A sample of an unknown substance is decomposed and is found to be 59% strontium (Sr), 8.0% carbon, and 33% oxygen. What is its empirical formula?

CStO3

Explanation:

What is need to be done here is to first divide the percentage compositions by the respective atomic masses. Atomic mass of strontium is 90, that of oxygen is 16 and that of carbon is 12.

Let's proceed with the divisions:

St = 59/90 = 0.656

O = 33/16 = 2.0625

C = 8/12 = 0.667

Now, we divide through all by the smallest answer which is 0.656:

St = 0.656/0.656 = 1

O = 2.0625/0.656 = 3.144 = 3

C = 0.667/0.656 = 1.0167 = 1

The empirical formula is thus CStO3