Just another Sig Fig Question

[natalie.krystal7]

Hey guys,

Was working through some random questions in prep for next year, and yet again as has been happening the whole time I've done Chem, I get confused with significant figures and their usage.

Here's something I don't quite get, and would love for someone to shed some light:

All these questions have to do with the simple concept of mols and mass.

One goes along the lines of:

13.4 / (58.7 + 2(35.5) = 0.103 mols
0.103 x 2 = 0.207
But the answer apparently, when taking sig figs into consideration is 2.07 x 10^-1. Why? Aren't both answers with 3 sig figs?

The next is:

1.2 x (27.0 + 3(35.5) = 160.2
= 1.6 x 10^2

But then the next question;

With the last part going 8.90 x 32 = 284.80g, is apparently 285g when with correct no. of sig figs.

Why wasn't it 2.8 x 10^2 or 2.85 x 10^2?


And lastly, when its said that sig fig calculations only occur at the end, what does this mean? In the question previously mentioned, with 8.90 x 32 as the last step, why isn't 2 the new number of sig figs required?

And where should we take the number of significant figures our answer should be from? The numbers given us in the original question or the ones we've worked out?

Sorry if this sounds really dumb, but all the sig fig questions I've asked my teacher, he's always replied with 'I don't know', or 'I'm not sure'.

Thanks for your help in advance ^^

[dptjandra]

"But the answer apparently, when taking sig figs into consideration is 2.07 x 10^-1. Why? Aren't both answers with 3 sig figs?"

This answer is better because it not only has the right number of significant figures, but is also in 'scientific notation', which is the norm, especially when dealing with moles.

"With the last part going 8.90 x 32 = 284.80g, is apparently 285g when with correct no. of sig figs.

Why wasn't it 2.8 x 10^2 or 2.85 x 10^2?"

2.85 x 10^2 g would be accepted, and is arguably 'better' than 285g...people tend to be a bit more slack when dealing with masses in grams, just by convention.

"And lastly, when its said that sig fig calculations only occur at the end, what does this mean? In the question previously mentioned, with 8.90 x 32 as the last step, why isn't 2 the new number of sig figs required?"

You haven't really specified where 32 came from...if it came from M(O2), which would be the most obvious guess...then the table actually has 16.0 + 16.0 = 32.0...still 3 sig figs.

"And where should we take the number of significant figures our answer should be from? The numbers given us in the original question or the ones we've worked out?"

Sig figs in the final answer correspond to the least accurate piece of data used to calculate the answer.  Eg. if you used a mass of 16.45g and temperature of 25.0 C and molar mass of 32.0 g mol-1, you would have a final answer of 3 sig figs.  If you used an answer from a previous part of the question with only 2 sig figs to get your next answer, then your answer in the next part should have only 2.  Note that scaling factors (eg. if the equation was something like 2X + Y --> X2Y) are not subject to sig figs...ie. if you multiplied your amount by 2, you would not need to put only 1 sig fig.  This is because it does not count as a "piece of experimental data".

[thushan]

Yup. Here are the general rules:

a + b = c: (or -)

c is rounded to the least number of decimal places of a or b:
Eg: 20.21 + 1.22 = 21.43 and 21.23 + 6.2 = 27.4

a x b = c: (or /)

this is the one most people are familiar with; c is rounded to least number of sig figs of a or b 
Eg: 12.0 x 4.00 = 48.0

a^b = c:

If b has x decimal places, c is rounded to x sig figs:
Eg 10^1.24 = 17. (the . is there for a reason, it signifies that both the 1 and the 7 are significant)

NB: If you're wide awake you would wonder what would happen if you get something like 4^3 or 10^5, with no decimal places; I did some experimenting and worked out that it's best to leave it as 10^x: so 4^3 = 10^2 ("no" significant figures) and 10^5 = 10^5. Try it out for yourself if you don't believe me.


log(base a) b = c


If b has y sig figs, then c is rounded to y decimal places:

Eg: [H+] = 0.00100 M => pH = -log(10)0.00100 = 3.000

[Romperait]

Thushan, do you believe vce chemistry would require us to use the last two rules you posted (to do with exponentials and logs)? I've never seen them before and putting them into practice could potentially be quite annoying...

[stonecold]

Thushan, do you believe vce chemistry would require us to use the last two rules you posted (to do with exponentials and logs)? I've never seen them before and putting them into practice could potentially be quite annoying...

Yeah, know them just in case, as it could come up in a acid base pH question.

[Romperait]

Thushan, do you believe vce chemistry would require us to use the last two rules you posted (to do with exponentials and logs)? I've never seen them before and putting them into practice could potentially be quite annoying...

Yeah, know them just in case, as it could come up in a acid base pH question.

Rightio. Writing notes for 'em now! Thanks Stonecold.

[Pixon]

Thushan's post is probably more concise and easy to follow than what you'll find in textbooks; in fact, most textbooks completely neglect to explain significant figures.

[natalie.krystal7]

Heyy, thanks so much for your replies guys , but just a couple of other quick questions:
 
These are parts of other questions, concerning the calculation of CO2 given the molar volume of gases at SLC in Heinemann Chem 2.

n(CO2) = V/Vm

26 881.7 = V/24.5 = 658 602.15 L
             = 659 000 L with 3 sig figs.

Why isn't it 6.59 x 10^5? Or is that acceptable as well?


And:

Calc. the mass of iron(III) oxide required.

Shown in solutions to be = 1 430 949.8g
Convert to kg; = 1 430 949.8 / 1000 = 1430.9 kg
                                                  = 1430 with the correct number of sig figs.

Firstly, is it assumed that we now take 1000 as the number of sig figs to use? But why :s

And secondly, why not 1431, if we have to make it 4 figures shouldn't we round up?

Thanks again!! Seriously alot..

[thushan]

Heyy, thanks so much for your replies guys , but just a couple of other quick questions:
 
These are parts of other questions, concerning the calculation of CO2 given the molar volume of gases at SLC in Heinemann Chem 2.

n(CO2) = V/Vm

26 881.7 = V/24.5 = 658 602.15 L
             = 659 000 L with 3 sig figs.

Why isn't it 6.59 x 10^5? Or is that acceptable as well?


And:

Calc. the mass of iron(III) oxide required.

Shown in solutions to be = 1 430 949.8g
Convert to kg; = 1 430 949.8 / 1000 = 1430.9 kg
                                                  = 1430 with the correct number of sig figs.

Firstly, is it assumed that we now take 1000 as the number of sig figs to use? But why :s

And secondly, why not 1431, if we have to make it 4 figures shouldn't we round up?

Thanks again!! Seriously alot..

Could I see the whole question?

[Romperait]

Um for the first one, using scientific notation or just writing the number (i.e. 659 000) are both fine.

For the second question, the 1000 isn't taken into account for significant figures because it's just converting the units (from g to kg). I think you need to look at the information provided at the start of the question to find the smallest number of significant figures...

[thushan]

Um for the first one, using scientific notation or just writing the number (i.e. 659 000) are both fine.

For the second question, the 1000 isn't taken into account for significant figures because it's just converting the units (from g to kg). I think you need to look at the information provided at the start of the question to find the smallest number of significant figures...

OK in conversion to units as we define EXACTLY 1000 g = 1 kg, we can say that factor of 1000 has an infinite number of significant figures.

[Romperait]

Um for the first one, using scientific notation or just writing the number (i.e. 659 000) are both fine.

For the second question, the 1000 isn't taken into account for significant figures because it's just converting the units (from g to kg). I think you need to look at the information provided at the start of the question to find the smallest number of significant figures...

OK in conversion to units as we define EXACTLY 1000 g = 1 kg, we can say that factor of 1000 has an infinite number of significant figures.

That's another way of saying that it isn't taken into account, yeah?

[thushan]

Yeah - but I just wanted to make it clear that it isn't 'taken into account' because it's an exact value

[natalie.krystal7]

Hmm, so where did they get the 4 significant figures from, and why wasn't the value rounded up? :s

And thushan, which question were you referring to previously which you wanted to see in full?


Also, here's something else which is annoying (I've put the question and whole working out down this time):

Calculate the amount of H20 in mols in a 20.0g sample of CuSO4`5H2O.

Step 1. Calc the molar mass of CuSO4`5H2O.

55.9 + 32.1 + 4(16.0) + 5(2.0+16.0) = 249.7

Step 2. Calc the amount of CuSO4`5H2O.

20.0 / 249.7 = 0.0801mol.

Step 3. Since 1 mol of CuSO4`5H2O contains 5 mol of water molecules, find the amount of H2O.

5 x 0.0801 = 0.4005 Firstly, why are we even putting the answer as 0.4005? Shouldn't it automatically be taken as 3 sig figs b'cos of the 0.0801?

Step 4. Express ans with correct number of sig figs.

n(H2O) = 5 x 0.0801 = 0.4005
                            = 0.401 mol < Why three sig figs here? Isn't the least accurate value 5? I understand one sig fig would be rather ridiculous; but how did they decide on 3?


And it's the same with this question as well (Calc the amount of Cl in mol in a 34 g sample of FeCl3):

Step 1. Calc the amount of FeCl3
          = 0.2095 mol

Step 2. As each mole of FeCl3 contains 3 mol of Cl- ions, calc the amount of Cl- ions.

n(Cl-) = 3 x 0.2095 = 0.628mol

Step 3. Correct number of sig figs.

n(Cl-) = 0.63mol < Why 2 sig figs? >.< Gaah.


Thank you for putting up with me but this is all so frustrating when you don't get it lol.

[thushan]

20.0g sample => 3 sig figs.

Oh don't worry about the question now, it just wasn't clear what you were talking about before in your solution. Now it's been explained.