Question

Hydrated sodium tetraborate is heated

to drive off the water. You find there are
0.01312 mol Na2B4O7 and 0.1311 mol
H2O in the sample. What is the formula
of the hydrate?
A. Na2B4O7 H₂O
B. Na2B4O7 2H₂O
C. Na2B4O7-10H₂O
D. Na2B4O7-13H₂O

Answer 1

Answer: C. Na2B4O7•10H2O

Step-by-step explanation:

Hope this helps :)

Answer 2

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.

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