Video: Determining the Empirical and Molecular Formulas of a Gas from Its Formula Mass and Percentage Composition

What are the empirical and molecular formulas for a gas molecule if its formula mass is 28 atomic mass units and elemental analysis revealed its percentage composition to be C = 86% and H = 14%? [A] CH₂, C₂H₄ [B] CH₃, C₂H₆ [C] CH, C₂H₂ [D] CH₂, C₃H₉ [E] CH, C₂H₄

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Video Transcript

What are the empirical and molecular formulas for a gas molecule if its formula mass is 28 atomic mass units and elemental analysis revealed its percentage composition to be C equals 86 percent and H equals 14 percent? A) CH₂ and C₂H₄, B) CH₃ and C₂H₆, C) CH and C₂H₂, D) CH₂ and C₃H₉, or E) CH and C₂H₄.

An empirical formula simply presents the atoms or ions of each element in the simplest ratio. For instance, in hydrogen peroxide, there is one oxygen atom for each hydrogen atom. So the empirical formula is simply HO. But there are two hydrogen atoms and two oxygen atoms in a molecule of hydrogen peroxide. So the molecular formula is H₂O₂. And this is the structure of a hydrogen peroxide molecule. So the molecular formula represents the composition of a single molecule.

The next thing the question tells us is the formula mass. In this context, the formula mass is simply the mass of a single molecule. For instance, with hydrogen peroxide, the formula mass is equal to two times the atomic mass of hydrogen, which is one, plus two times the atomic mass of oxygen, which is 16. Giving us 34 unified atomic mass units. You might see amu instead of u. They refer to the same property and give us the same result. So I’m going to use you throughout.

The next bit of information in the question refers to elemental analysis. This is the process where we work out how much of something is made up of each element. In our example of hydrogen peroxide, we can see that hydrogen contributes two out of the 34 unified atomic mass units that make up a molecule of hydrogen peroxide. So the mass contribution of hydrogen is 5.9 percent. This means that oxygen contribute 94.1 percent of the mass. Our job is to work backwards. We know this particular gas contains only carbon and hydrogen because their mass percentages sum to 100 percent.

So our formula must have this form, where we have a certain number of carbon atoms and a certain number of hydrogen atoms. And we know that the number of carbon atoms multiplied by the atomic mass of carbon plus the number of hydrogen atoms multiplied by the atomic mass of hydrogen must equal the formula mass of 28 unified atomic mass units. The contribution from carbon makes up 86 percent of the mass. And the contribution from hydrogen makes up 14 percent. We can work out the relative amount of carbon and hydrogen by simply dividing the percentage by the atomic mass.

Here, I’ve simply removed the units because we’re only interested in the magnitude of the number. What this means is roughly for every seven of carbon, we have 14 of hydrogen. But we don’t exactly have seven of carbon. We have seven and a sixth. Well, the truth is that the numbers you get from elemental analysis are often rounded. So we can dismiss any value that’s not quite a round number. So the ratio of carbon to hydrogen in our mystery compound is about seven to 14, which is one to two. So for every carbon atom we have, there must be two hydrogen atoms. This means our empirical formula is CH₂.

But if we add the atomic mass of carbon and twice the atomic mass of hydrogen. We only get 14 unified atomic mass units. That’s only half our formula mass. So we need two lots of these amounts of atoms. So our molecular formula is two lots of carbon and two lots of two hydrogen. So that’s C₂H₄. This means, for a gas molecule with a mass of 28 atomic mass units, which is 86 percent by mass of carbon and 14 percent by mass of hydrogen. The empirical formula is CH₂, and the molecular formula is C₂H₄.

Another way of doing this would be to simply take the formula mass and work out the contributions from each element. So the mass due to hydrogen is four unified atomic mass units. And the mass due to carbon is 24 unified atomic mass units. We can then divide by the atomic masses to give us the number of atoms per molecule, two carbons, four hydrogens. We could then simplify the molecular formula to give us the empirical formula. So this process goes the other way around, but it’s just as effective.

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