Question

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

Answer 1

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. 

Answer 2

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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