Determine the empirical formulas for the compound with the following percent composition: 36.2% aluminum and 63.8% 3.8% sulfur Al_(2)S_(4) Al_(2)S_(3) Al_(3)S_(2) AlS_(3) Multiple Choice 1 point

To determine the empirical formula for the compound with the given percent composition, we need to convert the percentages to grams and then to moles.

Given:
- Aluminum (Al): 36.2%
- Sulfur (S): 63.8%

Assume we have 100 grams of the compound. This means we have:
- 36.2 grams of Al
- 63.8 grams of S

Next, we convert the masses to moles using the molar masses of aluminum (Al) and sulfur (S):
- Molar mass of Al = 26.98 g/mol
- Molar mass of S = 32.06 g/mol

Calculate the moles of each element:
$$\text{Moles of Al} = \frac{36.2 \text{ g}}{26.98 \text{ g/mol}} \approx 1.34 \text{ moles}$$
$$\text{Moles of S} = \frac{63.8 \text{ g}}{32.06 \text{ g/mol}} \approx 1.99 \text{ moles}$$

Next, we find the simplest whole number ratio of the moles of Al to S:
$$\text{Ratio of Al to S} = \frac{1.34}{1.34} : \frac{1.99}{1.34} \approx 1 : 1.49$$

To get whole numbers, we can multiply both ratios by a common factor to get the smallest whole number ratio. In this case, multiplying by 2 gives:
$$1 \times 2 : 1.49 \times 2 \approx 2 : 3$$

Thus, the empirical formula for the compound is:
$$\text{Empirical formula} = \text{Al}_2\text{S}_3$$

So, the correct answer is:
$$\boxed{\text{Al}_2\text{S}_3}$$