Question

Question #6d48d

Answer 1

`"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.