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johnkellock

Mass of Na2CO3, then add 10 (molecular mass of water which is 18g)

So we have 106g+180g = 286g if we are looking for the percentage of water in the compound we do this 180g/286g= 62.9

 SO2 because sulfurs atomic mass is 32g and Oxygen atomic mass is 16g

Answer: The mass percent of water in \(Na_2CO_3.10H_2O\) is 63%.

Explanation:

In \(Na_2CO_3.10H_2O\), there are 2 sodium atoms, 1 carbon atom 13 oxygen atoms and 20 hydrogen atoms.

To calculate the mass percent of element in a given compound, we use the formula:

\(\text{Mass percent of water}=\frac{\text{Mass of water}}{\text{Molar mass of}Na_2CO_3.10H_2O}\times 100\)

Mass of water = \(10\times 18g/mol=180g\)

Mass of \(Na_2CO_3.10H_2O\)= 286 g

Putting values in above equation, we get:

\(\text{Mass percent of water}=\frac{180}{286}\times 100=63\%\)

Hence, the mass percent of water in \(Na_2CO_3.10H_2O\) is 63%

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