A robot spacecraft returned samples from the planetesimal 98765 ALEKS, located in the outer Solar System. Mass spectroscopic analysis produced the following
data on the isotopes of tin in these samples:
Isotope
e mass relative
(amu) abundance
112 Sn 111.9
93.5%
116 Sn 115.9 6.5%
Use these measurements to complete the entry for tin in the Periodic Table that would be used on 98765 ALEKS. Be sure your answers have the correct number
of significant digits.
Caution: your correct answer will have the same format but not necessarily the same numbers as the entry for tin in the Periodic Table we use here on Earth
x
5
?

2. You can assume that the relative abundance (93.5% and 6.5%) are exact to the extense that they do not affect the precision of the measures, so the result will contain the same number of significant digits as the isotope mass, i.e. 4

Entry (answer)

50 ← atomic number

Sn ← chemical symbol

112.2 ← atomic mass

Explanation:

Table of data

Mass spectroscopic analysis

Isotope mass relative abundance

(amu)

¹¹²Sn 111.9 93.5%

¹¹⁶Sn 115.9 6.5%

Then entry in the periodic table consists of the chemical symbol (Sn), the atomic number above the symbol, and the atomic mass below the symbol.

1) Atomic number

The atomic number is the same in any planet, as it is the number of protons and identifies uniquely the element. So, tin (Sn) has 50 protons in any place and the atomic number is 50.

2) Atomic mass

Atomic mass is the weigthed average of the atomic masses of the element in its nature state; so it depends of the relative abundance, and si calculated with this formula:

atomic mass = ∑ (relative abundance × atomic mass) of each isotope.

atomic mass Sn = %¹¹²Sn × relative abundance ¹¹²Sn + %¹¹⁶Sn × relative abundance ¹¹⁶Sn

atomic mass Sn = 93.5% × 111.9 amu + 6.5% × 115.9 amu = 112.16 ≈ 112.2 amu

To keep 4 significant figures the number 112.16 hast to be rounded the closest number with one decimal, which is 112.2 (112.16 is closer to 112.2 than to 112.1)