Question

problem

1. an unknown element x has three naturally occurring isotopes:

x - 35: mass = 34.96885 amu, % abundance = 75.76%
x - 37: mass = 36.96590 amu, % abundance = 24.24%
x - 39: mass = 38.96371 amu, % abundance = 0.20%
a. calculate the average atomic mass of x. show all work for full credit. (round to 5
isotope
% abundance
35(0.7576) 26.516
35.
\\( \frac { 35 ( 0.7576 ) } { 0.7576 } \\)
\\( \frac { 37 ( 0.2424 ) ( 8.9688 ) } { 0.2424 } \\)
\\( \frac { 39 ( 0.002 ) 0.078 } { 0.002 } \\)
35.5628
b. identify the unknown element with name and atomic number:
cl - 17

Explanation:

Step1: Convert percentage abundances to decimals

For X - 35: \(75.76\%=0.7576\)
For X - 37: \(24.24\% = 0.2424\)
For X - 39: \(0.20\%=0.002\)

Step2: Calculate the contribution of each isotope to the average atomic mass

Contribution of X - 35: \(34.96885\times0.7576=\frac{34.96885\times7576}{10000}\approx26.516\)
Contribution of X - 37: \(36.96590\times0.2424=\frac{36.96590\times2424}{10000}\approx8.9688\)
Contribution of X - 39: \(38.96371\times0.002 = 0.07792742\)

Step3: Sum up the contributions

\(26.516+8.9688 + 0.07792742=35.56272742\approx35.56\) (rounded to two decimal places)

Answer:

a. The average atomic mass of \(X\) is approximately \(35.56\) amu.
b. The element is Chlorine (\(Cl\)) with atomic number \(17\).