Doubts regarding Atomic mass,relative atomic mass and mass number.

rishch

I was reading my science textbook and they have introduced the concepts of atomic mass,relative atomic mass and mass number.Here are some doubts I have-

1)Is atomic mass a ratio ?
2)1/12th of a C-12 atom is half a proton,half a neutron and half an electron.Since mass of an electron is negligible and since 1 proton=1 neutron,1/12th of a C-12 atom is basically the mass of one proton or one neutron.A hydrogen atom has 1 proton so why didn't they take 1u = mass of 1 hydrogen atom?
3)Also,if 1u = mass of a proton = mass of a neutron aren't mass number and atomic mass the exact same thing ?
4)In my book the atomic mass of Chlorine is given as 35.5.How can an atom have half a neutron or proton ?
5)Can someone please explain what relative atomic mass is along with an example as in my book just the definition is given and I couldn't understand a thing.

Amok

It's mass of an atom in atomic mass units. So you could say it's the ratio of the mass of the atom to that of 1/12 of the mass of C-12.

I dunno, historical reasons I guess. Probably has to do with the fact that C-12 has the same number of neutrosn and protons. And 1/12 of the mass of C-12 is half a proton, half a neutron and 1 electron (not half an electron).

1u is not exactly the mass of a proton, you said so yourself. However, since neutrons and protons have similar masses and electrons have very small masses, mass numbers and atomic mass values do look alike.

It doesn't. That's the average mass of a chlorine atom in nature, taking into account isotopic abundance. Chlorine has basically two isotopes; Cl-35 (76% of all chlorine) and Cl-37 (24% of all chlorine). So the weighted average is:

35*0.76 + 37*0.24 = 35.5



I dunno what you mean by "relative atomic mass". The word "relative" probably comes from the fact that you measure the mass of an atom relative to the C-12 atom. If an atom has isotopes in nature, its mass is usually given as a weighted average of the masses of its isotopes (as I explained just above).