Question

model 3 - proposed average atomic mass calculations
marys method
(78.99)(23.9850 amu) + (10.00)(24.9858 amu) + (11.01)(25.9826 amu)
------------------------------ = ________
100
jacks method
(0.7899)(23.9850 amu) + (0.1000)(24.9858 amu) + (0.1101)(25.9826 amu) = ________
alans method
23.9850 amu + 24.9858 amu + 25.9826 amu
------------------------------ = ________
3

Explanation:

Step1: Calculate Mary's method

First, calculate each product:
$(78.99)(23.9850)=78.99\times23.9850 = 1904.57015$
$(10.00)(24.9858)=249.858$
$(11.01)(25.9826)=11.01\times25.9826 = 286.068426$
Then sum them up: $1904.57015 + 249.858+286.068426=2440.496576$
Finally, divide by 100: $\frac{2440.496576}{100}=24.40496576\approx24.405$ amu

Step2: Calculate Jack's method

Calculate each product:
$(0.7899)(23.9850)=0.7899\times23.9850 = 18.9457515$
$(0.1000)(24.9858)=2.49858$
$(0.1101)(25.9826)=0.1101\times25.9826 = 2.86068426$
Then sum them up: $18.9457515+2.49858 + 2.86068426=24.30501576\approx24.305$ amu

Step3: Calculate Alan's method

Sum the atomic - masses: $23.9850+24.9858 + 25.9826=74.9534$
Then divide by 3: $\frac{74.9534}{3}=24.984466\cdots\approx24.984$ amu

Answer:

Mary's method: $24.405$ amu
Jack's method: $24.305$ amu
Alan's method: $24.984$ amu