Question

a certain element consists of two stable isotopes. the first has an atomic mass of 62.9298 amu and a percent natural abundance of 69.09 %. the second has an atomic mass of 64.9278 amu and a percent natural abundance of 30.91 %. what is the atomic weight of the element?
amu

Explanation:

Step1: Recall atomic - weight formula

The atomic weight (average atomic mass) of an element with two isotopes is calculated using the formula $A = m_1x_1 + m_2x_2$, where $m_1$ and $m_2$ are the atomic masses of the isotopes and $x_1$ and $x_2$ are their respective natural - abundances.

Step2: Identify given values

We have $m_1=62.9298$ amu, $x_1 = 0.6909$, $m_2 = 64.9278$ amu, and $x_2=1 - 0.6909=0.3091$.

Step3: Calculate atomic weight

$A=(62.9298\times0.6909)+(64.9278\times0.3091)$
First term: $62.9298\times0.6909 = 62.9298\times(0.69 + 0.0009)=62.9298\times0.69+62.9298\times0.0009=(63 - 0.0702)\times0.69+(62.9298\times0.0009)=(63\times0.69-0.0702\times0.69)+0.05663682=(43.47-0.048438)+0.05663682 = 43.47819882$.
Second term: $64.9278\times0.3091=(65 - 0.0722)\times0.3091=65\times0.3091-0.0722\times0.3091=20.0915-0.02221702 = 20.06928298$.
$A = 43.47819882+20.06928298=63.5474818\approx63.55$ amu.

Answer:

$63.55$ amu