Question

on a distant planet, the fractional abundance of helium is given in the table below. what is the average atomic mass on this planet? solve to 5 significant figures. hint: the sum of fractions should equal 1.

isotope	isotopic mass (amu)	fraction
he-4	4.0026	?

Explanation:

Step1: Find fraction of He-4

The sum of fractions is 1. So fraction of He-4 = \(1 - 0.407 = 0.593\)

Step2: Calculate average atomic mass

Average atomic mass = (mass of He-3 × fraction of He-3) + (mass of He-4 × fraction of He-4)
= \((3.0160 \times 0.407) + (4.0026 \times 0.593)\)
First, calculate \(3.0160 \times 0.407 \approx 1.227512\)
Then, calculate \(4.0026 \times 0.593 \approx 2.3735418\)
Add them together: \(1.227512 + 2.3735418 \approx 3.6010538\)
Round to 5 significant figures: \(3.6011\) (or check calculation again for precision)
Wait, let's recalculate more precisely:
\(3.0160 \times 0.407\):
\(3.0160 \times 0.4 = 1.2064\), \(3.0160 \times 0.007 = 0.021112\), total \(1.2064 + 0.021112 = 1.227512\)
\(4.0026 \times 0.593\):
\(4.0026 \times 0.5 = 2.0013\), \(4.0026 \times 0.09 = 0.360234\), \(4.0026 \times 0.003 = 0.0120078\)
Sum: \(2.0013 + 0.360234 = 2.361534 + 0.0120078 = 2.3735418\)
Total average: \(1.227512 + 2.3735418 = 3.6010538\), which rounds to \(3.6011\) amu (5 significant figures)
Wait, maybe I made a mistake in significant figures. Let's check the given values:
He-3 mass: 3.0160 (5 sig figs), fraction: 0.407 (3 sig figs)
He-4 mass: 4.0026 (5 sig figs), fraction: 0.593 (3 sig figs)
When multiplying, the number of sig figs is determined by the least, but when adding, it's about decimal places. Wait, but the problem says to solve to 5 significant figures. Let's do the calculation with more precision:
\(0.407 \times 3.0160 = 0.407 \times 3 + 0.407 \times 0.0160 = 1.221 + 0.006512 = 1.227512\)
\(0.593 \times 4.0026 = 0.593 \times 4 + 0.593 \times 0.0026 = 2.372 + 0.0015418 = 2.3735418\)
Sum: \(1.227512 + 2.3735418 = 3.6010538\), which is \(3.6011\) when rounded to 5 significant figures (the fifth digit is 0, wait no: 3.6010538. Let's count: 3 (1), 6 (2), 0 (3), 1 (4), 0 (5), 5... So to 5 sig figs, it's 3.6011? Wait, 3.6010538: the first five significant figures are 3,6,0,1,0? Wait no, significant figures start from the first non-zero digit. So 3.6010538: digits are 3 (1), 6 (2), 0 (3), 1 (4), 0 (5), 5, 3, 8. Wait, no, 3.6010538 is 3.6010538. So the first five significant figures: 3,6,0,1,0? Wait, no, 3.6010538. Let's write it as 3.6010538. The first significant figure is 3, second 6, third 0, fourth 1, fifth 0? Wait, no, 3.6010538: the digits are 3 (1), 6 (2), 0 (3), 1 (4), 0 (5), 5 (6), 3 (7), 8 (8). Wait, maybe I miscalculated the sum. Let's do the multiplication again:

\(3.0160 \times 0.407\):
\(3.0160 \times 0.407 = 3.0160 \times (0.4 + 0.007) = 3.0160 \times 0.4 + 3.0160 \times 0.007 = 1.2064 + 0.021112 = 1.227512\)

\(4.0026 \times 0.593\):
\(4.0026 \times 0.593 = 4.0026 \times (0.5 + 0.09 + 0.003) = 4.0026 \times 0.5 + 4.0026 \times 0.09 + 4.0026 \times 0.003 = 2.0013 + 0.360234 + 0.0120078 = 2.0013 + 0.360234 = 2.361534 + 0.0120078 = 2.3735418\)

Now sum: \(1.227512 + 2.3735418 = 3.6010538\)

Now, 3.6010538. Let's count significant figures:

1. 3 (first non-zero)
2. 6
3. 0
4. 1
5. 0 (the fifth significant figure is 0, the next digit is 5, so we round up the fifth digit: 0 becomes 1? Wait, no: 3.6010538. The number is 3.6010538. To five significant figures, we look at the sixth digit to round. The fifth digit is 0 (position: 3.6010 538), so the sixth digit is 5. So we round the fifth digit up: 0 becomes 1. So 3.6011. Yes, that's correct. So 3.6011 amu.

Answer:

\(3.6011\) (amu)