# Question ca4d9

Apr 3, 2017

11.8 to three significant numbers

#### Explanation:

Multiple 23.5 % xx 10.2 = 2.39 amu

Multiple  76.5% xx 12.4 = 9.49 amu

Add the two values to find the average mass

$2.39 + 9.49 = 11.88 a \mu$ To three significant digits this is 11.8 amu.

Apr 3, 2017

${A}_{r}$ of $X = 11.9$ amu

#### Explanation:

In order to calculate an atomic mass you do the following:
Sum the products of mass of each isotope and its relative abundance (in percentage) and then divide that sum by 100.

E.g.  (X' × %X' + X'' × %X'' )/100 = A_r of X
(The percentages must always add up to 100.)

That procedure works for elements with more than just two isotopes, you simply keep adding to the sum part of the equation (see below for the equation for four isotopes). This works with mass spectra, which provides the mass of isotopes (the position of a peak on the x-axis) and their relative abundance (height of the peak). The equation can be used with any unit for the mass too, amu, or g. Solution to this problem
In this case the equation becomes:
${A}_{r}$ of X = (23.5 × 10.2 + 76.5 × 12.4) / 100 = 11.9  amu

Extension
For four isotopes the equation is:
E.g.  (X' × %X' + X'' × %X'' + X''' × %X''' + X'''' × %X'''' )/100 = A_r# of X