Why does $\mathrm{NO_3}$ have charge $1-$? [closed]

Question

I'm given the following problem:

Write the molecular equation for the reaction $\mathrm{Iron\;(III)\; Nitrate\; and\; Sodium \;Phosphate}$

I begin by attempting to determine the empirical formula of $\mathrm{Iron\;(III)\; Nitrate}$. In class, we were taught to memorize the following table:

Group     Charge
#1        +1
#2        +2
#3        +3
#5        -3
#6        -2
#7        -1

Additionally, we were asked to memorize the "formula" for Nitrate ($\mathrm{NO_3}$).

The $\mathrm{III}$ after Iron indicates that it has a $3+$ charge. From the table above, $\mathrm{N}$ has charge $3-$, and $\mathrm{O}$ has charge $2-$. There are $3$ $\mathrm{O's}$, giving a net charge of $-2 \cdot 3=-6$.

Unable to determine the empirical formula of $\mathrm{Iron\;(III)\; Nitrate}$, I looked it up to find $\mathrm{Fe(NO_3)_3}$.

If $\mathrm{NO_3}$ has charge $6-$, then $3$ $\mathrm{NO_3}$'s has charge $18-$. Meanwhile, $\mathrm{Fe}$ has only charge $3+$.

As it turns out, $NO_3$ actually has charge $1-$. How is this determined? I'm told its "just something that needs to be memorized" in order to progress in the course.

Answer 1

As it turns out, the simple "adding and subtracting" of charge values (from the table given in the question) we were taught doesn't always (ever?) work. Drawing the Lewis structure for $\mathrm{NO_3}$, I obtain:

...the charge is then clear.