Question

What is the percent composition of water in $N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$.

Answer 1

Decahydrate sodium sulfate ($N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$) is also known as mirabilis or Glauber's salt. The molecules of water in the compound are known as the water of crystallization. These water molecules are present inside the crystals of the solid molecule.

Complete answer:
The mass percentage composition is used to determine the composition of a constituent in a compound or a molecule.
Now, to determine the percent composition of water in decahydrate sodium sulfate ($N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$), we first need to find out the molecular mass of decahydrate sodium sulfate ($N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$).
The molecular mass of one mole of a compound can be calculated by taking the sum of atomic masses of all the elements present in the molecule.
The atomic mass of elements present in decahydrate sodium sulfate ($N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$) is as follows.
N = 22.99 u
S = 32.06 u
O = 15.99 u
H = 1.008 u
So, the molecular mass of decahydrate sodium sulfate ($N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$) will be

& {{M}_{N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O}}=(2\times {{M}_{Na}})+{{M}_{S}}+(4\times {{M}_{O}})+[10\times \\{(2\times {{M}_{H}})+{{M}_{O}}\\}] \\\ & {{M}_{N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O}}=(2\times 22.99)+32.06+(4\times 15.99)+[10\times \\{(2\times 1.008)+15.99\\}] \\\ & {{M}_{N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O}}=322.06\text{ g/mol} \\\ \end{aligned}$$ Since decahydrate sodium sulfate ($N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$) compounds contain 10 molecules of water of crystallization, the mass of water molecules in one mole of decahydrate sodium sulfate ($N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$) will be $$\begin{aligned} & {{M}_{10{{H}_{2}}O}}=10\times [(2\times {{M}_{H}})+{{M}_{O}}] \\\ & {{M}_{10{{H}_{2}}O}}=10\times [(2\times 1.008)+15.99] \\\ & {{M}_{10{{H}_{2}}O}}=180.06\text{ g/mol} \\\ \end{aligned}$$ Now, the formula to calculate the percentage composition of a constituent in a compound is given by: $$\text{percentage composition = }\dfrac{\text{mass of constituent in 1 mole of compound}}{\text{molar mass of compound}}\times 100$$ %$$\text{(}{{\text{H}}_{2}}\text{O)= }\dfrac{180.06}{322.06}\times 100$$ %$$({{H}_{2}}O)=55.91$$% So, the percent composition of water in decahydrate sodium sulfate ($N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$) will be 55.91%. **Note:** It should be noted that when aqueous solutions are used to form crystals of a substance, water is often incorporated in the crystals. Upon heating the substance, one could remove the water molecules of crystallization from the substance, but doing so would also lead to a loss in crystalline properties of the substance.