Question

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naturally occurring zirconium exists as five stable isotopes: $^{90}zr$ with a mass of 89.905 amu (51.45%); $^{91}zr$ with a mass of 90.906 amu (11.22%); $^{92}zr$ with a mass of 91.905 amu (17.15%); $^{94}zr$ with a mass of 93.906 amu (17.38%), and $^{96}zr$ with a mass of 95.908 amu (2.80%). calculate the average mass of zirconium.
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Explanation:

Step1: Convert percentages to decimals

$51.45\% = 0.5145$, $11.22\%=0.1122$, $17.15\% = 0.1715$, $17.38\%=0.1738$, $2.80\% = 0.0280$

Step2: Multiply mass of each isotope by its abundance

For $^{90}Zr$: $89.905\times0.5145 = 46.2561225$
For $^{91}Zr$: $90.906\times0.1122 = 10.1996532$
For $^{92}Zr$: $91.905\times0.1715 = 15.7617075$
For $^{94}Zr$: $93.906\times0.1738 = 16.3108628$
For $^{96}Zr$: $95.908\times0.0280 = 2.685424$

Step3: Sum up the products

$46.2561225+ 10.1996532+15.7617075+16.3108628+2.685424 = 91.21377$

Answer:

$91.21$ amu