Question

What is the average atomic mass of Titanium? Titanium has 5 isotopes: Ti-46 (8.0%) Ti-47(7.8%) Ti-48(73.4%) Ti-49(5.5%) Ti-50(5.3%)

Answer 1

`"47.923 u"`

Explanation:

When the problem doesn't provide you with the actual atomic mass of an isotope, `m_a`, you can use its mass number, `A`, as an approximation of its atomic mass.

In this case, you will have

`""^46"Ti " -> " "m_a ~~ "46 u"`

`""^47"Ti " -> " " m_a ~~ "47 u"`

`""^48"Ti " -> " " m_a ~~ "48 u"`

`""^49"Ti " -> " " m_a ~~ "49 u"`

`""^50"Ti " -> " " m_a ~~ "50 u"`

Now, the average atomic mass of titanium is calculated by taking the weighted average of the atomic masses of its stable isotopes.

Simply put, each isotope `i` will contribute to the average atomic mass of the element in proportion to its decimal abundance, which is simply the percent abundance divided by `100`.

`color(blue)(|bar(ul(color(white)(a/a)"avg. atomic mass" = sum_i m_"a i" xx "abundance of i"color(white)(a/a)|)))`

The decimal abundances for these five isotopes will be

`""^46"Ti: " 8.0/100 = 0.080`

`""^47"Ti: " 7.8/100 = 0.078`

`""^48"Ti: " 73.4/100 = 0.734`

`""^49"Ti: " 5.5/100 = 0.055`

`""^50"Ti: " 5.3/100 = 0.053`

The average atomic mass of titanium will thus be

`"avg. atomic mass " =`

`"46 u" xx 0.080 + "47 u" xx 0.078 + "48 u" xx 0.734 + "49 u" xx 0.055 + "50 u" xx 0.053`

`"avg. atomic mass " = color(green)(|bar(ul(color(white)(a/a)color(black)("47.923 u")color(white)(a/a)|)))`