What is the molecular weight of 429 g of
lead(II) nitrite? Answer in units of g=mol.
if you can show work and answer plz.
thanks for your time.
2006-12-09 04:27:00 · 3 answers · asked by dojorno5 2 in Science & Mathematics ➔ Chemistry
The molecular weight is intrinsic to a molecule, so forget about the mass. I think the problem probably just threw that in to confuse you.
If you look on your periodic table, you should be able to find the atomic mass of each atom. It's usually either at the bottom of the square representing that atom or in a separate table. The molecular mass of a molecule is just the sum of the atomic masses of the atoms which compose it. The units of atomic/molecular mass are already in g/mol, so we won't have to convert it.
In this case, the molecule is Pb(NO3)2 (I know there are two nitrate groups because lead(II) means the lead has a +2 charge, and each nitrate group has a -1 charge). That means its molecular mass is equal to the atomic mass of Pb + 2* the atomic mass of N + 6* the atomic mass of O. That's 207.2 + 2*14.0 + 6*16.0 = 207.2 + 28 + 96 = 331.2 g/mol.
2006-12-09 04:34:11 · answer #1 · answered by Amy F 5 · 0⤊ 0⤋
Pb(No3)
then the molecular weight of leadnitrite is--
207+(14+16*3)=269
2006-12-09 12:41:31 · answer #2 · answered by sanu 2 · 0⤊ 0⤋
I think what is being asked here is the number of moles of lead II nitrate in 429 g of the compound, so here is how you proceed:
1. find the mass of lead nitrate= Pb+ 2x (N +O3)=269.
this means that 269g of lead nitrate contains 1 mole of the compound; 269g is the molar mass of lead II nitrate.
2.then find the number of moles in 429 g:
269g - 1mole
therefore 429 g = 1/269 x 429 = 1.595 moles
2006-12-09 13:59:00 · answer #3 · answered by Crazygirl 3 · 0⤊ 0⤋