Question

isotopes and average atomic mass
part i: match the word with its definition.
name:
5 (out)
1)______ ion
a) equal to the number of protons
2)______ isotope
b) whole number equal to protons and neutrons
3)______ atomic number
c) weighted average atomic mass of all naturally - occurring isotopes of an element.
4)______ average atomic mass
d) atom of the same element with a different number of neutrons.
5)______ mass number
e) atom of the same element with a different number of electrons.
part ii: average atomic mass
elements come in a variety of isotopes, meaning they are made up of atoms with the same atomic number but different atomic masses. these atoms differ in the number of neutrons.
the average atomic mass is the weighted average of all the isotopes of an element.
example: a sample of cesium is 75% $^{133}$cs, 20% $^{132}$cs and 5% $^{134}$cs. what is its average atomic mass?
answer: 0.75 x 133 = 99.75
0.20 x 132 = 26.4
0.05 x 134 = 6.7
132.85 amu = average atomic mass
(remember: to change a percent to a decimal, divide by 100, or move the decimal point 2 places to the left.)
determine the average atomic mass of the following mixtures of isotopes. use 4 significant figures.
show all work!
1.) 80% $^{127}$i, 17% $^{126}$i, 3% $^{128}$i
2.) 15% $^{55}$fe, 85% $^{56}$fe

Explanation:

Step1: Convert percentages to decimals for 127I, 126I and 128I

$80\% = 0.8$, $17\%=0.17$, $3\% = 0.03$

Step2: Calculate the contribution of each isotope to the average atomic mass for the iodine - mixture

Contribution of 127I: $0.8\times127 = 101.6$
Contribution of 126I: $0.17\times126=21.42$
Contribution of 128I: $0.03\times128 = 3.84$

Step3: Sum up the contributions to get the average atomic mass of iodine

$101.6 + 21.42+3.84=126.86\approx126.9$

Step4: Convert percentages to decimals for 55Fe and 56Fe

$15\% = 0.15$, $85\%=0.85$

Step5: Calculate the contribution of each isotope to the average atomic mass for the iron - mixture

Contribution of 55Fe: $0.15\times55 = 8.25$
Contribution of 56Fe: $0.85\times56=47.6$

Step6: Sum up the contributions to get the average atomic mass of iron

$8.25 + 47.6=55.85$

Answer:

1. $126.9$ amu
2. $55.85$ amu