Question #240cc
1 Answer
Explanation:
The idea here is that all the sulfur that was present in the
Your starting point here will be to figure out the percent composition of sulfur in barium sulfate. To do that, use the molar mass of barium sulfate and the molar mass of elemental sulfur
#M_ ("M BaSO"_ 4) = "233.43 g mol"^(-1)#
#M_"M S" = "32.065 g mol"^(-1)#
To make the calculations easier, let's pick a sample of barium sulfate that contains exactly
#"% S" = (32.065 color(red)(cancel(color(black)("g"))))/(233.43color(red)(cancel(color(black)("g")))) xx 100% = 13.7365%#
This means that you get
#0.4813 color(red)(cancel(color(black)("g BaSO"_4))) * "13.7365 g S"/(100color(red)(cancel(color(black)("g BaSO"_4)))) = "0.066114 g S"#
Therefore, the percent composition of sulfur in the organic compound was
#"% S" = (0.066114 color(red)(cancel(color(black)("g"))))/(0.157color(red)(cancel(color(black)("g")))) xx 100% = color(darkgreen)(ul(color(black)(42.1%)))#
The answer is rounded to three sig figs.