The calculated masses (from III) of anhydrous salt and water, and the identity of the hydrated Use- compound (on the data slip supplied with the experiment), determine the number of hydrated waters in the hydrated compound of anhydrous salt. Formula for anhydrous salt mass = water lost. The moles of anhydrous salt and the moles of water are calculated. Determine the simplest mole “hydrated water (per mole) of compound and get the moles. Formula of hydrated compound calculation determines the formula for hydrated compound [I on data slip]. Slip number Data for hydrate (supplied of hydrated compound before heating 3ZS99 mass upon heating water 41260623 loss IeDS of residue after heating: anhydrous salt. Calculate the moles of anhydrous salt moles ullei. Determine the empirical formula of hydrated compound. Formula of hydrated compound.
The Correct Answer and Explanation is:
To determine the formula of a hydrated compound, follow these steps:
Given Data:
- Mass of hydrated compound (before heating): 3.2599 g
- Mass of anhydrous salt (after heating): 2.60623 g
- Mass of water lost = 3.2599 g – 2.60623 g = 0.65367 g
Step 1: Calculate the moles of anhydrous salt
To calculate moles of the anhydrous salt, we need its molar mass. Suppose from the data slip the anhydrous salt is CuSO₄, whose molar mass is:
- Cu = 63.55 g/mol
- S = 32.07 g/mol
- O₄ = 4 × 16.00 = 64.00 g/mol
- Total = 63.55 + 32.07 + 64.00 = 159.62 g/mol
Moles of CuSO₄=2.60623 g159.62 g/mol=0.01633 mol\text{Moles of CuSO₄} = \frac{2.60623\text{ g}}{159.62\text{ g/mol}} = 0.01633 \text{ mol}Moles of CuSO₄=159.62 g/mol2.60623 g=0.01633 mol
Step 2: Calculate the moles of water lost
- Molar mass of H₂O = 18.02 g/mol
Moles of H₂O=0.65367 g18.02 g/mol=0.03627 mol\text{Moles of H₂O} = \frac{0.65367\text{ g}}{18.02\text{ g/mol}} = 0.03627 \text{ mol}Moles of H₂O=18.02 g/mol0.65367 g=0.03627 mol
Step 3: Determine the mole ratio of water to salt
Mole ratio=0.036270.01633≈2.22\text{Mole ratio} = \frac{0.03627}{0.01633} ≈ 2.22Mole ratio=0.016330.03627≈2.22
This ratio is approximately 2.22 : 1, which is close to 2:1. Since empirical formulas require whole numbers, round to the nearest whole number.
So, 2 moles of water per mole of salt.
Step 4: Write the formula of the hydrated compound
If the anhydrous salt is CuSO₄, and there are 2 water molecules per formula unit, the formula of the hydrated compound is: CuSO₄\cdotp2H₂O\boxed{\text{CuSO₄·2H₂O}}CuSO₄\cdotp2H₂O
Final Answer:
Empirical formula of hydrated compound: CuSO₄·2H₂O
Explanation (300 words):
The empirical formula of a hydrated compound tells us the simplest whole-number ratio of the salt to water molecules in the compound. In this experiment, the mass of the hydrate before heating was measured to be 3.2599 g, and after heating, the mass was reduced to 2.60623 g. This mass difference is due to the loss of water during heating. The mass of the water lost is found by subtracting the mass after heating from the initial mass, resulting in 0.65367 g of water.
To determine how many moles of each component are present, divide the mass of each substance by its molar mass. The molar mass of copper(II) sulfate (CuSO₄) is 159.62 g/mol, and water’s molar mass is 18.02 g/mol. This gives us 0.01633 mol of CuSO₄ and 0.03627 mol of H₂O. To find the number of water molecules per salt formula unit, divide the moles of water by the moles of anhydrous salt. The ratio is approximately 2.22:1, which we round to 2:1 for simplicity and accuracy in chemistry.
Thus, for every mole of CuSO₄, there are 2 moles of water, leading to the hydrated formula CuSO₄·2H₂O. This empirical formula provides essential insight into the composition of the hydrated compound and is crucial in both laboratory and industrial settings, where knowing the exact hydration state affects reactivity, storage, and usage.
