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