Question 6f50c

Oct 1, 2015

$\text{50.281 u}$

Explanation:

You need to use titanium's atomic mass to figure out the mass and abundance of its fifth isotope.

• $\text{^46"Ti} \to$ $\text{45.953 u}$ and 8.0%
• $\text{^47"Ti} \to$ $\text{46.592 u}$ and 7.3%
• $\text{^48"Ti} \to$ $\text{47.948 u}$ and 73.8%
• $\text{^49"Ti} \to$ $\text{48.948 u}$ and 5.5%

Now take a look at a periodic table. Titanium's molar mass is known to be $\text{47.867 g/mol}$.

If you take into account the fact that one unified atomic mass unit, or $u$ (amu is a very old term that is sometimes used interchangeably with $u$, although it was not defined the same), is equivalent to $\text{1 g/mol}$, you can say that titanium has a relative atomic mass of

47.867color(red)(cancel(color(black)("g/mol"))) * "1 u"/(1color(red)(cancel(color(black)("g/mol")))) = "47.867 u"

The abundances of the five stable isotopes must add up 100%, which means that the percent abundance of the fifth isotope will be

%""^50"Ti" = 100% - (8.0 + 7.3 + 73.8 + 5.5)% = 5.4%

The relative atomic mass of titanium is the sum of each isotope's atomic mass multiplied by its abundance

"relative atomic mass" = sum_i("isotope"""_i * "abundance"""_i)#

This means that you have - I'll use fractional abundances, which are simply percent abundances divided by $100$

$45.953 \cdot 0.08 + 46.952 \cdot 0.073 + 47.948 \cdot 0.738 + 48.948 \cdot 0.055 + \textcolor{b l u e}{x} \cdot 0.054 = 47.867$

This means that $\textcolor{b l u e}{x}$, which represents the atomic mass of the fifth isotope, will be

$\textcolor{b l u e}{x} \cdot 0.054 = 47.867 - 45.1552$

$\textcolor{b l u e}{x} = \frac{2.71178}{0.054} = \text{50.218 u}$

SIDE NOTE THe actual atomic mass of titanium's fifth stable isotope, titanium-50, is $\text{49.945 u}$.

From what I can tell, the difference between this result and the actual value comes from the way the values given to you for the abundances have been rounded.