# Question f9d12

Jun 10, 2016

${\text{55.3 g mol}}^{- 1}$

#### Explanation:

The trick here is to realize that the density of the gaseous element is actually five times that of molecular oxygen, ${\text{O}}_{2}$.

Your strategy here will be to determine the

You can use the ideal gas law equation

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

Here you have

$P$ - 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

to show that the molecular mass of the molecule is five times that of atomic oxygen.

Under similar conditions for pressure and temperature, equal number of moles of gas occupy the same volume. If you take

$n = \frac{m}{M} _ M$

to be the number of moles of gas present in a sample, you can say that

$P V = \frac{m}{M} _ M \cdot R T$

Rearrange to get

$P \cdot {M}_{M} = \frac{m}{V} \cdot R T$

which is equivalent to

${\overbrace{\frac{m}{V}}}^{\textcolor{b l u e}{\text{density}}} = \frac{P}{R T} \cdot {M}_{M}$

As you know, the ratio between the mass of a substance and the volume it occupies gives you that substance's density. Assuming that you're dealing with ideal gases, the same number of moles of ${\text{X}}_{3}$ and of ${\text{O}}_{2}$ will occupy the same volume $V$.

Now, you have

${m}_{{O}_{2}} / V = \frac{P}{R T} \cdot {M}_{{\text{M O}}_{2}} \to$ for oxygen gas

and

${m}_{{X}_{3}} / V = \frac{P}{R T} \cdot {M}_{{\text{M X}}_{3}} \to$ for the triatomic molecule ${\text{X}}_{3}$

You know that

${\overbrace{{m}_{{X}_{3}} / \textcolor{red}{\cancel{\textcolor{b l a c k}{V}}}}}^{\textcolor{p u r p \le}{{\text{density of X"_ 3)) = 5 xx overbrace(m_ (O_2)/color(red)(cancel(color(black)(V))))^(color(darkgreen)("density of O}}_{2}}}$

$\textcolor{w h i t e}{a a a a} {m}_{{X}_{3}} = 5 \times {m}_{{O}_{2}}$

This means that the difference between these densities lies in the fact that each individual molecule of ${\text{X}}_{3}$ is five times heavier than a molecule of ${\text{O}}_{2}$.

Therefore, the molar mass of ${\text{X}}_{3}$ must be five times bigger than the molar mass of ${\text{O}}_{2}$.

${M}_{{\text{M X"_ 3) = 5 xx M_("M O}}_{2}}$

The molar mass of oxygen gas is approximately equal to ${\text{32 g mol}}^{- 1}$, which means that you have

M_ ("M X"_3) = 5 xx "32 g mol"^(-1) = "160 g mol"^(-1)

The molecule is triatomic, which of course means that it contains $3$ atoms of element $\text{X}$. Therefore, the molar mass of $\text{X}$ will be

M_"M X" = M_("M X"_3)/3 = "160 g mol"^(-1)/3 = color(green)(|bar(ul(color(white)(a/a)color(black)("55.3 g mol"^(-1))color(white)(a/a)|)))

I'll leave the answer rounded to three sig figs.

Jun 10, 2016

Atomic mass(relative) =55.3 (up to one decimal place)

#### Explanation:

For the sake of convenient discussion the Symbol of the gaseous element is considered as E .

Since it is tri-atomic one its molecular formula will be written as ${E}_{3}$

If its Relative atomic mass is a then its Relative molecular mass will be 3a .

Now by the given condition

Under the same condition of temperature and pressure

$\left(\text{The density of "E_3" gas")/("The density of " O_2 " gas }\right) = 5$

Hence we can write Mass per unit voume in place of density to represent the ratio.

$\left(\text{The Mass of unit volume of " E_3 " gas")/(" The Mass of unit volume of " O_2 " gas }\right) = 5$

Again by Avogadro's Law It is known that under the same condition of temperature and pressure equal volume of all gases contain same number of molecules. Here we consider that under the same condition of temperature and pressure unit volume of all gases contain n molecules.

So the relation becomes

$\left(\text{The Mass of n molecules of " E_3 " gas")/(" The Mass of n molecules of " O_2 " gas }\right) = 5$

$= \left(\text{The Mass of 1 molecule of " E_3 " gas")/(" The Mass of 1 molecules of " O_2 " gas }\right) = 5$

Dividing both numerator and denominator by

${\text{Mass of" 1/12 " part of " }}_{6}^{12} C$ we get

The ratio

$= \left(\left(\text{The mass of 1 molecule of " E_3" gas")/ ( "Mass of 1/12th part of ""_6^12C" ) )/(("The mass of 1 molecule of "O_2 " gas ")/( "Mass of 1/12th part of " ""_6^12C}\right)\right) = 5$

$\implies \left(\text{The Relative Molecular mass of " E_3" gas")/("The Relative Molecular mass of "O_2 " gas }\right) = 5$

Inserting

$\text{The Relative Molecular mass of " O_2 " gas } = 32$

$\text{ and Relative Molecular mass of } {E}_{3} = 3 a$

we get

=>(3a)/32"=5=>color(blue)(a =160/3~~55.3)#

Please Note that Relative Molecular mass,Relative Atomic mass are unit less quantities
As discussed here
http://digipac.ca/chemical/molemass/moles2.htm