A sample of strontium has a relative atomic mass of 87.7 and consists of three isotopes, 86Sr, 87Sr and 88Sr. In this sample, the ratio of abundances of the isotopes 86Sr: 87Sr is 1:1. Calculate the percentage abundance of the 88Sr isotope in this sample

Q. A sample of strontium has a relative atomic mass of 87.7 and consists of three isotopes, ⁸⁶Sr, ⁸⁷Sr and ⁸⁸Sr. In this sample, the ratio of abundances of the isotopes ⁸⁶Sr: ⁸⁷Sr is 1:1. Calculate the percentage abundance of the ⁸⁸Sr isotope in this sample.A. The Mr is the weighted mean of the percentage abundance of each isotope and its Ar. The first step would be to write out the full equation with the unknown input. Mr=87.7 Mr=[(186)+(187)+(x88)]/3 = 87.7 ---- 86+87+88x=87.73 ----- 173 +88x = 263.1 ---- 88x = 263.1-173 -> 88x=90.1 ----x=90.1/88 = 1.12625 =~1.13 percentage abundance. A tip would be to see if this is a sensible value; the relative atomic mass was 87.7 - closer to 88 than 87, therefore a higher abundance of ⁸⁸Sr than ⁸⁷Sr makes sense.