Question

The thallium (present as `Tl_2SO_4`) in a `9.486` g pesticide sample was precipitated as Thallium(I) iodide. Calculate the mass percent of `Tl_2SO_4` in the sample if `0.1824` g of `TlI` was recovered?

Help is much appreciated.

Answer 1

The mass percent of `"Tl"_2"SO"_4` in the sample is `1.465%`.

Explanation:

Step 1. Write an equation for the reaction

A partial equation for the reaction is

`M_text(r):color(white)(m)504.83color(white)(mmmmll)331.29`
`color(white)(mmm)"Tl"_2"SO"_4 + … → "2TlI" + …`

We don’t know what the other reactants and products are.

However, that doesn't matter as long as atoms of `"Tl"` are balanced.

Step 2. Calculate the moles of `"TlI"`

`"Moles of TlI" = 0.1824 color(red)(cancel(color(black)("g TlI"))) × "1 mol TlI"/(331.29 color(red)(cancel(color(black)("g TlI")))) = 5.5058 × 10^"-4"color(white)(l) "mol TlI"`

Step 3. Calculate the moles of `"Tl"_2"SO"_4`

`"Moles of Tl"_2"SO"_4 = 5.5058 × 10^"-4"color(red)(cancel(color(black)("mol TlI"))) × (1 "mol Tl"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol TlI"))))`

`= 2.7529 × 10^"-4"color(white)(l) "mol Tl"_2"SO"_4`

Step 4. Calculate the mass of `"Tl"_2"SO"_4`

`"Mass of Tl"_2"SO"_4 = 2.7529 × 10^"-4"color(red)(cancel(color(black)("mol Tl"_2"SO"_4))) × (504.83 "g Tl"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol Tl"_2"SO"_4)))) = "0.138 97 g Tl"_2"SO"_4`

Step 5. Calculate the mass percent of `"Tl"_2"SO"_4`

`"Mass %" = ("mass of Tl"_2"SO"_4)/"mass of pesticide" × 100 % = ("0.138 97" color(red)(cancel(color(black)("g"))))/(9.486 color(red)(cancel(color(black)("g")))) × 100 % = 1.465 %`

I hope that helps!