What is the empirical formula of this compound? If the molar mass of the molecular formula is 865.80 g/mol, what is the molecular formula of the compound ?

Question

A compound containing only silicon and fluorine is 64.89% silicon by mass.

Answer 1

Hi Luky,

Thanks for your question!

Part 1. The Empirical Formula

Imagine we had a 100 g sample of this silicon fluoride compound. This compound would be made up of

64.89 g silicon,

35.11 g fluorine.

Using the molar mass of each element, we can convert grams to moles:

(64.89 g silicon)*(1 mol silicon)/(28.0855 g silicon) = 2.310 mol silicon,

(35.11 g fluorine)*(1 mol fluorine)/(18.9984 g fluorine) = 1.848 mol fluorine.

Once we know the amount of moles of each element in our compound, we can conclude that the amount of silicon in our compound is roughly 2.310/1.848 ~ 1.250 times the amount of fluorine in our compound. Thus, we would be tempted to conclude that an empirical formula for the compound would be

Si_1.25F₁.

The only problem is that the indices in this formula are not counting numbers. We can remedy this problem by quadrupling the amount of Si and F present in the empirical formula, giving us

Si₅F₄.

Now, we have a formula with indices that are counting numbers and, most importantly, the ratio between the amount of silicon and amount of fluorine is still 1.25. Thus, our final answer to this first part is Si₅F₄.

Part 2. The Molecular Formula

We first calculate the molar mass of our empirical formula, which will be

M_Empirical = 5*(28.0855 g/mol) + 4*(18.9984 g/mol) = 216.4211 g/mol.

In contrast, we are told the molar mass of the molecular formula is M_Molecular = 865.80 g/mol. Dividing these two molar masses, we see that

M_Molecular/M_Empirical = (865.80 g/mol)/(216.4211 g/mol) = 4.0005 ~ 4.

So, we should expect to quadruple the amount of each element in our empirical formula, giving us a final molecular formula of Si₂₀F₁₆.