Download Atomic Mass and Isotopes: Problem Set 3 Solutions and more Lecture notes Reasoning in PDF only on Docsity!
Chemistry 121
Mines, Spring 2020 (Tro 5/e)
Answer Key, Problem Set 3 – Complete Version (with explanations)
- 2.7 4 ; 2. 2.7 6 ; 3. NT1 ; 4. 3.24 ADD part (ii); 5. 3.29 & 3.30 ADD part (ii); 6. NT
% abundance of isotopes and (average) atomic mass.
- 2.74. The atomic mass of copper is 63.546 amu. Do any copper isotopes [atoms] have a mass of 63.546 amu? Explain.
Answer : No. The value of 63.546 amu is a weighted average of the two naturally occurring isotopes on Earth (Cu-63 and Cu-65; See Example 2.5). So there are only two “kinds” of Cu atoms and thus (only) two isotopic masses—one smaller than the average and one larger than it. More specifically, all of the Cu- 63 atoms have a mass less than 63.546 amu and all of the Cu- 65 atoms have a mass greater than it. But no Cu atoms have the precise mass of the weighted average.
- 2.7 6. (i) An element has four naturally occurring isotopes with the (isotopic) masses and natural abundances given here. Find the atomic mass of the element and identify it.
Isotope Mass(amu) Abundance (%)
1 135.90714 0.
2 137.90599 0.
3 139.90543 88.
4 141.90924 11.
(ii) Also describe how you could predict which isotope’s isotopic mass the atomic mass would be closest to, and whether or not the atomic mass would be greater than or less than this isotope’s atomic mass, before doing the actual calculation of the atomic mass.
Answers:
(i) : 140. 1 amu; From a Periodic Table, this atomic mass matches that of cerium (symbol Ce; #58)
(ii) : Closest to isotope 3’s isotopic mass, but slightly greater than it (full answer below)
Work / Reasoning :
Let me address (ii) first. Let’s call the element X for now. One can predict that the atomic mass of X should be closest to the isotopic mass of isotope 3 because the vast majority of X atoms (over 88%) are atoms of isotope 3. Also, since the vast majority of the other atoms (in any sample of X) are of isotope 4 (11% vs less than 1 % for isotopes 1 and 2), whose isotopic mass is greater than that of isotope 3, one can predict that the atomic mass will be slightly larger than the isotopic mass of isotope 3. As such, I would predict that the atomic mass should be slightly larger than 140 amu.
To get the precise value for X’s atomic mass, multiply each isotopic mass by its fractional abundance (obtained by dividing the % abundance by 100) and sum:
(weighted average) atomic mass = 0. 0019 ( 135.90714 amu) + 0. 0025 ( 137.90599 amu) +
- 8843 ( 139.90543 amu) + 0.1113( 141.90924 amu)
= 0.2 582 … + 0. 3447 .. + 123. 718 … + 15.7 94 … = 140. 115 …= 140. 1 amu
- NT1. The isotopic mass of Cu-63 is 62.9396 amu and its natural abundance (on Earth) is 69.17%. The atomic mass of Cu (on Earth) is 63.546 amu. If 1 g = 6.022 x 10^23 amu (this is a mass “conversion” relationship—it has nothing to do with Cu per se), how many atoms would there be in a 1.000-gram (i.e., 6.022 x 10^23 - amu) sample of (a) Cu-63 (i.e., a sample of copper that has only Cu-63 atoms in it. This would be very expensive to produce here on Earth, but it could be done!)? (b) a “regular” sample of Cu on planet Earth?
(reasonable; a bit over 140 amu, as predicted)
(c) i) Do all the individual atoms in the sample in (b) above have different masses than the atoms in the sample in (a) above? Be specific. ii) What causes fewer atoms to be in the second sample than the first?
Answers : (a) 9.568 x 10^21 atoms; (b) 9.477 x 10^21 atoms;
(c) i) No, not even most of them are different! 69.17% of the atoms in Sample 2 have identical masses to all of the atoms in Sample 1. 30.83% of the atoms in Sample 2 have a greater mass.
ii) Since some of the atoms in Sample 2 have a greater mass, it will take fewer of them to add up to a certain mass (like 1.000 g). For example, it will take fewer bowling balls to reach a mass of 16 lb (i.e., one ball!) than the number of tennis balls needed to reach 16 lb (clearly way more than 32 since a tennis ball is nowhere near even half a pound in mass).
Work / Reasoning for (a) & (b) :
Each sample has a mass of 6.022 x 10^23 amu. Each atom in (a) has a mass of 62.9396 amu. Thus, the number of atoms in the first sample (to 4 SF) is just:
9.568 x 10 atoms
21 = 9.56 7 9...x 10 = 62.9396amu/atom
6.022 x 10 amu 21 23 (all Cu-63 atoms)
Since the (average) atomic mass of Cu is 63.546 amu, the number of atoms in the “regular” sample of Cu (with both isotopes present in the ratio of their natural abundances) is just:
9.477 x 10 atoms 21 = 9.47 6 5...x 10 = 63.546amu/atom
6.022 x 10 amu 21
23 (some Cu-63 and some Cu-65)
Ions vs. Ionic Compounds, Molecular vs. Ionic Compounds
- 3.24. (i) Determine the number of each type of atom in each formula [one formula unit of each substance]. (ADD part (ii): Determine the number of each type of ion in one formula unit of each substance.)
- This column was only added to try to clarify the literal meaning of subscripts in a formula; it is not necessary for you to have this in your homework papers.
Literal “Expansion” of formula (no charges)*
of each type of atom
(in one FU)
of each type of ion
(in one FU)
(a) Ca(NO 2 ) 2 Ca; NO 2 ; NO 2 one Ca; two N; four O one Ca2+; two NO 2 -
(b) CuSO 4 Cu; S; O; O; O; O one Cu; one S; four O one Cu2+; one SO 42 -
(c) Al(NO 3 ) 3 Al; NO 3 ; NO 3 ; NO 3 one Al; three N; nine O one Al3+; three NO 3 -
(d) Mg(HCO 3 ) 2 Mg; HCO 3 ; HCO 3 one Mg; two H; two C; six O one Mg2+; two HCO 3 -
iodine monochloride (remember to omit mono for 1st^ element)
(e) Pb 3 (PO 4 ) 2 metal first ionic compound; Pb is Type II, so look at anion first. PO 43 -
(phosphate) charge on Pb is 2+ (neutrality principle; total neg. charge = 2 x - 3 = - 6 so three Pb's must total +6)
lead(II) phosphate
(f) KIO 3 metal first ionic compound; K is Type I (1A), so K+. IO 3 must be - 1, IO 3 -^ is
iodate ion
potassium iodate
(g) H 2 CO 3 H first acid; CO 32 -^ is carbonate ion; - ates become "-ic acids"
acid name is carbonic acid
(h) Sr 3 N 2 metal first ionic compound; Sr is Type I (2A), so Sr2+. N^3 -^ is nitride ion
strontium nitride
(i) Al 2 (SO 3 ) 3 metal first ionic compound; Al is Type I (3A), Al3+; SO 3 must be 2-;
(neutrality principle) SO 32 -^ is sulfite ion
aluminum sulfite
(j) SnO 2 metal first ionic compound; Sn is Type II, so look at anion first. O^2 -^ (oxide)
charge on Sn is 4+ (neutrality principle; total neg. charge = 2 x - 2 = - 4 so one Sn must total +4)
tin(IV) oxide
(k) HClO H first acid; ClO-^ is hypochlorite ion; - ites become "-ous acids"
acid name is hypochlorous acid
(l) ammonium hydrogen phosphate NH 4 +, HPO 42 -^ (NH 4 ) 2 HPO 4 (neutrality principle)
(m) chromium(VI) sulfide Cr^6 +, S^2 -^ CrS 3 (neutrality principle)
(n) silicon dioxide SiO 2 (prefixes tell you the subscripts)
(o) sodium sulfite Na+, SO 32 -^ Na 2 SO 3 (neutrality principle)
(p) aluminum hydrogen sulfate Al3+, HSO 4 -^ Al(HSO 4 ) 3 (neutrality principle)
(q) nitrogen trichloride NCl 3 (prefixes tell you the subscripts)
(r) hydrobromic acid hydro___ic acid - ide anion bromide Br-^ HBr
(s) bromous acid ous acid - ite anion bromite BrO 2 -^ HBrO 2
(t) perbromic acid ic acid - ate anion perbromate BrO 4 -^ HBrO 4
(u) potassium hydrogen sulfide K+, HS-^ (sulfide is S^2 -^ + H+^ = HS-^ [hydrogen sulfide]) KHS (neutrality princ.)
