The density of #"CO"_2# at #0^@ "C"# is #"1.98 g/L"#. How many mols are found in #"88.0 L"# of #"CO"_2# at this temperature and #"1 atm"#?

2 Answers
Apr 8, 2016

Answer:

#"3.88 moles CO"_2#

Explanation:

The thing to remember about Standard Temperature and Pressure, STP, is that when an ideal gas is being kept under these conditions for pressure and temperature, one mole of this gas occupies #"22.7 L"#.

The volume occupied by one mole of an ideal gas at any conditions for pressure and temperature is called the molar volume.

Under STP conditions, which are currently defined as a pressure of #"100 kPa"# and a temperature of #0^@"C"#, you will be dealing with the molar volume of a gas at STP, which is equal to #"22.7 L"#.

So, you can use the molar volume of a gas at STP as a conversion factor to figure out how many moles of carbon dioxide, #"CO"_2#< would occupy #"88.0 L"#

#88.0 color(red)(cancel(color(black)("L"))) * overbrace("1 mole"/(22.7color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas aat STP")) = "3.87665 moles"#

Rounded to three sig figs, the answer will be

#"volume of CO"_2 = color(green)(|bar(ul(color(white)(a/a)"3.88 moles"color(white)(a/a)|)))#

SIDE NOTE More often than not, you'll see STP conditions being defined as a pressure of #"1 atm"# and a temperature of #0^@"C"#.

Under these conditions for pressure and temperature, one mole of any ideal gas occupies #"22.4 L"#.

If this is the value given to you for the molar volume of a gas at STP, simply redo the calculation using #"22.4 L"# instead of #"22.7 L"#.

Apr 8, 2016

Here's an alternative way to do this, without requiring the assumption that #"CO"_2# is ideal (though it is quite close).

If you use the density of #"CO"_2# (which is given as #"1.98 g/L"# at #0^@ "C"#), it is possible to solve for the exact number of #"mol"#s.

The main catch with using the ideal gas law is that it assumes all gases have the same molar volume at STP, and of course, the "more realistic" gases don't.

When you use the experimental density and convert that to the molar density, it circumvents that assumption because you'd be using all experimental data. Here's how you do it:

#1/(barV) = barrho = n/V = m/(VM_r) = \mathbf((rho)/M_r)#

where:

  • #V# is volume in #"L"#
  • #n# is #"mol"#s
  • #M_r# is relative molar mass in #"g/mol"#
  • #m# is mass in #"g"#
  • #rho# is the mass density in #"g/mol"#

The molar density has units of #"mol/L"#. Thus, simply multiply by the volume to get:

#"mol CO"_2 = ("1.98" cancel"g")/(cancel"L")xx("mol")/(44.009 cancel"g") xx 88.0 cancel"L"#

# = color(blue)("3.92 mol CO"_2)#

And this is larger but somewhat close to Stefan's answer of #"3.88 mol"# by about #1.2%#.

#Z = (PV)/(nRT) = 0.99435 ~~ 1# at #15^@ "C"# suggests that #"CO"_2# is actually easier to compress than an ideal gas.

This answer agrees with that statement because there are a greater number of #"mol"#s of #"CO"_2# than expected for an ideal gas for the same volume, meaning that it is more compressed and thus more dense than an ideal gas.