Question

In a laboratory experiment, pure lead metal reacted with excess sulfur to produce a lead sulfide compound. The following data was collected: Mass of empty evaporating dish: 25.000 g Mass of evaporating dish and lead metal: 26.927 g Mass of evaporating dish and lead sulfide: 27.485 g Solve for the empirical formula of the lead sulfide compound. Be sure to show, or explain, all of your calculations.

Answer 1

To get the empirical formula, we need to know the ratio of the moles of both lead and sulfur.

First, lets get the mass of each:
mass of lead = mass of evaporating dish and lead - mass of empty dish
= 26.927 - 25.000 = 1.927 gm of lead (Pb)
mass of sulfur = mass of dish and lead sulfide - mass of dish and lead
= 27.485 - 26.927 = 0.558 gm of sulfur (S)

The next step is to get the number of moles in 1.927 gm of Pb and in 0.558 gm of S:
From the periodic table:
molar mass of lead = 207.2 gm
molar mass of sulfur = 32 gm

number of moles = mass / molar mass
number of moles of Pb = 1.927 / 207.2 = 0.0093 moles
number of moles of S = 0.558 / 32 = 0.0174 moles

The third step is to get the ratio between the moles by dividing the number of moles of each by the smaller of the two:
1 mole of Pb (0.0093 / 0.0093) reacts with 0.0174 / 0.0093 = 1.99 (approximately 2 moles) of sulfur.

Final step is to write the empirical formula based on the ratio:
Lead sulfide compound is written as : PbS2