posted by Tran .
Heres the question:
"In determining the simplest formula of lead sulfide, 2.46 grams of lead are placed in a crucible with 2.00 grams of sulfur. When the reaction is complete, the product has a mass of 3.22 grams. What mass of sulfur should be used to in the simplest forumla calculation? Find the simplest formula of lead sulfide."
Could someone go through steps for me.
mass Pb = 2.46 g.
mass lead sulfide = 3.22 g.
mass S used = 3.22-2.46 = 0.76 g.
mols Pb = 2.46/207.20 = 0.01187
mols S = 0.76/32.066 = 0.02370
ratio: Divide the smallest number by itself thereby assuring you that will be 1.0000.
mols Pb = 0.01187/0.01187 = 1.000
mols S = 0.02370/0.01187 = 1.9967
rounded to whole numbers we have Pb = 1.0 and S = 2.0 so formula is PbS2.
Note: I have carried out the numbers to far more places than is allowed so round down to two (since 0.76 has two places) at the most. Also, PbS is the formula for lead sulfide but these numbers don't reflect that.
Check my work.
Thanks for the fast reply!