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.
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
