What is the empirical formula of a compound that contains 10.0 g of aluminum and 39.4 g of chlorine?

• A. AlCl_3
• B. Al_2Cl_3
• C. Al_2Cl_6
• D. AlCl

Answer: (A)

Verified step by step guidance

1. Determine the number of moles of each element by dividing the given mass by its molar mass. For aluminum (Al), use the molar mass approximately 26.98 g/mol, and for chlorine (Cl), use approximately 35.45 g/mol. Calculate moles of Al as \(\frac{10.0\ \text{g}}{26.98\ \text{g/mol}}\) and moles of Cl as \(\frac{39.4\ \text{g}}{35.45\ \text{g/mol}}\).
2. Find the simplest whole number mole ratio between aluminum and chlorine by dividing both mole values by the smaller number of moles obtained in the previous step.
3. If the mole ratio is not a whole number, multiply both ratios by the smallest factor that converts them into whole numbers (e.g., 2, 3, 4, etc.).
4. Write the empirical formula by using the whole number mole ratios as subscripts for each element. For example, if the ratio is 1:3 for Al:Cl, the empirical formula is \(\mathrm{AlCl_3}\).
5. Double-check that the subscripts are in the simplest whole number ratio and that the formula matches one of the given options.