# A component of protein called serine has an approximate molar mass of 100 g/mole. If the percent composition is as follows, what is the empirical and molecular formula of serine?

## The percent composition is: C=34.95% H = 6.844% O=46.56% and N=13.59%.

Mar 2, 2016

${\text{C"_3"H"_7"NO}}_{3}$

#### Explanation:

Here's a great example of a problem that allows you to use a bit of logic to find the molecular formula first, then backtrack to find the empirical formula.

Notice that you are given the molar mass of serine, which is said to be approximately equal to ${\text{100 g mol}}^{- 1}$.

That means that you can expect the actual molar mass of serine to be either a little smaller than ${\text{100 g mol}}^{- 1}$ or a little bigger than ${\text{100 g mol}}^{- 1}$.

You can pick a $\text{100-g}$ sample of serine and see how many moles of each element it contains. Remember, if you take the mass of one mole of a compound, the number of moles of each element must come out to be integers.

Why?

Because one molecule of a compound will always contain whole numbers of atoms of each constituent element.

So, if you were to pick a $\text{100-g}$ sample of serine, you would end up with

• $\text{34.95 g } \to$ carbon
• $\text{6.844 g } \to$ hydrogen
• $\text{46.56 g } \to$ oxygen
• $\text{13.59 g } \to$ nitrogen

Use the molar mass of each element to determine how may moles of each you get in one mole of serine

$\text{For C: " 34.95 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "2.91 moles C}$

$\text{For H: " 6.844color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "6.80 moles H}$

$\text{For O: "46.56color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "2.91 moles O}$

$\text{For N: " 13.59color(red)(cancel(color(black)("g"))) * "1 mole N"/(14.00674color(red)(cancel(color(black)("g")))) = "0.97 moles N}$

You can stop the calculations at this point and say that since these values must be rounded up to give integers, the molecular mass of serine is a little bigger than ${\text{100 g mol}}^{- 1}$.

Since you're still in the approximately equal to ${\text{100 g mol}}^{- 1}$, you can round these values to the nearest integer to get

$\text{For C: " 2.91 ~~ "3 moles C}$

$\text{For H: " 6.80 ~~ "7 moles H}$

$\text{For O: " 2.91 ~~ "3 moles O}$

$\text{For N: " 0.97 ~~ "1 mole N}$

Therefore, if one mole of serine contains that many moles of each element, you can say that its molecular formula is

$\textcolor{g r e e n}{{\text{C"_3"H"_7"N"_1"O"_3 implies "C"_3"H"_7"NO}}_{3}} \to$ molecular formula

Since $3 : 7 : 1 : 3$ is also the smallest whole number ratio that can exist between these values, this will also be serine's empirical formula

$\textcolor{g r e e n}{{\text{C"_3"H"_7"NO}}_{3}} \to$ empirical formula

If you want to test this result, simply divide the number of moles of each element you found in the $\text{100-g}$ sample by the smallest value of the group to get the compound's empirical formula.

Once you know the empirical formula, you can determine the molecular formula by using the $3 : 7 : 1 : 3$ ratio that exists between the elements.