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Author Topic: Sig figs when hydrogen is involved? Sig Figs rounding Q  (Read 629 times)  Share 

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horizon

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Sig figs when hydrogen is involved? Sig Figs rounding Q
« on: December 24, 2011, 10:13:58 pm »
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I've noticed that in answers to stoichiometry questions, they will often ignore the fact that hydrogen has 2 sig figs and round the answer to 3 sig figs instead (as it is the second lowest number of sig figs used in calculation)...
Why is this the case?
ARE THE ANSWERS ACTUALLY RIGHT? Or should we be rounding to 2 sig figs?
I've also seen the same thing applied when questions involve helium too- they do not round down to 2 sig figs, but rather, three.



Another question: if a calculation is split into parts and you must rely on your previous answer for each subsequent step, when do we round?
Should we write our answer to the correct number of sig figs for each step? Or should we use the calculator's figure (with lots and lots of decimals places) in our subsequent calculations and then only round off at the end?

In the latter case, if we are to only round off at the very end of the question, how would you present your answer for the parts in between- as in, would you still right it with the correct number of sig figs or with lots and lots of decimals trailing behind?

I hope VCAA has released a bulletin or statement to clarify this, anyone know?

Thanks in advance btw.

thushan

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Re: Sig figs when hydrogen is involved? Sig Figs rounding Q
« Reply #1 on: December 24, 2011, 10:22:12 pm »
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OK, I think you're talking about using hydrogen as part of calculating the amount of sth from molar mass?

Rules for addition/subtraction and multiplication/division are different.

In addition/subtraction, we look at the least number of DECIMAL PLACES.
In multiplication/division, we look at the least number of SIG FIGS.

So say we have to find out n(H2O) when m(H2O) = 0.129 g.

So, M(H2O) = 1.0 + 1.0 + 16.0 = 18.0 g/mol --> least accurate value to 1 dp, so answer to 1 dp. (we're adding)
Note that this answer also HAPPENS to be to 3 sigfigs.

So n(H2O) = 0.129/18.0 = 7.17E-3 mol. We're dividing, look at sigfigs, least accurate value is to 3 sf, so answer is to 3 sf.

Make sense?

Oh, about the calc qn, answer all qns to correct sig figs, but use the value on calc (the one to zillions of sig figs) for subsequent calcs.
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