Suppose the amu had been defined as 1/10th of the mass of a phosphorus atom. What would be the relative mass of carbon-12?

Oct 11, 2015

$\text{1.2011 u}$

Explanation:

So, let's say that you pick a phosphorus atom, more specifically a phosphorus-31 atom, since that is the only stable isotope of phosphorus, as the base for the unified atomic mass unit, or $u$.

The main idea here is that the relative atomic masses of two atoms do not change when the unified atomic mass unit changes because the atoms themselves do not change as a result of that.

This means that you can use the current definition of $u$ to find the ratio that must exist between a phosphorus-31 atom and a carbon-12 atom for different values of $u$.

So, the current definition of $u$ is $\frac{1}{12} \text{th}$ of the mass of a neutral carbon-12 atom. Since $u$ is defined as

$\text{1 u = 1 g/mol}$

you can use the current atomic masses of phosphorus-31 and carbon-12 to find the ratio that must exist between these two atoms

${A}_{\text{r phosphorus-31"/A_"r carbon-12" = "30.9738 u"/"12.011 u" = (30.9738 color(red)(cancel(color(black)("g/mol"))))/(12.011color(red)(cancel(color(black)("g/mol")))) = 2.578786 " " }} \textcolor{b l u e}{\left(1\right)}$

This means that a phosphorus-31 atom is $2.578786$ times heavier than a carbon-12 atom.

The atomic mass of a mole of phosphorus-31 atoms is known to be $\text{30.9738 g}$. Since one mole of any element contains exactly $6.022 \cdot {10}^{23}$ atoms of that element, you can say that the mass of one atom of $\text{^31"P}$ will be

$30.9738 \text{g"/color(red)(cancel(color(black)("mole"))) * (1color(red)(cancel(color(black)("mole"))))/(6.022 * 10^(23)"atoms of P") = 5.1434 * 10^(-23)"g/atom}$

Now, you need $\frac{1}{10} \text{th}$ of this value as the definition of the unified atomic mass unit, which will give you

$\text{1 u" = (5.1434 * 10^(-23)"g")/10 = 5.1434 * 10^(-24)"g}$

If $u$ is $\frac{1}{10} \text{th}$ of the mass of a single phosphorus-31 atom, then it follows that the atomic mass of one mole of phosphorus using the new definition of $u$ will be $\frac{1}{10} \text{th}$ of the old one.

${A}_{\text{r phosphorus-31" = 1/10 * "30.9738 u" = "3.09738 u}}$

This means that the relative atomic mass of carbon-12 will be, using equation $\textcolor{b l u e}{\left(1\right)}$

A_"r carbon-12" = "3.09738 u"/2.578786 = color(green)("1.2011 u")