What is the relative atomic mass of a hypothetical element that consists of the following isotopes in the indicated natural abundances? Isotope - Isotopic mass (amuamu) - Relative abundance (%%) 1 - 78.9 - 12.6 2 - 80.9 - 13.9 3 - 82.9 - 73.5

82.1

Explanation:

Isotopes of the same element are atoms of the same element (so, having same number of protons) having a different number of neutrons (and so, a different atomic mass).

In this problem, this element has 3 different isotopes, with atomic mass (in atomic mass units, a.m.u.) and relative abundance listed below:

1) 78.9 12.6%

2) 80.9 13.9%

3) 82.9 73.5%

In order to calculate the relative atomic mass unit of this element, we have to calculate the weighed average of the relative atomic mass of each isotope.

This means that we can calculate it as the sum of each atomic mass unit multiplied by the relative frequency:

where is the atomic mass of each isotope and the relative frequency. Therefore, we find:

h let me think

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.