Question

What is the percentage of water in the following compound?

Answer 1

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%

Answer 2

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