# Help with question 11 c guys ?

##### 1 Answer

**DISCLAIMER:** *LONG ANSWER!*

The idea here is to use two situations and conservation of mass to separately determine the mass percent of two out of three components in thiophene, which we denote as

- Find mass percent of
#"S"# in thiophene. - Find mass percent of
#"C"# in thiophene and use that to deduce mass percent of#"H"# in thiophene. - Assume
#"100 g"# of thiophene to find the mol ratios. - Check with the molecular mass to see if the empirical formula is any different from the molecular formula.

In the first reaction:

#"7.96 g C"_x"H"_y"S"_z harr "16.65 g CO"_2#

In the second set of reactions:

#"4.31 g C"_x"H"_y"S"_z harr "11.96 g BaSO"_4#

where we assume thiophene was the limiting reagent...

This begins part

STEP 1: MASS PERCENT OF SAs the quantity of sulfur must obey

conservation of mass, here is a nice trick we can use: we can convertfrom the mass of one substance to anotherusing theratio of their molar masses.

#"11.96 g " cancel("BaSO"_4) xx (32.065 cancel"g/mol" " S")/(233.38 cancel"g/mol" cancel("BaSO"_4))#

#=# #"1.643 g S"# thatthiophene,came fromassuming

no other reactantscontained sulfur.That means we can use the thiophene mass from the second set of reactions to find the

mass percentof#"S"# in thiophene.

#(1.643 cancel"g" " S")/(4.31 cancel("g") " thiophene") xx 100% = color(green)ul(38.13%"w/w S")# in thiophene

STEP 2: MASS PERCENT OF C & HFrom the first reaction, notice that thiophene is the only source of carbon for

#"CO"_2# . Thus, we can similarly find the mass of#"C"# that is in#"CO"_2# , and by conservation of mass, that was the mass in thiophene.

#"16.65 g " cancel("CO"_2) xx (12.011 cancel"g/mol" " C")/(44.009 cancel"g/mol" cancel("CO"_2))#

#=# #"4.544 g C"# thatthiophene.came fromThis means the

mass percentof#"C"# in thiophene is:

#(4.544 cancel"g" " C")/(7.96 cancel"g" " thiophene") xx 100% = color(green)ul(57.09% "w/w C")# in thiopheneThis leaves a hydrogen atom

mass percentof:

#100.00% - 38.13% - 57.09%#

#= color(green)ul(4.78_7% "w/w H")# in thiophene(where the subscript indicates the digit past the last significant digit.)

STEP 3: MOL RATIOS & EMPIRICAL FORMULACurrently our mass percents are only relative. If we

achoose#"100 g"# mass of thiophene, then in that mass, we have absolute masses:

#4.78_7# #"g H"# #"57.09 g C"# #"38.13 g S"# Now, if we have everything in mols, they can be compared in a ratio. So, divide by their molar masses to get the

mol quantitiesof each:

#(4.787 cancel"g H")/(1.0079 cancel"g H""/""mol H") = "4.749 mols H"#

#(57.09 cancel"g C")/(12.011 cancel"g C""/mol C") = "4.753 mols C"#

#(38.13 cancel"g S")/(32.015 cancel"g S""/mol S") = "1.191 mols S"# We see that the mols of

#"H"# and#"C"# are about equal, and are also about four times the mols of#"S"# . To proceed, divide by thesmallestmol quantity:

#(1.191 cancel"mols" "S")/(1.191 cancel"mols") = bb1# equiv.of#"S"#

#(4.753 cancel"mols" "C")/(1.191 cancel"mols") ~~ (4.749 cancel"mols" "H")/(1.191 cancel"mols") = 3.991#

#~~ bb4# equivs.of#"C"# and#"H"# eachSo, we construct an

empirical formulaof:

#color(blue)("C"_4"H"_4"S")# This is currently the most "reduced" form of thiophene's chemical formula, i.e. the mol ratios are as small as they can be with integer subscripts.

And now for part

STEP 4: MOLECULAR FORMULAKnowing that its

molecular massis#"84 amu"# , or#"84 g/mol"# , we compare with the molecular mass we would get using its empirical formula to see if we have the proper molecular formula yet.

#M_("C"_4"H"_4"S") = 4 xx "12.011 g/mol" + 4 xx "1.0079 g/mol" + "32.065 g/mol"#

#=# #"84.141 g/mol"# #~~# #"84 g/mol"#

So, our empirical formula IS the molecular formula.