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## Q. 5.10

Chemical analysis of a gaseous compound showed that it contained 33.0 percent silicon (Si) and 67.0 percent fluorine (F) by mass. At 35°C, 0.210 L of the compound exerted a pressure of 1.70 atm. If the mass of 0.210 L of the compound was 2.38 g, calculate the molecular formula of the compound. Si2F6

Strategy This problem can be divided into two parts. First, it asks for the empirical formula of the compound from the percent by mass of Si and F. Second, the information provided enables us to calculate the molar mass of the compound and hence determine its molecular formula. What is the relationship between empirical molar mass and molar mass calculated from the molecular formula?

## Verified Solution

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  $(SiF_{3})_{2} or Si_{2}F_{6}$.