Strontium in the natural environment $($Earth$)$ has an isotopic distribution indicated in the first table below. A sample created in the laboratory was designed to have a different isotopic distribution than naturally occurring strontium. Using the data in the second table below, calculate the atomic mass of the lab sample of Strontium.

Natural Strontium:

$\backslash$table$[[$Isotope$,$Exact Mass $($amu$),\backslash$table$[[$Percent$],[$Abundance$],[(\%)]]],{[}^{84}Sr,83\mathrm{.}913,0\mathrm{.}56],{[}^{86}Sr,85\mathrm{.}909,9\mathrm{.}86],{[}^{87}Sr,86\mathrm{.}909,7\mathrm{.}00],{[}^{88}Sr,87\mathrm{.}906,82\mathrm{.}58]]$

Lab sample:

$\backslash$table$[[$Isotope$,$Exact Mass $($amu$),\backslash$table$[[$Percent$],[$Abundance$],[(\%)]]],{[}^{84}Sr,83\mathrm{.}913,52\mathrm{.}00],{[}^{86}Sr,85\mathrm{.}909,28\mathrm{.}00],{[}^{87}Sr,86\mathrm{.}909,8\mathrm{.}00],{[}^{88}Sr,87\mathrm{.}906,12\mathrm{.}00]]$

The atomic mass of naturally occurring $Sr$ is amu $($hint $-$ where can we find this?$)$

The atomic mass of the Lab created sample of $Sr$ is

$\_$amu