Question

atomic symbols
a is the mass number: total number of protons + neutrons
z is the atomic number: the number of protons
q is the overall charge: number of electrons - number of protons

sample problems
complete the following table for isotopes:

(p^{+})	(n^{0})	mass number	symbol
30	34	64	(^{64}zn)
13	14	27	(^{27}al)

complete the following table for ions:

(p^{+})	(e^{-})	charge	symbol
37			(s^{2 -})

boron is 19.9% b - 10 (10.0129 amu) and 80.1% b - 11 (11.0093 amu). what is the atomic mass of boron?
10.81 amu

an element is 90.48% isotope a, with a mass of 19.9924 amu, and 0.27% isotope b, with a mass of 20.9938 amu. there is also a third isotope with a mass of 21.9914 amu. what is the element?
ne

gallium has two stable isotopes. one is ga - 69, with a mass of 68.9256 amu, and the other is ga - 71 with a mass of 70.9247 amu. what is the percent abundance of each?
60.1% ga - 69; 39.9% ga - 71

Explanation:

Step1: Recall atomic number and mass - number relationships

The mass number \(A\) of an isotope is the sum of the number of protons (\(p^{+}\)) and neutrons (\(n^{0}\)), i.e., \(A=p^{+}+n^{0}\). The atomic symbol is written as \(\ce{^{A}_{Z}X^{q}}\), where \(Z = p^{+}\)

Step2: Calculate ion charges

The charge \(q\) of an ion is given by \(q=p^{+}-e^{-}\)

Step3: Calculate atomic mass of boron

Use the formula for average atomic mass \(M=\sum_{i}x_im_i\), where \(x_i\) is the fractional abundance and \(m_i\) is the mass of the isotope

Step4: Identify the element

First find the abundance of the third isotope, then calculate the average atomic mass using the formula \(M=\sum_{i}x_im_i\) and compare with known atomic masses

Step5: Solve for gallium isotope abundances

Set up an equation for the average atomic mass of gallium using the abundances and masses of the two isotopes and solve for the unknown abundance.

Answer:

1. For the isotope table:

• For the first row with \(p^{+}=6\) and \(n^{0}=6\), the mass number \(A=p^{+}+n^{0}=6 + 6=12\), and the symbol is \(\ce{^{12}_6C}\)
• For the second - row with \(p^{+}=30\) and \(n^{0}=34\), the mass number \(A = 30+34 = 64\), and the symbol is \(\ce{^{64}_{30}Zn}\)
• For the third - row with \(p^{+}=13\) and \(n^{0}=14\), the mass number \(A=13 + 14=27\), and the symbol is \(\ce{^{27}_{13}Al}\)

1. For the ion table:

• For \(p^{+}=26\) and \(e^{-}=23\), the charge \(q=p^{+}-e^{-}=26 - 23=3+\), and the symbol is \(\ce{Fe^{3 + }}\)
• For the ion with symbol \(\ce{S^{2-}}\), \(p^{+}=16\) (since sulfur has atomic number 16), and if \(q=- 2\), then \(-2=16 - e^{-}\), so \(e^{-}=18\)
• For \(p^{+}=37\), for a common ion of rubidium (\(\ce{Rb}\)), a common ion is \(\ce{Rb^{+}}\), so \(q = 1+\), and \(e^{-}=36\)

1. For the atomic mass of boron:

• The atomic mass \(M\) of boron is calculated using the formula \(M=\sum_{i}x_im_i\), where \(x_i\) is the fractional abundance and \(m_i\) is the mass of the isotope.
• \(x_1 = 0.199\), \(m_1=10.0129\) amu, \(x_2 = 0.801\), \(m_2 = 11.0093\) amu
• \(M=(0.199\times10.0129)+(0.801\times11.0093)\)
• \(0.199\times10.0129 = 1.9925671\)
• \(0.801\times11.0093=8.8184493\)
• \(M=1.9925671 + 8.8184493=10.8110164\approx10.81\) amu

1. For the unknown element:

• Let the abundance of the third isotope be \(x_3=1-(0.9048 + 0.0027)=0.0925\)
• The average atomic mass \(M=(0.9048\times19.9924)+(0.0027\times20.9938)+(0.0925\times21.9914)\)
• \(0.9048\times19.9924 = 18.09912352\)
• \(0.0027\times20.9938 = 0.05668326\)
• \(0.0925\times21.9914=2.0341045\)
• \(M=18.09912352+0.05668326 + 2.0341045=20.19\) amu, which is close to the atomic mass of neon (\(\ce{Ne}\))

1. For the percent - abundance of gallium isotopes:

• Let the abundance of \(\ce{Ga - 69}\) be \(x\), then the abundance of \(\ce{Ga - 71}\) is \(1 - x\)
• The average atomic mass of gallium is \(69.723\) amu.
• \(69.723=68.9256x+70.9247(1 - x)\)
• \(69.723=68.9256x+70.9247-70.9247x\)
• \(69.723-70.9247=(68.9256 - 70.9247)x\)
• \(-1.2017=-1.9991x\)
• \(x=\frac{1.2017}{1.9991}\approx0.601 = 60.1\%\), so the abundance of \(\ce{Ga - 69}\) is \(60.1\%\) and the abundance of \(\ce{Ga - 71}\) is \(39.9\%\)