Final answer:
The ratio of the relative abundances of the isotopes of indium with m/z = 113 to m/z = 115 is 1:3 or 25% to 75%, calculated using the definition of relative atomic mass.
Step-by-step explanation:
To calculate the ratio of the relative abundances of the two isotopes of indium with m/z = 113 and m/z = 115, given the relative atomic mass of 114.5, you can use the following formula, which is derived from the definition of relative atomic mass:
Relative Atomic Mass = (Abundance of Isotope 1 × Mass of Isotope 1) + (Abundance of Isotope 2 × Mass of Isotope 2)
In this case, let the abundance of isotope with m/z = 113 be x, and the abundance of isotope with m/z = 115 be (1-x), since the total abundance must equal 100% or 1 in fraction form. Using the given relative atomic mass:
114.5 = (x × 113) + ((1-x) × 115)
Solve for x to find the abundance of the isotope with m/z = 113:
114.5 = 113x + 115 - 115x
114.5 = 113x - 115x + 115
114.5 = -2x + 115
x = (115 - 114.5) / 2
x = 0.5 / 2
x = 0.25 or 25%
Therefore, the ratio of the relative abundances of the isotopes m/z = 113 to m/z = 115 is 1:3 or 25% to 75%.