Question

Was the density you determined accurate, taking into account the total instrument uncertainty?

I got 0.778 g/ml
the theoretical value is 0.789 g/ml
the instrumental uncertainty was 0.2%

How do I know if its accurate ??
please help

Answer 1

(.789-.778)/.778=1.4 percent error.

Answer 2

does that mean is accurate?

Answer 3

It means your answer (experimental result) is 1.4% above the accepted value. To call this 'accurate' based on a %Error calculation would be unwise unless there is a published accepted %Error 'range' for the experiment instruments used to collect the raw data. That is, your answer would be published as (Experimental Density) ± 1.4%(Experimental Density) = 0.778 g/ml ± 1.4%(0.778)g/ml giving an experimental range of [0.778 – (1.4% of 0.778)] to [0.778 + (1.4% of 0.778)] => [(0.778 – 0.011)g/ml to (0.778 + 0.011)g/ml] => [0.767g/ml to 0.789g/ml] for your experimental results. Now, this is based on your instructor (or note in the procedures section of your experiment) giving an ‘accepted’ %Error. You would then be able to determine the ‘expected’ range of reliability of the lab data collected based on an accepted value and accepted %Error. For example, let’s say you are given a 2% Error as an accepted %Error. Then your answer of 0.778g/ml would be ‘accurate’ with respect to the published %Error. That is, Density = Accepted Density ± 2% of accepted density = 0.789g/ml ± 2%(0.789g/ml) = (0.789 ± 0.016)g/ml => [(0.773g/ml) to (0.805g/ml)]. Since your experimental results falls within the range based on accepted value variation of data, then your answer can be said to be accurate. However, if the instructor/reference manual says the accepted %error is, say 1%, then this would give an accepted range of (0.789 ± 0.008)g/ml => [(0.781)g/ml to (0.797)g/ml]. Your answer of 0.778g/ml would NOT be accurate b/c it was not within the ‘accepted’ range of %Error for the instrument used to obtain your data. %Error is generally the ‘Accuracy’ factor as it indicates variation of data about an accepted value, but the accepted value and the standard %Error for the instrument used needs to be given.

Answer 4

woow thank you so much