Question #6d48d

1 Answer

Answer:

#"C"_2"H"_6"O"#
#"45.88 g/mol"#

Explanation:

So, you know that your organic compound, #"C"_x"H"_y"O"#, contains #34.87%# percent composition by mass oxygen.

Moreover, you know that the chemical formula of the compound contains one mole of oxygen, #x# moles of carbon, and #y# moles of hydrogen.

To find the molar mass of the compound, let's say #z#, use oxygen's known percent composition

#(1 xx 16.0color(red)(cancel(color(black)("g/mol"))))/(zcolor(red)(cancel(color(black)("g/mol")))) xx 100 = 34.87#

This means that #z# will be equal to

#z = (16.0 xx 100)/34.87 = "45.88 g/mol"#

The molar mass of the compound is equal to #"45.88 g/mol"#. To find the values of #x# and #y#, use a sample of #"45.88 g"# of the compound.

Since this represents one mole of the compound, it will contain one mole of oxygen, the equivalent of #"16.0 g"#.

The remaining mass will be the carbon and the hydrogen

#m_"C" + m_"H" + m_"O" = m_"compound"#

#m_"C" + m_"H" = "45.88 g" - "16.0 g" = "29.88 g"#

Use carbon and hydrogen's molar masses to write these mases using the number of moles of each you get per mole of compound, which is of course #x# and #y#, respectively

#overbrace(12.011"g"/color(red)(cancel(color(black)("mol"))) * xcolor(red)(cancel(color(black)("moles"))))^(color(blue)("mass of carbon")) + overbrace(1.008"g"/color(red)(cancel(color(black)("mol"))) * ycolor(red)(cancel(color(black)("moles"))))^(color(red)("mass of hydrogen")) = "29.88 g"#

#12.011 * x + 1.008 * y = 29.88#

At this point, it becomes obvious that #x=2# and #y=6#.

Think of it like this. You're dealing with one mole of the compound, so #x# and #y# are whole numbers.

Since you cannot have #x>2#, i.e. #x = 3, 4, 5...# since that would not satisfy the above equation, you can deduce that the formula can either contain

  • one mole of carbon, #x=1#

This would imply that

#y = (29.88 - 12.011 xx 1)/1.008 = 17.72 ~~ 18#

This is not avalid option because you cannot have one carbon atom attached to an oxygen atom and #18# hydrogen atoms.

  • two moles of carbon

This will get you

#y = (29.88 - 2 xx 12.011)/1.008 = 5.81 ~~ 6#

This is not a very clean result, but it will have to do. The compound will thus be #"C"_2"H"_6"O"#, which is either ethanol or methoxymethane.

SIDE NOTE Using a percent composition by mass of oxygen equal to #34.78%# will actually produce better results.

In this case, the molar mass will be #"46.003 g/mol"#.

For #x=2# you will get

#y = (30.003 - 2 xx 12.011)/1.008 = 5.93 ~~ 6#

The values are cleaner this time.