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Unformatted text preview: $oiw‘tions To Homework 11.
Sef.3.2o\° EXP—miss; 2 81 5 Bardsz 1 :~ 6. (M) We compute the mass of the reactants and compare that with the mass of the products
to answer this question.
reactant mass = mass of calcium carbonate + mass of hydrochloric acid solution 1.148g 1 mL soln
= 10.00 g calcium carbonate +1 14.8 g solution = 124.8 g reactants = 10.00 g calcium carbonate +1000 1111. soln x product mass = mass of solution + mass of carbon dioxide (M)
(a) (13) 1.9769 E ngw
= 120.40 g soln + 4.39 g carbon dioxide = 120.40 g soln + 2.22 L gasx The same mass within experimental error,
= 124.79 g products Thus, the law of conservation of mass obeyed / We can determine that carbon dioxide has a ﬁxed composition by ﬁnding the % C
in each sample. (In the calculations below, the abbreviation “cmpd” is short for compound.)
3.62 g C %C= —————— x 100% == 27.3% C % =: 7———,'— XI00% = 27.3% C
‘ 13.26 g cmpd 21.66 g cmpd
%C~= Jig—Cx100% 27.3% C
25.91 g cmpd Since all three samples have the same percent ot‘carbon. these data do establish that
carbon dioxide has a ﬁxed composition. Carbon dioxide contains only carbon and oxygen. As determined above, carbon dioxidq, 27.3 °/o C by mass. The percent of oxygen in carbon dioxide is obtained by difference.
%0 = 100.0 % (i27.3 %C) = 72.7 %0 Page .1 Exauise 2 16. 23. my 26. (M)
(a) As with the previous problem, one of the two elements must have the same mass in all of the compounds. This can be most readily achieved by setting the mass of iodine in all four compounds to 1.000 g. With this approach we only need to manipulate the data for compounds B and C. To normalize the amount of iodine in compound B to 1.000 g, we need to multiply the masses of both iodine and fluorine by 2. To accomplish the analogOus normalization of compound C, we must multiply by 4/3 (1.333). Cmpd. B: “normalized” mass of iodine = 0. 500 g 1 x 2 = 1.000 g 1
“normalized” mass of ﬂuorine = 0.2246 g F x 2 = 0.4492 g F Cmpd. C: “normalized” mass ofiodine = 0.750 g 1 x 1.333 = 1.000 g1
“normalized” mass of ﬂuorine = 0.5614 g F x 1.333 = 0.7485 g F Next we divide the mass of ﬂuorine in each compound by the smallest mass of ﬂuorine,
namely, 0.1497 g. This gives 1.000 for compound A, 3.001 for compound B, 5.000 for
compound C, and 7.001 for compound D. The ratios of the amounts of ﬂuorine in the
four compounds A : E: : C : D is 1 : 3 : 5 : 7. These results are consistent with the law
of multiple proportions because for a ﬁxed amount of iodine (1.000 g), the masses of
ﬂuorine in the four compounds are in the ratio of small whole numbers. (b) As with the preceding problem, we can ﬁgure out the empirical formulas for the four iodine—ﬂuorine containing compounds from the ratios of the amounts of ﬂuorine that
were determined in 16(a): Cmpd A: IF Cmpd 8: [F3 Cmpd C: IFS Cmpd D: 1F7 E i  6° '
r. )(a, cobalt 60 .430 (b)phosphorus32 1351) (c) Iron—.59 gzre (d) radium—226 3369.21 38
(E) _
(a) Since all of these species are neutral atoms, the numbers of electrons are the atomic” numbers, the subscript numbers. The symbols must: be arranged in order of increasing ' 0
value of these subscripts. ‘ISAr < 13K < :3 Co <‘ 33,01 < 1,2,ng < 1;(2)Sn < 12522Te i (b) The number of neutrons is given by the difference between the mass number and the
atomic number, AZ. This is the difference between superscripted and subscripted
values and is provided (in parentheses) after each element in the following list. §3K(20) < ngrm) < §3Cu(30) < §§Co(31)< ”.28n(62) < 12522T6(70) < ‘§§Cd(72) 50 (c) Here the nuclides are arranged by increasing mass number, given by the superscripts.
39 40 58 , 59 , 112 120 22
19K<13Ar< ”Cos 29Cu <~ Sn< Cd< ‘.Te Barely. 1 28. (E)
(a) The atomic number of Ra is 88 and equals the number of protons in the nucleus. The
ion’s charge is 2+ and, thus, there are two more protons than electrons: no.
protons = no. electrons + 2 = 88; no. electrons = 88 ~— 2 = 86. The mass number (228) is the sum of the atomic number and the number of neutrons: 228 = 88 + no.
neutrons; Hence, the number of neutrons = 228 — 88 = 140 neutrons. mass of isot(£e_ 2 228.030 u ‘ 16 = 14.2524
mass of 0 15.9994 u (b) The‘massof360i5159994u. ratio: 35 (M) We will use the same type of strategy and the same notation as we used previously in
Equation 35 to come up with the answer.
n = p + 1 There is one more neutron than the number of protons.
n + p = 9 X 3 = 27 The mass number equals nine times the ion’s charge of 3+. Substitute the ﬁrst relationship into the second, and solve for p.
27=(p+1)+p=2p+1 ‘
27 — 1 =———=13
p 2 Thus this is the +3 cation of the isotope Al27 —> ,2; A1”. 5.1.; (E)
(a) Ge is in group 14 and in the fourth period. (b) Other elements in group 16(6A) are similar to S: 0, Se, and Te. Most of the elements
in the periodic table are unlike S, but particularly metals such as Na, K, and Rb. (c) The alkali metal (group 1), in the fifth period is Rb.
(d) The halogen (group 17) in the sixth period is At. 44. (E) To determine the average atomic mass, we use the following expression:
average atomic mass = 2(isotopic mass x fractional natural abundance) Each of the three percents given is converted to a fractional abundance by dividing it by 100. Cr atomic mass = (49.9461x 0.0435) + (51.9405 x 0.8379) +(529407 x 0.0950) +(53.9389 >< 0.0236) = 2.17u +43.52u+5.03u+1.27u = 51.99 u
If all digits are carried and then the answer is rounded at the end, the answer is 52.00 ii. Pages \ Bemise. 7 66. 16. M)
(a) 23 g Na = 1 mol with a. density ~ 1 g/cm3. 1 mole = 23 g, so volume of 25.5 mol ~ 600 cu; (b) Liquid bromine occupies 725 mL or 725 cm3 (given). .
(c) 1.25X 1025 atoms Cr is ~20 moles. At ~50 g/mol, this represents approxrmately 1000 g. Given the density of 9.4 g/cm3, this represents about 1, 00 cm33 of volume.
((1) 2150 g solder at 9.4 g/cm3 represents approximately 200 cm . From this we can see that the liquid bromine would occupy the largest volume. (E) Since the molar mass oi’nitrogen is 14.0 g/mol, 25.0 g N is almost two moles (1.79 mol N), while 6.022 ><1023 Ni atoms is about one mole, and 52.0 g Cr (52.00 g/mol Cr) is also almost one mole. Finally, 10.0 cm3 Fe (55.85 g/mol Fe) has a mass ofabout 79 g. and contains about 1.4 moles ofatoms. Thus, 25.0 g N contains the greatest number of
atoms. Note: Even if you take nitrogen as N2, the answer is the same. E\ excl se. 5 (E) The greatest number of N atoms is found in the compound with the greatest number of
moles of N. The molar mass ofN2O = (12 mol Nx 14.0g N)+(lmol 0x16.0g O) = 44.0g/mol N20. Thus, 50.0 g N20 is slightly more than 1 mole of N20, and contains slightly more than 2 moles 0%
Each mole of N2 contains 2 moles ofN. The molar mass of NH3 is 17.0 g. Thus, there is I M
of NH3 present, which contains 1 mole of N. ‘ 5 The molar mass ofpyridine is(5 mol C x12.0g C) +(5mol H x 1.01g H)+14.0g N = 791g/mu;
Because each mole of pyridine contains 1 mole of N, we need slightly more than 2 moles of ;
pyridine to have more N than is present in the N20. But that would be a mass of about 158 g pyridine, and 150 mL has a mass of less than 150 g. Thus, the greatest number of N atoms is
present in 50.0 g N20. (E) For sorbic acid, CGHSOZ, (a) FALSE The C:H:O mole ratio is 3:411, but the mass ratio differs because moles of
different elements have different molar masses. (b) TRUE Since the two compounds have the same empirical formula, they have the
same mass percent composition. (c) TRUE Aspidinol, CmHmO“ and sorbic acid have the same empirical formula, C3H40.
(d) TRUE The ratio of H atoms to O atoms is 8:2 = 4:1 .Thus, the mass ratio is
(4mol Hxlg H):(lmol Oxl6.0gO)=4g H : 16g O=lg H14g 0. W4 Emmi sz 3 22. (E) Determine the mass ofa mole of Cr(NO3 )3 9HZO , and then the mass of water in a
mole. molar mass Cr(Nog)So 9H20 = 51.9961g Cr +(3x14.0067g N)+(18><15.9994g o)
+(18><:1.00794 g H) = 400.148g/mol Cr(NO3)3 x 9HZO 9molI—[ZO X18.0153gHzO mass H20 = ————— ——
1molCr(NO3 )3 X9HZO lmolHZO = 162.14gH20/m01Cr(NO3)39HZO V
162.14g HZO/molCr(NO3)39H,O '—100‘V=40_ 0 ﬁ
400.148g/molCr(N03)3.9HZO X o 52 /oH_O 32. (M) We base our calculation on 100.0 g of monosodium glutamate. 13.6g Naxﬂl—N—a 0.592 mol Na +0592 —>1.00 mol Na
22.99g Na
1mol C ,
35.5ng—=2.96m01C +0.592—>5.00mol(,
12.01g C
4.8ng1m01H =4.8molH +0.592—>8.1molH
1.01g H
8.3g NXM=059molN +0.592——>1.0molN
14.01gN
\lmol O . . ‘ .
37.8g 0 x m = 2.36 mol 0 +0592 ——> 3.99 mol 0 Empmcal formula : Nanga 4 533*
. g 46. (M) Thiophene contains only carbon, hydrogen, and sulfur, so there is no need to determine
the mass of oxygen by difference. We simply determine the amount of each element from the
mass of its combustion product. 1molCO2 x 1molC
X____,.__ __
2 44.010gC02 1molco2 lmol. H20 _2 mol H X 18.0153g HZO 1mol H20 2.7224 g CO = 0.061859 mol C —:0.01548 —> 3.996 mol C 0.5575 g H20 x = 0.06189 mol H + 0.01548 ——) 3.999 mol H 1mol 80:. 1 mol S ..——x— 0.9915g SO, = 0.01548mol S + 0.01548 —> 1.00 mol S X___.._—._
" 64.0648gSO2 1molSO2 The empirical formula of thiophene is C 4H 4S . 'Va315 ...
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 Winter '07
 ZAX,D
 Chemistry

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