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I'm having trouble understanding the answer to this question:

Most chemistry laboratories have balances that can weigh to the nearest milligram, would it be possible to weigh $5.64\times10^{18}$ molecules of octadecane, $\ce{C18H38}$, on such a balance?

I worked through the equation and my end result was $\pu{0.00238 g}$ which is equal to $\pu{2.38 mg}$ so I thought that the answer was that it could, but the book states that the answer is that it cannot. Why can't it? Am I misunderstanding how weighing to the nearest $\pu{mg}$ works?

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    $\begingroup$ significant figures. Weighing $5.6\cdot10^{18}$ molecules is not the same as weighing $5.64\cdot10^{18}$ molecules. $\endgroup$
    – MaxW
    Feb 5, 2017 at 18:14
  • $\begingroup$ So I made a rounding error? $\endgroup$ Feb 5, 2017 at 18:54
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    $\begingroup$ The balance can only weigh 23 or 24 mg, not 23.8 mg. $\endgroup$
    – MaxW
    Feb 5, 2017 at 19:00
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    $\begingroup$ Ah, I think I get it. So, it only measures in whole milligrams. Thanks for your help. $\endgroup$ Feb 5, 2017 at 19:03

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Couldn't agree less, but check my math.

First $\ce{C18H38}$ has a molecular weight of $254.5~\mathrm{g/mol}$.

Second $\mathrm{5.64\times 10^{18} ÷ 6.02214 \times 10^{23} = 9.3654 \times 10^{-6}}$ and so is $2.4\ \mathrm{mg}$.

A balance with an ability to weigh to the nearest mg would, in a perfect world give you a mass of 2 or possibly 3 mg. This wouldn't be very useful, but it wouldn't give you zero information, either. So, since I don't understand the context in which your book asked the question, I'd have to say it doesn't seem to be completely correct if it states "it cannot" weigh that amount. It certainly cannot weigh that amount with adequate precision for most purposes I can think of. But if, say, you were trying to determine whether there was enough of a sample to use in a mass spec., or to calculate the yield of a reactor-on-a-chip, then it would suffice.

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  • $\begingroup$ Your math is correct. The OP should have gotten 2.38 mg not 23.8. // We'll just have to agree to disagree on how significant figures impact the answer. $\endgroup$
    – MaxW
    Feb 5, 2017 at 20:40
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    $\begingroup$ Thanks for the correction, I had a bit of a mix-up with my decimals when copying from my work to the screen it seems. $\endgroup$ Feb 5, 2017 at 22:19
  • $\begingroup$ "Would it be possible to weight this amount?" should, I think, be interpreted as "Would it be possible to weigh out this amount?" It's not asking if the balance can give you an answer for the weight of a sample; rather, it's asking if the balance can be used to tell you how much stuff to take out of hte bottle to make 2.4mg. $\endgroup$ Feb 6, 2017 at 10:43
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I would like to expand upon the other answer a bit and claim that the balance indeed cannot weight 2.38 mg.

First of all, the question cannot be interpreted literally. The question must be interpreted as either:

  • Given this sample of $C_{18}H_{38}$ and this balance, can you conclude that it has $5.64 \times 10^{18}$ molecules.

or

  • Can you use this balance to make a sample that has $5.64 \times 10^{18}$ molecules.

Now, $5.64 \times 10^{18}$ has 3 significant figures. What it actually means is that you want between $(5.64\pm 0.01) \times 10^{18}$ or $(5.64 \pm 0.005) \times 10^{18}$ molecules. This converts to $2.38$ mg (I am simply assuming you did your math correctly), again with 3 significant digits.

Now, all measuring devices are living creature made and imperfect $^\text{[citation needed]}$. A measuring device that is accurate to the nearest unit could be off by up to that unit (or half, depending on who you ask.) For example, if you use a ruler accurate up to mm, then a measurement you make on that ruler can be off up to 1 or 0.5 mm, depending on who you ask. If you measure a line, and the line ends in between the 6mm notch and the 7mm notch, it does not make sense to say that the line is 6.5mm. The 6mm notch and the 7mm notch could be off up to a millimeter, after all. Maybe it depends on the season.

As such, you cannot use a measuring device that is accurate up to mg to conclude that something has 2.38 mg, or measure out 2.38 mg of something.

As an addendum, I would like to note that the question asks about a balance.

This is a balance

A balance's being accurate to the nearest mg could mean a couple of things.

  • One side is up to 1 mg heavier than the other side.

  • There is some friction in the balance, and it simply would not move, unless there is a difference of more than 1 mg between the two sides.

  • Where you put the items on the plate, and what the initial position of the balance is, could have an effect on the weighing, but no more than 1 mg worth.

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  • $\begingroup$ A "balance" doesn't have to be of that form. The question is probably about a microblance (google link) which would be a frictionless design. Spring balances (wikipedia) may not be common in chemistry labs but are another use of the term. $\endgroup$
    – Chris H
    Feb 6, 2017 at 9:49
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    $\begingroup$ +1 but the term "balance" is used to refer to any weighing scale. $\endgroup$ Feb 6, 2017 at 10:40
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    $\begingroup$ FYI: The "balance" with the digital readout that you see in a chem lab probably does work on the same principle as the balance shown above, except that instead of comparing the unknown weight to a set of known masses, it compares the unknown weight to a force supplied by an electromagnet. The readout is proportional to the amount of current that must flow in the magnet in order to exactly balance the unknown weight on the pan. $\endgroup$ Feb 6, 2017 at 11:31
  • $\begingroup$ "The 6mm notch and the 7mm notch could be off up to a centimeter" Surely, you mean "millimeter", right? I really hope my millimeter precision ruler isn't off by a centimeter! $\endgroup$
    – Mike Caron
    Feb 6, 2017 at 15:45
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    $\begingroup$ The significant figure factor is about precision not accuracy. $\endgroup$
    – MaxW
    Feb 6, 2017 at 16:41
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Accepting what is the "wrong" answer bugs me. Although I hope the point about significant figures has been made adequately, let me go into more detail.

QUESTION

Most chemistry laboratories have balances that can weigh to the nearest milligram, would it be possible to weigh $5.64×10^{18}$ molecules of octadecane, $\ce{C18H38}$, on such a balance?

Li Zhi correctly points out that octadecane has a molecular weight of 254.5 g/mole and that $5.64×10^{18}$ molecules would weigh 2.38 milligrams considering the significant figures.

Li Zhi makes the correct assessment when he says "It certainly cannot weigh that amount with adequate precision for most purposes I can think of."

Li Zhi is also correct when he says that "A balance with an ability to weigh to the nearest mg would, in a perfect world give you a mass of 2 or possibly 3 mg." But this is accuracy not precision.

Thus a balance has two factors to consider. Accuracy and precision. A balance which weighs to 1 mg has neither the accuracy nor the precision to weigh out 2.38 milligrams.

Accuracy: On average 2 mg will be 0.38 mg too low and 3 mg will be 0.62 mg too high. So either weight is biased.

Now if a chemist weighs out 2 mg 62% of the time and 3 mg 38% of the time, then on average the chemist weighs out 2.38 mg. (See how stupid this is?)

Let's look at the reverse too. On average 2 mg will contain $4.73\times10^{18}$ molecules and 3 mg will on average contain $7.10\times10^{18}$ molecules.

Precision: $5.64×10^{18}$ molecules of octadecane implies +/- 1 part in 564. If we assume the balance rounds, which is reasonable, then 2 mg is precise to 1 part in 4, not nearly enough.

The point here overall is that chemists make assumptions all the time to simply problems using significant figures. Using significant figures has to become innate to a chemist so that one can cut through the Gordian knot offered by so many chemistry problems.

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