Find the empirical formula of each of the following compounds: (a) 0.063 mol of chlorine atoms combined with \(0.22 \mathrm{~mol}\) of oxygen atoms; (b) \(2.45 \mathrm{~g}\) of silicon combined with \(12.4 \mathrm{~g}\) of chlorine; (c) 27.3 mass \% carbon and 72.7 mass \% oxygen

The mole ratio is a fundamental concept in chemistry used to determine the simplest ratio between elements in a compound. This ratio is found by dividing the number of moles of each element by the number with the lowest value of moles in the mixture.

For example, to find the mole ratio of chlorine and oxygen in part (a) of the exercise, we start with their given moles: 0.063 mol for chlorine and 0.22 mol for oxygen.

To simplify, we divide both numbers by the smaller value (0.063 mol):
\(\frac{0.063}{0.063} = 1\) for chlorine and \(\frac{0.22}{0.063} \approx 3.5\) for oxygen.
Since 3.5 is not a whole number, it's common to multiply both ratios by 2 to get whole numbers: \(1 \times 2 = 2 \) and \(3.5 \times 2 = 7 \).

Thus, the mole ratio of chlorine to oxygen is 2:7, making the empirical formula \text{Cl}\_2\text{O}\_7\. This showcases how mole ratios simplify the representation of compounds.

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Knowing the molar mass of elements is essential for calculating the number of moles from a given mass.

In part (b) of the exercise, we need to convert the masses of silicon and chlorine to moles. We use their molar masses: 28.0855 g/mol for silicon and 35.453 g/mol for chlorine.

The number of moles is found using the formula:
\(\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}\).
For silicon: \(\frac{2.45 \text{ g}}{28.0855 \text{ g/mol}} \approx 0.0872 \text{ mol}\), and for chlorine: \(\frac{12.4 \text{ g}}{35.453 \text{ g/mol}} \approx 0.350 \text{ mol}\).
This allows us to compare the amounts in moles, further simplifying to the simplest whole number ratio to find the empirical formula.

Mass percent composition shows the percentage by mass of each element in a compound. This helps when converting between mass of elements and their moles.

For instance, in part (c) of the exercise, we're given the mass percent composition of carbon (27.3%) and oxygen (72.7%). Assuming a total mass of 100 g makes calculations simpler: 27.3 g of carbon and 72.7 g of oxygen.

We convert these masses to moles using their respective molar masses (Carbon: 12.011 g/mol, Oxygen: 15.999 g/mol):
\(\frac{27.3 \text{ g}}{12.011 \text{ g/mol}} \approx 2.27 \text{ mol}\) for carbon and \(\frac{72.7 \text{ g}}{15.999 \text{ g/mol}} \approx 4.54 \text{ mol}\) for oxygen.
Finally, dividing by the smallest number of moles gives us a ratio of 1 carbon atom to 2 oxygen atoms, yielding the empirical formula \text{CO}\_2\.

The empirical formula represents the simplest whole number ratio of the elements in a compound. It’s derived from mole ratios, giving insight into compound structure without considering the actual mass quantities.

To find an empirical formula, follow these steps:

• Convert masses of elements to moles using their molar masses.
• Divide by the smallest number of moles among the elements to get the simplest ratio.
• If necessary, multiply these ratios by a common factor to obtain whole numbers.

Let’s look again at part (b). We have 0.0872 mol of silicon and 0.350 mol of chlorine. Dividing both by the smallest value (0.0872 mol):

\(\frac{0.0872}{0.0872} = 1\) for silicon and \(\frac{0.350}{0.0872} \approx 4\) for chlorine. This results in a 1:4 ratio, giving us the formula \text{SiCl}\_4\.

The empirical formula connects directly to experimental data, making it a vital tool for understanding chemical compositions.