Question

What is the percentage of water in the following compound?

Sodium carbonate decahydrate, Na2CO3 • 10H2O

% by mass H2O

Answer 1

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 2

Answer: The mass percent of water in
`Na_2CO_3.10H_2O` is 63%.

Step-by-step 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}* 100`

Mass of water =
`10* 18g/mol=180g`

Mass of
`Na_2CO_3.10H_2O`= 286 g

Putting values in above equation, we get:

`\text{Mass percent of water}=(180)/(286)* 100=63\%`

Hence, the mass percent of water in
`Na_2CO_3.10H_2O` is 63%