We follow the procedure in Example 3.9 (Section 3.5) to calculate the empirical formula by assuming that we have 100 g of the compound, so the percentages are converted to grams. The number of moles of Si and F are given by

n_{Si}=3.0  \cancel{g  Si}\times \frac{1  mol  Si}{28.09  \cancel{g  Si}}=1.17  mol  Si

 

n_{F}=67.0  \cancel{g  F}\times \frac{1  mol  F}{19.00  \cancel{g  F}}=3.53  mol  F

Therefore, the empirical formula is Si_{1.17}F_{3.53} , or, dividing by the smaller subscript (1.17), we obtain  SiF_{3} .

To calculate the molar mass of the compound, we need first to calculate the number of moles contained in 2.38 g of the compound. From the ideal gas equation

n=\frac{PV}{RT}

 

=\frac{(1.70  atm)(0.210  L)}{(0.0821  L  .  atm/K  .  mol)(308  K)} =0.0141  mol

Because there are 2.38 g in 0.0141 mole of the compound, the mass in 1 mole, or the molar mass, is given by

M=\frac{2.38  g}{0.0141  mol}=169  g/mol

The molar mass of the empirical formula SiF_{3} is 85.09 g. Recall that the ratio (molar mass/empirical molar mass) is always an integer (169⁄85.09 ≈ 2). Therefore, the molecular formula of the compound must be  (SiF_{3})_{2}   or   Si_{2}F_{6}.