Molar Mass of Lead(II) nitrate

Pb(NO3)2 — 331.208 g/mol, worked out element by element.

Molar mass of Lead(II) nitrate, Pb(NO₃)₂

331.208 g/mol

Molecular weight 331.208 · 9 atoms per formula unit

Also known as: pb(no3)2, lead nitrate.

One of the few soluble lead salts, which is what makes it useful for precipitation reactions.

How 331.208 g/mol is worked out

Mass contribution of each element
ElementAtomsAtomic massMass% of total
Lead1207.2207.20062.56%
Oxygen615.99995.99428.98%
Nitrogen214.00728.0148.46%
Total9331.208100%

M = 1 × 207.200 + 6 × 15.999 + 2 × 14.007 = 331.208 g/mol

Grams to moles for Lead(II) nitrate

Mass to moles conversions
MassMolesFormula units
1 g0.0030 mol1.818 × 10^21
10 g0.0302 mol1.818 × 10^22
25 g0.0755 mol4.546 × 10^22
50 g0.1510 mol9.091 × 10^22
100 g0.3019 mol1.818 × 10^23
250 g0.7548 mol4.546 × 10^23
331.208 g1 mol6.022 × 10^23

For any other amount, use the molar mass calculator, or the molarity calculator to make up a solution of a given concentration.

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Frequently Asked Questions

What is the molar mass of Lead(II) nitrate?

The molar mass of Lead(II) nitrate (Pb(NO3)2) is 331.208 g/mol. That is the sum of the standard atomic weights of all 9 atoms in one formula unit: 1 × 207.200 + 6 × 15.999 + 2 × 14.007 = 331.208.

What is the molecular weight of Lead(II) nitrate?

331.208 — the same number as the molar mass, but as a dimensionless ratio rather than in g/mol. The two terms describe the same arithmetic, so a molecular weight of 331.21 and a molar mass of 331.21 g/mol are the same answer.

How many moles are in 100 g of Lead(II) nitrate?

100 ÷ 331.208 = 0.3019 mol. To convert any mass to moles, divide by the molar mass; to go the other way, multiply.

What is the percent composition of Lead(II) nitrate?

By mass, Lead(II) nitrate is 62.56% lead, 28.98% oxygen, 8.46% nitrogen. Each element's share is its total mass in the formula divided by 331.208.

How many atoms are in one formula unit of Lead(II) nitrate?

9: 1 lead, 6 oxygen, 2 nitrogen. One mole contains 6.02214076 × 10²³ formula units, so a mole of Lead(II) nitrate contains 9 × 6.022 × 10²³ atoms in total.