**Empirical formula**

**09/30/16**

*“How can we obtain the structure of a chemical just by knowing the percentage of it’s components?”*

Wouldn’t it be really cool if just by knowing the percentage composition of the different atoms in a compound, you could **obtain the chemical formula**? Well, let’s figure out how to do it by completing an example. Let’s say that you obtain a chemical with percentage 58.64% Carbon [C], 8.16% Hydrogen [H], and 43.20% Oxygen [O]. The first step we must take is to **change this percentage into something more realistic**, such as mass. To make the math easy, let’s use a mass sample of 100g. This means that this compound will have 58.64 grams of carbon[C], 8.16 grams of hydrogen [H], and 43.20 g of oxygen [O]. The second step will be to take these masses and change it **into moles**. After doing the math, we will end up with 4.409 moles of carbon [C], 8.905 moles of hydrogen [H], and 2.7 moles of oxygen [O]. if you notice, all of the moles are in “messy” values, so we need to simplify this somehow. We can accomplish this by **dividing the moles by the lowest number **(In this case, the lowest number is 2.7)**. **After we do the math, we will end up with 1.5 moles of carbon [C], 3 moles of hydrogen [H], and 1 mole of oxygen [O]. Now, we simply have to make everything a whole number. We could do this by multiplying all of mole values by 2, giving us 3 moles of carbon [C], 6 moles of hydrogen [H], and 2 moles of oxygen [O]. Our final compound formula will be C3H6O2, which just so happens to be the chemical formula for **Acetate. **Since this process will always give you the formula with the simplest amount of proportions, it is called the **empirical formula.**