Question 425a5

Jun 18, 2017

${\text{C"_8"H"_6"O}}_{4}$

Explanation:

The first thing to do here is to figure out the empirical formula of the unknown carboxylic acid.

Start by converting the percent composition to masses by picking a $\text{100-g}$ sample of this compound.

You will have

• $\text{57.83 g C}$
• $\text{3.64 g H}$
• $\text{38.52 g O}$

Next, use the molar masses of the three elements to convert the masses to moles

$\text{For C: " 57.83 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "4.815 moles C}$

$\text{For H: " 3.64 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.008color(red)(cancel(color(black)("g")))) = "3.611 moles H}$

$\text{For O: " 38.52 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "2.408 moles O}$

Next, divide all three values by the smallest one to get the mole ratio that exists between the three elements in the compound

"For C: " (4.815 color(red)(cancel(color(black)("moles"))))/(2.408color(red)(cancel(color(black)("moles")))) = 1.9996 ~~ 2

"For H: " (3.611 color(red)(cancel(color(black)("moles"))))/(2.408color(red)(cancel(color(black)("moles")))) = 1.4996 ~~ 1.5

"For O: " (2.408color(red)(cancel(color(black)("moles"))))/(2.408color(red)(cancel(color(black)("moles")))) = 1

Now, in order to get the empirical formula of the compound, you need the smallest whole number ratio that exists between its constituent elements.

$\text{C : H : O = } 2 : 1.5 : 1$

To get the smallest whole number ratio, multiply all three values by $2$. You will have

$\text{C : H : O = } 4 : 3 : 2$

You can thus say that the empirical formula of the compound is

${\text{C"_4"H"_3"O}}_{2}$

Now, the thing to keep in mind about the molecular formula is that it is always a multiple of the empirical formula.

$\text{molecular formula" = color(blue)(n) * "empirical formula}$

This means that the molar mass of the compound is equal to a multiple of the molar mass of the empirical formula.

The molar mass of the empirical formula is

$4 \cdot {\text{12.011 g mol"^(-1) + 3 * "1.008 g mol"^(-1) + 2 * "15.9994 g mol}}^{- 1}$

$= {\text{83.07 g mol}}^{- 1}$

This means that you have

${\text{molar mass" = color(blue)(n) * "83.07 g mol}}^{- 1}$

Since you know that you have

"molar mass" in ["158 g mol"^(-1), "167 g mol"^(-1)]

you can say that $\textcolor{b l u e}{n}$ will vary between

$\textcolor{b l u e}{n} = \left(158 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g mol"^(-1)))))/(83.07color(red)(cancel(color(black)("g mol}}^{- 1}}}}\right) = 1.902$

and

$\textcolor{b l u e}{n} = \left(167 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g mol"^(-1)))))/(83.07color(red)(cancel(color(black)("g mol}}^{- 1}}}}\right) = 2.01$

Since it's obvious that the only whole number present in this interval is

$\textcolor{b l u e}{n = 2}$

you can say that the molecular formula will be

"molecular formula" = color(blue)(2) * "C"_4"H"_3"O"_2 = color(darkgreen)(ul(color(black)("C"_8"H"_6"O"_4)))#