example 1 point two isotopes of rubidium occur naturally. isotopes of rubidium rubidium - 85 rubidium - 87…

example 1 point two isotopes of rubidium occur naturally. isotopes of rubidium rubidium - 85 rubidium - 87 percent abundance 72.15 % 27.85 % atomic mass 84.908 86.9167 calculate the average atomic mass to four decimal places. answer

example 1 point two isotopes of rubidium occur naturally. isotopes of rubidium rubidium - 85 rubidium - 87 percent abundance 72.15 % 27.85 % atomic mass 84.908 86.9167 calculate the average atomic mass to four decimal places. answer

Answer

Explanation:

Step1: Convert percentages to decimals

$72.15% = 0.7215$, $27.85%=0.2785$

Step2: Use average - atomic - mass formula

The average atomic mass formula is $m = m_1\times x_1 + m_2\times x_2$, where $m_1$ and $m_2$ are the atomic masses of the isotopes and $x_1$ and $x_2$ are their fractional abundances. $m=(84.9117\times0.7215)+(86.9092\times0.2785)$

Step3: Calculate the first product

$84.9117\times0.7215 = 84.9117\times\frac{7215}{10000}=61.25389155$

Step4: Calculate the second product

$86.9092\times0.2785 = 86.9092\times\frac{2785}{10000}=24.2042122$

Step5: Sum the two products

$m = 61.25389155+24.2042122 = 85.45810375\approx85.4581$

Answer:

$85.4581$