<blockquote>
<p>The abundance of an isotope, for example, 84Sr, is calculated as: (84Sr/88Sr)/( 84Sr/88Sr + 86Sr/88Sr + 87Sr/88Sr + 88Sr/88Sr) where 87Sr/88Sr can be calculated as 86Sr/88Sr x 87Sr/86Sr and 88Sr/88Sr is, of course, 1.</p>
</blockquote>
<p>It would very nice if you could add a reference to that calculation.</p>
<p>The abundance of an isotope, an isotope of strontium for example, is calculated as the ratio of the atomic mass of the isotope in some sample to the naturally present atomic mass of the element, averaged over many samples.</p>
<p>Strontium has four naturally occurring isotopes (<span class="math-container">$^{84}\text{Sr}$</span>, <span class="math-container">$^{86}\text{Sr}$</span>, <span class="math-container">$^{87}\text{Sr}$</span>, and <span class="math-container">$^{88}\text{Sr}$</span>), one somewhat long-lived isotope, <span class="math-container">$^{90}\text{Sr}$</span>, with a half-life of 28.9 years, and several very short-lived isotopes that don't occur in nature. Even ignoring that one somewhat long-lived isotope (it is but a tiny, tiny trace isotope, so it is safe to ignore it), all four remaining isotopes must be taken into account. Three of those four isotopes are presumably primordial. The fourth, <span class="math-container">$^{87}\text{Sr}$</span>, is a mix of primordial and radioactive decay from <span class="math-container">$^{87}\text{Rb}$</span>. As an aside, this feature is what makes rubidium–strontium dating important.</p>
<p>Suppose you have broken down a sample of strontium into its isotopes, by mass. Then the relative abundance of any one of those isotopes is calculated, by atomic mass, as the ratio of the atomic mass of the isotope in question to the atomic mass of the sample as a whole. This will, by design, add to one (or 100%).</p>

CC BY-SA 4.0
2024-10-29T10:01:02.493Z
2024-10-29T10:01:02.493Z
David Hammen
<p>I've found, embarrasingly, that I'm unable to explain this when probed although I know it does works. In sum, how does the following work, exactly, in simple mathematical terms:</p>
<p>The abundance of an isotope, for example, 84Sr, is calculated as: (84Sr/88Sr)/( 84Sr/88Sr + 86Sr/88Sr + 87Sr/88Sr + 88Sr/88Sr) where 87Sr/88Sr can be calculated as 86Sr/88Sr x 87Sr/86Sr and 88Sr/88Sr is, of course, 1.<br />
In this way, we calculate the following abundances:</p>
<p>88Sr: 0.474%</p>
<p>87Sr: 8.381%</p>
<p>86Sr: 20.952%</p>
<p>84Sr: 70.192%</p>

CC BY-SA 4.0
2024-10-26T23:32:33.033Z
I'm not sure what you are asking. If you're asking how these sum to 1 (or 100%), that's by design. Those four isotopes are the only stable (or nearly stable in the case of $^{87}\text{Sr}$, which has such a ridiculously long half life that it might as well be considered to be stable). All of the other isotopes have ridiculously short half lives. If you're asking why the calculation is designed that way, that's a different question. If you're asking what those isotope ratios mean, that too is a different question.
David Hammen
CC BY-SA 4.0
2024-11-08T00:11:32.767Z
how does the fractional abundance of a ratio give the abundance of a particular isotope in a ratio?
Josiah
<geology><geochemistry><geochronology><isotope>
How can the ratios of some elements isotopes provide the fractional abundance?
Josiah
# How can the ratios of some elements isotopes provide the fractional abundance?

(Asked by: username_0 on 2024-10-26)

I've found, embarrasingly, that I'm unable to explain this when probed although I know it does works. In sum, how does the following work, exactly, in simple mathematical terms:

The abundance of an isotope, for example, 84Sr, is calculated as: (84Sr/88Sr)/( 84Sr/88Sr + 86Sr/88Sr + 87Sr/88Sr + 88Sr/88Sr) where 87Sr/88Sr can be calculated as 86Sr/88Sr x 87Sr/86Sr and 88Sr/88Sr is, of course, 1.  
In this way, we calculate the following abundances:

88Sr: 0.474%

87Sr: 8.381%

86Sr: 20.952%

84Sr: 70.192%

**Tags:** geology, geochemistry, geochronology, isotope

> **Comments:**
> **username_1**: I'm not sure what you are asking. If you're asking how these sum to 1 (or 100%), that's by design. Those four isotopes are the only stable (or nearly stable in the case of $^{87}\text{Sr}$, which has such a ridiculously long half life that it might as well be considered to be stable). All of the other isotopes have ridiculously short half lives. If you're asking why the calculation is designed that way, that's a different question. If you're asking what those isotope ratios mean, that too is a different question.  
> **username_0**: how does the fractional abundance of a ratio give the abundance of a particular isotope in a ratio?  

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## 1 answers

### Answer by username_1 on 2024-10-29. Score: 1 | Accepted answer

> The abundance of an isotope, for example, 84Sr, is calculated as: (84Sr/88Sr)/( 84Sr/88Sr + 86Sr/88Sr + 87Sr/88Sr + 88Sr/88Sr) where 87Sr/88Sr can be calculated as 86Sr/88Sr x 87Sr/86Sr and 88Sr/88Sr is, of course, 1.

It would very nice if you could add a reference to that calculation.

The abundance of an isotope, an isotope of strontium for example, is calculated as the ratio of the atomic mass of the isotope in some sample to the naturally present atomic mass of the element, averaged over many samples.

Strontium has four naturally occurring isotopes ($^{84}\text{Sr}$, $^{86}\text{Sr}$, $^{87}\text{Sr}$, and $^{88}\text{Sr}$), one somewhat long-lived isotope, $^{90}\text{Sr}$, with a half-life of 28.9 years, and several very short-lived isotopes that don't occur in nature. Even ignoring that one somewhat long-lived isotope (it is but a tiny, tiny trace isotope, so it is safe to ignore it), all four remaining isotopes must be taken into account. Three of those four isotopes are presumably primordial. The fourth, $^{87}\text{Sr}$, is a mix of primordial and radioactive decay from $^{87}\text{Rb}$. As an aside, this feature is what makes rubidium–strontium dating important.

Suppose you have broken down a sample of strontium into its isotopes, by mass. Then the relative abundance of any one of those isotopes is calculated, by atomic mass, as the ratio of the atomic mass of the isotope in question to the atomic mass of the sample as a whole. This will, by design, add to one (or 100%).


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