Determine the simplest formula of a compound with the following composition: Cr=26.52%, S=24.52%, O=48.96%, Atomic Weights: Cr=52, S=32

We need to find the ratio of atoms, so we need the atomic weight of each element,
which we can find from a periodic chart.
The atomic weight ofCris 52.0, that of O is 16.0 and S is 32.0
For the first compound, 100 grams of it contain 26.52gm. of Cr, 48.96 gm. of O and
24.52 gm. of S. Dividing each of those by its corresponding weight we obtain:
For Cr, 26.52/52 = 0.51 and for O, 48.96/16 = 3.06, and for S, 24.52/32 = 0.766.
Comparing these numbers, we see that the Cr to S ratio is about 2 to 3, so for each
two Cr atoms we have 3 S atoms, and the ratio of Cr to O is 1 to 6, or 2 to 12 if we
eliminate fractions. So the ratio is Cr2, S3, O12, which gives us an empirical formula
of Cr2S3O12 - it's probably chromium(III) sulfate, Cr2(SO4)3.
For the second, we have 0.657 gm. of magnesium whose atomic weight is 24.3, so
its atomic ratio is 0.657/24.3 = 0.027; for S, 0.865/32 = 0.027; for O, 1.73/16 = 0.108
These numbers are in a ratio of 1 to 1 to 4, so we have 1 atom of Mg to 1 atom of S
to 4 atoms of O, so our compound is MgSO4.

Thank You
Ruchi
Askiitians Faculty