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Chapter 5

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?

Step-by-Step

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}.