Question

What is the mass percent of nitrogen in lead (ii) nitrate (pb(no3)2)?

Answer 1

The mass percent of nitrogen in lead (II) nitrate (Pb(NO3)2) is calculated by dividing the total mass of nitrogen by the molecular mass of lead (II) nitrate and multiplying by 100%, which is approximately 8.46%.

Step-by-step explanation:

To determine the mass percent of nitrogen in lead (II) nitrate (Pb(NO3)2), you need to find the total mass of nitrogen in the compound and divide it by the molecular mass of lead (II) nitrate. The molar mass of lead (II) nitrate is the sum of the mass of one lead atom, two nitrogen atoms, and six oxygen atoms. Thus, the molar mass of Pb(NO3)2 is 207.2 (mass of Pb) + 2(14.007) (mass of N) + 6(15.999) (mass of O) = 331.209 g/mol. The total mass of nitrogen is 2 × 14.007 g/mol = 28.014 g/mol.

To find the mass percent, we use the formula:

Mass percent of N = (Total mass of N / Molar mass of Pb(NO3)2) × 100%

Mass percent of N = (28.014 g/mol / 331.209 g/mol) × 100% ≈ 8.46%

Answer 2

In order to calculate the mass percent of nitrogen in Pb(NO3)2, first, list the molar mass of all elements.

Pb = 207.2 g/mol
N = 14 g/mol
O = 16 g/mol

Then, multiply each element to the number of each atom in the compound to get the mass contribution of each element, in this case:

Pb = (207.2 x 1) = 207.2
N = (14 x 2) = 28
O = (16 x 6) = 96

Add the resulting products to obtain the total molecular mass of the compound. This is equal to 331.2 g/mol. Finally, to get the mass percent of nitrogen, divide its mass contribution by the total molecular mass, as in, (28/331.2) x 100%. Thus, the mass percent of nitrogen in Pb(NO3)2 is 8.45%.