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Degger [83]
3 years ago
6

Percent of water in Na2CO3 • 10H2O

Chemistry
1 answer:
Troyanec [42]3 years ago
5 0
Hey there! 

In order to solve for the percentage of water in the compound, you will first need to find its total molar mass. You can do this by adding up the molar masses of each individual element in the compound. Then, you will divide the mass that you find of the water molecules by the total mass to get the percentage. 

→ Na₂CO₃ ×<span> 10 H</span>₂<span>O

</span>→ Na₂ = 22.9898 × 2 = 45.9796
→ C = 12.0107
→ O₃ = 15.999 × 3 = 47.997
→ 10 H₂O = 18.015 × 10 = 180.15

Now, just add all of those numbers up for the total molar mass. 

→ 45.9796 + 12.0107 + 47.997 + 180.15 = <span>286.1373
</span>
The last step is to divide the molar mass of the 10 water molecules by the total mass. 

→ 180.15 ÷ 286.1373 = <span>0.62959 </span>≈ 0.63

Your answer will be about 63%. 

Hope this helped you out! :-)
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