# Question f0d05

Mar 31, 2016

${\text{C"_4"H"_6"O}}_{6}$

#### Explanation:

My guess here would be that you already have the empirical formula of tartaric acid, and that you must find its molecular formula by using its molar mass.

Assuming that the starting point here is the empirical formula of tartaric acid looks like this

${\text{C"_2"H"_3"O}}_{3} \to$ empirical formula

Now, a compound's empirical formula tells you the smallest whole number ratio that exists between the atoms of its constituent elements.

In this case, you know that the molecular formula, which tells you the exact number of atoms that make up a molecule of said compound, will contain

color(black)("2 atoms of C" color(red)(" for every ") {("3 atoms of H"),("3 atoms of O") :}

In essence, you're looking for a multiple of the empirical formula.

Tartaric acid is said to have a molar mass of ${\text{150 g mol}}^{- 1}$. Your goal here will be to determine the molar mass of its empirical formula, then use this value to determine the compound's molecular formula.

In this case, you will have

$2 \times \text{12.011 g mol"^(-1) " } \textcolor{b l u e}{+}$
$3 \times {\text{1.00794 g mol}}^{- 1}$
$3 \times {\text{15.9994 g mol}}^{- 1}$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a}}$
${\text{75.044 g mol}}^{- 1} \to$the molar mass of the empirical formula

So, if the empirical formula has a molar mass of ${\text{75.044 g mol}}^{- 1}$, and the molar mass of tartaric acid is ${\text{150 g mol}}^{- 1}$, it follows that you'll need to multiply the empirical formula by

$\left(150 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g mol"^(-1)))))/(75.044color(red)(cancel(color(black)("g mol}}^{- 1}}}}\right) = 1.999 \approx \textcolor{red}{2}$

to get the molecular formula. This means that the molecular formula of tartaric acid will be

("C"_2"H"_3"O"_3)_color(red)(2) implies color(green)(|bar(ul(color(white)(a/a)"C"_4"H"_6"O"_6color(white)(a/a)|)))#

Here's how a molecule of tartaric acid looks like