# 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.

Sep 8, 2017

The mass percent of ${\text{Tl"_2"SO}}_{4}$ in the sample is 1.465%.

#### Explanation:

Step 1. Write an equation for the reaction

A partial equation for the reaction is

${M}_{\textrm{r}} : \textcolor{w h i t e}{m} 504.83 \textcolor{w h i t e}{m m m m l l} 331.29$
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 $\text{Tl}$ are balanced.

Step 2. Calculate the moles of $\text{TlI}$

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

Step 3. Calculate the moles of ${\text{Tl"_2"SO}}_{4}$

"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 ${\text{Tl"_2"SO}}_{4}$

${\text{Mass of Tl"_2"SO"_4 = 2.7529 × 10^"-4"color(red)(cancel(color(black)("mol Tl"_2"SO"_4))) × (504.83 "g Tl"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol Tl"_2"SO"_4)))) = "0.138 97 g Tl"_2"SO}}_{4}$

Step 5. Calculate the mass percent of ${\text{Tl"_2"SO}}_{4}$

"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!