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"#?

1 Answer
Dec 27, 2017

Answer:

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#.