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Chapter 6: Problem 37

What is the molecular formula of a gaseous fluoride of sulfur containing \(70.4 \%\) F and having a density of approximately \(4.5 \mathrm{g} / \mathrm{L}\) at \(20^{\circ} \mathrm{C}\) and \(1 \mathrm{atm} ?\)

Short Answer

Expert verified
The molecular formula of the gaseous fluoride of sulfur is SF4.

Step by step solution

01

Calculate mass of sulfur

Given that the gaseous fluoride contains 70.4% fluorine (F), it means that it must contain 100% - 70.4% = 29.6% sulfur (S). If we consider 100g as a total, there are 70.4g of fluorine and 29.6g of sulfur.
02

Calculate moles of F and S

We calculate the moles by dividing the mass of each element by their respective atomic masses. The atomic mass of F = 19.00 g/mol and of S = 32.07 g/mol. Thus, the moles of F = 70.4g / 19.00 g/mol = 3.7 mol and the moles of S = 29.6g / 32.07 g/mol = 0.92 mol.
03

Determine the ratio of moles

To find the simplest whole-number ratio, divide each number of moles by the smallest number of moles calculated. In this case, divide both F and S moles by 0.92. So, ratio of moles of F and S becomes F:S = 3.7 / 0.92 : 0.92 / 0.92 = 4 : 1. This implies the empirical formula is SF4.
04

Calculate molecular formula

Given that the molecular mass of a gaseous fluoride of sulfur (M) can be calculated by using the ideal gas law equation \(M = dRT / P\), where d is density (4.5 g/L), R = 0.0821 L.atm/K.mol (ideal gas constant), T is temperature in Kelvin (20°C = 293.15 K), P is the pressure in atm (1 atm). Substituting these values into the ideal gas law equation gives M = 4.5g/L * 0.0821 L.atm/K.mol * 293.15 K / 1 atm = 108.15 g/mol.
05

Identify the molecular formula

The empirical formula SF4 has the molar mass as 32.07g + 4x19.00g = 108 g/mol, which is exactly the molar mass calculated in previous step. Therefore, the molecular formula is the same as the empirical formula - SF4.

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