Chapter 4, Problem 71A

Sulfur Show that the atomic mass of the element sulfur is 32.065 amu.

Interpretation Introduction

Interpretation:

The atomic mass of S needs to be determined using the given information of its isotopes.

Concept introduction:

Atomic mass of an element is the weighted average mass of the isotopes of that element.

Answer to Problem 71A

Atomic mass of sulfur is $32.065amu$.

Explanation of Solution

Given Information:

$S\left(31.972amu,94.99\%\right)$, $S\left(32.971amu,\text{ 0}\text{.75}\%\right)$, $S\left(33.968amu,\text{ 4}\text{.25}\%\right)$, $S\left(35.967amu,\text{ 0}\text{.01}\%\right)$

Calculation:

Atomic mass of an element is the sum of mass contribution of the isotopes of that element.

$S\left(31.972amu,94.99\%\right)$

Atomic mass: $31.972amu$

Percent Abundance: 94.99%

$\text{Mass contribution }(M_1):(31.972)\left(94.99\%\right)=30.3702$

$S\left(32.971amu,\text{ 0}\text{.75}\%\right)$

Atomic mass: $32.971amu$

Percent Abundance: 0.75%

$\text{Mass contribution }(M_2):(32.972)\left(0.75\%\right)=0.2472$

$S\left(33.968amu,\text{ 4}\text{.25}\%\right)$

Atomic mass: $33.968amu$

Percent Abundance: 4.25%

$\text{Mass contribution }(M_3):(33.968)\left(4.25\%\right)=1.4436$

$S\left(35.967amu,\text{ 0}\text{.01}\%\right)$

Atomic mass: $35.967amu$

Percent Abundance: 0.01%

$\text{Mass contribution }(M_4):(35.968)\left(0.01\%\right)=0.0036$

$\begin{array}{l}\text{Atomic mass of sulfur }= M_1 +{\text{ M}}_2{\text{ + M}}_3+ M_4 \\ =30.3702+0.2472+1.4436+0.0036 \\ =\mathrm{32.064632.065}\end{array}$

Atomic mass of sulfur will be 32.065

Conclusion

Thus,

Mass contribution of $S$ is 30.3702

Mass contribution of $S$ is 0.2472

Mass contribution of $S$ is 1.4436

Mass contribution of $S$ is 0.0036

$\text{Atomic mass of sulfur = Sum of mass contribution of isotopes of sulfur = 32}\text{.065}$