A sample of a compound contains 3.21 g of sulfur and 11.4 g of fluorine. Which of the following repr — Stoichiometry Chemistry Question

STEPS:

1. Understand the concept of an empirical formula: The empirical formula represents the simplest whole-number ratio of the atoms of each element present in a compound. To find this ratio, we must first convert the given masses of sulfur and fluorine into moles.
2. Identify the molar masses of the elements: Using the periodic table:
* The molar mass of Sulfur (S) is approximately $32.06\text{ g/mol}$ (commonly rounded to $32.1\text{ g/mol}$ on AP exams).
* The molar mass of Fluorine (F) is approximately $19.00\text{ g/mol}$.
3. Calculate the moles of each element in the sample:
* Moles of S:
$$n_{\text{S}} = \frac{3.21\text{ g}}{32.06\text{ g/mol}} \approx \mathbf{0.100\text{ mol}}$$
* Moles of F:
$$n_{\text{F}} = \frac{11.4\text{ g}}{19.00\text{ g/mol}} = \mathbf{0.600\text{ mol}}$$
4. Determine the simplest whole-number ratio: Divide the number of moles of each element by the smallest number of moles calculated ($0.100\text{ mol}$):
* S: $\frac{0.100}{0.100} = 1$
* F: $\frac{0.600}{0.100} = 6$
5. Formulate the empirical formula: The mole ratio is 1 sulfur atom to 6 fluorine atoms, which gives the empirical formula $\text{SF}_6$. This matches Option E.

*

WHY_OTHERS_WRONG:

• A is incorrect: An empirical formula of $\text{SF}_2$ corresponds to a 1:2 mole ratio. This ratio would require only $0.200\text{ mol}$ of fluorine (which is $3.80\text{ g}$ of F) for every $3.21\text{ g}$ of sulfur.
• B is incorrect: An empirical formula of $\text{SF}_3$ corresponds to a 1:3 mole ratio. This ratio would require only $0.300\text{ mol}$ of fluorine (which is $5.70\text{ g}$ of F) for every $3.21\text{ g}$ of sulfur.
• C is incorrect: An empirical formula of $\text{SF}_4$ corresponds to a 1:4 mole ratio. This ratio would require only $0.400\text{ mol}$ of fluorine (which is $7.60\text{ g}$ of F) for every $3.21\text{ g}$ of sulfur.
• D is incorrect: An empirical formula of $\text{SF}_5$ corresponds to a 1:5 mole ratio. This ratio would require only $0.500\text{ mol}$ of fluorine (which is $9.50\text{ g}$ of F) for every $3.21\text{ g}$ of sulfur.