# Question 240cc

Mar 1, 2017

42.1%

#### Explanation:

The idea here is that all the sulfur that was present in the $\text{0.157 g}$ sample of organic compound will now be present in the $\text{0.4813 g}$ of barium sulfate.

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)

${M}_{\text{M S" = "32.065 g mol}}^{- 1}$

To make the calculations easier, let's pick a sample of barium sulfate that contains exactly $1$ mole of this salt. Since you know that every mole of barium sulfate contains $1$ mole of sulfur, you can say that the percent composition of sulfur in barium sulfate is equal to

"% 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 $\text{13.7365 g}$ of sulfur for every $\text{100 g}$ of barium sulfate. Consequently, your sample will contain

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.