Can somebody help me with understanding the question A set of scales come with n weights each of which is an integer number of grams. In combination, on one side of the scale, the weights may be used to balance an equal mass. Explain why at most $2^{n}-1$ masses may be balanced. 
Does this question mean that let's say $n=4$, then each of the weights are $1,2,3,4$g and $15$g can be balanced using these $4$ weights?
 A: No, this question does not mean what you suggested. As one of the comments explains it's not about a particular weight of masses you can balance, but it's about how many different weights can you "produce" by combining different sets of the masses available. So for example is you have weighs $1g$, $2g$, and $4g$, you can produce:
$1$, $2$, and $4$ (by selecting only one weight each time), 
$1+2= 3$, $1+4=5$, $2+4=6$ (by selecting two weights each time) 
and $1+2+4 = 7$ (by selecting all three weights). 
This is a total of 7 different weights/masses (and we can check that $2^3-1 = 7$)
So the maximum number of weighs/masses we can produce is the number of all non-empty subsets of the set of weights available. How do we find this number then?
The easiest approach is probably the following: For every element of our set (our individual weights we start with) we have two choices: either include it in the subset we are building or not. So we have $2$ choices for the first element, times $2$ choices for the second etc.
$$\underbrace{2\times 2 \times \cdots \times 2}_{n \text{ times}} = 2^n$$
One of these combinations is the empty set (choosing for all elements not to include them). Since we do not want to include this case we simple subtract $1$ from the result: $2^n -1$
Here's another way to count all the different subsets. Let's assume that we build all subsets of exactly $i$ elements out of the $n$ possible. How many are these? The number of combinations of choosing $i$ elements out of a set of $n$ elements is written as $n\choose{i}$ and read "n choose i"
So we are looking for the result of this sum: $$\sum_{i=1}^n{n\choose i} $$
It is a relatively well-known fact that this number is $2^n-1$. One way to prove this is take $2^n$ and write it as $(1+1)^2$. Then use the binomial expansion of $(x+y)^n$ which is $\sum_{i=0}^n{n\choose{i}}x^i y^{n-i}$ 
$$2^n=(1+1)^n = \sum_{i=0}^n{n\choose{i}}1^i 1^{n-i} =\sum_{i=0}^n{n\choose{i}} =  \sum_{i=\color{red}{1}}^n{n\choose{i}} + {n\choose{0}}= \sum_{i=1}^n{n\choose{i}} + 1 \implies$$
$$ \sum_{i=1}^n{n\choose{i}}=2^n-1$$
