Question

Tin has 10 stable isotopes. The heaviest, 124Sn, make up 5.80% of naturally occurring tin atoms. How many atoms of 124Sn are present in 93.0 g of naturally occurring tin?

Answer 1

Answer: 2.735 x 10²¹

Step-by-step explanation:

Given:

Atomic mass of tin
`(M_{\text{Sn}})` = 118.71 g/mol

Percent of Sn present = 5.80%

Mass of Sn
`(m_{\text{Sn}})` = 93.0 g

To calculate moles:

`\text{Moles of Sn} = \frac{\text{Given mass}}{\text{Atomic mass}}`

`\text{Moles of Sn} = \frac{93.0 \text{g}}{118.71 \text{g/mol}} = 0.783 \ \text{mol}`

One mole
`(N_A)` contains Avogadro's number
`6.023 * 10^{23` of atoms.

Therefore, 0.23 moles of Sn contains:

`0.783 * 6.023 * 10^(23) = 4.716 * 10^(23) \ $atoms`

The total number of atoms in naturally occurring Sn can be calculated using the percentage of the heaviest isotope:

Percentage of heaviest isotope of tin (
`^(124)\text{Sn}`) = 5.80%

Total number of atoms in tin (
`N_{\text{total}}`) =
`(5.80)/(100) * 4.716 * 10^(23) \ $atoms`

Total number of atoms of (
`^(124)\text{Sn}`) in naturally occurring tin (
`N_(124)`) is therefore =
`2.735 * 10^(21) \ $atoms`