# Percent error and significant figures?

For an elementary chemistry class percent error is calculated as: [actual-experimental]/actual x 100% edit:bracket represents absolute value

So my question is if the experimental is close enough to the actual that the subtraction would give zero significant figures, would then the percent error then be zero?
EX: measured density of 0.997g/ml and actual density of 0.997171g/ml then, 0.997171-0.997=0.000171 which would be rounded to 0.000 due to sig. figs. -> there are no significant figures so this would mean the calculated 0.0171% error would be zero correct. If I remember my analytical correctly we will do away with sig. figs. and work with the propagation of error within each measurement but I have not thought about how this would work in a problem like this.

Actually the % error is calculated as:

$\frac{\text{experimental} - \text{actual}}{\text{actual}}\cdot100\%$

That way

• if experimental < actual then the error is negative
• if experimental > actual then the error is positive

So my question is if the experimental is close enough to the actual that the subtraction would give zero significant figures, would then the percent error then be zero?

Not exactly zero...

EX: measured density of 0.997g/ml and actual density of 0.997171g/ml then, 0.997171-0.997=0.000171 which would be rounded to 0.000 due to sig. figs

Correct for subtraction you have used. Addition and subtraction round to the the last common significant decimal place of all the measurements. That would be the 8 in 10.98 in the following example.

So $10.9\bar8 + 0.6754 + 0.4795 = 12.1349 \approx 12.13$

In actually you should have used the reverse
0.997 - 0.99717 = -0.000171

there are no significant figures so this would mean the calculated 0.0171% error would be zero correct.

Now this gets sticky. You could write that the error was 0% but that would indicate only one part in 100 precision. If you write 0.00% it isn't entirely clear if that is two or three significant figures. But you have claimed that the hundredth place in the % is significant, which it is not.

Remember that the minimum nonzero error possible was +/- 0.001 which yields an error of 0.1%.

So the error would be best written as 0.0% indicating that the error is 0 to 1 part in a thousand.

Reference: Significant Figures at Wikipedia

I'll leave the propagation of errors for a different exercise.

I believe that you are correct in saying that the percent error is zero in this case, which is what I tell my gen chem students who do not know how to statistically evaluate uncertainty. As the previous answer-er pointed out, "when subtracting, Addition and subtraction round to the the last common significant decimal place of all the measurements." The last common sig DP between 0.997 and 0.99717 is the third decimal place, so the result of your subtraction, when rounded to the correct number of sig figs is zero.

Here is the more important part and the rational behind the sig fig rules... 0.997 g/mL implies a precision of +/- 0.001g/ml. The accepted value of 0.997171 g/mL falls within this range. So, due to the limitations of the method used to determine the density, the experimental value and accepted value are not distinguishable from each other and that a percent error cannot be determined unless the experiment is redone in a way which increases precision.

FYI, my experience is the most Gen Chem teachers let sig fig errors slide with percent error since this concept can be difficult to understand.