# Question 6d48d

Oct 14, 2015

$\text{C"_2"H"_6"O}$
$\text{45.88 g/mol}$

#### Explanation:

So, you know that your organic compound, $\text{C"_x"H"_y"O}$, 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

$\left(1 \times 16.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g/mol"))))/(zcolor(red)(cancel(color(black)("g/mol}}}}\right) \times 100 = 34.87$

This means that $z$ will be equal to

$z = \frac{16.0 \times 100}{34.87} = \text{45.88 g/mol}$

The molar mass of the compound is equal to $\text{45.88 g/mol}$. To find the values of $x$ and $y$, use a sample of $\text{45.88 g}$ of the compound.

Since this represents one mole of the compound, it will contain one mole of oxygen, the equivalent of $\text{16.0 g}$.

The remaining mass will be the carbon and the hydrogen

${m}_{\text{C" + m_"H" + m_"O" = m_"compound}}$

${m}_{\text{C" + m_"H" = "45.88 g" - "16.0 g" = "29.88 g}}$

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"

$12.011 \cdot x + 1.008 \cdot y = 29.88$

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

$y = \frac{29.88 - 12.011 \times 1}{1.008} = 17.72 \approx 18$

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

$y = \frac{29.88 - 2 \times 12.011}{1.008} = 5.81 \approx 6$

This is not a very clean result, but it will have to do. The compound will thus be $\text{C"_2"H"_6"O}$, 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 $\text{46.003 g/mol}$.

For $x = 2$ you will get

$y = \frac{30.003 - 2 \times 12.011}{1.008} = 5.93 \approx 6$

The values are cleaner this time.