How To Find The Empirical Formula 11 Steps With Pictures

A compound that is made up of 40. 92% Carbon, 4. 58% hydrogen, and 54. 5% Oxygen would have an empirical formula of C3H4O3 (we will go through an example of how to find the EF of this compound in Part Two).

In a chemistry lab, to find the percentage composition, the compound would be examined through some physical experiments and then quantitative analysis. Unless you are in a lab, you will not need to actually do these experiments.

For example, let’s say that we have a compound that is made up of 40. 92% carbon. The atomic mass of carbon is 12 so our equation would be 40. 92 / 12 = 3. 41.

Let’s say that we are working with a compound that has three gram atoms: 1. 5, 2 and 2. 5. The smallest gram atom out of those three numbers is 1. 5. So to find the atomic ratio, you must divide all of the numbers by 1. 5 and then separate them with the symbol for ratio :. 1. 5 / 1. 5 = 1. 2 / 1. 5 = 1. 33. 2. 5 / 1. 5 = 1. 66. So your atomic ratio is 1 : 1. 33 : 1. 66.

Try 2. Multiply the numbers in your atomic ratio (1, 1. 33, and 1. 66) by 2. You get 2, 2. 66, and 3. 32. These are not whole numbers so 2 doesn’t work. Try 3. You get 3, 4, and 5 when you multiply 1, 1. 33, and 1. 66 by 3. Therefore, your atomic ratio of whole numbers is 3 : 4 : 5.

X3Y4Z5

Let’s say that the assignment asks you to look at a sample of vitamin C. It lists 40. 92% Carbon, 4. 58% hydrogen 54. 5% Oxygen—this is the percent composition. 40. 92% of the vitamin C is made up of carbon, while the rest is made up of 4. 58% hydrogen and 54. 5% oxygen.

Number of gram atoms of carbon = 40. 92 / 12 = 3. 41 Number of gram atoms of hydrogen = 04. 58 / 01 = 4. 58 Number of gram atoms of oxygen = 54. 50 / 16 = 3. 41

Carbon: 3. 41 / 3. 41 = 1 Hydrogen: 4. 58 / 3. 41 = 1. 34 Oxygen: 3. 41 / 3. 41 = 1 The atomic ratio is 1 : 1. 34 : 1.

1 x 3 = 3 (this works because 3 is a whole number). 1. 34 x 3 = 4 (4 is also a whole number). 1 x 3 = 3 (again, 3 is a whole number). Our whole number ratio is therefore Carbon(C) : Hydrogen(H) : Oxygen(O) = 3 : 4 : 3

C3H4O3