# 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

#### 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,

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) -># themole ratio isequalto 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 **mole** of the compound will produce

#2.0# molesof carbon dioxide#3.5# molesof water#0.50# molesof 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, **whole numbers**, so this will also be the compound's *empirical formula*

#"C"_2"H"_7"N -># molecularandempirical formula