Q127CP - Chemistry: Molecular Nature Of Matter And Change

Question

Prior to 1961, the atomic mass standard was defined as of the mass of ¹⁶O. Based on that standard: (a) What was the mass of carbon-12, given the modern atomic mass of oxygen is 15.9994 amu? (b) What was the mass of potassium-39, given its modern isotopic mass is 38.9637 amu?

Step-by-Step Solution

Answer

The mass of carbon-12 is $\mathbf{12}\mathbf{.}\mathbf{00045}\mathbf{ }\mathbf{amu}\mathbf{.}$ The mass of potassium-39 is$\mathbf{38}\mathbf{.}\mathbf{9652}\mathbf{ }\mathbf{amu}$ 

Step 1: Finding the mass of carbon-12

The carbon mass on old scale can be calculated as:

$$\begin{gathered} \frac{\mathbf{modern}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}{ }^{16}\mathbf{O}}{\mathbf{modern}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}{ }^{12}\mathbf{C}}=\frac{\mathbf{mass}\mathbf{ }\mathbf{of}{ }^{16}\mathbf{O}\mathbf{ }\mathbf{on}\mathbf{ }\mathbf{old}\mathbf{ }\mathbf{scale}}{\mathbf{mass}\mathbf{ }\mathbf{of}{ }^{12}\mathbf{C}\mathbf{ }\mathbf{on}\mathbf{ }\mathbf{old}\mathbf{ }\mathbf{scale}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\frac{\mathbf{15}\mathbf{.}\mathbf{9994}}{\mathbf{12}}\mathbf{=}\frac{\mathbf{16}}{\mathbf{m}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{m}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{00045}\mathbf{ }\mathbf{amu} \end{gathered}$$

Step 2: Finding the mass of potassium-39

The atomic ratio of modern isotopic mass of potassium-39 can be denoted as,

$$\begin{gathered} \frac{\mathbf{modern}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}{ }^{16}\mathbf{O}}{\mathbf{modern}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}{ }^{39}\mathbf{K}}=\frac{\mathbf{mass}\mathbf{ }\mathbf{of}{ }^{16}\mathbf{O}\mathbf{ }\mathbf{on}\mathbf{ }\mathbf{old}\mathbf{ }\mathbf{scale}}{\mathbf{mass}\mathbf{ }\mathbf{of}{ }^{39}\mathbf{K}\mathbf{ }\mathbf{on}\mathbf{ }\mathbf{old}\mathbf{ }\mathbf{scale}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\frac{\mathbf{15}\mathbf{.}\mathbf{9994}}{\mathbf{38}\mathbf{.}\mathbf{9637}}\mathbf{=}\frac{\mathbf{16}}{\mathbf{m}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{m}\mathbf{=}\mathbf{38}\mathbf{.}\mathbf{9652}\mathbf{ }\mathbf{amu} \end{gathered}$$

On solving the isotopic mass,

$$\begin{gathered} \mathbf{Isotopic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}{\mathbf{ }}^{39}\mathbf{K}\mathbf{=}\mathbf{measured}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{ratio}\mathbf{ }\mathbf{\times }\mathbf{mass}\mathbf{ }\mathbf{of}{ }^{16}\mathbf{O} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{435}\mathbf{\times }\mathbf{21}\mathbf{.}\mathbf{328}\mathbf{ }\mathbf{amu} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{51}\mathbf{.}\mathbf{938}\mathbf{ }\mathbf{amu} \end{gathered}$$