Question

There are four naturally occurring isotopes of the element chromium. The relative abundance of each is 50Cr - 4.345% 49.946044 amu 52Cr - 83.789% 51.940508 amu 53Cr - 9.501% 52.940649 amu 54Cr - 2.365% 53.93880 amu Find the average atomic mass of chromium. A. 52.061 amu B. 52.978 amu C. 51.996 amu D. 53.2503 amu

Answer 1

The average molar mass of an atom is equal to the summation of the molar mass of the isotopes multiplied by their relative abundance. In this case, we substitute and use the formula:

MW ave = 49.946044 amu*0.04345 + 51.940508 amu * 0.83789 +  52.940649 amu * 0.09501 + 53.93880 amu * 0.02365 = 51.966 amu. 

Answer is C

Answer 2

Answer : The correct option is, (C) 51.996 amu

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Mass of $^{50}Cr$ = 49.946044 amu

Percentage abundance of $^{50}Cr$ = 4.345 %

Fractional abundance of $^{50}Cr$ = 0.04345

Mass of $^{52}Cr$ = 51.940508 amu

Percentage abundance of $^{52}Cr$ = 83.789 %

Fractional abundance of $^{52}Cr$ = 0.83789

Mass of $^{53}Cr$ = 52.940649 amu

Percentage abundance of $^{53}Cr$ = 9.501 %

Fractional abundance of $^{53}Cr$ = 0.09501

Mass of $^{54}Cr$ = 53.93880 amu

Percentage abundance of $^{54}Cr$ = 2.365 %

Fractional abundance of $^{54}Cr$ = 0.02365

Now put all the given values in above formula, we get:

$\text{Average atomic mass of element}=\sum[(49.946044\times 0.04345)+(51.940508\times 0.83789)+(52.940649\times 0.09501)+(53.93880\times 0.02365)]$

$\text{Average atomic mass of element}=51.996amu$

Therefore, the atomic weight of the element is, 51.996 amu