Molar mass of barium nitrate (Ba(NO3)2)

The molar mass of barium nitrate (Ba(NO3)2) is 261.338 g/mol.

Step-by-step calculation

Element	Atoms	Atomic mass	Subtotal
Barium	× 1	137.33	= 137.33
Nitrogen	× 2	14.007	= 28.014
Oxygen	× 6	15.999	= 95.994
Total			261.338 g/mol

Percent composition by element

Barium 52.55%
Nitrogen 10.72%
Oxygen 36.73%

Try another formula

H₂O NaCl CO₂ C₆H₁₂O₆ H₂SO₄ Ca(OH)₂ CuSO₄·5H₂O C₈H₁₀N₄O₂

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Related compounds

AgNO₃ Silver nitrate NaNO₃ Sodium nitrate NH₄NO₃ Ammonium nitrate Pb(NO₃)₂ Lead(II) nitrate C₈H₁₀N₄O₂ Caffeine HNO₃ Nitric acid

FAQ

What is the molar mass of Ba(NO3)2?

The molar mass of barium nitrate (Ba(NO3)2) is 261.338 g/mol.

How do you calculate the molar mass of Ba(NO3)2?

Multiply the atomic mass of each element by the number of its atoms in Ba(NO3)2, then add the results: 1 × 137.33 (Ba) + 2 × 14.007 (N) + 6 × 15.999 (O) = 261.338 g/mol.

What percentage of Ba(NO3)2 is Barium?

Barium makes up 52.55% of the mass of barium nitrate (Ba(NO3)2).