# Question bed32

Jan 28, 2018

$\text{CH"_4"N"_2"O}$

#### Explanation:

The problem tells you that in this unknown compound, hydrogen, carbon, oxygen, and nitrogen are present in a $1 : 3 : 4 : 7$ mass ratio.

This means that the masses of each element will always be in that ratio regardless of the total mass of the sample.

So, for example, you can pick a sample that contains exactly $\text{1 g}$ of hydrogen and use the mass ratio to say that this sample will also contain

• $3 \times \text{1 g" = "3 g C}$
• $4 \times \text{1 g" = "4 g O}$
• $7 \times \text{1 g" = "7 g N}$

Now, to find the compound's empirical formula, you must figure out the smallest whole number mole ratio that exists between the elements.

To do that, use their respective molar masses to convert the number of grams in this sample to moles.

$\text{For H: " 1 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "0.9921 moles H}$

$\text{For C: " 3color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.0122color(red)(cancel(color(black)("g")))) = "0.2498 moles C}$

$\text{For O: " 4 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.2500 moles O}$

$\text{For N: " 7 color(red)(cancel(color(black)("g"))) * "1 mole N"/(14.0067color(red)(cancel(color(black)("g")))) = "0.4998 moles N}$

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

"For H: " (0.9921 color(red)(cancel(color(black)("moles"))))/(0.2498color(red)(cancel(color(black)("moles")))) = 3.972 ~~ 4

"For C: " (0.2498color(red)(cancel(color(black)("moles"))))/(0.2498color(red)(cancel(color(black)("moles")))) = 1

"For O: " (0.2500color(red)(cancel(color(black)("moles"))))/(0.2498color(red)(cancel(color(black)("moles")))) = 1.001 ~~ 1

"For N: " (0.4998 color(red)(cancel(color(black)("moles"))))/(0.2498color(red)(cancel(color(black)("moles")))) = 2.001 ~~ 2

Since $4 : 1 : 1 : 2$ is the smallest whole number mole ratio that can exist between the elements, you can say that the compound's empirical formula will be

$\text{C"_1"H"_4"N"_2"O"_1 => "CH"_4"N"_2"O}$

As an interesting fact, this is also the empirical formula of urea, "CO"("NH"_2)_2#, for which the empirical formula is also the molecular formula.