Switching glucose and formaldehyde would be really bad.
Switching glucose and formaldehyde would be really bad (0:42)! So let’s make sure we know the difference between empirical and molecular formulas! Our episode starts with a definition of empirical formulas (1:20) and the comparison of ionic and molecular compounds (1:49). It briefly recaps the significance of a chemical formula (2:16) and then recaps the steps to calculate an empirical formula using glucose as an example (2:50). But what if you do not have a whole-number ratio right away (5:30) and how do you go from Empirical to Molecular Formula (6:14)? One specific type of calculation is the combustion analysis (7:11).
Question: What is the first step, if I have given grams of elements as part of the substance instead of % composition?
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Hi and welcome to the APsolute Recap: Chemistry Edition. Today’s episode will recap Empirical and Molecular Formulas. This is our first episode of season 2! Looking for more chemistry recap? Go back and check out our other 46 episodes from season 1.
Lets Zoom out:
Unit 1 - Atomic Structure and Properties
Topic 1.3 Elemental Composition of Pure Substances
Big idea - Scale, Proportion and Quantity
Glucose has the well-known molecular formula C6H12O6. Its empirical formula is CH2O. That sounds familiar... That’s formaldehyde! Ugh, yes, the one they use to conserve specisms. It would be REALLY bad to mix those two up… I guess, we’ll better take a closer look at empirical and molecular formulas!
Lets Zoom in:
An empirical formula is the lowest whole number ratio of atoms or ions present in the compound, which is expressed by the subscripts in the formula. All ionic compounds are empirical formulas. The chemical formulas of molecular compounds can be empirical formulas, like ammonia, NH3, but they don’t have to be, like with glucose. Whereas the chemical formula for ionic compounds is the simplest ratio of the ions in the ionic lattice, the molecular formula actually tells us exactly how many atoms are bonded in the compound. Therefore, it is not necessarily the simplest ratio. We do have to take this into account when we are discussing the calculations of empirical and molecular formulas.
The chemical formula cannot only represent the individual particles, but it can, of course, also be seen as a ratio of moles. If I have NH3, then I have one nitrogen atom and three hydrogen atoms. But at the same time, I have 6.022 x 1023 nitrogen atoms and three times as many hydrogen atoms. And that is a key point in the calculation of empirical formulas.
So let’s dive in: often, these problems start with the % of elements, e.g. glucose has a percent composition by mass of 39.99% of carbon, 6.73% of hydrogen and 53.28% of oxygen. Our first step is to assume we have a 100g sample, and therefore can convert those percentages by mass values to g: 39.99 grams of carbon, 6.73 grams of hydrogen and 53.28 grams of oxygen. In the second step, since the empirical formula can be seen as the ratio of moles, we can convert these grams to moles by dividing by the molar masses. As an example: 39.99 grams of carbon times 1 mole of Carbon over 12.01 g of carbon is equal to 3.33 moles of carbon. For hydrogen this would be 6.67 moles of hydrogen and 3.33 moles of oxygen. Now, in this example the ratio is pretty easy to see. But it is not always this simple. So, the third step is to divide by the smallest number of moles. Why? Because then you get a ratio based on this element, which automatically becomes 1. In our case, we divide by 3.33 and get a ratio of 1:2:1 for Carbon to Hydrogen to Oxygen. And this is our empirical formula, CH2O.
Sometimes, after dividing by the smallest number of moles, you will end up with numbers with decimals, like 1 to 1.5, for example for a compound with aluminum and oxygen. Of course, we cannot write the chemical formula to be Al1O1.5, since we need a whole-numbered ratio. We have to multiply BOTH subscripts by 2 to get the formula of Al2O3. Pro tip: it can help to convert the decimal to a fraction to determine the factor: for example, if you have 2.33, which is 2 ⅓, you need to multiply it by 3 to get it to a whole number.
Now back to the candy - uhm, to the glucose: we have our empirical formula. To calculate the molecular formula we need one more piece of information: The molar mass of the molecular compound. Glucose has a molar mass of 180 g/mol. When we calculate the molar mass of the empirical formula, we get 30.0 g/mol. And so, the molecular formula has to be the empirical formula times six, which gets us to (drum roll please) C6H12O6. If it would indicate a molar mass of 30.0 g/mol, we would know the empirical formula is the molecular formula - and we would actually have formaldehyde and not glucose!
There is one special case for empirical and molecular formulas: combustion analysis. During combustion analysis, we have compounds containing at least carbon and hydrogen, sometimes also oxygen and nitrogen, that react with oxygen to form carbon dioxide and water. In those problems, we are given the mass of the carbon dioxide and water formed and have to use it to calculate the chemical formula of the original compound. But, that’s only a teaser and material for another episode!
To recap……
An empirical formula is the simplest whole-numbered ratio of atoms or ions present in the compound, which is expressed by the subscripts in the formula. We can also interpret it as a ratio of moles of constituent elements. Therefore we can calculate the empirical formula by converting percent composition to grams, grams to moles and dividing by the smallest number of moles. This will give us the ratio. To calculate the molecular formula from the empirical formula we use a factor derived from the molar masses.
Coming up next on the Apsolute RecAP Chemistry Edition: gravimetric analysis
Today’s Question of the day is about empirical formulas.
Question: What is the first step, if I have given grams of elements as part of the substance instead of % composition?
Answer: