# Question f2961

Nov 21, 2015

Here's what I got.

#### Explanation:

The Lewis formula is another term used for the Lewis structure, which as you know is a diagram that shows the bonding of the atoms and the lone pairs in a molecule.

Now, every time you're dealing with percent composition figures, your tool of choice will be the $\text{100.0-g}$ sample.

Picking a $\text{100.0-g}$ sample of the compound will allow you to convert the mass percentages of the elements that make up said compound to grams.

So, if you have a $\text{100.0-g}$ sample of this compound, you can say that it will contain

• $\text{25.0 g } \to$ carbon
• $\text{2.1 g } \to$ hydrogen
• $\text{39.6 g } \to$ fluorine
• $\text{33.3 g } \to$ oxygen

Your goal now is to determine how many moles of each element you get in that sample. This will allow you to determine the compound's empirical formula.

To do that, use the four elements' respective molar masses.

25.0color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "2.0814 moles C"

2.1color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "2.0835 moles H"

39.6color(red)(cancel(color(black)("g"))) * "1 mole F"/(19.0color(red)(cancel(color(black)("g")))) = "2.0842 moles F"

33.3color(red)(cancel(color(black)("g"))) * "1 mole O"/(16.0color(red)(cancel(color(black)("g")))) = "2.0813 moles O"

Next, divide these values by the smallest one to get the mole ratio that exists between the elements in the compound. Even without doing any calculations, it's clear that you will get a $1 : 1 : 1 : 1$ mole ratio between the four elements

"For C: " (2.0814color(red)(cancel(color(black)("moles"))))/(2.0813color(red)(cancel(color(black)("moles")))) = 1

"For H: " (2.0835color(red)(cancel(color(black)("moles"))))/(2.0813color(red)(cancel(color(black)("moles")))) = 1.001 ~~ 1

"For F: " (2.0842color(red)(cancel(color(black)("moles"))))/(2.0813color(red)(cancel(color(black)("moles")))) = 1.001 ~~ 1

"For O: " (2.0813color(red)(cancel(color(black)("moles"))))/(2.0813color(red)(cancel(color(black)("moles")))) = 1

The empirical formula, which tells you what the smallest whole number ratio between the elements in the compound is, will thus be

$\text{C"_1"H"_1"F"_1"O"_1 implies "CHFO}$

To get the molecular formula of the compound, compare the molar mass of the empirical formula with the molar mass of the compound, $\text{48.0 u}$.

The molar mass of the empirical formula will be

$\text{19.0 u" + "16.0 u" + "1.00794 u" + "12.011 u" = "48.0189 u}$

This means that you have

overbrace("48.0189 u")^(color(red)("empirical formula")) xx color(blue)(n) = "48.0 u"#

$\textcolor{b l u e}{n} = \left(48.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{u"))))/(48.0189color(red)(cancel(color(black)("u}}}}\right) = 0.9996 \approx 1$

Therefore ,the molecular formula of the compound will be $\text{CHFO}$.

Now, in order to be able to draw its Lewis structure, you must first determine the total number of valence electrons found in the molecule.

Each element will contribute its specific number of valence electrons, so that the molecule will have

• four valence electrons from carbon
• one valence electron from hydrogen
• seven valence electrons from fluorine
• six valence electrons from oxygen

This means that the molecule will have a total of $18$ valence electrons.

Now, the carbon atom will be the central atom for this molecule. The hydrogen and the fluorine atoms will form single bonds to the central carbon atom.

This will ensure that they both complete their octet (duet in the case of hydrogen). These bonds will account for $4$ of the $18$ valence electrons.

The oxygen atom will form a double bond to the central carbon atom. This will ensure that both the oxygen and the carbon atom will have complete octets.

The double bond will account for $4$ valence electrons, which means that you'll be left with $10$ valence electrons to place as lone pairs.

Fluorine will get three lone pairs of electrons and oxygen will get two lone pairs of electrons. The Lewis structure for this molecule will thus look like this