Question 66004

Mar 23, 2016

${\text{34 g mol}}^{- 1}$

Explanation:

The idea here is that you need to use the ideal gas law equation to find a relationship between the density of the gas and its molar mass.

As you know, the ideal gas law equation looks like this

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

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

Now, the number of moles can be expressed using the mass, $m$, and the molar mass, ${M}_{M}$, of the gas

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{n = \frac{m}{M} _ M} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug this into the ideal gas law equation to get

PV = m/M_M * RT" " " "color(red)("(*)")

Density is defined as mass per unit of volume. If you take $m$ to be the mass of the gas and $V$ the volume it occupies, you can say that

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\rho = \frac{m}{V}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Notice that you can rearrange equation $\textcolor{red}{\text{(*)}}$ by multiplying both sides by ${M}_{M}$

$P V \cdot {M}_{M} = \frac{m}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{M}_{M}}}}} \cdot R T \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{M}_{M}}}}$

Now divide both sides by $V$ to get

$\frac{P \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{V}}} \cdot {M}_{M}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{V}}}} = \frac{m}{V} \cdot R T$

Finally, isolate ${M}_{M}$ on one side of the equation and plug in the density of the gas

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{M}_{M} = \rho \cdot \frac{R T}{P}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now, STP conditions are defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$. In order to use the above equation, make sure to convert the temperature to Kelvin and the pressure from kPa to atm by using the conversion factor

$\text{1 atm " = " 101.325 kPa}$

Plug in your values to get

M_M = 1.5"g"/color(red)(cancel(color(black)("L"))) * (0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))#

${M}_{M} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{34 g mol}}^{- 1} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ rounded to two sig figs

SIDE NOTE Many sources still use STP conditions as a pressure of * $\text{1 atm}$ and a temperature of* ${0}^{\circ} \text{C}$.

If this is the definition of STP given to you, simply redo the calculations using these values for pressure and temperature. In this particular case, the answer will be the same because it must be rounded to two sig figs.