Question #ed628

1 Answer
Apr 1, 2016

Answer:

#"C"_6"H"_6"OS"#

Explanation:

Your strategy here will be to use the molar masses of carbon dioxide, #"CO"_2#, water, #"H"_2"O"#, and barium sulfate, #"BaSO"_4#, to determine how many moles of carbon, hydrogen, and sulfur were present in the initial sample.

This will also allow you to determine the mass, and subsequently the number of moles of oxygen present in the initial sample.

Start with carbon dioxide. You know that one mole of carbon dioxide contains one mole of carbon, so you can say that the #"0.969 g"# sample will contain

#0.969color(blue)(cancel(color(black)("g"))) *(1color(red)(cancel(color(black)("mole CO"_2))))/(44.01color(blue)(cancel(color(black)("g")))) * "1 mole C"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.02202 moles C"#

Do the same for water. In this case, one mole of water will contain two moles of hydrogen, so you can say that the #"0.198 g"# sample will contain

#0.198color(blue)(cancel(color(black)("g"))) * (1color(red)(cancel(color(black)("mole H"_2"O"))))/(18.015color(blue)(cancel(color(black)("g")))) * "2 moles H"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "0.02198 moles H"#

Now focus on sulfur. You know that the second experiment produced #"0.859 g"# of barium sulftate. You also know that one mole of barium sulfate contains one mole of sulfur, which means that the initial sample contained

#0.859color(blue)(cancel(color(black)("g"))) * (1color(red)(cancel(color(black)("mole BaSO"_4))))/(233.39color(blue)(cancel(color(black)("g")))) * "1 mole S"/(1color(red)(cancel(color(black)("mole BaSO"_4)))) = "0.003681 moles S"#

Now, in order to find the number of moles of oxygen present in your #"0.464 g"# sample of this unknown compound, you must first determine the mass of oxygen it contained.

To do that, use the molar masses of carbon, hydrogen, and sulfur.

#0.02202color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "0.2645 g C"#

#0.01099color(red)(cancel(color(black)("moles H"))) * "1.00794 g"/(1color(red)(cancel(color(black)("mole H")))) = "0.02215 g H"#

#0.003681color(red)(cancel(color(black)("moles S"))) * "32.065 g"/(1color(red)(cancel(color(black)("mole S")))) = "0.1180 g S"#

The mass of oxygen will thus be

#m_(O) = m_"compound" - (m_(C) + m_(H) + m_(S))#

#m_(O) = "0.464 g" - ("0.2645 g" + "0.02215 g" + "0.1180 g")#

#m_(O) = "0.05935 g O"#

This will be equivalent to

#0.05935color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.003710 moles O"#

To get the mole ratio that exists between carbon, hydrogen, sulfur, and oxygen in the compound, divide all values by the smallest one

#"For C: " (0.02202color(red)(cancel(color(black)("moles"))))/(0.003710color(red)(cancel(color(black)("moles")))) = 5.94 ~~ 6#

#"For H: " (0.02198color(red)(cancel(color(black)("moles"))))/(0.003710color(red)(cancel(color(black)("moles")))) = 5.92 ~~ 6#

#"For S: " (0.003681color(red)(cancel(color(black)("moles"))))/(0.003710color(red)(cancel(color(black)("moles")))) = 0.992 ~~ 1#

#"For O: " (0.003710color(red)(cancel(color(black)("moles"))))/(0.003710color(red)(cancel(color(black)("moles")))) = 1#

As you know, a compound's empirical formula tells you the smallest whole number ratio that exists between the atoms that make up said compound.

In this case, a ratio of #6:6:1:1# cannot be reduced to a smaller whole number ration, which means that your unknown compound's empirical formula will be

#color(green)(|bar(ul(color(white)(a/a)"C"_6"H"_6"OS"color(white)(a/a)|)))#