# An organic compound crystallizes in an orthorhombic system with 2 molecules per unit cell. If the density of the crystal is "1.419 g/cm"^3, what is the molar mass of the compound?

## The unit cell dimensios are $12.05 \cdot {10}^{- 8} \text{cm}$, $15.05 \cdot {10}^{- 8} \text{cm}$, and $2.69 \cdot {10}^{- 8} \text{cm}$.

May 14, 2017

${\text{208 g mol}}^{- 1}$

#### Explanation:

The first thing you need to do here is to figure out the volume of a single unit cell by using

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{V = L \times l \times h}}}$

• $L = 15.05 \cdot {10}^{- 8}$ $\text{cm}$
• $l = 2.69 \cdot {10}^{- 8}$ $\text{cm}$
• $h = 12.05 \cdot {10}^{- 8}$ $\text{cm}$

The volume of a single unit cell will thus be

$V = 15.05 \cdot {10}^{- 8} \textcolor{w h i t e}{.} \text{cm" * 2.69 * 10^(-8)color(white)(.)"cm" * 12.05 * 10^(-8)color(white)(.)"cm}$

$V = 4.878 \cdot {10}^{- 22}$ ${\text{cm}}^{3}$

Now, the compound's molar mass tells you the mass of exactly $1$ mole of the compound. As you know, a mole is defined as

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 mole" = 6.022 * 10^(23)color(white)(.)"molecules}}}} \to$ Avogadro's constant

Since you know that each unit cell holds $2$ molecules, you will need to calculate the volume occupied by

6.022 * 10^(23) color(red)(cancel(color(black)("molecules"))) * "1 unit cell"/(2color(red)(cancel(color(black)("molecules")))) = 3.011 * 10^(23) $\text{unit cells}$

To do that, use the volume of a single unit cell

3.011 * 10^(23)color(red)(cancel(color(black)("unit cells"))) * (4.878 * 10^(-22)color(white)(.)"cm"^3)/(1color(red)(cancel(color(black)("unit cell")))) = 146.9 ${\text{cm}}^{3}$

So, you know that $1$ mole of molecules occupies ${\text{146.9 cm}}^{3}$ and that each ${\text{1 cm}}^{3}$ of this compound has a mass of $\text{1.419 g}$.

This means that the mass of $1$ mole will be

146.9 color(red)(cancel(color(black)("cm"^3))) * overbrace("1.419 g"/(1color(red)(cancel(color(black)("cm"^3)))))^(color(blue)("the density of the compound")) = "208.45 g"

Therefore, you can say that this organic compound has a molar mass of

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molar mass = 208 g mol}}^{- 1}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for one of the dimensions of the unit cell.