Question #cc97c

The idea here is that you need to determine how many electrons you get in a neutral lithium-6 atom, then compare the total mass of these electrons to the mass of the lithium-6 atom itself.

So, lithium-6 is one of the two stable isotopes of lithium, #"Li"#, an element that has an atomic number equal to #3#.

This means that in order for an atom to be an atom of lithium, it must contain #3# protons in its nucleus and #3# electrons surrounding its nucleus.

This means that a lithium-6 atom will have #3# electrons surrounding its nucleus. Since the mass of one electron is said to be equal to #1/1836"u"#, you can say that the combined mass of these three electrons will be

#3 color(red)(cancel(color(black)("electrons"))) * (1/1836"u")/(1color(red)(cancel(color(black)("electron")))) = 3/1836"u"#

Now, the atomic mass of a lithium-6 atom can be found here

https://en.wikipedia.org/wiki/Isotopes_of_lithium#Table

So, you know that one atom of lithium-6 has a mass of approximately #"6.015123 u"#.

This values represents the mass of

#{: ("3 protons"), ("3 neutrons") :}} -># present in the nucleus

#"3 electrons " -># surrounding the nucleus

You can get the percent contribution of the electrons to the mass of the atom by dividing their combined mass by the total mass of the atom and multiplying the result by #100#

In your case, you will have

#"% electrons" = (3/1836color(red)(cancel(color(black)("u"))))/(6.015123color(red)(cancel(color(black)("u")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"0.0272%"color(white)(a/a)|)))#

ALTERNATIVELY

You can solve this problem without looking for the actual mass of a lithium-6 atom. All you need to know is that one nucleon, i.e. one proton or one neutron, has a mass of #"1 u"#.

In this case, you would have

#color(purple)(|bar(ul(color(white)(a/a)color(black)("mass of"color(white)(a) ""^6"Li atom" = "mass of nucleons" + "mass of electrons")color(white)(a/a)|)))#

This will get you

#"mass of" color(white)(a)""^6"Li atom" = 6 xx "1 u" + 3/1836"u" = "6.001634 u"#

Once again, you will have

#"% electrons" = (3/1836color(red)(cancel(color(black)("u"))))/(6.001634color(red)(cancel(color(black)("u")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"0.0272%"color(white)(a/a)|)))#