Determine the empirical formula of a compound containing 1.71 g of silicon and 8.63 g of chlorine.

The answer to your question is: SiCl₄

Explanation:

Data

amount of Si      1.71 g

amount of Cl     8.63 g

MW Si = 28 g

MW Cl = 35.5

Process (rule of three)

For Si                                                        For Cl

        28 g of Si 1 mol                      35.5 g of Cl 1 mol

          1.71g of Si     x                              8.63 g of Cl  x

         x = 1.71 x 1 / 28 = 0.06 mol                          x = 8.63 x 1 / 35.5 = 0.24 mol

Now, divide both results by the lowest of them.

Si = 0.06 mol / 0.06 = 1 molecule of Si     Cl = 0.24 / 0.06 = 4 molecules of Cl

Finally

                     Si₁ Cl₄ or SiCl₄

the answer is b & c i believe : )

the answer is 75.0%.