Significant figures in calculations?

[Aar0n_101]

So I know this must sound kinda dumb, but in equations, like thermochemical equations, that require multiple steps and calculations, how do we use significant figures? Do we round significant figures with each step or round at the final answer? E.g, The molar heat of combustion of ethanol is tabulataed as -1364 kJ mol. If the density of  ethanol is 0.790 g ML, calculate the energy evolved in MJ when 1.00L of ethanol is burned. Thanks

[sweetiepi]

So I know this must sound kinda dumb, but in equations, like thermochemical equations, that require multiple steps and calculations, how do we use significant figures? Do we round significant figures with each step or round at the final answer? E.g, The molar heat of combustion of ethanol is tabulataed as -1364 kJ mol. If the density of  ethanol is 0.790 g ML, calculate the energy evolved in MJ when 1.00L of ethanol is burned. Thanks

Hey!
1) You use the lowest amount of sig figs supplied by VCAA in the final answer
2) You shouldn't round off your answer until you get to the answer, else your answer could be off by a ittle/a lot
Hope this helps!

[Aar0n_101]

Hey!
1) You use the lowest amount of sig figs supplied by VCAA in the final answer
2) You shouldn't round off your answer until you get to the answer, else your answer could be off by a ittle/a lot
Hope this helps!

As in, we would have to round to significant figures when calculating the mol in the formula n=m/M, and then use that to find something else. So we would have to use significant figures at EVERY point where we are multiplying and dividing? I say this because in my textbook they first find the mass of a compound via the density formula, and then to to find the mol they use m/M, that is mass over molar mass, which in this case was 790/46.0. However they gave the answer as 17.17, which is 4 significant figures, while the lowest amount of sig figs in the eqaution had only 3 sig. Is this a mistake or am I missing something

[cookiedream]

You use the lowest amount of sig figs supplied by VCAA in the final answer

My Chemistry teacher told me that in a calculation if we had to work out molecular formula of a substance, and hydrogen is involved, our final answer must be in two significant figures. She said that was because the molar mass of hydrogen provided by the VCAA Chemistry Data Booklet is 1.0, which is 2 sig figs.
Would VCAA accept this, or do we have to rely on only the values given in the question itself to determine the minimum number of significant figures?

[Somedudelmaoy]

Why is it in VCAA's own answers they dont follow this rule? Question 9biii in the 2016 exam it rounds off to 2 sig figs even though it should be 3

[Phenomenol]

My Chemistry teacher told me that in a calculation if we had to work out molecular formula of a substance, and hydrogen is involved, our final answer must be in two significant figures. She said that was because the molar mass of hydrogen provided by the VCAA Chemistry Data Booklet is 1.0, which is 2 sig figs.
Would VCAA accept this, or do we have to rely on only the values given in the question itself to determine the minimum number of significant figures?

For addition/subtraction I believe you are actually meant to round off to the lowest number of decimal places in the numbers you use. So the molar mass of anything you calculate from the periodic table will be to 1 decimal place, which could be any number of significant figures depending on the number of digits.

Having said that I am also not sure if the molar mass you manually calculate is actually something you need to restrict the significant figures of the final answer with.

As in, we would have to round to significant figures when calculating the mol in the formula n=m/M, and then use that to find something else. So we would have to use significant figures at EVERY point where we are multiplying and dividing? I say this because in my textbook they first find the mass of a compound via the density formula, and then to to find the mol they use m/M, that is mass over molar mass, which in this case was 790/46.0. However they gave the answer as 17.17, which is 4 significant figures, while the lowest amount of sig figs in the eqaution had only 3 sig. Is this a mistake or am I missing something

The significant figures of the final answer is the only one that matters. Try to keep as many decimal places in your calculator, without rounding, as you go. It doesn't really matter what you write for the rest of your working.

Why is it in VCAA's own answers they dont follow this rule? Question 9biii in the 2016 exam it rounds off to 2 sig figs even though it should be 3

According to the exam report, the 2 sig figs come from the 0.10 that is used towards the end of the calculations. In part b) ii. 10.00 - 9.90 = 0.10 which is consistent with what I have said in the topmost response.

[Aar0n_101]

Thanks guys turns out its not even on the study guide lmao just spent 2 hours revising it as well. This happens when you dont pay attention in class and try to cram a years worth of knowledge into a weekend.

[Somedudelmaoy]

According to the exam report, the 2 sig figs come from the 0.10 that is used towards the end of the calculations. In part b) ii. 10.00 - 9.90 = 0.10 which is consistent with what I have said in the topmost response.

But VCAA gave the volume of both the container (10.00) and used NaOH (9.90) to more than 2 sig figs. Don't we use sig figs since our result cannot be more accurate than our given values? If so, we know accurately the amount of NaOH left - 0.100 to 3 sig figs. Basically what im getting at is all the values given in the question are 3 sig figs or more, we calculate that 0.01 ourselves so i dont really get why we restrict the answer to 2 sig figs

[Phenomenol]

But VCAA gave the volume of both the container (10.00) and used NaOH (9.90) to more than 2 sig figs. Don't we use sig figs since our result cannot be more accurate than our given values? If so, we know accurately the amount of NaOH left - 0.100 to 3 sig figs. Basically what im getting at is all the values given in the question are 3 sig figs or more, we calculate that 0.01 ourselves so i dont really get why we restrict the answer to 2 sig figs

But you don't know the volume of NaOH left as accurately as 0.100, only as 0.10 because for addition/subtraction you don't carry through minimal sig figs, you carry through minimal decimal places... as I have said in a reply above.

After all, if I have measured an empty beaker to be 10000.0 g, then I add some compound and the new mass is 10001.2 g, am I going to say the mass of compound added is 1.20000 g? Highly unlikely. Better to say 1.2 g.

Second point to make: I believe each calculation must be done successively in terms of sig figs/decimals - you consider the sig figs of the immediate values you are using and carry that through to determine the sig figs for the answer in that step of the calculation, not the sig figs of numbers from the very beginning.

Because of the way that addition/subtraction combined with multiplication/division can change the number of sig figs of an answer from what you began with, I think the sig figs must be carried through step by step in this manner...

...I think.

If you are purely doing multiplication/division throughout a calculation then this does not matter - you can simply look at the values you started with since for purely multiplication/division in every step, sig figs do not fluctuate.

So I believe by the subtraction calculation our amount of NaOH is 0.10 to highest precision, and the resulting division in part iii. to calculate the concentration gives an answer limited to 2 sig figs due to this.

This is overall just a means as to justify why VCAA might be right in this situation. Hope this helps.

[Somedudelmaoy]

But you don't know the volume of NaOH left as accurately as 0.100, only as 0.10 because for addition/subtraction you don't carry through minimal sig figs, you carry through minimal decimal places... as I have said in a reply above.

After all, if I have measured an empty beaker to be 10000.0 g, then I add some compound and the new mass is 10001.2 g, am I going to say the mass of compound added is 1.20000 g? Highly unlikely. Better to say 1.2 g.

Second point to make: I believe each calculation must be done successively in terms of sig figs/decimals - you consider the sig figs of the immediate values you are using and carry that through to determine the sig figs for the answer in that step of the calculation, not the sig figs of numbers from the very beginning.

Because of the way that addition/subtraction combined with multiplication/division can change the number of sig figs of an answer from what you began with, I think the sig figs must be carried through step by step in this manner...

...I think.

If you are purely doing multiplication/division throughout a calculation then this does not matter - you can simply look at the values you started with since for purely multiplication/division in every step, sig figs do not fluctuate.

So I believe by the subtraction calculation our amount of NaOH is 0.10 to highest precision, and the resulting division in part iii. to calculate the concentration gives an answer limited to 2 sig figs due to this.

This is overall just a means as to justify why VCAA might be right in this situation. Hope this helps.

Hmmm yeah i get what you mean. So would VCAA penalise me for giving the answer to 3 sig figs?

[Phenomenol]

Hmmm yeah i get what you mean. So would VCAA penalise me for giving the answer to 3 sig figs?

Only if they picked that particular question to be the one they judge for correct sig figs. 1 mark would be taken off from the entire exam if the sig figs in that one question weren't right then. It would be extremely evil to make it that question really... haha

To be honest, I have also heard from other people there is a 1 sig fig lenience - i.e. if the correct sig figs are 2 but your answer has either 1 or 3 then that will be acceptable. But I wouldn't bank on this claim.