Elemental Analysis

Using atomic absorption, the following percent composition data were obtained: 40.9 % Carbon (C), 4.57 % Hydrogen (H), and 54.5 % Oxygen (O). What is the empirical formula of the sample?

Our first step is to convert from a percentage to grams. We can assume that the total gram amount is 100 g. Technically, we can choose any amount since the ratio will be the same, but this method is simpler. Doing this conversion, we get:

\(100\,g*40.9\%=40.9\,g\,C\)

\(100\,g*4.57\%=4.57\,g\,H\)

\(100\,g*54.5\%=54.5\,g\,O\)

Next, we need to convert to moles. This is because the empirical formula (and all chemical formulas) are in units of moles. To do this, you divide the number of moles by each element's atomic mass, which is the amount of grams of the element per mole. It is also the weight listed under the element on the periodic table.

\(\frac{40.9\,g}{\frac{12.01\,g}{mol}}=3.41\,mol\,C\)

\(\frac{4.57\,g}{\frac{1.01\,g}{mol}}=4.52\,mol\,H\)

\(\frac{54.5\,g}{\frac{16.00\,g}{mol}}=3.41\,mol\,O\)

Our last step is to divide each molar amount by the smallest variable. Doing so will give us the ratio between elements.

\(\frac{3.41\,mol\,C}{3.41}=1\,C\)

\(\frac{4.52\,mol\,H}{3.41}=1.33\,H\)

\(\frac{3.41\,mol\,O}{3.41}=1\,O\)

We have a ratio of 1:1.33:1; however, chemical formulas are in whole numbers. So what do we do? Well, we multiply our ratio by the smallest number possible to get all whole numbers. In this case, that number is 3.

\(1\,C*3=3\,C\)

\(1.33\,H*3=4\,H\)

\(1\,O*3=3\,O\)

So our final formula is:

\(C_3H_4O_3\)

Using atomic absorption, the following percent composition data were obtained: 85.6 % Carbon (C) and 14.4 % Hydrogen (H). What is the molecular formula of the sample if the total mass is 84.16 g?

Since we know the actual mass, we will multiply the percentages by that mass:

\(84.16\,g*85.6\%=72.1\,g\,C\)

\(84.16\,g*14.4\%=12.1\,g\,H\)

Now we convert from grams to moles:

\(\frac{72.1\,g}{\frac{12.01\,g}{mol}}=6\,mol\,C\)

\(\frac{12.1\,g}{\frac{1.01\,g}{mol}}=12\,mol\,H\)

So the molecular formula is:

\(C_{6}H_{12}\)

An unknown gaseous sample was analyzed using a mass spectrometer, giving the data below. What is the molecular formula of the sample?

Fig. 2 - Mass spectrometry data for an unknown sample.

Our first step is to understand what this data is telling us. Mass spectrometry breaks things into fragments, not necessarily elements, so each peak is a fragment or an element. The peak that has 100% intensity is the peak for the molecule itself, so we know that the entire molecule has a mass of 44 g.

This means that the other peaks are either elements or fragments (i.e., smaller molecules). Let's label each peak, going from left to right.

The first peak is 12 g, which means it is carbon, which has an atomic mass of 12 g/mol. The second peak is 16 g, so it is oxygen (oxygen has an atomic mass of 16 g/mol).

Now for the 28 g peak. It could be either silicon or carbon monoxide (CO), which both have masses of 28 g/mol. We know that this sample is a gas, so it is more likely that this is CO and not silicon (which is usually a solid). This also makes sense since the total mass of the sample is 44 g, and the mass of C+O+Si is 56 g.

Now that we know our peaks, we can determine the identity of the sample. So we know that the molecule contains two elements: C and O, so we can subtract those masses from the total mass.

\(44-(12.00+16.00)=16.00\)

Since there are 16 g "left over", that means there is another oxygen atom present, so the complete formula is \(CO_2\)