Find the empirical formula for a compound which contains 32.8% chromium and 67.2% chlorine

67.2% Cl = 67.2g Cl x 1 molCl/ 35.45g Cl= 1.89 /0.63 = 3 Cl

32.8% Cr = 32.8g Cr x 1 molCr/51.99g Cr= 0.63 /0.63 = 1 Cr

The empirical formula is CrCl3

First convert the mass in grams to moles, then divide both by the smallest number.

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abhishekprakash2625 abhishekprakash2625 · Answer 2 · 2021-11-26T11:41:31-05:00

The empirical formula of a chemical compound is the simplest whole-number ratio of atoms present in a compound and a chemical formula that gives the total number of atoms of each element in each molecule of a substance

According to the question.

To convert the mass in grams to moles, we have to divide both by the smallest number.

\(67.2\ percentage\ = \frac{67.2g Cl * 1 molCl}{35.45g Cl}= \frac{1.89}{0.63} = 3 Cl\)

\(32.8\ percentage\ Cr = \frac{32.8g Cr * 1 molCr}{51.99g Cr}= \frac{0.63 }{0.63} = 1 Cr\)

Hence, after joining the compound the empirical formula is \(CrCl_3\)

The empirical formula is \(CrCl_3\)

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