Question

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"`?

Answer 1

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.

`color(blue)(ul(color(black)(PV = nRT)))`

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("atm L")/("mol K")`
• `T` is the absolute temperature of the gas

Now, you know that when kept at a temperature of `-120^@"C"` and a pressure of `"320 mmHg"`, this compound has a density of `"1.809 g L"^(-1)`, which means that under these conditions for pressure and temperature, `"1 L"` of this compound has a mass of `"1.809 g"`.

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

`PV = nRT implies n = (PV)/(RT)`

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

`n = (320/760 color(red)(cancel(color(black)("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"))))`

`n = "0.03349 moles"`

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

• `2` moles of element `"X"`, `2 xx "X"`
• `1` mole of oxygen, `1 xx "O"`

Since elemental oxygen gas a molar mass of about `"16.0 g mol"^(-1)`, you can say that the mass of element `"X"` in `1` mole of `"X"_2"O"` is

`"mass X" = "54.16 g " - " 16.0 g"`

`"mass X"` `=` `"38.16 g"`

This is the mass of exactly `2` moles of element `"X"`, which implies that the molar mass of `"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 `"X"` would be

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

The closest match that you have for the identity of element `"X"` is fluorine, `"F"`, which has a molar mass of

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

This would make the unknown compound oxygen difluoride, `"OF"_2`.