On combustion, 1.0 L of a gaseous compound of hydrogen, carbon, and nitrogen gives 2.0 L of #CO_2#, 3.5 L of #H_2O# vapor, and 0.50 L of #N_2# at STP. What is the empirical formula of the compound?

1 Answer
Jan 4, 2016

Answer:

#"C"_2"H"_7"N"#

Explanation:

The most important thing to realize here is that for gases that react under the same conditions for pressure and temperature, the mole ratio is equivalent to the volume ratio.

This can be proven using the ideal gas law equation for two gases that are part of a chemical reaction at the same conditions for pressure, #P#, and temperature, #T#.

You can say that

#P * V_1 = n_1 * RT -># for the first gas

#P * V_2 = n_2 * RT -># for the second gas

Divide these two equations to get

#(color(red)(cancel(color(black)(P))) * V_1)/(color(red)(cancel(color(black)(P))) * V_2) = (n_1 * color(red)(cancel(color(black)(RT))))/(n_2 * color(red)(cancel(color(black)(RT))))#

Therefore,

#color(blue)(n_1/n_2 = V_1/V_2) -># the mole ratio is equal to the volume ratio

Now, since the unknown compound contains carbon, hydrogen, and nitrogen, let's represent its molecular formula, not its empirical formula, as #"C"_x"H"_y"N"_z#.

The unbalanced chemical equation for this combustion reaction will thus look like this

#"C"_x"H"_y"N"_text(z(g]) + "O"_text(2(g]) -> "CO"_text(2(g]) + "H"_2"O"_text((g]) + "N"_text(2(g])#

You know that at STP, #"1.0 L"# of this compound will produce

  • #"2.0 L"# of carbon dioxide
  • #"3.5 L"# of water
  • #"0.50 L"# of nitrogen gas

Since we've shown that the mole ratio is equivalent to the volume ratio, you can say that #1.0# mole of the compound will produce

  • #2.0# moles of carbon dioxide
  • #3.5# moles of water
  • #0.50# moles of nitrogen gas

From this point on, it all comes down to using mass conservation to help you find how many moles of each element were present in one mole of compound.

You know that you get

  • one mole of carbon for every one mole of carbon dioxide
  • two moles of hydrogen for every one mole of water
  • two moles of nitrogen for every one mole of nitrogen gas

This means that one mole of the original compound contained a total of

  • #x = 2 xx 1 = 2# moles of carbon
  • #y = 3.5 xx 2 = 7# moles of hydrogen
  • #z = 0.5 xx 2 = 1# mole of nitrogen

Therefore, the molecular formula for this compound is

#"C"_2"H"_7"N" -># ethylamine or dimethylamine

Now, the empirical formula gives you the smallest whole number ratio that exists between the elements that make up a compound.

In this case, #2:7:1# is the smallest ratio that can formed with whole numbers, so this will also be the compound's empirical formula

#"C"_2"H"_7"N -># molecular and empirical formula