# Find the empirical formula of the compound.A 2.203 g sample of an organic compound was extracted from a plant. When it was burned in oxygen, the hydrogen in the compound was converted to 1.32 g of...

Find the empirical formula of the compound.

A 2.203 g sample of an organic compound was extracted from a plant. When it was burned in oxygen, the hydrogen in the compound was converted to 1.32 g of water and the carbon was oxidised to 3.23 g of carbon dioxide.

txmedteach | Certified Educator

We'll start by considering the combustion equation:

Sample + 02 -> CO2 + H20

This tells us that the only products are going to be CO2 and H20, which is important because there is no other place for the carbon or hydrogen to go (let's assume complete combustion, so no CO is produced)!

Amount of Carbon

First, let's figure out how much carbon we started off with by determining what mass of the carbon dioxide is carbon. To do this, we'll need to figure out what proportion of the mass of CO2 is carbon based on molar masses given on the periodic table in the following way:

% mass Carbon = Molar Mass Carbon/Molar Mass CO2

% mass Carbon = 12.0 / (12.0 + 2*16.0) = 27.3 %

This means that 27.3% of our mass of carbon dioxide is carbon:

Mass Carbon = % mass Carbon * mass CO2

Mass Carbon = 0.273 * 3.23 = 0.882 g Carbon

Now, we continue with Hydrogen

Finding the mass of hydrogen

We will find the mass of hydrogen in the same way we found the mass of carbon above.

We'll start by finding our percentage of hydrogen in the water produced by the reaction. There is one difference, though. In water, we have 2 hydrogen atoms per molecule, so we'll be doubling the numerator:

% mass Hydrogen = 2 * Molar mass Hydrogen/Molar Mass Water

% mass Hydrogen = 2 * 1.01/(2*1.01 + 16.0) = 11.2%

Now, we know 11.2% of the mass of water is hydrogen, so we'll figure out our actual mass of hydrogen by multiplying our % hydrogen by the mass of water:

mass Hydrogen = % mass Hydrogen * mass water

mass Hydrogen = 0.112 * 1.32 g = 0.148 g Hydrogen

Now, we'll see whether or not we've accounted for all the mass in our sample

Find out if we're missing anything

Now, we know the mass of carbon and hydrogen we started off with. We also know the mass of our original sample. If only carbon and hydrogen were present, the combined mass of carbon and hydrogen should equal the mass of the sample. Let's see if this is the case:

Mass Carbon + Mass Hydrogen ?= Mass Sample

0.882 + 0.148 ?= 2.203

1.030 != 2.203

Looks like we're missing some mass! In many biological organic macromolecules, you have a lot of oxygen. In fact, in 2 out of the 4 types of macromolecules, pretty much the only types of atoms you have are Carbon, Hydrogen, and Oxygen.

Because nitrogen and phosphorus don't fit into our regular combustion equation, we'll assume the rest of the mass in the sample is Oxygen, and see whether we get a reasonable answer. If we assume the rest of the mass is oxygen, we get the following result:

Mass Sample - (Mass Carbon + Mass Hydrogen) = Mass Oxygen

2.203 - 1.030 = 1.173 g

To determine if this is reasonable, let's figure out the number of moles of each element and then calculate the molar ratios. To do this, we'll divide our mass of each element by their respective molar mass:

Moles O = Mass O / Molar Mass O = 1.173 / 16.0 = 0.073 mol

Moles H = Mass H / Molar Mass H = 0.148 / 1.01 = 0.147 mol

Moles C = Mass C / Molar Mass C = 0.882 / 12.0 = 0.074 mol

To get a sense of the ratio of moles O to moles H to moles C, we'll just divide each number by the element with the smallest number of moles (in our case, 0.073 mol Oxygen)

Moles O/Moles O = 1

Moles H/Moles O = 2.01

Moles C/Moles O = 1.01

This gives us the ratio of each atom to each other atom. This means that for 1 oxygen atom, we have about 2 hydrogen atoms and 1 carbon atom. This, incidentally is our Empirical Formula:

CH(2)O

This makes sense based on biology, because it is the canonical empirical formula for carbohydrates!