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The formula of the anhydrous compound was determined to be CuSO 4. Calculate the formula of the hydrated compound. Solution: 1) Assume 100. g of the hydrate is present. That means this: CuSO 4---> 74.7 g H 2 O ---> 25.3 g. 2) Change to moles: CuSO 4---> 74.7 g / 159.607 g/mol = 0.468 mol H 2 O ---> 25.3 g / 18.0 g/mol = 1.406 mol. 3) Divide by ...
- Calculate Empirical Formula When Given Percent Composition Data
Generally speaking, in empirical formula problems, C = 12, H...
- Calculate Empirical Formula When Given Percent Composition Data
Formula of a Hydrate (\(\text{Anhydrous Solid}\ce{*}x\ce{H2O}\)) The formula of a hydrate can be determined by dehydrating a known mass of the hydrate, then comparing the masses of the original hydrate and the resulting anhydrous solid.
30 Απρ 2018 · Divide the mass of your anhydrous (heated) salt sample by the molar mass of the anhydrous compound to get the number of moles of compound present. In our example, 16 grams / 160 grams per mole = 0.1 moles.
The formula of the anhydrous compound was determined to be CuSO 4. Calculate the formula of the hydrated compound. Example #12: 0.572 grams of a hydrate is heated to dryness, ending with 0.498 grams of anhydrous compound. What is the percentage of water by mass in the hydrate?
1) Determine the mass of the anhydrate and of the water that was lost: CuCl 2 ---> 17.917 − 16.221 = 1.696 g H 2 O ---> 18.360 − 17.917 = 0.443 g. 2) Determine moles of each: CuCl 2 ---> 1.696 g / 134.452 g/mol = 0.012614 mol H 2 O ---> 0.443 g / 18.015 g/mol = 0.02459 mol. 3) Divide by smallest:
17 Ιουν 2023 · A hydrate contains a definite number of water molecules bound to each ionic compound (also called the anhydrous salt). The formula of the hydrate is represented by the formula of the anhydrous salt followed by a dot and x H 2 O, where x is the number of moles of water per mole of the anhydrous salt. The hydrate is named by naming the anhydrous ...
Use the figures above to calculate the degree of hydration in Al(NO 3) 3 ·n H 2 O. Mass of anhydrous salt = 54.80 – 52.67 = 2.13 g Mass of water lost = 56.42 – 54.80 = 1.62 g