The formula is tetrasodium borate-Na2B4O7.10H2O
Given- 0.0132 mol Na2B4O7 and 0.1311 mol H2O
Step 1
Using the molar mass of the anhydrous Na2B4O7 and its mass percentage, we can calculate the molar mass of the hydrate (if we look at it as 100% of the mass) by stoichiometry.
Molar mass of Na = (22.990 g/mol)
Molar mass of B = (10.811 g/mol)
Molar mass of O = (15.999 g/mol)
Molar mass of Na2B4O7= 2⋅22.990 g/mol +4⋅10.811 g/mol +7⋅15.999 g/mol = 201.217 g/mol
201.217g/mol : 52.8%=x g/mol : 100%
x g/mol = 201.217 g/mol⋅100%÷52.8 %
x g/mol= 381.093 g/mol
Step 2
In 381.093 g of hydrate, we have 201.217 g of anhydrous Na2B4O7 , the rest of the mass is water.
381.093g−201.217g= 179.876 g of water
Molar mass of H = 1.008 g/mol
Molar mass of O = 15.999 g/mol
Molar mass of H2O= 1.008 g/mol+ 15.999g/mol = 18.015 g/mol
179.876g ÷18.015 = 9.98= 10 moles of water per mole of hydrate.