Question #eb15a

1 Answer
Aug 20, 2016

Answer:

#"C"_4"H"_4"S"#

Explanation:

!! LONG ANSWER !!

The first thing to do here is to use the mass of carbon dioxide produced by the combustion of your

#7.96 color(red)(cancel(color(black)("mg"))) * "1 g"/(10^3color(red)(cancel(color(black)("mg")))) = 7.96 * 10^(-3)"g"#

sample of thiophene. To do that, use the fact that every mole of carbon dioxide, #"CO"_2#, contains one mole of carbon, and that carbon dioxide has a molar mass of #"44.01 g mol"^(-1)#.

So, the sample of carbon dioxide produced by the combustion reaction will be equivalent to

#16.65 color(red)(cancel(color(black)("mg"))) * (1 color(red)(cancel(color(black)("g"))))/(10^3color(red)(cancel(color(black)("mg")))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = 3.783 * 10^(-4)"moles CO"_2#

Use the fact that carbon has a molar mass of #"12.011 g mol"^(-1)# to calculate the mass of carbon present in the sample of thiophene that underwent combustion

#3.783 * 10^(-4)color(red)(cancel(color(black)("moles CO"_2))) * (1color(red)(cancel(color(black)("mole C"))))/(1color(red)(cancel(color(black)("mole CO"_2)))) * "12.011 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = 4.54 * 10^(-3)"g C"#

The percent composition of carbon in thiophene will thus be

#(4.54 * color(red)(cancel(color(black)(10^(-3)"g"))) )/(7.96 * color(red)(cancel(color(black)(10^(-3)"g")))) xx 100 = "57.04% C"#

Now, you know that another sample of thiophene was subjected to a series of reactions that resulted in the formation of barium sulfate, #"BaSO"_4#.

This compound is said to contain all the sulfur that was initially present in the sample of thiophene

#4.31 color(red)(cancel(color(black)("mg"))) * "1 g"/(10^3color(red)(cancel(color(black)("mg")))) = 4.31 * 10^(-3)"g"#

Once again, use the mass of barium sulfate produced by the reactions to figure out the percent composition of sulfur in thiophene.

Barium sulfate has a molar mass of #"233.43 g mol"^(-1)#, and one mole of barium sulfate contains one mole of sulfur, which has a molar mass of #"32.065 g mol"^(-1)#.

You will thus have

#11.96 color(red)(cancel(color(black)("mg"))) * (1color(red)(cancel(color(black)("g"))))/(10^3color(red)(cancel(color(black)("mg")))) * "1 mole BaSO"_4/(233.43 color(red)(cancel(color(black)("g")))) = 5.123 * 10^(-5)"moles BaSO"_4#

This means that the sample of thiophene contained

#5.123 * 10^(-5)color(red)(cancel(color(black)("moles BaSO"_4))) * (1color(red)(cancel(color(black)("mole S"))))/(1color(red)(cancel(color(black)("mole BaSO"_4)))) * "32.065 g"/(1color(red)(cancel(color(black)("mole S")))) = 1.643 * 10^(-3)"g"#

of sulfur. You can thus say that the percent composition of sulfur in thiophene is equal to

#(1.643 color(red)(cancel(color(black)( * 10^(-3)"g"))))/(4.31 color(red)(cancel(color(black)( * 10^(-3)"g")))) xx 100 = "38.12% S"#

Now, you didn't mention this in the question, but I'm fairly certain that the problem given to you does mention the fact that thiophene is made up of carbon, sulfur, and hydrogen.

The percent composition of hydrogen in thiophene will thus be

#"% H" = 100% - ("% S" + "% C")#

#"% H" = 100% - (57.04% +38.12%) = "4.84% H"#

At this point, all you have to do is use the percent composition of the compound to find its empirical formula.

Let's pick the first sample of thiophene, #7.96 * 10^(-3)"g"#. This sample contains

#4.54 * 10^(-3)"g C " -># calculated earlier

#7.96 * 10^(-3)color(red)(cancel(color(black)("g thiophene"))) * "4.84 g H"/(100color(red)(cancel(color(black)("g thiophene")))) = 3.853 * 10^(-4)"g H"#

#7.96 * 10^(-3) color(red)(cancel(color(black)("g thiophene"))) * "38.12 g S"/(100color(red)(cancel(color(black)("g thiophene")))) = 3.034 * 10^(-3)"g S"#

Use the molar masses of the three elements to calculate how many moles of each you have in the sample

#"For C: " 3.783 * 10^(-4)"moles C " -># calculated earlier

#"For S: " 3.034 * 10^(-3) color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.065color(red)(cancel(color(black)("g")))) = 9.462 * 10^(-5)"moles S"#

#"For H: " 3.853 * 10^(-4)color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = 3.823 * 10^(-4)"moles H"#

Divide all values by the smallest one to get the mole ratio that exists between carbon, hydrogen, and sulfur in thiophene

#"For C: " (3.783 * 10^(-4) color(red)(cancel(color(black)("moles"))) )/(9.462 * 10^(-5)color(red)(cancel(color(black)("moles")))) = 3.998 ~~ 4#

#"For H: " (3.823 * 10^(-4)color(red)(cancel(color(black)("moles"))))/(9.462 * 10^(-5)color(red)(cancel(color(black)("moles")))) = 4.04 ~~ 4#

#"For S: " (9.462 * 10^(-5)color(red)(cancel(color(black)("moles"))))/(9.462 * 10^(-5)color(red)(cancel(color(black)("moles")))) = 1#

Since #4:4:1# is the smallest whole number ratio that can exist here, you can say that the empirical formula of thiophene is

#color(green)(|bar(ul(color(white)(a/a)color(black)("C"_4 "H"_4"S"_1 implies "C"_4"H"_4"S")color(white)(a/a)|)))#