 mol 0.35462 mol C6 H6 2 O2 needed which is less than what is actually present. Therefore the limiting reactant must be C6 H6 . FW of CO2 is 44.0095 g/mol The limiting reactant (C6 H6 ) will yield 2.12772 mol CO2 which is equal to 93.64 g of CO2 .

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