Let's look at where these numbers come from. For carbon, the main isotopes are carbon-12 and carbon-13. Carbon-13 has an abundance of 1.109%.
Consider a molecule with $n$ carbons in it. The ratio of the $M_{1}$ peak to the $M_{0}$ peak is related to this above fraction, but you have to account for the fact that there are $n$ ways to create this molecule with one carbon as carbon-13 and the rest as carbon-12. This is basically one of the binomial cofficients:
$$\binom{n}{1} = n$$
You can take the total distribution of $p + (1-p)$, where $p$ is abundance of carbon-13. If you have $n$ carbons, we have:
$$\left(p + (1-p)\right)^{n}=(1-p)^{n} + \binom{n}{1}(1-p)^{n-1}p+\ldots$$
Each of $n+1$ terms corresponds to the relative height of one of your MS peaks. The ratio between peak 0 and peak 1 is:
$$\frac{\binom{n}{1}(1-p)^{n-1}p}{(1-p)^{n}}=\frac{np}{1-p}$$
Since $p=0.01109$, this expression comes to approximately:
$$n\cdot\frac{1.1}{100}$$
This means if you have $n$, you can estimate the height of the first peak relatively to the zero-th order peak. Technically, given the method I provided, you can estimate an higher order peak.
Order, if you have the base peak and the first peak, you can estimate $n$. This isn't supposed tobe perfect if you're using 2 significant figures, but your measurement of the MS peak intensity is probably similar in terms of error, but you should find that even in your example, $1/99$ is a lot closer to $1.1/100$ than $n\cdot (1.1/100)$ for values of $n$ other than 1.
