terminology - Significant Figures Interpretation

Please correct me if I’m wrong, but I’ve read that the reason why we use significant figures is to avoid making the result of a calculation more accurate than the starting values prior to the calculation. For example, if we were seeking the ratio of two weights, lets call them A and B, using a cheap scale and an expensive scale we could get the following varied results:

Cheap Scale $$A = 34 \space oz$$ $$B = 23 \space oz$$

Expensive Scale $$A = 34.0000 \space oz$$ $$B = 23.0000 \space oz$$

Computing the ratio $\frac{A}{B}$ using a calculator we get the following value: $$\frac{34}{23} = 1.47826$$ Now this is a correct ratio depending on the scale used (ie the cheap scale could not have produced such an accurate measurement given the starting values) This leads to the following question:

Why are the zeros behind a decimal point ignored if they are just place holders?

For example, the value 0.000023 oz only has two significant figures, but isn’t that value something only a very fine tuned scale could measure (keeping in line with the previous weight example). So why are these zeros ignored?

Answer

This is part of the reason I really don't like the way we teach significant figures. What's really important isn't the number of digits you have, but rather the uncertainty in your measurements. I think significant figures are taught as an approximation to the proper uncertainty analysis, but it really seems to be not worth the trouble.

What's really going on is this: when you say $A = 34$ oz, you're actually saying that $A$ is closer to 34 oz than it is to either 35 oz or to 33 oz. In other words, you mean $A = (34 \pm 1)$ oz. Likewise, when you say $A = 34.0$ oz, you're really saying $A = (34.0 \pm 0.1)$ oz. And in the final example you give, when you say $A = 0.000023$ oz $= 2.3 \times 10^{-5}$ oz, you're actually saying $A = (2.3 \pm 0.1)\times 10^{-5}$ oz.

The more precise way to do things is to use the uncertainty you have for each measurement, and use the rules for propagation of uncertainties to get the uncertainty in your final answer. One advantage of using the uncertainties is that your uncertainty won't always be 1 unit of your least significant digit, but the rules for propagating significant digits essentially imply that. Using sig figs usually gets close to the answer you would get from properly using uncertainties, but there are some cases where the result is really weird.

To answer your main question directly, you ignore those placeholder zeros because they don't affect the fractional uncertainty of the quantity you're looking at. Keeping them would give you an answer very different from the proper uncertainty analysis.

but isn’t that value something only a very fine tuned scale could measure?

Not necessarily. The cost of an instrument is much more closely related to the relative precision of its output, rather than the absolute precision. If all you're trying to do is weigh things with masses between $1 \times 10^{-5}$ and $1 \times 10^{-4}$ oz, then you don't care at all if your scale can measure 1 oz. But if you have objects that are 1 oz and $1 \times 10^{-5}$ oz, and you care about the difference, then you'll need a really precise scale.