Corrected exercises: measurement in the laboratory, significant figures and uncertainty (202-SN1-RE)

$0.00450$ g: the zeros placed BEFORE the first nonzero digit only position the decimal point, and they never count. What remains is $4$, $5$ and the final zero, which counts because it is written after a nonzero digit and after the decimal point: 3 significant figures. $100.0$ mL: the zeros are either sandwiched between digits or follow the decimal point, so all of them count: 4 significant figures. $2.50\times 10^{3}$ mL: in scientific notation only the coefficient counts, never the power of ten: 3 significant figures.

$0.08070$ mol: the leading zero does not count, the zero between 8 and 7 counts, the final zero counts: 4 significant figures. $6.022\times 10^{23}$: the coefficient $6.022$ carries 4, and the exponent 23 contributes none. Avogadro's constant as tabulated here therefore has 4 significant figures.

Neither is a measurement. The coefficient 2 comes from balancing the equation: it is a counting number, exact by construction, so its precision is unlimited. The equality $1$ L $=1000$ mL is a definition of units, also exact. In practice, neither of them ever limits the number of significant figures of a result. This is the classic trap of stoichiometry calculations, where students think they must round to one figure because of the coefficient 2.

$0.00450$ g $=4.50\times 10^{-3}$ g. The coefficient keeps its three significant figures, final zero included: writing $4.5\times 10^{-3}$ would throw away information about the precision of the weighing. For $2500$ mL measured to the nearest millilitre: $2.500\times 10^{3}$ mL, so the exponent is 3, and this time the coefficient carries 4 significant figures.

The last digit written is the uncertain digit, give or take one unit. $100.0$ mL: $\dfrac{0.1}{100.0}\times 100=0.10\,\%$. $2.50\times 10^{3}$ mL: the uncertain digit is the hundredths place of the coefficient, that is $0.01\times 10^{3}=10$ mL, so $\dfrac{10}{2500}\times 100=0.40\,\%$. $0.00450$ L: $\dfrac{0.00001}{0.00450}\times 100=0.22\,\%$. So $100.0$ mL is the most precise, at $0.10\,\%$, even though it is the volume whose notation contains the fewest zeros. The number of significant figures, not the length of the number, decides the precision.

$M=40.08+12.01+3(16.00)=40.08+12.01+48.00=100.09$ g/mol. Addition rule: line up on the number of DECIMAL PLACES of the term that has the fewest. Here every term has two decimal places, so the result keeps two: $100.09$ g/mol, which is 5 significant figures. Note that the factor 3 is an exact number from the formula and limits nothing.

$m=52.3874-48.1236=4.2638$ g. Both weighings have four decimal places, so the result keeps four. It still has 5 significant figures, exactly like the original data.

$m=52.39-48.12=4.27$ g. Two decimal places are kept, but only 3 significant figures remain, whereas each weighing carried 4. Subtracting two close numbers destroyed a significant figure: this is loss of significant figures by cancellation, the one case where a calculation loses precision without any rule being broken. Weighing by difference should therefore be avoided on a coarse balance when the sample is small compared with its container.

Analytical balance: $n=\dfrac{4.2638}{100.09}=0.0425997$ mol. Division rule: keep the smallest number of SIGNIFICANT FIGURES, here 5 for both factors, so $n=0.042600$ mol. Top-loading balance: $n=\dfrac{4.27}{100.09}=0.0426616$ mol, limited to 3 significant figures by the mass: $n=0.0427$ mol.

Analytical balance: $\dfrac{0.0001}{4.2638}\times 100=0.0023\,\%$. Top-loading balance: $\dfrac{0.01}{4.27}\times 100=0.23\,\%$. A factor of 100 between the two, for the same procedure and the same chemical. That is the exact cost of the wrong instrument, and it is then paid on everything calculated from this mass.

Volumetric pipette: $\dfrac{0.03}{25.00}\times 100=0.12\,\%$. Beaker: $\dfrac{5}{100}\times 100=5.0\,\%$. The graduations on a beaker are not a measuring instrument, they are fill marks.

$\dfrac{5.0}{0.12}=42$. The pipette is about 42 times more precise in relative terms. Yet the beaker is much bigger: the precision of an instrument cannot be read from its capacity.

The volumetric pipette. It is the only instrument on the list designed to DELIVER a single, fixed volume with high precision, which is exactly what an aliquot requires. The volumetric flask, on the other hand, is used to CONTAIN a volume, not to deliver one: the diluted solution is prepared in it, nothing is withdrawn from it.

The beaker, or at a pinch the graduated cylinder. A volume that goes into no calculation does not need to be known: measuring it precisely makes the result neither more accurate nor more precise. Tying up the burette for this means occupying the most fragile piece of glassware and the slowest to fill, and it would have to be rinsed with the titrant before use anyway.

$25.00+50.0=75.0$ mL, a single decimal place. The addition rule is about DECIMAL PLACES, and it is the graduated cylinder, with its single decimal place, that decides. Writing $75.00$ mL would claim a precision the cylinder does not have: the two digits of the pipette were lost as soon as the liquids were poured into the same container. This is the thread of the whole set: the weakest link fixes the last digit.

Set A: $\dfrac{0.789+0.788+0.790+0.789}{4}=\dfrac{3.156}{4}=0.789$ g/mL. Set B: $\dfrac{3.204}{4}=0.801$ g/mL. Set C: $\dfrac{3.156}{4}=0.789$ g/mL. Set D: $\dfrac{3.080}{4}=0.770$ g/mL.

Set A: $0.790-0.788=0.002$. Set B: $0.802-0.800=0.002$. Set C: $0.805-0.770=0.035$. Set D: $0.800-0.740=0.060$. The range measures PRECISION, that is, how well the measurements agree with one another, and says absolutely nothing about their agreement with the true value.

Same mean, $0.789$ g/mL, so the same apparent accuracy. But the range of C is $0.035$ against $0.002$ for A: C is accurate only because large random errors cancel out, whereas A is accurate AND precise. Publish A. The mean of C would land somewhere else if the pair made four more measurements, while that of A is reproducible: a mean that is right by luck is not a result.

Set B. Its four measurements are very precise, with a range of $0.002$ like A, but all of them are shifted by $+0.012$ g/mL from the true value. A constant, reproducible offset is the signature of a systematic error. Plausible causes: a pycnometer not properly dried after the previous experiment, a balance that was not tared, or a meniscus that was always read from above.

Set B can be rescued: since the offset is constant, calibrating the balance or redoing the zero removes it entirely, and the data already collected can even be corrected afterwards by subtracting $0.012$. Set D cannot be rescued: it is both offset and scattered, there is no constant offset to remove, and only a complete repeat of the experiment, with a corrected measuring technique, will give anything useful. Remember the hierarchy: precision is improved by technique, accuracy by calibration.

$n=\dfrac{m}{M}=\dfrac{2.4530}{58.44}=0.0419747$ mol. The mass carries 5 significant figures, the molar mass 4: the smaller one decides, so $n=0.04197$ mol.

$C=\dfrac{n}{V}=\dfrac{0.0419747}{0.2500}=0.1678987$ mol/L. Keep 4 significant figures: $C=0.1679$ mol/L. Note that the UNROUNDED value of $n$ was used for this calculation: rounding happens only at the end, otherwise rounding errors pile up.

Weighing: $\dfrac{0.0001}{2.4530}\times 100=0.0041\,\%$. Flask: $\dfrac{0.15}{250.0}\times 100=0.060\,\%$. These two numbers, not the raw tolerances, are what gets added: a tolerance of $0.15$ mL can only be compared with a tolerance of $0.0001$ g once each is expressed relative to the quantity measured.

The concentration is a quotient, so the RELATIVE uncertainties add: $0.0041+0.060=0.064\,\%$. Absolute uncertainty: $0.00064\times 0.1679=0.000108$ mol/L, that is $0.0001$ mol/L after rounding to one significant figure, as is always done with an uncertainty. The result is written $C=(0.1679\pm 0.0001)$ mol/L. The uncertainty falls on the fourth digit: the four significant figures claimed in b) are therefore justified, no more and no fewer.

The flask dominates, by a factor $\dfrac{0.060}{0.0041}=15$. A counterintuitive and useful conclusion: buying a more precise balance would change nothing in the result, whereas switching to a tighter-tolerance class A flask, or simply preparing a larger volume, would gain almost everything. The general rule: in a chain of measurements, only the dominant link is worth improving, and it is found from the relative uncertainties, never by eye.

Mass of water: $49.8654-24.3172=25.5482$ g. Volume: $V=\dfrac{m}{\rho}=\dfrac{25.5482}{0.99820}=25.5943$ mL. This is the whole point of the method: a volume is never measured, it is DEDUCED from a weighing and a tabulated density, because a balance is a hundred times more precise than any glassware.

Mass of the liquid: $44.5021-24.3172=20.1849$ g. The flask is the same, so the volume is the same: this is the only fact the entire experiment rests on, which is why the pycnometer must be filled up to the capillary, with no bubble, at the same temperature.

$\rho=\dfrac{20.1849}{25.5943}=0.78865$ g/mL. Both factors carry 6 significant figures; we keep 5, out of caution about the third decimal place of the mass of water. Note that the volume was not needed to reason this out: $\rho=\rho_{water}\times\dfrac{m_{liquid}}{m_{water}}$ gives the same result directly, and that is the form to remember.

$0.78865$ g/mL is $0.00035$ from ethanol, $0.0024$ from acetone, $0.015$ from propan-1-ol and $0.130$ from hexane. It is ethanol, and the gap to the second candidate is seven times larger than the gap to the first: the conclusion holds. Be careful, an identification by density alone is never proof; it is acceptable here because the list of candidates is given and none other lies within a thousandth.

Each mass is a difference of two weighings, so its absolute uncertainty is $2\times 0.0001=0.0002$ g. Relative: $\dfrac{0.0002}{25.5482}=7.8\times 10^{-4}\,\%$ for the water, $\dfrac{0.0002}{20.1849}=9.9\times 10^{-4}\,\%$ for the liquid, and $\dfrac{0.00001}{0.99820}=1.0\times 10^{-3}\,\%$ for the tabulated density. Total: $2.8\times 10^{-3}\,\%$, that is an absolute uncertainty of $2.8\times 10^{-5}\times 0.78865=2.2\times 10^{-5}$ g/mL. The uncertainty therefore falls on the fifth digit: $\rho=(0.78865\pm 0.00002)$ g/mL. The five figures are justified, and that is what makes it possible to decide between ethanol and acetone, which three significant figures would have confused.

Slope $=\dfrac{1.1478-0.9982}{20.00-0}=\dfrac{0.1496}{20.00}=0.007480$ (g/mL) per mass percent. The three intermediate points stray from this line by at most $0.002$ g/mL (the largest gap is at 5 %), which is small next to the $0.15$ g/mL span of the data: the relation is close enough to linear over the interval studied, so it can be used.

The y-intercept is $0.9982$ g/mL, which is directly the point measured at $0\,\%$. Its physical meaning is immediate: it is the density of pure water at 20 °C. A calibration whose y-intercept lands far from the expected value signals a handling error before the line has even been used: it is the cheapest check of the whole session.

$\rho=0.007480\,C+0.9982$, with $\rho$ in g/mL and $C$ as a mass percent.

Invert the relation: $C=\dfrac{\rho-0.9982}{0.007480}=\dfrac{1.0525-0.9982}{0.007480}=\dfrac{0.0543}{0.007480}=7.259$, that is $C=7.26\,\%$. Only three significant figures: the subtraction $1.0525-0.9982$ leaves $0.0543$, three significant figures out of five to start with. Loss of significant figures by cancellation once again, and it is the fundamental limit of the calibration method, not a slip in the calculation.

Because $1.25$ g/mL lies outside the interval over which the line was established, $[0, 20]\,\%$. A calibration line is worthless outside its range: nothing guarantees that linearity continues there, and here the chemistry settles it regardless of any statistics, since the solubility of sodium chloride levels off around $26\,\%$ by mass at 20 °C. A $33.7\,\%$ solution does not exist, so the reported result describes a liquid that is not in the beaker. A well-fitted model is still wrong outside the window where it was fitted.

TRUE. A stoichiometric coefficient is a counting number, produced by balancing and not by a measurement: it is exact, with unlimited precision. The same goes for the subscripts of a chemical formula, for definitions of units and for the number of times an operation is repeated. The opposite error exists too: believing that a tabulated constant is exact. $N_{A}=6.022\times 10^{23}$ as written here carries 4 significant figures and can limit a result.

FALSE. It carries three: $4$, $5$ and the final zero. Zeros placed before the first nonzero digit are only position markers and never count, which scientific notation makes obvious: $4.50\times 10^{-3}$ g. The correct sentence: the mass $0.00450$ g carries three significant figures.

FALSE, twice over. First, the rule is misnamed: in an addition or a subtraction, one reasons on DECIMAL PLACES, not on significant figures. Second, the count is wrong: $52.39-48.12=4.27$ keeps two decimal places, but only three significant figures, whereas each datum carried four. The correct sentence: the result keeps two decimal places, therefore three significant figures, the subtraction of two close numbers having destroyed one.

TRUE. Precision measures how well the measurements agree with one another, accuracy how well they agree with the true value: the two are independent qualities. A balance that is off by $12$ mg gives very reproducible weighings that are all wrong by the same amount. It is even the most dangerous case, because reproducibility inspires confidence and no repetition of the measurement will reveal the problem: only a calibration will.

FALSE. Pipette: $\dfrac{0.03}{25.00}\times 100=0.12\,\%$. Flask: $\dfrac{0.15}{250.0}\times 100=0.060\,\%$. The flask is twice as precise in relative terms, even though its tolerance is five times larger in absolute terms. The correct sentence: the $250.0$ mL volumetric flask is twice as precise in relative terms, because a tolerance can only be compared once it is expressed relative to the quantity measured.

$n=\dfrac{1.0432}{158.03}=6.6013\times 10^{-3}$ mol, then $C_{0}=\dfrac{6.6013\times 10^{-3}}{0.1000}=6.6013\times 10^{-2}$ mol/L. The mass carries 5 significant figures, the molar mass 5, the flask 4: $C_{0}=0.06601$ mol/L, four significant figures.

Dilution conserves the amount of substance: $C_{1}V_{1}=C_{0}V_{0}$, so $C_{1}=C_{0}\times\dfrac{10.00}{250.0}=0.066013\times 0.04000=2.6405\times 10^{-3}$ mol/L, that is $C_{1}=2.641\times 10^{-3}$ mol/L. The unrounded value of $C_{0}$ is kept to carry on, and rounding is done only for display.

$C_{2}=C_{1}\times\dfrac{25.00}{500.0}=2.6405\times 10^{-3}\times 0.05000=1.3203\times 10^{-4}$ mol/L, that is $C_{2}=1.320\times 10^{-4}$ mol/L.

First dilution: factor $\dfrac{250.0}{10.00}=25.00$. Second: $\dfrac{500.0}{25.00}=20.00$. Overall factor $25.00\times 20.00=500.0$. Check: $\dfrac{C_{0}}{C_{2}}=\dfrac{6.6013\times 10^{-2}}{1.3203\times 10^{-4}}=500.0$. Dilution factors MULTIPLY, they do not add; believing in a factor of $45$ is the most common error in this calculation, and it gives a concentration eleven times too large.

It is not the balance. It gives 5 significant figures, one more than everything else: it is the only instrument in the chain that limits nothing. The limit comes from the five pieces of volumetric glassware, each carrying exactly 4 significant figures, $100.0$, $10.00$, $250.0$, $25.00$ and $500.0$ mL. The result is therefore given to 4 significant figures, and no finer balance would change that. This is the practical lesson of the set: before improving one step, find out which one is in charge, and it is rarely the most impressive.

$V=23.17-0.45=22.72$ mL. Two READINGS are made on the burette, the starting one and the endpoint one, and each is good to $\pm 0.05$ mL: the absolute uncertainties of a difference ADD, so $\Delta V=0.05+0.05=0.10$ mL. Forgetting the initial reading, or believing that it cancels out, is the most common reasoning error in a titration.

$n(NaOH)=C\times V=0.1043\times 0.02272=2.3697\times 10^{-3}$ mol. Four significant figures on both sides, so we keep four.

The equation is in a $1:1$ ratio, so $n(HCl)=n(NaOH)=2.3697\times 10^{-3}$ mol. $C(HCl)=\dfrac{2.3697\times 10^{-3}}{0.02500}=0.094788$ mol/L, that is $0.09479$ mol/L before the uncertainty is examined.

Burette: $\dfrac{0.10}{22.72}\times 100=0.44\,\%$. Pipette: $\dfrac{0.03}{25.00}\times 100=0.12\,\%$. Titrant: $\dfrac{0.0001}{0.1043}\times 100=0.096\,\%$. The concentration sought is a product of quotients, so the relative uncertainties add: $0.44+0.12+0.096=0.66\,\%$. Absolute uncertainty: $0.0066\times 0.09479=6.2\times 10^{-4}$ mol/L. The uncertainty falls on the fourth decimal place, so we publish $C=(0.0948\pm 0.0006)$ mol/L, three significant figures. The $0.09479$ on the calculator claimed a fifth decimal place that the procedure cannot support.

The burette dominates, with $0.44\,\%$, two thirds of the total. The remedy is not a better instrument but a better procedure: titrating $50.00$ mL of acid instead of $25.00$ mL doubles the volume delivered, to about $45.4$ mL, while $\Delta V$ stays $0.10$ mL. The relative uncertainty of the burette falls to $\dfrac{0.10}{45.44}\times 100=0.22\,\%$ and, the other two contributions being unchanged, the total to about $0.44\,\%$ ($0.22+0.12+0.096$, assuming the other two contributions stay as they were, a cautious assumption for the pipette), a third less uncertainty without buying anything. This is why a titration is always aimed at an endpoint near three quarters of the burette rather than its first quarter.