The mass of an atom may be determined by adding the masses of the protons and neutrons in the nucleus and then subtracting the mass defect. Why is it unnecessary to include the electrons when determining the mass of an atom?

Question

The mass of an atom may be determined by adding the masses of the protons and neutrons in the nucleus and then subtracting the mass defect.  Why is it unnecessary to include the electrons when determining the mass of an atom?

Henry Antrim · Accepted Answer

Because the electrons have so little mass.

Neutrons and protons have masses that are nearly equal, with a proton having roughly 1836 times the mass of an electron.

Peyton Adams · Answer

Well, because electrons are really light... only about 1/1836 the mass of a proton or neutron... You get about #1000# times worse of a percent difference with respect to the literature, but you still get within #0.05%# typically... For light atoms, it is the most acceptable.  For heavy atoms, the absolute difference in #"amu"# when calculating the mass defect, if you exclude the electrons, does increase by a factor of about #10# sometimes. LIGHT ATOM EXAMPLE Consider, for example, the #bb("Mg"-25)# isotope. Here is a table listing the (real, not calculated) isotopic mass, and for that isotope,  #m_p" "" "" "m_n" "" "" "" "m_e" "" "" "" ""mass defect"#
#ul("(amu)"" "color(white)(/)"(amu)"" "" "" ""(amu)"" "" "" ""(amu)"" "" ")#
#12.087312color(white)(//.)13.112632color(white)(//)0.00658296color(white)(//.)0.220707# #ul("Isotopic mass" ("amu"))#
#24.985837# Below is the calculated isotopic mass #"w/o  e"^(-)" "" ""w/  e"^(-)" "" "" ""% diff"#
#ul("(amu)"" "" "color(white)(/)"(amu)"" "" "" "" "" "" "" "" "" ")#
#24.979237color(white)(//.)24.985820color(white)(//.)-6.80385 xx 10^(-5)# The percent difference with respect to the true atomic mass without electrons included is #-2.64150 xx 10^(-2)%#, so we are still within #0.05%# error, albeit 3 orders of magnitude worse. For general chemistry, that is still perfectly acceptable. The electron mass is only about #0.0263%# of the total isotopic mass, and the calculated mass defect by excluding the electrons is #"0.214107 amu"#, only about #"0.0066 amu"# less than what it should be. HEAVY ATOM EXAMPLE Consider, for example, the #bb("U"-238)# isotope. Here is a table listing the (real, not calculated) isotopic mass, and for that isotope,  #m_p" "" "" "" "m_n" "" "" "" "m_e" "" "" "" ""mass defect"#
#ul("(amu)"" "" "color(white)(/)"(amu)"" "" "" ""(amu)"" "" "" ""(amu)"" "" ")#
#92.669392color(white)(//.)147.264944color(white)(//)0.05046936color(white)(//.)1.934192# #ul("Isotopic mass" ("amu"))#
#238.050788# Below is the calculated isotopic mass #"w/o  e"^(-)" "" ""w/  e"^(-)" "" "" ""% diff"#
#ul("(amu)"" "" "color(white)(/)"(amu)"" "" "" "" "" "" "" "" "" ")#
#238.000144color(white)(//.)238.050613color(white)(//.)-7.44520 xx 10^(-5)# The percent difference with respect to the true atomic mass without electrons included is #-2.12761 xx 10^(-2)%#, so we are still within #0.05%# error, albeit 3 orders of magnitude worse. For general chemistry, again, that is still perfectly acceptable. The electron mass is only about #0.0212%# of the total isotopic mass. However, the calculated mass defect by excluding the electrons is #"1.883548 amu"#, about #"0.0506 amu"# less than what it should be. It doesn't seem like a big difference, but you do get almost an order of magnitude difference in the mass defect, which would be enough to give you a wrong answer in an online homework system.

Jeremiah Adams · Answer

undefined Since electrons are much less massive than protons and neutrons, their mass cannot be taken into consideration when calculating the mass of an atom because it is so small.