ITS A CHEMISTRY QUESTION I NEED HELP **Question:**When a sample of a hydrated CaBr\(_2\) was heated to a constant weight, 3.24 g of the anhydrous (dry) CaBr\(_2\) was obtained and 1.75 g of water was released. What is the empirical formula of the hydrate?**Options:**1. CaBr\(_2 \cdot\) 2H\(_2\)O2. CaBr\(_2 \cdot\) 6H\(_2\)O3. None of these choices is correct4. CaBr\(_2 \cdot\) 12H\(_2\)O**Explanation:**The problem involves determining the empirical formula of a hydrate of calcium bromide. The sample was heated, and the water of hydration was removed, resulting in a certain mass of anhydrous CaBr\(_2\) and a known mass of water released. To find the empirical formula of the hydrate, one must calculate the molar ratio of CaBr\(_2\) to H\(_2\)O.**Steps:**1. Calculate moles of anhydrous CaBr\(_2\): \[ \text{Moles of CaBr\(_2\)} = \frac{3.24 \, \text{g}}{199.9 \, \text{g/mol}} \]2. Calculate moles of water: \[ \text{Moles of H\(_2\)O} = \frac{1.75 \, \text{g}}{18.02 \, \text{g/mol}} \]3. Determine the ratio of moles of water to moles of CaBr\(_2\).4. Use this ratio to determine which hydrate formula matches the data.

The sample is given as a hydrated CaBr2. Weight of dry CaBr2 = 3.24 g Weight of water released = 1.75 gSo the total weight of hydrated sample will be the sum of dry and water released. That is, Hydrated sample weight = 3.24+1.75 = 4.99 g Hence we can calculate the percentage of water in the compound. Water loss % = (1.75/4.99)× 100 = 35Molecular mass of CaBr2 = 200 g So, We can assume for 35%, there must be 5-7 water molecules should be there and in option 6H2O is there. If 6 molecules of water is there, 6×18 = 108 g % of water in it = (108/(200+108))×100 = 35 % So there will be 6 molecules of water in it. Hence the emperical formula of the hydrate compound will be CaBr2.6H2O.