# Significant Figures Calculations and Precise Instruments [closed]

I understand the rules for significant figures and how to read instruments, except for one thing.

I was told that our digital scale is particularly precise to the 1,000ths decimal place.

So in multiplication/division, the answer should have the same number of significant figures as the term with the least number of significant figures.

So if I weigh something on a balance accurate to the 1000th's place that happens to read 0.300g. Then I should record that weight as 0.300g.

The confusion for me is that when we determine significant figures from abstract numbers we first have to determine the precision(significant figures) of each term. Well, 0.300g is just one sig-fig. So if I were to then divide it by a volume to find the density and the volume is 0.111 Liters the result would be 0.03g/ml.

What was the point in writing it down initially to the precision of the instrument (0.300g instead of 0.3g) if in terms of calculations we treat it as a low resolution(imprecise/low-sig-fig) number? OR am I doing it wrong, and the significant figures are determined via the precision of the instrument.

My understanding of the purpose convention of significant figures it to let people know the quality of the measurements. Well if you have a high precision instrument, and happen to weigh something that has only one-sig fig isn't this misleading?

## closed as off-topic by Mithoron, A.K., Jon Custer, Karl, Todd MinehardtOct 27 '18 at 20:13

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• I think you are doing it wrong. Given 0.300g, I would read that as having 3 sig figs while 0.3g would be 1 sig fig. It sometimes helps to write these in scientific notation to see the difference (i.e. $3.00\times10^{-1}$ vs $3\times10^{-1}$) – Tyberius Oct 24 '18 at 18:14
• If you are saying 0.300, then you are expressing certainly about that second zero. But then it's significant, isn't it? – Zhe Oct 24 '18 at 20:04
• I am doing it wrong. I was mixing the rules for decimal vs no decimal. Woops. I appreciate everyone's answers. Since my question was malformed I'm not going to credit an answer but I will upvote everyone's participation here. – Xzila Oct 31 '18 at 3:24

$$\dfrac{0.300\text{ grams}}{0.111\text{ liters}\times 1000\text {ml/liter}} \approx 0.0027027\text{ g/ml}\ce{->[rounding]}2.70\times10^{-3}\text{ g/ml}$$