Question

When a sample of hydrate is heated, water of hydration is lost.
If you start with 60.0-gram sample of Na₂CO₃⦁10H₂O, what will be the mass of the sample (in grams) after all of the water is removed by heating?

Answer 1

To find the mass of the sample after all of the water is removed by heating, we first need to determine the molar mass of Na₂CO₃·10H₂O (sodium carbonate decahydrate) and then the mass of the anhydrous form (Na₂CO₃).

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 H (hydrogen) = 1.01 g/mol

Now calculate the molar mass:

For Na₂CO₃:
\[
\text{Molar mass of Na}_2\text{CO}_3 = 2(\text{Na}) + 1(\text{C}) + 3(\text{O}) = 2(22.99) + 12.01 + 3(16.00)
\]
\[
= 45.98 + 12.01 + 48.00 = 105.99 \text{ g/mol}
\]

For 10H₂O:
\[
\text{Molar mass of } 10\text{H}_2\text{O} = 10(2(\text{H}) + 1(\text{O})) = 10(2(1.01) + 16.00) = 10(2.02 + 16.00) = 10(18.02) = 180.20 \text{ g/mol}
\]

Total molar mass of Na₂CO₃·10H₂O:
\[
105.99 + 180.20 = 286.19 \text{ g/mol}
\]

2. **Calculate the mass of anhydrous Na₂CO₃:**

The mass fraction of Na₂CO₃ in Na₂CO₃·10H₂O is given by:
\[
\text{Mass of Na}_2\text{CO}_3 = 105.99 \text{ g/mol}
\]
\[
\text{Mass fraction of Na}_2\text{CO}_3 = \frac{105.99}{286.19}
\]

Now we can compute this mass fraction:
\[
\text{Mass fraction of Na}_2\text{CO}_3 \approx 0.3705
\]

3. **Calculate the mass of anhydrous Na₂CO₃ from the 60.0 g sample:**
\[
\text{Mass of Na}_2\text{CO}_3 = 60.0 \text{ g} \times 0.3705 \approx 22.23 \text{ g}
\]

Thus, the mass of the sample after all of the water is removed by heating is approximately **22.23 grams**.