# Question a065d

Jul 21, 2017

The molecular formula is a) ${\text{MCl}}_{4}$.

#### Explanation:

Step 1. Calculate the molar mass of the compound

To do this, we can use the Ideal Gas Law:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Since $n = \frac{m}{M}$. we can write

$p V = \frac{m}{M} R T$

And we can rearrange this to

$M = \frac{m R T}{p V}$

STP is defined as 0 °C and 1 bar, so

M = ("16.96 g" × "0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar"))) × 1 color(red)(cancel(color(black)("L")))) = "385.16 g·mol"^"-1"#

Step 2. Calculate the formula of the compound

Your compound contains 37.321 % $\text{Cl}$.

(a) Calculate the mass of $\text{Cl}$ in 1 mol of the compound.

$\text{Mass of Cl" = 385.16 color(red)(cancel(color(black)("g compound"))) × "37.321 g Cl"/(100 color(red)(cancel(color(black)("g compound")))) = "143.74 g Cl}$

(b) Calculate the moles of $\text{Cl}$

$\text{Moles of Cl" = 143.74 color(red)(cancel(color(black)("g Cl"))) × "1 mol Cl"/(35.45 color(red)(cancel(color(black)("g Cl")))) = "4.055 mol Cl" ≈ "4 mol Cl}$

∴ The formula of the compound is ${\text{MCl}}_{4}$.