To calculate the percentage by mass of lead in 1 mole of Pb(NO3)2, we first need to determine the molar mass of the compound.
Pb(NO3)2 consists of one lead atom (Pb), two nitrate ions (NO3), and each nitrate ion contains one nitrogen atom (N) and three oxygen atoms (O).
The molar mass of lead (Pb) is 207.2 g/mol.
The molar mass of nitrate (NO3) is calculated as follows:
Nitrogen (N): 1 atom x 14.01 g/mol = 14.01 g/mol
Oxygen (O): 3 atoms x 16.00 g/mol = 48.00 g/mol
Total molar mass of nitrate (NO3): 14.01 g/mol + 48.00 g/mol = 62.01 g/mol
Now, let's calculate the molar mass of Pb(NO3)2:
1 atom of lead (Pb): 1 x 207.2 g/mol = 207.2 g/mol
2 nitrate ions (NO3): 2 x 62.01 g/mol = 124.02 g/mol
Total molar mass of Pb(NO3)2: 207.2 g/mol + 124.02 g/mol = 331.22 g/mol
To find the percentage by mass of lead in 1 mole of Pb(NO3)2, we divide the molar mass of lead by the molar mass of the compound and multiply by 100%:
(207.2 g/mol / 331.22 g/mol) x 100% ≈ 62.53%
Therefore, the percentage by mass of lead in 1 mole of Pb(NO3)2 is approximately 62.53%.
Calculate percentage by mass lead in 1 mole of Pb(NO3)2
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