Question

How many moles of `(CH_3)_3NH^+` are in 6.0 g of `(CH_3)_3NH^+` ?

Answer 1

`"0.10 mole"`

Explanation:

The thing to keep in mind here is that the charge of an ion does not affect its molar mass, so don't be confused by the fact that your polyatomic ion carries a `1+` overall charge.

So, you're dealing with the trimethylammonium cation, `("CH"_3)_3"NH"^(+)`, which is formed when trimethylamine, `("CH"_3)_3"N"`, accepts a proton, `"H"^(+)`.

Your starting point here will be to figure out the molar mass of the trimethylammonium cation by using the molar mass of trimethylamine, which is listed as

`M_("M" ("CH"_3)_3"N") = "59.11026 g mol"^(-1)`

So, the molar mass of the trimethylammonium cation must account for the extra proton that is added to the neutral compound. The molar mass of a proton is equal to the molar mass of a neutral hydrogen atom

`M_("M H") = "1.00794 g mol"^(-1)`

This means that the molar mass of the trimethylammonium cation will be

`M_("M"("CH"_3)_3"NH"^(+)) = "59.11026 g mol"^(-1) + "1.00794 g mol"^(-1)`

`M_("M"("CH"_3)_3"NH"^(+)) = "60.11820 g mol"^(-1)`

Now, the molar mass of a substance tells you the mass of one mole of that substance. In this case, one mole of trimethylammonium cations has a mass of `"60.11820 g"`.

This means that your `"6.0 g"` sample will contain

`6.0 color(red)(cancel(color(black)("g"))) * ("1 mole"color(white)(a)("CH"_3)_3"NH"^(+))/(60.11820color(red)(cancel(color(black)("g")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.10 moles"color(white)(a)("CH"_3)_3"NH"^(+))color(white)(a/a)|)))`

The answer is rounded to two sig figs, the number of sig figs you have for the mass of trimethylammonium cations.