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.

1 Answer

Answer:

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!