A 22.60 gram sample of chromium is heated in the presence of excess chlorine. A metal chloride is formed with a mass of 68.83 g. Determine the empirical formula of the metal chloride.

CrCl₃.

Explanation:

From the question given above, the following data were obtained:

Mass of chromium (Cr) = 22.60 g.

Mass of metal chloride = 68.83 g.

Empirical formula =.?

Next, we shall determine the mass of chlorine (Cl) in the metal chloride. This can be obtained as follow:

Mass of chromium (Cr) = 22.60 g.

Mass of metal chloride = 68.83 g.

Mass of chlorine (Cl) =.?

Mass of chlorine (Cl) = (Mass of metal chloride) – (Mass of chromium)

Mass of chlorine (Cl) = 68.83 – 22.60

Mass of chlorine (Cl) = 46.23 g

Finally, we shall determine the empirical formula for the metal chloride as follow:

Cr = 22.60 g.

Cl = 46.23 g

Divide by their molar mass

Cr = 22.60 / 52 = 0.435

Cl = 46.23 / 35.5 = 1.302

Divide by the smallest

Cr = 0.435 /0.435 = 1

Cl = 1.302 /0.435 = 3

The empirical formula for the metal chloride is CrCl₃

it is a solid

explanation:

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole