# 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

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## 1 Answers

ruchi yadav
askIITians Faculty 27 Points
10 years ago
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 YouRuchiAskiitians Faculty`

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