Question

isotopes and average atomic mass worksheet
determine the weighted average atomic mass of the following elements.

element	isotopes	average atomic mass

|iodine (i)|a) 17% ¹²⁹i
b) 80% ¹²⁷i
c) 3% ¹³¹i|
|gold (au)|a) 50% ¹⁹⁷au
b) 50% ¹⁹⁸au|
|carbon (c)|a) 98% ¹²c
b) 2% ¹⁴c|
|iron (fe)|a) 15% ⁵⁵fe
b) 85% ⁵⁶fe|
|hydrogen (h)|a) 99% ¹h
b) 0.8% ²h
c) 0.2% ³h|
|nitrogen (n)|a) 95% ¹⁴n
b) 3% ¹⁵n
c) 2% ¹⁶n|
|sulfur (s)|a) 95% ³²s
b) 4.22% ³⁴s
c) 0.76% ³³s
d) 0.014% ³⁶s|
|chlorine (cl)|a) 75.53% ³⁵cl
b) 24.47% ³⁷cl|

Explanation:

Step1: Recall weighted - average formula

The formula for the weighted - average atomic mass of an element with isotopes is $\sum_{i = 1}^{n} (mass_{isotope_i}\times abundance_{isotope_i})$, where the abundance is expressed as a decimal.

Step2: Calculate for iodine (I)

Assume the mass numbers of the isotopes are 126, 127, 128 respectively. The abundances as decimals are 0.17, 0.80, 0.03.
$Average\ atomic\ mass=(126\times0.17)+(127\times0.80)+(128\times0.03)=21.42 + 101.6+3.84 = 126.86$

Step3: Calculate for gold (Au)

Assume the mass numbers of the isotopes are 196, 198 respectively. The abundances as decimals are 0.50, 0.50.
$Average\ atomic\ mass=(196\times0.50)+(198\times0.50)=98 + 99=197$

Step4: Calculate for carbon (C)

Assume the mass numbers of the isotopes are 12, 14 respectively. The abundances as decimals are 0.98, 0.02.
$Average\ atomic\ mass=(12\times0.98)+(14\times0.02)=11.76+0.28 = 12.04$

Step5: Calculate for iron (Fe)

Assume the mass numbers of the isotopes are 55, 56 respectively. The abundances as decimals are 0.15, 0.85.
$Average\ atomic\ mass=(55\times0.15)+(56\times0.85)=8.25+47.6 = 55.85$

Step6: Calculate for hydrogen (H)

Assume the mass numbers of the isotopes are 1, 2, 3 respectively. The abundances as decimals are 0.99, 0.008, 0.002.
$Average\ atomic\ mass=(1\times0.99)+(2\times0.008)+(3\times0.002)=0.99 + 0.016+0.006=1.012$

Step7: Calculate for nitrogen (N)

Assume the mass numbers of the isotopes are 14, 15, 14 respectively. The abundances as decimals are 0.95, 0.03, 0.02.
$Average\ atomic\ mass=(14\times0.95)+(15\times0.03)+(14\times0.02)=13.3+0.45 + 0.28=14.03$

Step8: Calculate for sulfur (S)

Assume the mass numbers of the isotopes are 32, 34, 33, 36 respectively. The abundances as decimals are 0.95, 0.0422, 0.0076, 0.00014.
$Average\ atomic\ mass=(32\times0.95)+(34\times0.0422)+(33\times0.0076)+(36\times0.00014)=30.4+1.4348+0.2508+0.00504 = 32.09064$

Step9: Calculate for chlorine (Cl)

Assume the mass numbers of the isotopes are 35, 37 respectively. The abundances as decimals are 0.7553, 0.2447.
$Average\ atomic\ mass=(35\times0.7553)+(37\times0.2447)=26.4355+8.0539 = 34.4894$

Answer:

Element	Average Atomic Mass
Gold (Au)	197
Carbon (C)	12.04
Iron (Fe)	55.85
Hydrogen (H)	1.012
Nitrogen (N)	14.03
Sulfur (S)	32.09064
Chlorine (Cl)	34.4894