Molecular Formula

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Image courtesy of freefoto.com

Image courtesy of freefoto.com

 

I imagine that some of you have probably worked at a fast food restaurant at some point in your life. Fast food restaurants put their food together in a manner that is similar to an assembly line in a factory. It is a very ordered and predictable process.

Let’s say that you are working at a fast food restaurant and you assemble the burgers. Imagine that you have been assembling these burgers for so long that you can practically assemble them in your sleep. You are so familiar with the process that you have completely memorized the recipe (it’s pretty simple, 1 hamburger bun and 1 beef patties) and you even know the mass of a completely assembled burger, 100 grams.

 

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Burger Recipe

Burger Recipe

Burger Recipe

We can see from the recipe above that the ratios of hamburger buns (B) to meat patties (P) is 1:1. So we can write this ratio as a formula, BP.

 

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As you are preparing burgers for customers, you are always concerned with this recipe and the ratio of hamburger buns to meat patties. If a customer asked you for 3 complete burgers what would be the ratio of hamburger buns to meat patties for this order?

Triple Burger Recipe

Triple Burger Recipe

 

For this particular order we would have a 3:3 ratio. We can also write this ratio as a formula, B3P3.

What if you had a really smart customer order 400 grams worth of burgers? What would be the ratio of hamburger buns to meat patties? Well since a complete burger is 100 grams we would need 4 times the amount of this basic recipe. So our ratio would go from a 1:1 ratio to a 4:4 ratio. We can also write this ratio as a formula, B4P4

In chemistry, we have a similar method to understanding different multiples of a simple ratio. We learned in the previous lesson (empirical formula) that an empirical formula is the simplest whole number ratio of the atoms in a chemical compound. That is kind of like our basic burger recipe with the 1:1 ratio. The molecular formula on the other hand is the actual number of atoms in a chemical formula.

For example, the molecular formula of glucose is C6H12O6. This tells us that there are 6 carbon atoms, 12 hyrodgen atoms, and 6 oxygen atoms in this molecular. This is the molecular formula for glucose. What would the empirical formula of glucose be? Well we need to look at the simplest ratio of Carbon to hydrogen to oxygen. I can see that all of the subscript numbers are divisible by 6, so if I reduce this ratio by dividing everything by 6 I get, CH2O. This is the empirical formula for glucose.

 

Here’s another example,

The empirical formula for a particular chemical compound is CH2O. If the actual chemical compound has a molar mass of 180 g/mol, what is the molecular formula for this compound?

 

Here are the steps to solve a molecular formula problem like this,

Step 1  – Determine the mass of the empirical formula

Step 2 – Determine how many times larger the actual compound is from the empirical formula compound by dividing the molar mass of the actual compound by the molar mass of the empirical formula

step 3 – Multiple all subscripts in the empirical formula by the number you determined in step 2

 

Step 1

Step 1

Step 1

 

Step 2

Step 2

Step 2

 

Step 3

Step 3

Step 3

 

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