Example: Calculation of moles and grams of boron in boric acid

1. Molar mass of boric acid (H₃BO₃)

  • Hydrogen (H):3×1,008=3,024 g/mol
  • Boron (B):1×10.81=10.81 g/mol
  • Oxygen (O):3×16.00=48.00 g/mol
  • In total: 3,024+10.81+48.00=61,834 g/mol

2. Amount of boric acid in 1 gram

n H3BO3 = 1 g 61.834 g/mol ≈ 0.01617 mol

3. Amount of boron (B)

1 molecule of H₃BO₃ contains 1 boron atom:

nB = n H3BO3 = 0.01617 mol

4. Concentration in mol/liter

0.0162 mol/L boron (B)
(when dissolving 1 g of H₃BO₃ in 1 liter)

 

Calculation of the mass of boron (B) in boric acid (H₃BO₃)

1. Amount of boron in moles

nB=0.01617moles

(already from the calculation of the amount of substance of H₃BO₃)

 

2. Molar mass of boron

MB=10.81g/mol

 

3. Calculating the boron mass

mB=nB×MB=0.01617moles×10.81g/mol≈0.175G

Mass of boron:0.175g B

(contained in 1 g H₃BO₃)

 

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Category: Analysis
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