Answer to Question #239433 in General Chemistry for Ellen

Question

On earth, tin has 10 different stable isotopes. The heaviest, \ce{^{124}Sn}

X

124

​X

2

2

124

​Sn makes up 5.80\%

5.80% of naturally occurring tin atoms. How many atoms of \ce{^{124}Sn}

X

124

​X

2

2

124

​Sn are present in 18.3

18.3 g of a natural sample of tin?

Considering \ce{^{124}Sn}

X

124

​X

2

2

124

​Sn has an isotopic mass of \ce{{123.905} amu}

123.905 amu , calculate the mass of this isotope in the sample.

Answer 1

The average atomic mass of Sn is 118.71 g/mol

the percentage of heaviest Sn is 5.80%

the given mass of Sn is 82g

The total moles of Sn will be = mass / atomic mass = 82/118.71=0.691

Total atoms of Sn in 82g = 6.023× 10²³× 0.691= 4.16×10²³

the percentage of heaviest Sn is 5.80%

So the total atoms of Sn¹²⁴ = 5.80% X 4.16× 10²³

Total atoms of Sn¹²⁴ = 2.41× 10²² atoms

the mass of Sn¹²⁴ will be =$\frac{2.41× 10^{22} ×124}{6.023×10^{23}}$ = 4.96g