Question

When a sample of hydrate is heated, water of hydration is lost.
If you start with 60.0-gram sample of Na₂CO₃⦁10H₂O, how much water (in grams) will be lost?

Answer 1

To find out how much water of hydration will be lost when heating a sample of sodium carbonate decahydrate (Na₂CO₃·10H₂O), we first need to determine the molar mass of the compound and the water it contains.

1. **Calculate the molar mass of Na₂CO₃·10H₂O**:
- Molar mass of Na (Sodium) = 22.99 g/mol
- Molar mass of C (Carbon) = 12.01 g/mol
- Molar mass of O (Oxygen) = 16.00 g/mol
- Molar mass of H₂O (Water) = 18.02 g/mol

Now, calculate the total for Na₂CO₃ and 10H₂O:

- Molar mass of Na₂CO₃ = 2(22.99) + 12.01 + 3(16.00)
= 45.98 + 12.01 + 48.00
= 105.99 g/mol

- Molar mass of 10H₂O = 10 × 18.02 = 180.20 g/mol

Therefore, the total molar mass of Na₂CO₃·10H₂O is:
\[
105.99 + 180.20 = 286.19 \text{ g/mol}
\]

2. **Calculate the mass fraction of water in Na₂CO₃·10H₂O**:
The mass of water in one mole of Na₂CO₃·10H₂O is 180.20 g (from the 10H₂O part).

To determine the fraction of the hydrate that is water:
\[
\text{Mass fraction of water} = \frac{180.20 \text{ g}}{286.19 \text{ g}} \approx 0.629
\]

3. **Determine the amount of water lost from 60.0 g of Na₂CO₃·10H₂O**:
Multiply the mass fraction of water by the mass of the hydrate sample:
\[
\text{Mass of water lost} = 60.0 \text{ g} \times 0.629 \approx 37.74 \text{ g}
\]

Thus, when heating the 60.0-gram sample of Na₂CO₃·10H₂O, approximately **37.74 grams of water** will be lost.