An unknown element "X" with oxygen forms a gas compound "X"_2"O" whose density at -120^@"C" and a pressure of "320 mmHg" is "1.809 g/L". What is element "X"?

Dec 27, 2017

Here's what I got.

Explanation:

Your starting point here will be to use the ideal gas law equation and the density of this compound to find its molar mass.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the initial pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the mixture
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Now, you know that when kept at a temperature of $- {120}^{\circ} \text{C}$ and a pressure of $\text{320 mmHg}$, this compound has a density of ${\text{1.809 g L}}^{- 1}$, which means that under these conditions for pressure and temperature, $\text{1 L}$ of this compound has a mass of $\text{1.809 g}$.

To make the calculations easier, let's assume that we have exactly $\text{1 L}$ of this gas. Use the ideal gas law equation to find the number of moles of gas present in this sample

$P V = n R T \implies n = \frac{P V}{R T}$

Plug in your values to find--do not forget to convert the temperature of the gas to Kelvin and its pressure to atmospheres!

$n = \left(\frac{320}{760} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 1 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + (-120)) color(red)(cancel(color(black)("K}}}}\right)$

$n = \text{0.03349 moles}$

Now, this sample has a mass of $\text{1.809 g}$ and contains $0.003349$ moles of $\text{X"_2"O}$, so you can say that the molar mass of the compound, i.e. the mass of exactly $1$ mole of $\text{X"_2"O}$, will be

M_ ("M X"_2"O") = "1.809 g"/"0.03349 moles" = "54.16 g mol"^(-1)

Notice that the chemical formula of the compound suggests that every $1$ mole of $\text{X"_2"O}$ contains

• $2$ moles of element $\text{X}$, $2 \times \text{X}$
• $1$ mole of oxygen, $1 \times \text{O}$

Since elemental oxygen gas a molar mass of about ${\text{16.0 g mol}}^{- 1}$, you can say that the mass of element $\text{X}$ in $1$ mole of $\text{X"_2"O}$ is

$\text{mass X" = "54.16 g " - " 16.0 g}$

$\text{mass X}$ $=$ $\text{38.16 g}$

This is the mass of exactly $2$ moles of element $\text{X}$, which implies that the molar mass of $\text{X}$ is

M_ ("M X") = "38.16 g"/"2 moles" = "19.08 g mol"^(-1)

Rounded to two sig figs, the molar mass of element $\text{X}$ would be

M_ ("M X") = "19 g mol"^(-1)

The closest match that you have for the identity of element $\text{X}$ is fluorine, $\text{F}$, which has a molar mass of

M_ ("M F") = "18.998 g mol"^(-1)

This would make the unknown compound oxygen difluoride, ${\text{OF}}_{2}$.