# Why is the empirical formula and chemical formula of "C"_12"H"_22"O"_11 the same?

Sep 15, 2017

Here's why that is the case.

#### Explanation:

The thing to remember about a compound's molecular formula is that you can determine it by multiplying its empirical formula by a whole number, let's say $n$.

color(blue)(ul(color(black)("molecular formula" = n xx "empirical formula")))" " " "color(darkorange)("(*)")

Moreover, you should know that a compound's empirical formula must show the smallest whole number ratio that exists between the elements that make up the compound.

In your case, you have ${\text{C"_12"H"_22"O}}_{11}$ as the empirical formula because the ratio that exists between carbon, $\text{C}$, hydrogen, $\text{H}$, and oxywen, $\text{O}$, is the smallest whole number ratio that can exist for this formula.

In other words, you can divide $12$, $22$, and $11$ by a whole number and still get whole numbers for all three elements.

Now, in order to see why this also represents the compound's molecular formula, you need to compare the molar mass of sucrose and the molar mass of the empirical formula.

M_ ("M sucrose") = "342.30 g mol"^(-1)

https://en.wikipedia.org/wiki/Sucrose

For the empirical formula, you have

overbrace(12 xx "12.011 g mol"^(-1))^(color(blue)("for carbon")) + overbrace(22 xx "1.008 g mol"^(-1))^(color(blue)("for hydrogen")) + overbrace(11 xx "16.0 g mol"^(-1))^(color(blue)("for oxygen"))

$\approx {\text{342.30 g mol}}^{- 1}$

Now use equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ to find the value of $n$

overbrace("342.30 g mol"^(-1))^(color(blue)("the molar mass of sucrose")) = n xx overbrace("342.30 g mol"^(-1))^(color(blue)("the molar mass of the empirical formula"))

As you can see, you have

$n = 1$

which means that ${\text{C"_12"H"_22"O}}_{11}$ is both the empirical and the molecular formula.