What is the percentage of water in Na2CO3 and 10H2O?

Answer 1

overall weight of a molecule = 23*2+12+16*3+10*18=286

weight of H2O per molecule =10*18=180

% H2O =(180/286)100=62.9

Answer 2

The formula given shows that there are ten water molecules consisting of 20 hydrogen atoms and 10 oxygen atoms in the water content of the hydrated salt. The remainder of the formula shows that there are 2 sodium atoms, 1 carbon atom, and 3 oxygen atoms in the remainder of the formula unit.

The gram* atomic masses of the elements involved are: H, 1.00794; C, 12.011; O, 15.9994; and Na, 22.9898. The fraction by mass of water in the hydrated salt therefore is [20(1.00794) + 10(15.9994)]/[20(1.00794) + 10(15.9994) + 2(22.9898) + 12.011 + 3(15.9994)] or 0.62959. The pounds of water in 100 pounds of the hydrated salt therefore is 63.0, to the justified number of significant digits.**

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*Because the question is about proportions, any unit for atomic mass may be used, as long as the unit is consistent for all the atoms involved.

** All of the integers appearing before a left parenthesis in the arithmetic expression given for calculation are exact, but the number of significant digits justified is limited by the value of "100 pounds".

Answer 3

The percentage of water in Na2CO3·10H2O is 63.90%.