Problem Set 7.
1.
K
(39K
40
K
41
K
Determine the average atomic mass of
mass(amu)
%
38.9637074 (12) 93.2581 (44)
39.9639992 (12) 0.0117 (1)
40.9618254 (12) 6.7302 (44))
Br
79
Br
81
Br
mass (amu)
%
78.9183361 (26) 50.69 (7)
80.916289 (6)
49.31 (7)
Ni:
58
Ni
60
Ni
61
Ni
62
Ni
64
Ni
mass(amu)
%
57.9353462 (16) 68.0769 (89)
59.9307884(16) 26.2231 (77)
60.9310579 (16)
1.1399 (6)
61.9283461 (16)
3.6345 (17)
63.9279679 (17)
0.9256 (9)
Atomic mass of potassium K:
m(K) = 38.9637074 amu x 93.2581 / 100 + 39.9639992 amu x 0.0117 / 100 + 40.9618254 amu x 6.7302 / 100
= 36.3368 amu + 0.00468 amu + 2.7568 amu = 39.0983 amu
Note that in the first product the least number of significant digits is 6 so I rounded off that product to 6 digits; in the second product
one of the factors has three digits and so I rounded off the product to three digits; in the last product one of the factors has 5 significant
digits and so I rounded off the product to 5 digits.
When adding the doubtful digits in the fourth decimal place and so I rounded off the sum to 4 decimal places.
Atomic mass of bromine Br:
m(Br) = 78.9183361 amu x 50.69 / 100 + 80.916289 amu x 49.31 / 100 = 40.00 amu + 39.90 amu = 79.90 amu
Note that in the first product the least number of significant digits is 4 so I rounded off that product to 4 digits; in the second product
one of the factors has 4 digits and so I rounded off the product to 4 digits. No rounding off was necessary for the sum.
Atomic mass of nickel Ni:
m(Ni) = 57.9353462 amu x 68.0769 / 100 + 59.9307884 amu + 26.2231 / 100 + 60.9310579 amu x 1.1399 / 100 +
61.9283461 amu x 3.6345 / 100 + 63.9279679 amu x 0.9256 / 100
= 39.4406 amu + 15.7157 amu + 0.69455 amu + 2.2507 amu + 0.59172 amu = 58.6935 amu
2.
The atomic mass of rubidium (Z=37)is 85.4678(3) amu; rubidium has two naturally occurring isotopes. Their atomic masses
are 84.911794 (3) amu and 86.909187 (3) amu. Use this data to figure out how to compute the percentage distribution.
Let the fraction of the first isotope (m1 = 84.911794 amu) be f1 and the fraction of the second isotope (m2 = 86.909187 amu) be f2 . Then
the atomic mass of the rubidium can be computed as follows:
m(Rb) = f1 x m1 + f2 x m2
The sum of the fractions is equal to 1
f1 + f2 = 1
We can write the two equations for the factors as follows
85.4678 = f1 x 84.911794 + f2 x 86.909187
To solve we will isolate f1 from the first equation
f1 = 1 - f2
Substitute into the second equation we write
85.4678 = (1 - f2) x 84.911794 + f2 x 86.909187
Expand the left-hand side of the above equation
85.4678 = 84.911794 - f2 x 84.911794 + f2 x 86.909187
Factor f2 on the left
85.4678 = 84.911794 + f2 x (- 84.911794 + 86.909187)
85.4678 = 84.911794 + f2 x 1.997393
85.4678 - 84.911794 = f2 x 1.997393
0.5501 = f2 x 1.997393
0.2784 = f2
Consequently f1 = 1 - 0.2784 = 0.7216. The percentage distribution is obtained by multiplying by 100. There is 72.16% of isotope of
mass 84.911794 amu and 27.84% of isotope of mass 86.909187 amu.
3.
Determine the number of significant digits: in the table below:
Number
signif. digits
4.
45.26 cm
0.109 m
0.00025 kg
163 mL
0.60 g
62.700 cm
2.360x103 g
5.90x10-3kg
4
3
2
3
2
5
4
3
Round off to three significant digits:
Number
1.2472 cm
0.03605 L
2.103 cm
4.585 m
round-off
1.25 cm
0.0360 L
2.10 cm
4.59 m
5.
Complete the following operations stating a correct number of significant digits :
a.
8.16x106 + 5.32x105 = (8.16x101 + 5.32)x105 = (81.6 + 5.32)x105 = 86.9 x105 = 8.69 x106
b.
3.245 - 0.38 = 2.86 (second decimal place)
c.
1.2 x 2.68 = 3.2 (the least number of significant digits is 2)
d.
0.34 x10.100 = 3.4 (the least number of significant digits is 2)
e.
2.754 / 0.00230 = 1.20 x 103 (the least number of significant digits is 3; We need to use scientific notation to show
the three significant digits because the doubtful digit is zero)
6.
Determine the symbol of X element in the following nuclear reactions and determine the energy released in these reaction
using the mass-energy equivalence 1 amu = 1.5 x 10-10 J. Also identify whether the reaction is a fusion reaction or a fission
reaction. Note that you must state the results with appropriate number of significant digits.
a.
m = 44.9559100 amu + 7.0160030 amu - 51.9405098 amu = 0.00314032 amu
b.
m = 239.05216 amu - (9.0121822amu + 230.033126amu) = 0.00682amu
c.
Sorry the mass of this isotope of Xenon is not known.
d.
m = 11.0093054 amu + 4.00260324 amu - 15.00010897 amu = 0.01179967 amu
e.
m = 43.9554806 amu + 12.0000000 amu - 55.9349393 amu = 0.0205413 amu