Rubidium has two common isotopes, $""^85"Rb"$ and $""^87"Rb"$. If the abundance of $""^85"Rb"$ is 72.2% and the abundance of $""^87"Rb"$ is 27.8%, what is the average atomic mass of rubidium?

87.71 amu (I am assuming degrees of significance here...) In order to determine the average of an element, we take the weighted average of all of the of that element....

The abundance of $""_8^16"O"$ is 99.762 %, and the abundance of $""_8^18"O"$ is 0.201 %. Assume you have 100 000 atoms of O. Then you have 37 atoms of $""_8^17"O"$...

The idea with that have two naturally occurring is that the percent abundances of those two must add up to give $100%$. In calculations, it is often easier to...

$""^63Cu$ has $69.2%$ abundance. $""^65Cu$ has $30.8%$ abundance. So, the weighted average is $62.93xx69.2%$ $+$ $64.93xx30.8%$ $=$ $63.55$ $"amu"$. If we look at , copper metal (a mixture of isotopes...

Multiply the atomic mass of each isotope times its percent abundance in decimal form. Add them together. $"Average atomic mass of Br"$$=$$(78.92xx0.5069)+(80.92xx0.49331)="79.92 u"$

We know that: $x_(1)106.9059 + x_(2)108.9047 = 107.8682$ $(i)$ $x_(1) + x_(2) = 1$ $(ii)$ We assume that silver has only the 2 , which for a first approximation...

So, we take the weighted atomic mass of $""^87Rb$ and $""^85Rb$. $(85xx80.2% + 87xx19.8%)"amu" = 85.40$ $"amu"$. Check to see if this atomic mass is close to that listed on...

The average of an element is determined by taking the weighted average of the atomic masses of its naturally occurring . Now, weighted average simply means that each...

As you know, the average of an element is calculated by taking the weighted average of the atomic masses of its stable . More specifically, the average atomic...

And thus the $"weighted average"$ is: $(62.93xx69.2%+64.93xx30.8%)*"amu"$ $=$ $63.55$ $"amu"$