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What is the percentage of water in the following compound? Answer using one decimal place.

Sodium carbonate decahydrate, [tex]Na_2CO_3 \cdot 10H_2O[/tex]

[tex]\boxed{\ \% \text{ by mass } H_2O\ }[/tex]

Sagot :

GinnyAnswer
To find the percentage of water in sodium carbonate decahydrate ([tex]\(Na_2CO_3 \cdot 10H_2O\)[/tex]), we need to follow a step-by-step process:

1. Determine the molar masses of the individual components:

- Sodium ([tex]\(Na\)[/tex]): 22.99 g/mol
- Carbon ([tex]\(C\)[/tex]): 12.01 g/mol
- Oxygen ([tex]\(O\)[/tex]): 16.00 g/mol
- Hydrogen ([tex]\(H\)[/tex]): 1.01 g/mol

2. Calculate the molar mass of anhydrous sodium carbonate ([tex]\(Na_2CO_3\)[/tex]):

- Molar mass of [tex]\(Na_2CO_3\)[/tex]:
- [tex]\(2 \times 22.99\)[/tex] (for 2 sodium atoms) = 45.98 g/mol
- [tex]\(1 \times 12.01\)[/tex] (for 1 carbon atom) = 12.01 g/mol
- [tex]\(3 \times 16.00\)[/tex] (for 3 oxygen atoms) = 48.00 g/mol

- Adding these contributions gives:
- Total molar mass of [tex]\(Na_2CO_3\)[/tex] = 45.98 + 12.01 + 48.00 = 105.99 g/mol

3. Calculate the molar mass of 10 water molecules ([tex]\(10 \times H_2O\)[/tex]):

- Molar mass of one water molecule ([tex]\(H_2O\)[/tex]):
- [tex]\(2 \times 1.01\)[/tex] (for 2 hydrogen atoms) = 2.02 g/mol
- [tex]\(1 \times 16.00\)[/tex] (for 1 oxygen atom) = 16.00 g/mol

- Molar mass of one [tex]\(H_2O\)[/tex] = 2.02 + 16.00 = 18.02 g/mol
- Molar mass of [tex]\(10 \times H_2O\)[/tex] = 10 \times 18.02 = 180.2 g/mol

4. Calculate the molar mass of sodium carbonate decahydrate ([tex]\(Na_2CO_3 \cdot 10H_2O\)[/tex]):

- Molar mass of hydrated compound = Molar mass of [tex]\(Na_2CO_3\)[/tex] + Molar mass of [tex]\(10H_2O\)[/tex]
- Total molar mass = 105.99 + 180.2 = 286.19 g/mol

5. Calculate the percentage by mass of water in the compound:

- Mass percentage of water = [tex]\(\left(\frac{\text{Molar mass of } 10H_2O}{\text {Molar mass of } Na_2CO_3 \cdot 10H_2O}\right) \times 100\)[/tex]
- Mass percentage of water = [tex]\(\left(\frac{180.2}{286.19}\right) \times 100 \approx 62.97\%\)[/tex]

Rounding to one decimal place, the percentage by mass of water is approximately:
- [tex]\[ \boxed{63.0\%} \][/tex]
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