Question

Question #6f50c

Answer 1

`"50.281 u"`

Explanation:

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

So, start with what you know

• `""^46"Ti" ->` `"45.953 u"` and `8.0%`
• `""^47"Ti" ->` `"46.592 u"` and `7.3%`
• `""^48"Ti" ->` `"47.948 u"` and `73.8%`
• `""^49"Ti" ->` `"48.948 u"` and `5.5%`

Now take a look at a periodic table. Titanium's molar mass is known to be `"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 `"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 * 0.08 + 46.952 * 0.073 + 47.948 * 0.738 + 48.948 * 0.055 + color(blue)(x) * 0.054 = 47.867`

This means that `color(blue)(x)`, which represents the atomic mass of the fifth isotope, will be

`color(blue)(x) * 0.054 = 47.867 - 45.1552`

`color(blue)(x) = 2.71178/0.054 = "50.218 u"`

SIDE NOTE THe actual atomic mass of titanium's fifth stable isotope, titanium-50, is `"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.