Day: September 30, 2016

Empirical formula

Empirical formula

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.