Question

A 25.37 g sample of a compound is found to contain 8.24 g carbon, 0.92 g hydrogen and 16.21 g topoints chlorine. What is the empirical formula for this compound? When entering the formula place the elements in the following order: C,H,Cl. For subscripts, write the number after the element it corresponds to. For example, H₂O should be written as H₂ O.

Answer 1

The empirical formula for the compound can be determined by calculating the mole ratios of the elements present in the sample. For carbon, the mole ratio is 8.24 g C / 12.01 g/mol C = 0.686 mol C. The empirical formula is C0.686H0.911Cl0.457.

Step-by-step explanation:

The empirical formula for the compound can be determined by calculating the mole ratios of the elements present in the sample. To calculate the mole ratios, divide the mass of each element by the molar mass of that element. In this case, we have 8.24 g of carbon, 0.92 g of hydrogen, and 16.21 g of chlorine. The molar masses of carbon, hydrogen, and chlorine are 12.01 g/mol, 1.01 g/mol, and 35.45 g/mol, respectively.

For carbon, the mole ratio is 8.24 g C / 12.01 g/mol C = 0.686 mol C

For hydrogen, the mole ratio is 0.92 g H / 1.01 g/mol H = 0.911 mol H

For chlorine, the mole ratio is 16.21 g Cl / 35.45 g/mol Cl = 0.457 mol Cl

The empirical formula is then C0.686H0.911Cl0.457