Chem 151
Honors Chemistry
Fall 2007
Ass 1
1
1: 4, 9, 17, 23, 28 2: 3, 5, 12, 19
4. Since 1800 nearly 200 sincere but erroneous reports of the discovery of new chemical
elements have been made. Why have mistaken reports of new elements been so
numerous? Why is it relatively easy to prove that a material is not a chemical
element, but difficult to prove absolutely that a material is an element?
A substance is an element if it cannot be further divided. In some instances, very
stable compound exist, and these can be mistaken for elements until just the right
conditions are found for the compound’s decomposition. If it can be divided it is not
an element, so as soon as such a reaction is found, one doesn’t have to look further.
But proving the negative (ie, this is not a compound) is difficult. Some of the more
recent difficulties in finding new elements arise because the transuranium elements
are all short lived, and only a few nuclei are produced, making chemical
transformation challenging.
9. Vanadium and Oxygen for a series of compound with the following compositions:
Mass % V
Mass % O
76.10
23.90
67.98
32.02
61.42
38.58
56.02
43.98
What are the relative numbers of atoms of oxygen in the compounds for a given mass
of vanadium?
In 100 g of the compound there are the following
Grams V Moles V
Grams O
Moles O
76.10
76.10/50.942=1.495
23.90
23.90/16= 1.493
67.98
67.98/50.942=1.334
32.02
32.02/16=2
61.42
61.42/50.942 = 1.21
38.58
38.58/16=2.41
56.02
56.02/50.942 = 1.10
43.98
43.98/16 = 2.75
The simplest formulae satisfying these conditions are VO, V2O3,VO2,and V2O5
O/V
1
1.5
2
2.5
21. Only two isotopes of boron (B) occur in nature; their atomic masses and abundances
are given in the following table. Complete the table by computing the relative atomic
mass of 11B to 4 significant figures, taking the tabulated relative atomic mass of natural
boron as 10.811.
isotope
10
B
11
B
%Abundance
19.61
80.39
Atomic Mass
10.013
?
2
The average atomic mass is the atomic mass of the two isotopes, weighted by the
abundance:
AverageMass=Mass(10)*Abundance(10)+Mass(11)*Abundance(11)
10.811= 10.013*.1961+Mass(11)*0.8039
10.811 ! 10.013* 0.1961
so Mass(11) =
=11.01
0.8039
23. Soft wood chips weighing 17.2 kg are placed in an iron vessel and mixed with 150.1
kg of water and 22.43 kg of sodium hydroxide. A steel lid seals the vessel which is then
placed in an oven at 250 °C for 6 hours. Much of the wood decomposes, but the iron
vessel and lid do not react.
a) Classify each of the materials mentioned as a substance or mixture.
Subclassify the substances as elements or compounds.
The iron vessel and lid are clearly elements and substances.
The wood chips are a mixture of compounds, the water and sodium hydroxide is a
substance and compounds. The whole works is a mixture of compounds.
b) Determine the mass of the contents of the iron vessel after the reaction.
Mass is conserved. The initial mass is 17.2 + 150.1 + 22.43 kg = 189.73k = final mass.
28. Naturally occurring Rb consists of 85Rb and 87Rb. The atomic mass of the mixture
found in nature is 85.4678. Calculate the percentage of the two isotopes found in nature.
The apparent atomic mass of the naturally occurring Rb = 85.4678 =M85F85 + M87F87,
where M85 is the atomic mass of the 85 isotope, and F85 is the fraction of the F85 isotope.
Clearly F85 + F87 =1, so you have two equations and 2 unknowns.
Thus 85.4678 =M85F85 + M87 (1- F85) = 84.9117F85 + 86.9092*(1-F85)
F85 = 0.7216; F87 = 0.2784
---------Chapter 2------3. Compute the relative molecular masses of the following compounds on the 12C scale:
a) P4O10.
MW = 4*30.97+10*16=283.88
b) BrCl.
MW=79.90+35.45=115.35
c) Ca(NO3)2
MW=40.08+2*(14+3*16)=164.08
d) KMnO4
MW=39.1+54.94 + 4*16=158.04
e) (NH4)2SO4 MW=2*(14.01+4*1.01)+32.07+4*16=132.17
5. Suppse that a person counts out gold atoms at the rate of one each second for the
entire span of an 80 year life. Has the person counted enough atoms to be detected with
an ordinary balance? Explain
3
One atom/s for 80 years is 80*365*24*60*60 atoms = 2.52*109.
This is 2.52x109/6.02x1023 moles = 4.2x10-15 moles which weighs .82 picograms.
A really sensitive balance might weigh micrograms, so this is about 106 too small.
12. Calculate the number of atoms of silicon (Si) in 415 cm3 of the colorless gas disilane
at 0°C and atmospheric pressure where its density is .00278 g/cc. The molecular formula
of disilane is Si2H6.
The density is mass/volume, so the mass of Si2H6 is
density*volume = .00278 g/cc*415 cc=1.154 g
the molecular weight of = 28.09*2+6*1.01=62.24
the number of moles of Si2H6 = wt/mw = 1.154/62.24 = 18.5 mmoles
the number of molecules = 18.5*10-3 * 6.02x1023=1.1159e+22
the number of Si atoms is twice this 2.23x1022
19. Zinc phosphate is used as a dental cement. A 50.00 mg sample is broken down into
its constituent elements and gives 16.58 mg oxygen, 8.02 mg phosphous and 25.40 mg
zinc. Determine the empirical formula of zinc phosphate.
Convert weight of each element into the number of moles of each element:
16.58 mg oxygen = 16.58/15.999 mmoles = 1.04 mmoles
8.02 mg phosphorous = 8.02/30.974 mmoles =0.259mmoles
25.40mg zinc = 25.40/65.407 = 0.388 mmoles
We divide all by the smallest number, that for phosphorous (thereby getting a formula
with one P in it),
P= 1
O = 1.04/0.259 = 4.02
Zn = 0.388/.259 = 1.5
Now multiply everybody by 2 to get integers: Zn3(PO4)2