Post by: ed_saifa on September 23, 2009, 12:08:01 pm
Addition/subtraction
The result has the same number of decimal places as the measurement with the fewest decimal places
Multiplication/division
the result contains the same number of significant figures as the measurement with the fewest significant figures
Logarithms/exponentials
In a logarithm, the answer has the same number of decimal places as the number of significant figures in the input value. In an exponential, the result has the same number of significant figures as the number of decimal places in the input value.
Post by: kendraaaaa on September 23, 2009, 12:41:25 pm
Post by: Mao on September 23, 2009, 07:22:26 pm
Post by: IntoTheNewWorld on September 23, 2009, 07:52:10 pm
Stuck. Because it is quite significant
*slaps* =P
Post by: monokekie on September 23, 2009, 08:05:28 pm
Post by: Mao on September 23, 2009, 10:48:00 pm
do we need to consider sig fig for ANY calculation in chem? i was told before by some people that it is only the ones associated with concentrations that counts, but what about the heat of combustions? calibration factors? heat contents? and electricity currents? should they count as well?You should consider the sig fig of every number you use in calculations, and adjust your answers accordingly.
Post by: NE2000 on October 24, 2009, 04:59:43 pm
Post by: nels on October 24, 2009, 05:14:27 pm
Post by: simpak on November 03, 2009, 07:34:17 pm
I'm totally unsure as to whether that makes sense or not.
I'll demonstrate...
So say you were calculating, idk...
Q=It.
And you had 23.0 seconds and 5.667 Amps, for our sake.
Your answer would be...130.341.
And you're going to go on to use Q in the next equation in order to find the number of mol of electrons for instance.
Would you round there, to 130 because you just did a calculation in which the value with the lowest number of sig figs had 3?
Or would you carry the whole 130.341 into the next equation and only round up/down at the end, when you give the final result? Because you could end up with a significantly different result depending on the values you use.
Argh, I hope it made sense that way. Thanks.
Post by: nels on November 11, 2009, 02:46:31 pm
Post by: dav132 on November 11, 2009, 05:52:23 pm
i know that we shouldn't round values before you use them in the next calculations,
otherwise you won't get the right final result...
but when u actually write down the in between steps, do you write them with the right sig figs?
e.g. If you have to work out the calibration factor of a bomb calorimeter (3 mark question),
and you need to work out the number of mole of a substance to do that,
would you write down n=0.031 or n=0.0306111120090210.....?
Post by: kenhung123 on December 17, 2009, 02:39:22 pm
i still get confused with the in between steps...I would like to know too.
i know that we shouldn't round values before you use them in the next calculations,
otherwise you won't get the right final result...
but when u actually write down the in between steps, do you write them with the right sig figs?
e.g. If you have to work out the calibration factor of a bomb calorimeter (3 mark question),
and you need to work out the number of mole of a substance to do that,
would you write down n=0.031 or n=0.0306111120090210.....?
Also, when people are saying carry it to the next calculation do you mean like just do it to about 5 decimal places? As sometimes it is difficult to carry the values across in a scientific calculator. For E.g. I got 1.0031241252 and I need to divide 5 by 231 and multiply 1.0031241252, I can't use "Ans" button as its not stored as the ans after doing a later calculation.
Otherwise I will just write down the intermediate answer to 5 decimal places and ensure its quite accurate and forget about the calculator problems haha.
Post by: Edmund on December 18, 2009, 05:55:43 pm
Writing it down to 5 decimal places is fine. What you could do is:i still get confused with the in between steps...I would like to know too.
i know that we shouldn't round values before you use them in the next calculations,
otherwise you won't get the right final result...
but when u actually write down the in between steps, do you write them with the right sig figs?
e.g. If you have to work out the calibration factor of a bomb calorimeter (3 mark question),
and you need to work out the number of mole of a substance to do that,
would you write down n=0.031 or n=0.0306111120090210.....?
Also, when people are saying carry it to the next calculation do you mean like just do it to about 5 decimal places? As sometimes it is difficult to carry the values across in a scientific calculator. For E.g. I got 1.0031241252 and I need to divide 5 by 231 and multiply 1.0031241252, I can't use "Ans" button as its not stored as the ans after doing a later calculation.
Otherwise I will just write down the intermediate answer to 5 decimal places and ensure its quite accurate and forget about the calculator problems haha.
Post by: jimmy999 on December 29, 2009, 03:42:20 pm
Post by: jayfan on January 07, 2010, 12:58:25 pm
The closest reference on this from VCAA is within the 2008 Unit 4 Examiner's Report (http://www.vcaa.vic.edu.au/vcaa/vce/studies/chemistry/assessreports/2008/chemistry2_assessrep_08.pdf), although its not strictly relevant to what we are discussing:
"The question often arises about the impact of relative atomic masses on significant figures, particularly H given as 1.0
in the data book. It may be appropriate to quote the accepted rules for the use of significant figures in calculations.
1. For multiplication and division, the result contains the same number of significant figures as the measurement
with the fewest significant figures. For example, the mass of 0.251 mol H2 = 0.251 x 2.0 = 0.50.
2. For addition and subtraction, the result has the same number of decimal places as the measurement with the fewest decimal places. For example, M(C2H6O) = 2 x 12.0 + 6x1.0 + 1x16.0 = 46.0. For example, 34.652 – 2.36 = 32.29."
At our school, we are taught to round off to the number of significant figures of the least accurate data used in a calculation at the end of THAT calculation. This means you round off each step, and then use that rounded off answer in your next step.
In another word, we are taught NOT to keep results in our calculators all the way.
I think this reasoning makes sense in the real world, because when you are doing experiments, your results can only be as accurate as your least accurate apparatus, which would THEN influence all your subsequent calculations. - i.e. There's no point in keeping your results in your calculator, even though it would be more "mathematically" correct.
Post by: kenhung123 on January 07, 2010, 01:01:56 pm
Post by: stonecold on June 02, 2010, 06:05:04 pm
I was always confused about significant figures right up until the week before my Unit 3 exam. Then it became so clear and easy to do
A comment from someone who did so well warms my heart a little. :)
And just to clarify, you don't allow for relative atomic mass when determining sig figs yeah?
Post by: kyzoo on June 02, 2010, 06:41:40 pm
Post by: stonecold on June 02, 2010, 06:45:52 pm
Post by: kyzoo on June 02, 2010, 06:59:05 pm
m(H2) = 20.0g
So n (H2) = m/M = /20.0/2.0 = 10 mol
Post by: stonecold on June 02, 2010, 07:00:20 pm
Post by: Michael0007 on June 06, 2010, 10:17:53 pm
could someone please clarify point no.2 of the Vcaa report aswell?
Post by: brainbloopers on November 06, 2012, 12:21:32 pm
could someone please clarify point no.2 of the Vcaa report aswell?
Point 2 is confusing basically they are trying to say that with multiplication you always use the lowest sig fig but with addition and subtraction it's the lowest decimal point in the whole calculation.
2 (1s.f) *12.0 (3 s.f) = 20 1.sf (very unintuitive I know ! ) + 1 (1 s.f) * 6.0 (2 significant figures) = 6 1 s.f + 2 (1 s.f) * 16.0 (3 s.f) = 20 (1.sf) then 20+6+20 = 46 BUT SINCE YOU ADDED you use the lowest decimal place number in the calculation which was one therefore it's 46.0
Hope that helped.
Post by: Reckoner on November 06, 2012, 12:48:19 pm
If you then work out the n(e-), and your write
n(e-)= 5.18/96500
=5.36 x 10^-5 (using the "exact" 5.175 from the calculator)
Is that ok? because the value of 5.18/96500 is 5.37 x 10^-5.
Post by: k0k0p0p on November 09, 2012, 03:49:35 pm
When writing in the middle of a calculation, you can write down a few more figures, for example if the question calls for 3 significant figures then write 3.14152 or something similar. You could write 3.14 for a step in a 3 significant figure question, but keep the numbers in your calculator. I would NEVER write a calculation with fewer figures than are called for in the answer, ie. do not write 3.1 for the above value.
With the final figure, ensure significant figures are perfect. Convert to scientific notation if there could be any dispute over the number of significant figures, ie. 0.00314 would be 3.14 *10^-3 to 3 significant figures.
For pH, the numbers after the decimal point are significant, ie. 3.14 would be to 2 significant figures (not 3).
There will always be at least one mark that depends on using the correct significant figures within the exam, but this could be any of the questions so ensure all answers use the correct significant figures.
Essentially, keep the figures in your calculator and use these between calculations. Also, be careful with the acidity constants in the data book as these are only to 2 significant figures! (gotta love pi!)
Post by: Shyam995 on November 25, 2013, 08:28:06 am
Stuck. Because it is quite significant
lol love the pun
Post by: NathanJ on December 24, 2013, 11:53:24 am
eg.
A student determined the water content of a sample of jam. The following
measurements were obtained:
Mass of evaporating dish: 20.22 g
Mass of jam and evaporating dish
before heating: 30.95 g
after heating: 27.22 g
after more heating: 26.50 g
after more heating: 26.49 g
What was the percentage, by mass, of water in the jam?
A5.
Step 1 Find the mass of the moist jam by subtracting mass of evaporating dish.
Mass of moist jam = 30.95 – 20.22 g
= 10.73 g
Step 2 Find the mass of water.
Mass of water = 30.95 – 26.49 g
= 4.46 g
Step 3 Find the percentage of water in moist jam.
% (H2O) =
10.73
4.46 × 100
= 41.57%
Step 4 Express the answer with the correct number of significant figures. (this is where my question comes into play.)
Is the % (H2O) = 41.6% or does it equal 41.57% because of the 4 sig digs in the question ignoring the 3 that happens during the working out.
Cheers,