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209 Chemistry Chapter 3: Homework #3 answers 14 P4 units have molecular weight 4 30.9738 = 123.8952. The density of the P4 unit is given as 1.823 g/cm3 and we are given a cylinder 6.50 cm in length and 1.22 cm in diameter. The volume of a cylinder is given by d2 V = r2 h = h 4 2 3 for our sample V = 3.14159 0.61 6.50 cm = 7.598 cm3 ; and given the density d = 1.823 we know that we have a total mass of P4 m = V d = 7.598 cm3 1.823 g = 13.852 g cm3 (a) Given the molecular weight calculated above, that corresponds to n moles of P4 given by n = 13.852 grams sample 1 mole = 0.1118 moles P4 123.8952 g (b) This one is easy; one mole contains NA = 6.0221 1023 molecules; so 0.1118 moles contains 0.1118 6.0221 1023 = 6.733 1022 P4 molecules. 18 Given the compound Ge[S(CH2 )4 CH3 ]4 , we are asked to figure out what we have. First, let's start with a picture of the molecule, or at least an artist's conception of what might be the molecule: CH3 CH2 CH2 CH2 CH2 S | CH3 CH2 CH2 CH2 CH2 S - Ge - SCH2 CH2 CH2 CH2 CH3 | SCH2 CH2 CH2 CH2 CH3 (a) How many atoms are found in one formula unit? Well, there are four equivalent groupings of atoms, each containing 1 S atom, 4 CH2 groups, and a single CH3 group. By my count that corresponds to 1 + 4 3 + 4 = 17 atoms in each supgrouping; there are four of them, plus the central Ge atom, fo I count a total of 69 atoms per formula unit. (b) How many C and/or H atoms? They are found in CH2 and CH3 groups; there are 4 4 of the former, and 4 1 of the latter. So the total count of carbon atoms is 20; the total count of hydrogen atoms is 44; and the ratio of C atoms to H atoms is 20:44; or in simplified numbers, 5:11. (c) The ratio by mass of Ge to S; each Ge weighs 72.61 (on average, assuming that we have a natural abundance distribution of isotopes and that there are sufficient atoms present so that the average reasonably represents any distribution; and there are 4 S atoms per Ge, each of which weighs (again, on average) 32.066. So the ratio of Ge mass to S mass is 72.61 : (4 32.066) which is about 0.5661:1. (d) 1 mole of the compound contains 4 moles of S atoms, each of which contains 32.066 g of sulfur atoms. So 1 mole of the compound contains 4 moles S atoms 32.066 g/mole = 128.264 g sulfur 1 (e) We are given 33.10 grams of sample; we should calculate how many moles of sample we have which requires that we find the formula weight of the compound. That requires that we do the detailed countings of atoms; that's done above; and the formula weight is therefore 1 72.61 + 4 32.066 + 20 12.011 + 44 1.00794 = 485.444. 33.10 grams contains 33.10 g 1 mole/485.444 g = 6.8185 10-2 moles of our compound; each mole contains 20 C atoms, so the number of C atoms in our 33.10 gram sample is 6.8185 10-2 moles 20 atoms C/mole 6.0221 1023 = 8.212 1023 atoms of C 32 The analyzed % by mass of MSG is 13.6% Na, 35.5% C, 4.8% H, 8.3% N, and 37.8% O. We require the empirical formula for this material; that requires that we translate from mass units to mole units. Choose a representative sample of, say, 100 g MSG (this could be done for any size sample, we will ultimately ignore the absolute quantities anyway) and find 13.6 g Na 1 mole Na = 0.5916 moles Na 22.9898 g Na 1 mole C = 2.9566 moles C 12.011 g C 1 mole H = 4.7622 moles H 4.8 g H 1.00794 g H 1 mole Na = 0.5926 moles N 8.3 g N 14.0067 g N 1 mole Na = 2.3626 moles O 37.8 g O 15.9994 g O We wish to find an empirical formula; let's work on the expectation that there are one unit of both Na and N in the empirical formula, so the other atoms are found in quantities that are likely to be integer multiples of the number of moles of Na found in our sample; for example, if there is a single Na atom in our compound represented by the 0.5916 moles calculated above, then the 2.9566 moles of C in our sample represent 2.9566/0.5916 = 5.00 moles of C; similarly, we have 8.05 moles of H and 3.99 moles of O. We're not quite done; remember that (1) there are only 2 or 3 significant figures in the measurements provided, and (2) there is a preference strong for integer numbers of atoms in the empirical formula (or, perhaps, halves or thirds indicating that the formula contains multiple atoms of what we thought was the smallest contributor. So our empirical formula is likely to be NaC4 H8 NO4 ; the ordering of the atoms is largely irrelevant. 35.5 g C 46 1.3020 g of thiophene is burned, and yields 2.7224 g CO2 ; 0.5575 g H2 O, and 0.9915 g SO2 . Each molecule of CO2 represents one C atom that was found in thiophene; similarly, each molecule of SO2 represents one S atom in thiophene. Each molecule of H2 O found represents two H atoms that were originally in the thiophene; after we have accounted for all the atoms of C, H and S the remaining mass in the thiophene sample must represent O atoms that found their way to one of our products. So let's count moles; first, of C in the thiophene: 2.7224 g CO2 0.5575 g H2 O 1 mole CO2 1 mole C = 6.186 10-2 moles C 44.0103 g CO2 1 mole CO2 2 mole H 1 mole H2 O = 6.189 10-2 moles H 18.01528 g H2 O 1 mole H2 O 2 1 mole SO2 1 mole S = 1.548 10-2 moles S 64.0648 g SO2 1 mole SO2 This would appear to represent a molecule with the empirical formula C4 H4 SOn ; where n is as yet unknown; and we have 1.548 10-2 moles of the molecule. If 1.3020 g thiophene represents 1.548 10-2 moles of thiophene, then the formula weight of the molecule would appear to be 1.3020 g/1.548 10-2 = 84.13 g/mol; so far, our molecular fragment weighs 4 12.0115 + 4 1.00794 + 32.066 = 84.14 g/mol; therefore, there is no oxygen to worry about! Thiophene has the empirical formula C4 H4 S. 0.9915 g SO2 52 (a) Net charge on the unit is -2; each O atom is assigned an oxidation state of -2, so the total assigned to the 3 oxygen atoms is -6; that leaves the S with an oxidation state of +4 (as the sum of all the oxidation states gives the net charge; +4 + (-6) = -2). (b) Net charge is -2; each O atom is assigned an oxidation state of -2, so the two S atoms have to total +4, and each S atom is assigned an oxidation state +2. (c) Net charge is -2; each O atom is assigned an oxidation state of -2, so the two S atoms have to total +14 between them, and each S atom is assigned an oxidation state +7. (d) Net charge is -1; each O atom is assigned an oxidation state of -2, and the H atom is normally assigned +1 (unless attached to a metal atom). So the S atom must account for +6 units of charge, which is its oxidation state. (e) Net charge is -2; each O atom is assigned an oxidation state of -2, so the four S atoms have to total +10 between them, and each S atom is assigned an oxidation state +2.5. A little peculiar, we'll suggest what this might mean at some later date. 80 Note that we must calculate the Cu content by difference; it isn't measured directly in our experiment. We need to calculate the moles of each of the other metals found in the various salts, and convert those mole quantities back to masses so that we can account for the total of 1.1713 g of sample. SnO2 has formula weight 150.7088; 0.245 grams of the tin salt (stannic oxide, or tin (IV) oxide) represent 1.626 10-3 moles of SnO2 , which represents 1.626 10-3 moles 118.710 g/mole = 0.1930 g Sn metal. PbSO4 has formula weight 303.28; 0.115 grams of the lead salt (plumbous sulphate, or lead (II) sulphate) represent 0.115 g 1 mole/303.28 g PbSO4 = 3.792 10-4 moles of PbSO4 , which represents 3.792 10-4 moles 207.2 g/mole = 0.07857 g Pb metal. Zn2 P2 O7 has formula weight 304.7234 g. Our 0.246 g sample of the salt corresponds to 0.246 g 1 mole Zn salt/304.7234 g Zn salt = 8.073 10-4 moles; that represents 8.073 10-4 moles 2 moles Zn/mole salt 65.39 g/mole Zn = 0.1056 g Zn. The total mass of metal accounted for so far is 0.1930 + 0.07857 + 0.1056 = 0.37717 g; the last digit is probably not reliable at any level, so call it 0.3772 g. That lets us account for the Cu which represents 1.1713 - 0.3772 = 0.7941 g Cu; in percentages, that corresponds to: 0.7941 100% = 67.80% Cu 1.1713 0.1930 100% = 16.48% Sn 1.1713 0.07857 100% = 6.71% Pb 1.1713 0.1056 100% = 9.02% Zn 1.1713 3

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