Find the percentage of water in Na2CO3*2H2O?
(2*molar mass H2O)/(molar mass Na2CO3.2H2O)*100 =
To find the percentage of water in Na2CO3*2H2O, we need to calculate the ratio of the mass of water to the total mass of the compound and then multiply by 100 to express it as a percentage.
First, we need to determine the molar mass of the entire compound, Na2CO3*2H2O.
The molar mass of Na2CO3 is:
2(Na) + 1(C) + 3(O) = 46.0 g/mol + 12.0 g/mol + 48.0 g/mol = 106.0 g/mol
The molar mass of H2O is:
2(H) + 1(O) = 2.0 g/mol + 16.0 g/mol = 18.0 g/mol
Therefore, the total molar mass of Na2CO3*2H2O is:
106.0 g/mol + 2(18.0 g/mol) = 142.0 g/mol
Next, we need to identify the mass of water in the compound, which is 2H2O. Since the molar mass of water is 18.0 g/mol, the mass of 2H2O is 2(18.0 g/mol) = 36.0 g/mol.
Finally, to find the percentage of water in Na2CO3*2H2O:
(36.0 g/mol / 142.0 g/mol) x 100 = 25.35%
Therefore, Na2CO3*2H2O contains approximately 25.35% of water.