Chemistry · Book 2 · Bachelor Year 1

University Chemistry — Year 1

University Chemistry — Year 1 · Bachelor Year 1

29Lab Techniques I: Safety, Measurement and Separation

Two students recrystallise the same batch of aspirin and measure its melting point on the same heated bench. One reads 134 ∘C134\,{}^{\circ}\mathrm{C}, the other 136 ∘C136\,{}^{\circ}\mathrm{C}. Do they disagree? Is the product pure? Neither question can be answered from the two numbers alone: a measurement is worth something only with its uncertainty, and a comparison only with a rule. This last chapter gathers the practical side of the year: how to read the hazards of what is on the bench, how to state a measured value and its uncertainty and compare it with another, and how the classic operations of the organic laboratory — extraction, filtration, recrystallisation, distillation — separate a product from what surrounds it, then check that it is pure.

You already know

Solubility and miscibility from intermolecular forces, polar and apolar solvents (Chapter 4); burettes, pipettes and the end point of a titration (Chapter 15). The school volume (grade 10) introduced liquid–liquid extraction, recrystallisation, thin-layer chromatography and the yield of a synthesis; here they are given their numbers.

Hot filtration during a recrystallisation: goggles, gloves and a lab coat; the hot solution passes through a fluted filter paper, which keeps back the insoluble impurities. Needles of a recrystallised product wait in the dish on the right.
Hot filtration during a recrystallisation: goggles, gloves and a lab coat; the hot solution passes through a fluted filter paper, which keeps back the insoluble impurities. Needles of a recrystallised product wait in the dish on the right.

29.1 Hazards and their labels

Definition 29.1 (Hazard and risk)

A hazard is the intrinsic property of a substance (or of a situation) that can cause harm: flammability, corrosiveness, toxicity. The risk is the likelihood that harm actually occurs, and its severity, under the given conditions of use: it depends on the hazard and on the exposure (quantity, concentration, duration, route, protection).

A hazard cannot be changed without changing the substance; a risk is reduced by working on the exposure: a smaller quantity, a more dilute solution, a fume cupboard, gloves and goggles, or a less hazardous reagent for the same job.

Definition 29.2 (GHS labelling)

The labels of chemicals follow the Globally Harmonized System (GHS).

  • A hazard pictogram is a black symbol on a white diamond with a red border; there are nine, GHS01 to GHS09.
  • The signal word is Danger for the more severe hazard categories and Warning for the less severe; a label carries one only, the more severe.
  • A hazard statement is a standard phrase, coded H followed by three digits, describing a hazard: H2xx physical hazards, H3xx health hazards, H4xx environmental hazards.
  • A precautionary statement, coded P and three digits, gives a measure to take: P1xx general, P2xx prevention, P3xx response, P4xx storage, P5xx disposal.
  • The safety data sheet is the document the supplier provides with a product, in standard sections: identification, hazards, composition, first aid, fire fighting, handling and storage, exposure controls and protection, physical and chemical properties, stability, toxicity, disposal and transport.

GHS01
explosive

GHS02
flammable

GHS03
oxidising

GHS04
gas under pressure

GHS05
corrosive

GHS06
acute toxicity

GHS07
harmful, irritant

GHS08
health hazard

GHS09
environment
The nine GHS pictograms: exploding bomb, flame, flame over circle, gas cylinder, corrosion, skull and crossbones, exclamation mark, health hazard (long-term effects: cancer, mutations, reproduction, sensitisation of the airways, aspiration) and environment.

Example 29.3 (Reading the label of ethanol)

A bottle of ethanol carries the pictograms GHS02 and GHS07, the signal word Danger, and the statements H225 (highly flammable liquid and vapour, a physical hazard, whose category calls for Danger) and H319 (causes serious eye irritation, a health hazard, Warning on its own). Its precautionary statements include P210 (keep away from heat, sparks and open flames; no smoking), P233 (keep the container tightly closed), P280 (wear gloves and eye protection), P305+P351+P338 (if in eyes: rinse cautiously with water for several minutes, remove contact lenses, continue rinsing) and P403+P235 (store in a well-ventilated place, keep cool). The practical consequences: no flame on the bench where ethanol is used, goggles, a closed bottle.

Method 29.4 (Preparing a session from the safety data sheets)

For each substance used or formed:

  1. read its pictograms, signal word and H statements (section 2 of the sheet), and note the physical hazards (fire, pressure, reactivity) apart from the health and environmental ones;
  2. for each operation (weighing, heating, transferring, filtering), estimate the exposure: quantity, state (a volatile liquid or a fine powder reaches the lungs), duration;
  3. choose the measures: substitution by a less hazardous reagent, smaller scale, fume cupboard, then personal protection (P2xx statements);
  4. plan the response to an incident (P3xx: eyes, skin, swallowing, fire) and where each waste goes (P5xx).

In the lab — Waste

Nothing goes down the sink by default. Organic solvents are collected in two containers, halogenated (dichloromethane, chloroform) and non-halogenated (ethanol, propanone, cyclohexane, ethyl ethanoate), since they are treated differently; aqueous acid and base solutions are neutralised before disposal; solutions of heavy-metal ions (chromium, copper, silver) have their own containers; broken glass and contaminated solids are kept apart from ordinary rubbish. The statement P273, “avoid release to the environment”, on a label is a reminder that the waste container, not the drain, is the end of the experiment.

29.2 Measurement and uncertainty

A measured value is never the exact value of the quantity measured: repeat the measurement and the result changes a little; read the scale and the last digit is a guess. The uncertainty says how far from the measured value the true value may reasonably lie.

Definition 29.5 (Measurement uncertainty)

  • The measurement uncertainty is a parameter that characterises the dispersion of the values that can reasonably be attributed to the quantity measured.
  • The standard uncertainty u(x)u(x) is that uncertainty expressed as a standard deviation.
  • A type A evaluation obtains uu by the statistical analysis of a series of repeated measurements; a type B evaluation obtains it by any other means: the graduation of an instrument, the tolerance stated by its maker, a calibration certificate, the number of digits of a tabulated value.
  • The expanded uncertainty U=k uU = k\,u is the standard uncertainty multiplied by a coverage factor kk, usually k=2k = 2; for a normal distribution the interval x±2ux \pm 2u contains the true value with a probability of about 95 %95\,\%.
  • The relative uncertainty is u(x)/∣x∣u(x)/|x|, often given in per cent.

Proposition 29.6 (Type A evaluation)

If nn independent measurements x1,…,xnx_1, \dots, x_n of the same quantity are made, the best estimate is their mean xˉ\bar x; the experimental standard deviation is

s=1n−1∑i=1n(xi−xˉ)2,s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar x)^2},

and the standard uncertainty of the mean is u(xˉ)=s/nu(\bar x) = s/\sqrt{n}.

Proof. Admitted at this level; the statistics of repeated measurements, including the Student coefficient used when nn is small, are treated in the Year 2 volume. ∎

Left: twenty readings of the same titration end point (illustrative data), read to 0.05\, mL, with the normal curve of the same mean V = 12.455\, mL and standard deviation s = 0.071\, mL. Right: the rectangular distribution of a type B evaluation, a value known only to lie between x_0 - a and x_0 + a; its standard deviation is a/√3. Left: twenty readings of the same titration end point (illustrative data), read to 0.05\, mL, with the normal curve of the same mean V = 12.455\, mL and standard deviation s = 0.071\, mL. Right: the rectangular distribution of a type B evaluation, a value known only to lie between x_0 - a and x_0 + a; its standard deviation is a/√3.
Left: twenty readings of the same titration end point (illustrative data), read to 0.05 mL0.05\,\mathrm{mL}, with the normal curve of the same mean Vˉ=12.455 mL\bar V = 12.455\,\mathrm{mL} and standard deviation s=0.071 mLs = 0.071\,\mathrm{mL}. Right: the rectangular distribution of a type B evaluation, a value known only to lie between x0−ax_0 - a and x0+ax_0 + a; its standard deviation is a/3a/\sqrt3.

Example 29.7 (Twenty end points)

For the twenty readings of the figure, Vˉ=12.455 mL\bar V = 12.455\,\mathrm{mL}, s=0.071 mLs = 0.071\,\mathrm{mL} and u(Vˉ)=0.071/20=0.016 mLu(\bar V) = 0.071/\sqrt{20} = 0.016\,\mathrm{mL}. One reading is uncertain by about ss; the mean of twenty by about s/20s/\sqrt{20}, four and a half times less. The result is V=12.46 mLV = 12.46\,\mathrm{mL} with U=2u=0.03 mLU = 2u = 0.03\,\mathrm{mL}.

Proposition 29.8 (Type B evaluation for a rectangular distribution)

If all that is known of a quantity is that it lies, with equal probability, anywhere between x0−ax_0 - a and x0+ax_0 + a, its standard uncertainty is

u=a3.u = \frac{a}{\sqrt3}.

Proof. The probability density is 1/(2a)1/(2a) on [x0−a,x0+a][x_0 - a, x_0 + a] and zero elsewhere; its mean is x0x_0 by symmetry and its variance is

u2=∫x0−ax0+a(x−x0)2dx2a=12a[t33]−aa=12a⋅2a33=a23.u^2 = \int_{x_0-a}^{x_0+a} (x - x_0)^2 \frac{\mathrm{d}x}{2a} = \frac{1}{2a} \left[\frac{t^3}{3}\right]_{-a}^{a} = \frac{1}{2a}\cdot\frac{2a^3}{3} = \frac{a^2}{3}. \qedhere

∎

Example 29.9 (A reading and a tolerance)

A burette graduated every 0.1 mL0.1\,\mathrm{mL} is read to half a division: the reading lies within ±0.05 mL\pm0.05\,\mathrm{mL} of the value noted, so u=0.05/3=0.029 mLu = 0.05/\sqrt3 = 0.029\,\mathrm{mL}. A 20 mL20\,\mathrm{mL} pipette whose maker states a tolerance of ±0.03 mL\pm0.03\,\mathrm{mL} delivers its volume with u=0.03/3=0.017 mLu = 0.03/\sqrt3 = 0.017\,\mathrm{mL}. A tabulated value given as 135 ∘C135\,{}^{\circ}\mathrm{C} is known to ±0.5 ∘C\pm0.5\,{}^{\circ}\mathrm{C}: u=0.29 ∘Cu = 0.29\,{}^{\circ}\mathrm{C}.

Proposition 29.10 (Combining uncertainties)

For independent quantities: if y=x1+x2y = x_1 + x_2 or y=x1−x2y = x_1 - x_2, then u(y)2=u(x1)2+u(x2)2u(y)^2 = u(x_1)^2 + u(x_2)^2; if y=x1x2y = x_1 x_2 or y=x1/x2y = x_1/x_2, then

(u(y)y)2=(u(x1)x1)2+(u(x2)x2)2.\left(\frac{u(y)}{y}\right)^2 = \left(\frac{u(x_1)}{x_1}\right)^2 + \left(\frac{u(x_2)}{x_2}\right)^2 .

Several sources of uncertainty on the same quantity (a type A part and type B parts) combine in the same way, as a sum of squares.

Proof. Admitted at this level. ∎

The squares, not the uncertainties themselves, add: two independent errors rarely push in the same direction at full size. A titrated volume is a difference of two burette readings, so its reading uncertainty is 2×0.029 mL=0.041 mL\sqrt2 \times 0.029\,\mathrm{mL} = 0.041\,\mathrm{mL}; and one large source of uncertainty dominates the others, which is where an improvement should be sought. The general rule, with partial derivatives, belongs to the Year 2 volume.

Method 29.11 (Reporting a result)

  1. List the sources of uncertainty; evaluate each as a standard uncertainty (type A from a series, type B as a/3a/\sqrt3 from a half-width).
  2. Combine them as a sum of squares (Proposition 29.10).
  3. Give the expanded uncertainty U=2uU = 2u with one or two significant figures, and round the value to the same decimal place: x=(value±U)x = (\text{value} \pm U) unit, k=2k = 2.

Definition 29.12 (Normalised deviation)

The normalised deviation (or zz-score) between two independent values x1x_1 and x2x_2 of the same quantity, of standard uncertainties u1u_1 and u2u_2, is

z=∣x1−x2∣u12+u22.z = \frac{|x_1 - x_2|}{\sqrt{u_1^2 + u_2^2}} .

The two values are said to be compatible when z≤2z \le 2; a measured value is compatible with a reference value under the same condition.

Example 29.13 (The two students)

The two melting points of the opening, 134 ∘C134\,{}^{\circ}\mathrm{C} and 136 ∘C136\,{}^{\circ}\mathrm{C}, were each obtained with a standard uncertainty of 0.8 ∘C0.8\,{}^{\circ}\mathrm{C} (reading and calibration of the bench). Then z=2/0.82+0.82=1.8≤2z = 2/\sqrt{0.8^2 + 0.8^2} = 1.8 \le 2: the students agree. Their mean, 135 ∘C135\,{}^{\circ}\mathrm{C}, is also compatible with the tabulated melting point of aspirin, 135 ∘C135\,{}^{\circ}\mathrm{C} under rapid heating. A tabulated value given to the unit has u=0.5/3=0.29 ∘Cu = 0.5/\sqrt3 = 0.29\,{}^{\circ}\mathrm{C}, so a single reading with u=0.8 ∘Cu = 0.8\,{}^{\circ}\mathrm{C} is compatible with it while it lies within 20.82+0.292=1.7 ∘C2\sqrt{0.8^2 + 0.29^2} = 1.7\,{}^{\circ}\mathrm{C} of it: a reading below about 133 ∘C133\,{}^{\circ}\mathrm{C} would have pointed to an impure product.

29.3 Separating

Definition 29.14 (Partition coefficient)

When a solute A is shared, at equilibrium, between two immiscible solvents, an organic phase and an aqueous phase, the ratio of its concentrations

K=[A]org[A]aqK = \frac{[\mathrm{A}]_{\mathrm{org}}}{[\mathrm{A}]_{\mathrm{aq}}}

is, for dilute solutions at a given temperature, a constant: the partition coefficient of A between the two solvents.

Proposition 29.15 (Several extractions are better than one)

A volume VaV_a of aqueous solution is extracted nn times, each time with a fresh volume VoV_o of organic solvent. The fraction of A left in the aqueous phase is

fn=(1+KVoVa)−n.f_n = \left(1 + K \frac{V_o}{V_a}\right)^{-n}.

For a fixed total volume of solvent, nn small portions extract more than one large portion.

Proof. Let mm be the amount of A in the aqueous phase before an extraction, and m′m' after. At equilibrium the organic phase holds m−m′m - m', and K=(m−m′)/Vom′/VaK = \dfrac{(m - m')/V_o}{m'/V_a}, so m−m′=KVoVam′m - m' = K\dfrac{V_o}{V_a} m' and m′=m/(1+KVo/Va)m' = m\big/\bigl(1 + K V_o/V_a\bigr). Each extraction multiplies the amount left by the same factor; after nn of them, fnf_n is the nn-th power. With a total volume VV split in nn portions V/nV/n, ln⁡fn=−nln⁡(1+KV/(nVa))\ln f_n = -n \ln(1 + KV/(nV_a)); since ln⁡(1+t)/t\ln(1+t)/t decreases with tt, nln⁡(1+t/n)n \ln(1 + t/n) increases with nn for t=KV/Va>0t = KV/V_a > 0, and fnf_n decreases, towards the limit e−KV/Va\mathrm{e}^{-KV/V_a}. ∎

Example 29.16 (One portion or three)

A solute with K=4K = 4 is extracted from 100 mL100\,\mathrm{mL} of water with 60 mL60\,\mathrm{mL} of ethyl ethanoate. In one portion, f1=1/(1+4×0.6)=0.29f_1 = 1/(1 + 4 \times 0.6) = 0.29: 71 %71\,\% is extracted. In three portions of 20 mL20\,\mathrm{mL}, f3=(1+4×0.2)−3=0.17f_3 = (1 + 4 \times 0.2)^{-3} = 0.17: 83 %83\,\%. Ethyl ethanoate (ρ=0.90 g/cm3\rho = 0.90\,\mathrm{g}/\mathrm{cm}^{3}) is the upper layer in the separating funnel; dichloromethane (ρ=1.33 g/cm3\rho = 1.33\,\mathrm{g}/\mathrm{cm}^{3}) would be the lower one.

Filtration separates a solid from a liquid. By gravity, through a fluted paper in a funnel, it keeps back an insoluble impurity from a hot solution that must not cool on the way (hot filtration, as in the photograph at the head of the chapter). Under reduced pressure, in a Büchner funnel on a side-arm flask (as drawn in the school volume), it collects a solid quickly and leaves it almost dry; the solid is washed on the filter with a little cold solvent.

Definition 29.17 (Recrystallisation)

Recrystallisation purifies a solid by dissolving it in the minimum of a hot solvent in which it is very soluble hot and little soluble cold, then letting it crystallise on cooling; the impurities either stay dissolved in the cold solvent (the mother liquor) or, being insoluble, are removed beforehand by hot filtration.

Method 29.18 (Recrystallising a solid)

  1. Choose the solvent: the product very soluble hot and little soluble cold, the soluble impurities soluble even cold, no reaction with the product, a boiling point below the product’s melting point; a mixture of two miscible solvents (ethanol and water) can be tuned to the product.
  2. Dissolve the crude solid in the minimum of boiling solvent, added in small portions, under reflux if the solvent is volatile or flammable.
  3. If an insoluble impurity remains, filter hot.
  4. Let the solution cool slowly, then in ice: slow cooling gives larger, purer crystals, which trap less mother liquor.
  5. Filter under reduced pressure, wash with a little ice-cold solvent, dry, weigh; check the purity (melting point, TLC).

Example 29.19 (The price of purity)

Aspirin dissolves in water at about 3.3 g/L3.3\,\mathrm{g}/\mathrm{L} at 25 ∘C25\,{}^{\circ}\mathrm{C} and 10 g/L10\,\mathrm{g}/\mathrm{L} at 37 ∘C37\,{}^{\circ}\mathrm{C}, much more near the boiling point. If 40 mL40\,\mathrm{mL} of mother liquor and 10 mL10\,\mathrm{mL} of washing water leave the filter at 25 ∘C25\,{}^{\circ}\mathrm{C}, they carry away about 0.050 L×3.3 g/L=0.17 g0.050\,\mathrm{L} \times 3.3\,\mathrm{g}/\mathrm{L} = 0.17\,\mathrm{g} of aspirin, whatever its purity. Every millilitre of solvent beyond the minimum costs yield.

A liquid is purified, or two liquids of well-separated boiling points are separated, by simple distillation: the mixture is boiled, the vapour, richer in the more volatile component, is condensed and collected. The temperature at the side arm is that of the vapour that distils; while it stays constant, a pure substance is passing over. Liquids of close boiling points need fractional distillation, a column between the flask and the head, treated with the liquid–vapour diagrams of the Year 2 volume.

Simple distillation. The thermometer reads the temperature of the vapour entering the condenser; cooling water enters at the low end of the condenser so that the jacket stays full; the boiling stones prevent violent bumping; the apparatus is open to the air at the receiver.
Simple distillation. The thermometer reads the temperature of the vapour entering the condenser; cooling water enters at the low end of the condenser so that the jacket stays full; the boiling stones prevent violent bumping; the apparatus is open to the air at the receiver.

Safety

A distillation apparatus is never closed: heated, a sealed system bursts. Boiling stones are added to the cold liquid, never to a liquid already hot (it may boil over at once). A flammable distillate is collected away from any flame, the receiver if needed in ice.

29.4 Identifying and checking purity

Definition 29.20 (Retention factor)

In thin-layer chromatography (TLC), a spot of the sample is deposited on a baseline near the foot of a plate coated with a stationary phase (silica); the plate stands in a closed tank in a little eluent, which rises through the coating by capillarity and carries the species at different rates. The retention factor of a species is

Rf=distance travelled by the spotdistance travelled by the eluent front,R_f = \frac{\text{distance travelled by the spot}} {\text{distance travelled by the eluent front}},

both measured from the baseline; 0≤Rf≤10 \le R_f \le 1, and for given stationary phase, eluent and temperature it characterises the species.

On polar silica, a polar species is held back more strongly than a less polar one and has the smaller RfR_f; a more polar eluent moves every spot further. Two species with the same RfR_f on one plate may still differ; two different RfR_f prove them different. A pure product gives a single spot.

A developed TLC plate: references A and B and a mixture M, deposited on the baseline. R_f( A) = d_ A/d_ front = 0.30, R_f( B) = 0.65; the mixture shows both spots, at the same heights.
A developed TLC plate: references A and B and a mixture M, deposited on the baseline. Rf(A)=dA/dfront=0.30R_f(\mathrm{A}) = d_{\mathrm{A}}/d_{\mathrm{front}} = 0.30, Rf(B)=0.65R_f(\mathrm{B}) = 0.65; the mixture shows both spots, at the same heights.

Method 29.21 (Running a TLC)

  1. Draw the baseline in pencil about 1 cm1\,\mathrm{cm} from the foot; deposit small spots of dilute solutions of the sample and of the references side by side.
  2. Put a few millimetres of eluent in the tank (below the baseline), close it, let the atmosphere saturate; stand the plate in it.
  3. Remove the plate when the front is about 1 cm1\,\mathrm{cm} from the top; mark the front at once; dry.
  4. Reveal colourless spots (ultraviolet lamp on a fluorescent plate, iodine vapour, a stain); circle them in pencil.
  5. Measure the distances from the baseline and compute each RfR_f; compare the sample with the references on the same plate.
A heated bench (Kofler bench): a metal strip heated at one end, with a temperature gradient along it; the solid, sprinkled on the strip, melts beyond a sharp line, read on the scale with the pointer. The box holds reference substances of known melting point, used to calibrate it. Photo: HBR, CC BY-SA 3.0, Wikimedia Commons.
A heated bench (Kofler bench): a metal strip heated at one end, with a temperature gradient along it; the solid, sprinkled on the strip, melts beyond a sharp line, read on the scale with the pointer. The box holds reference substances of known melting point, used to calibrate it. Photo: HBR, CC BY-SA 3.0, Wikimedia Commons.

Method 29.22 (Measuring a melting point on a heated bench)

  1. Switch the bench on well in advance: the gradient takes time to settle.
  2. Calibrate it with a pure reference substance melting close to the expected value, sprinkled on the strip; set the pointer so that the scale reads its melting point on the line.
  3. Sprinkle a few crystals of the dry sample from the cold end towards the hot end; push them back and forth with the spatula and set the pointer on the limit beyond which the solid melts at once.
  4. Read, repeat, clean the strip; take the mean of the readings and state the uncertainty (Method 29.11).

A pure crystalline substance melts sharply at its melting point; an impure one starts to melt lower and melts over a range of temperatures, because the impurity lowers the temperature at which the solid and the liquid coexist (the reason, the chemical potential of a solvent lowered by a solute, is in the Year 2 volume). A melting point well below the tabulated value, or a broad melting range, therefore reveals impurities; a value compatible with the tabulated one, in the sense of the normalised deviation, supports purity without proving it.

In the lab — Calibrating the bench

The references are chosen to bracket the expected value. Acetanilide, which melts at 114.3 ∘C114.3\,{}^{\circ}\mathrm{C}, and benzoic acid, at 122.4 ∘C122.4\,{}^{\circ}\mathrm{C}, are classic standards for products melting between 110 ∘C110\,{}^{\circ}\mathrm{C} and 130 ∘C130\,{}^{\circ}\mathrm{C}; for aspirin, a standard near 135 ∘C135\,{}^{\circ}\mathrm{C} is better still. Aspirin itself decomposes as it melts, so its observed melting point depends on how fast it is heated: tables give 135 ∘C135\,{}^{\circ}\mathrm{C} for rapid heating, which is what a bench does.

A liquid is identified, and its purity checked, by its refractive index nDn_D, the ratio of the speed of light in vacuum to its speed in the liquid, measured with the yellow D line of sodium at a stated temperature (usually 20 ∘C20\,{}^{\circ}\mathrm{C}), to four decimal places. Water has nD=1.333n_D = 1.333, ethanol 1.3611 and cyclohexane 1.4266 at 20 ∘C20\,{}^{\circ}\mathrm{C}. The index falls slightly as the temperature rises, so the instrument is thermostatted.

An Abbe refractometer: a drop of liquid is spread between two prisms; the eyepiece shows a field half light and half dark, and the boundary, set on the cross-hairs, gives the refractive index on the scale. Photo: Foreade, CC BY-SA 4.0, Wikimedia Commons.
An Abbe refractometer: a drop of liquid is spread between two prisms; the eyepiece shows a field half light and half dark, and the boundary, set on the cross-hairs, gives the refractive index on the scale. Photo: Foreade, CC BY-SA 4.0, Wikimedia Commons.

29.5 Exercises

Exercise 29.1 ★

A bottle of propanone carries the pictograms GHS02 and GHS07 and the statements H225, H319 and H336. Which signal word does it carry? Which statements describe a physical hazard, which a health hazard? Name two precautions they call for.

Solution

Solution of Exercise 29.1.

Danger: H225 (highly flammable liquid, category 2) calls for it, and the label carries the more severe word. H225 is a physical hazard; H319 (serious eye irritation) and H336 (drowsiness or dizziness) are health hazards. Precautions: no flame or spark nearby, a closed bottle; goggles; work in a ventilated place or a fume cupboard.

Exercise 29.2 ★

Hazard or risk? (a) Concentrated sulfuric acid causes severe burns. (b) Weighing 50 mg50\,\mathrm{mg} of a toxic powder in a fume cupboard with gloves exposes the operator little. (c) Sodium releases hydrogen with water. (d) The same solvent is more dangerous used hot in an open beaker than cold in a closed bottle.

Solution

Solution of Exercise 29.2.

(a) A hazard, a property of the acid. (b) A risk, small because the exposure is small. (c) A hazard. (d) A risk: same hazard, larger exposure (vapour, splashes) when hot and open.

Exercise 29.3 ★

On a TLC plate the eluent front is 6.0 cm6.0\,\mathrm{cm} above the baseline; the sample gives spots at 2.1 cm2.1\,\mathrm{cm} and 4.2 cm4.2\,\mathrm{cm}. Compute their RfR_f. A reference deposited on the same plate rises to 4.2 cm4.2\,\mathrm{cm}: what can and cannot be concluded?

Solution

Solution of Exercise 29.3.

Rf=2.1/6.0=0.35R_f = 2.1/6.0 = 0.35 and 4.2/6.0=0.704.2/6.0 = 0.70. The second spot has the same RfR_f as the reference: the sample may contain it (it is consistent), but equal RfR_f on one eluent do not prove identity; a second eluent, or a co-spot of sample and reference that stays a single spot, strengthens the conclusion. The spot at 0.35 is certainly another species.

Exercise 29.4 ★

A burette graduated every 0.1 mL0.1\,\mathrm{mL} is read to half a division. Compute the standard uncertainty of one reading, then of a delivered volume, the difference of two readings.

Solution

Solution of Exercise 29.4.

Half-width 0.05 mL0.05\,\mathrm{mL}, rectangular: u=0.05/3=0.029 mLu = 0.05/\sqrt3 = 0.029\,\mathrm{mL}. For a difference of two independent readings, u=2×0.029=0.041 mLu = \sqrt{2} \times 0.029 = 0.041\,\mathrm{mL}.

Exercise 29.5 ★★

Five weighings of the same sample give 0.2512 g0.2512\,\mathrm{g}, 0.2508 g0.2508\,\mathrm{g}, 0.2515 g0.2515\,\mathrm{g}, 0.2510 g0.2510\,\mathrm{g} and 0.2505 g0.2505\,\mathrm{g}. Compute the mean, the experimental standard deviation, the standard uncertainty of the mean, and report the result with k=2k = 2.

Solution

Solution of Exercise 29.5.

Mean 0.251 00 g0.251\,00\,\mathrm{g}; deviations +2+2, −2-2, +5+5, 0, −5-5 (in 10−4 g10^{-4}\,\mathrm{g}), sum of squares 58×10−8 g258 \times 10^{-8}\,\mathrm{g}^{2}, s=58×10−8/4=3.8×10−4 gs = \sqrt{58 \times 10^{-8}/4} = 3.8 \times 10^{-4}\,\mathrm{g}; u=s/5=1.7×10−4 gu = s/\sqrt5 = 1.7 \times 10^{-4}\,\mathrm{g}; U=3.4×10−4 gU = 3.4 \times 10^{-4}\,\mathrm{g}: m=(0.25100±0.00034)m = (0.25100 \pm 0.00034) g, k=2k = 2.

Exercise 29.6 ★★

A solute with partition coefficient K=3K = 3 between an organic solvent and water is extracted from 50 mL50\,\mathrm{mL} of water with 30 mL30\,\mathrm{mL} of solvent in total. Compute the fraction extracted in one portion, in two portions of 15 mL15\,\mathrm{mL}, and in three of 10 mL10\,\mathrm{mL}.

Solution

Solution of Exercise 29.6.

One portion: f=1/(1+3×30/50)=0.357f = 1/(1 + 3 \times 30/50) = 0.357, 64.3 %64.3\,\% extracted. Two of 15 mL15\,\mathrm{mL}: f=1.9−2=0.277f = 1.9^{-2} = 0.277, 72.3 %72.3\,\%. Three of 10 mL10\,\mathrm{mL}: f=1.6−3=0.244f = 1.6^{-3} = 0.244, 75.6 %75.6\,\%.

Exercise 29.7 ★★

A titration gives a concentration of 0.1012 mol/L0.1012\,\mathrm{mol}/\mathrm{L} with standard uncertainty 0.0006 mol/L0.0006\,\mathrm{mol}/\mathrm{L}; the solution was prepared at 0.1000 mol/L0.1000\,\mathrm{mol}/\mathrm{L} with standard uncertainty 0.0002 mol/L0.0002\,\mathrm{mol}/\mathrm{L}. Are the two values compatible? What if the titration uncertainty had been 0.0004 mol/L0.0004\,\mathrm{mol}/\mathrm{L}?

Solution

Solution of Exercise 29.7.

z=0.0012/0.00062+0.00022=0.0012/0.00063=1.9≤2z = 0.0012/\sqrt{0.0006^2 + 0.0002^2} = 0.0012/0.00063 = 1.9 \le 2: compatible, just. With 0.0004 mol/L0.0004\,\mathrm{mol}/\mathrm{L}: z=0.0012/0.00045=2.7z = 0.0012/0.00045 = 2.7, not compatible: the more precise titration reveals a real difference (an error in the preparation, or in the titration).

Exercise 29.8 ★★

The solubilities (illustrative values, g\mathrm{g} per 100 mL100\,\mathrm{mL}) of a solid in three solvents are: solvent P, 1.5 cold and 2.0 boiling; solvent Q, 0.3 cold and 12 boiling; solvent R, 9 cold and 25 boiling. Which solvent suits a recrystallisation, and why not the other two? With the right solvent, what is the largest fraction of 5.0 g5.0\,\mathrm{g} of the solid that can be recovered, if the minimum of boiling solvent is used and the mixture cooled?

Solution

Solution of Exercise 29.8.

Q: little soluble cold, very soluble hot. P dissolves the solid hardly better hot than cold (nothing crystallises on cooling); R keeps too much dissolved cold (large losses). With Q, 5.0 g5.0\,\mathrm{g} needs at least 5.0/12×100=42 mL5.0/12 \times 100 = 42\,\mathrm{mL} of boiling solvent, which keeps 0.3×0.417=0.13 g0.3 \times 0.417 = 0.13\,\mathrm{g} dissolved cold: at most 4.9 g4.9\,\mathrm{g}, 97.5 %97.5\,\%, can be recovered (before any loss on the filter).

Exercise 29.9 ★★

A colourless liquid, thermostatted at 20 ∘C20\,{}^{\circ}\mathrm{C}, has nD=1.3612n_D = 1.3612. Which of water, ethanol and cyclohexane can it be? Why is the temperature stated? What would a value of 1.350 suggest?

Solution

Solution of Exercise 29.9.

Ethanol (nD=1.3611n_D = 1.3611 at 20 ∘C20\,{}^{\circ}\mathrm{C}); water (1.333) and cyclohexane (1.4266) are excluded. The index varies with temperature, so a value means nothing without it. A value of 1.350, between water and ethanol, suggests a mixture, such as ethanol containing water.

Exercise 29.10 ★★★

With the data of Exercise 29.6, show that however finely the 30 mL30\,\mathrm{mL} of solvent is divided, at most a fraction 1−e−KV/Va1 - \mathrm{e}^{-KV/V_a} of the solute can be extracted, and compute it. How many portions reach 99 %99\,\% of that limit?

Solution

Solution of Exercise 29.10.

With t=KV/Vat = KV/V_a and nn portions, ln⁡fn=−nln⁡(1+t/n)\ln f_n = -n\ln(1 + t/n); as n→∞n \to \infty, nln⁡(1+t/n)→tn \ln(1 + t/n) \to t (since ln⁡(1+ε)∼ε\ln(1 + \varepsilon) \sim \varepsilon), and fnf_n decreases towards e−t\mathrm{e}^{-t} (Proposition 29.15). Here t=3×30/50=1.8t = 3 \times 30/50 = 1.8, e−1.8=0.165\mathrm{e}^{-1.8} = 0.165: at most 83.5 %83.5\,\%. Three portions already give 75.6 %75.6\,\%, 91 %91\,\% of the limit; reaching 99 %99\,\% of it takes 32 portions of less than 1 mL1\,\mathrm{mL}: beyond two or three portions, the gain is not worth the work.

Exercise 29.11 ★★★

A quantity is known to lie between x0−ax_0 - a and x0+ax_0 + a, values near x0x_0 being more likely: its density is triangular, (a−∣x−x0∣)/a2(a - |x - x_0|)/a^2. Check that it is normalised and show that its standard uncertainty is a/6a/\sqrt6. Compare with the rectangular case.

Solution

Solution of Exercise 29.11.

With t=x−x0t = x - x_0: ∫−aa(a−∣t∣)/a2 dt=2∫0a(a−t)/a2 dt=2(a2/2)/a2=1\int_{-a}^{a} (a - |t|)/a^2 \,\mathrm{d}t = 2 \int_0^a (a - t)/a^2 \,\mathrm{d}t = 2 (a^2/2)/a^2 = 1. The mean is x0x_0 by symmetry; the variance is 2∫0at2(a−t)/a2 dt=2a2(a43−a44)=a262 \int_0^a t^2 (a - t)/a^2 \,\mathrm{d}t = \frac{2}{a^2}\left(\frac{a^4}{3} - \frac{a^4}{4}\right) = \frac{a^2}{6}, so u=a/6=0.41 au = a/\sqrt6 = 0.41\,a, against a/3=0.58 aa/\sqrt3 = 0.58\,a for the rectangular case: knowing that the middle is more likely reduces the uncertainty.

Exercise 29.12 ★★★

A standard solution is made by dissolving the sample of Exercise 29.5 (molar mass 204.2 g/mol204.2\,\mathrm{g}/\mathrm{mol}, its uncertainty negligible) in a 100.00 mL100.00\,\mathrm{mL} volumetric flask whose volume has a standard uncertainty of 0.06 mL0.06\,\mathrm{mL}. Compute the concentration and its relative and expanded uncertainties. Which source dominates?

Solution

Solution of Exercise 29.12.

c=m/(MV)=0.25100/(204.2×0.10000)=0.012 292 mol/Lc = m/(MV) = 0.25100/(204.2 \times 0.10000) = 0.012\,292\,\mathrm{mol}/\mathrm{L}. Relative uncertainties: mass 1.7×10−4/0.2510=6.8×10−41.7 \times 10^{-4}/0.2510 = 6.8 \times 10^{-4}, volume 0.06/100.00=6.0×10−40.06/100.00 = 6.0 \times 10^{-4}; combined (6.82+6.02)×10−4=9.1×10−4\sqrt{(6.8^2 + 6.0^2)} \times 10^{-4} = 9.1 \times 10^{-4}, that is 0.091 %0.091\,\%; u(c)=1.1×10−5 mol/Lu(c) = 1.1 \times 10^{-5}\,\mathrm{mol}/\mathrm{L}, U=2.2×10−5 mol/LU = 2.2 \times 10^{-5}\,\mathrm{mol}/\mathrm{L}: c=(0.012292±0.000022)c = (0.012292 \pm 0.000022) mol/L, k=2k = 2. The weighing dominates, slightly; the two sources are of the same size.

29.6 Problem: Purifying Aspirin

Problem 29.1

Weekend problem — the hazards of a synthesis, the recrystallisation of crude aspirin and its yield, a melting point with its type A and type B uncertainties, and the normalised deviation from the tabulated value

A student makes aspirin by heating 2.00 g2.00\,\mathrm{g} of salicylic acid with an excess of ethanoic anhydride, then recrystallises the crude product. Molar masses (g/mol\mathrm{g}/\mathrm{mol}): salicylic acid CX7HX6OX3\ce{C7H6O3} 138.0, aspirin CX9HX8OX4\ce{C9H8O4} 180.0, ethanoic anhydride CX4HX6OX3\ce{C4H6O3} 102.0. Labels: salicylic acid GHS05, GHS07, H302, H318; ethanoic anhydride GHS02, GHS05, GHS06, GHS07, H226, H302, H314, H332; aspirin GHS07, H302. Solubility of aspirin in water: 3.3 g/L3.3\,\mathrm{g}/\mathrm{L} at 25 ∘C25\,{}^{\circ}\mathrm{C}. Tabulated melting point of aspirin (rapid heating): 135 ∘C135\,{}^{\circ}\mathrm{C}.

Part I — Hazards.

  1. What does each pictogram on the label of ethanoic anhydride mean?
  2. Which signal word does the label of salicylic acid carry? And that of aspirin?
  3. Ethanoic anhydride has the more severe hazards. Explain how the risk of handling 5 mL5\,\mathrm{mL} of it can nevertheless be made small.
  4. Which of the anhydride’s statements call for goggles and gloves, and which for the fume cupboard and the absence of flames?
  5. What is done at once if a drop of anhydride reaches an eye?
  6. The filtrate of the recrystallisation contains ethanoic acid and a little ethanol in water. Where does it go?

Part II — Synthesis and recrystallisation.

  1. Write the equation of the synthesis, with ethanoic acid CX2HX4OX2\ce{C2H4O2} as by-product, and check that it is balanced.
  2. Compute the amount of salicylic acid and the maximum mass of aspirin.
  3. The crude dry product weighs 2.41 g2.41\,\mathrm{g}. Why can the recrystallisation not be skipped although this is 92 %92\,\% of the maximum?
  4. Aspirin is recrystallised from a little ethanol completed with hot water. Why is the solubility in water at 25 ∘C25\,{}^{\circ}\mathrm{C}, compared with that near the boiling point, the property that matters?
  5. Why is the crude solid dissolved in the minimum of hot solvent?
  6. Why is the solution cooled slowly, then in ice?
  7. 40 mL40\,\mathrm{mL} of filtrate at 25 ∘C25\,{}^{\circ}\mathrm{C} leave the funnel, then 5 mL5\,\mathrm{mL} of washing water. Estimate the mass of aspirin lost with them.
  8. The pure dry product weighs 1.95 g1.95\,\mathrm{g}. Compute the yield.

Part III — Melting point.

  1. Six readings on the bench, calibrated just before, give 134, 135, 133, 135, 134 and 136 ∘C136\,{}^{\circ}\mathrm{C}. Compute the mean.
  2. Compute the experimental standard deviation.
  3. Deduce the type A standard uncertainty of the mean.
  4. Each reading is made to the nearest degree. Compute the type B uncertainty of reading.
  5. The calibration of the bench is guaranteed to within ±1 ∘C\pm1\,{}^{\circ}\mathrm{C}. Compute the corresponding type B uncertainty, then the combined standard uncertainty.
  6. Report the melting point with its expanded uncertainty (k=2k = 2).
  7. A classmate’s crude product melts between 120 and 128 ∘C128\,{}^{\circ}\mathrm{C}. What does this indicate?

Part IV — Comparison with the reference.

  1. The tabulated value is given to the unit. What standard uncertainty does that imply?
  2. On a TLC plate (front 6.0 cm6.0\,\mathrm{cm}), the crude product shows spots at 2.3 cm2.3\,\mathrm{cm} and 3.3 cm3.3\,\mathrm{cm}, salicylic acid one spot at 2.3 cm2.3\,\mathrm{cm}, the recrystallised product one spot at 3.3 cm3.3\,\mathrm{cm}. Compute the RfR_f values and conclude.
  3. Why does the melting point of aspirin depend on the heating rate?
  4. Compute the normalised deviation zz between the measured and the tabulated melting points, and say whether they are compatible.
Solution

Solution of Problem 29.1.

1. GHS02 flammable, GHS05 corrosive, GHS06 acute toxicity, GHS07 harmful or irritant. 2. Salicylic acid: Danger, because of H318 (serious eye damage, category 1). Aspirin: Warning (H302 only). 3. The hazard is fixed, the exposure is not: a small volume, measured and used in a fume cupboard, with gloves, goggles and a lab coat, the bottle closed at once, no flame nearby. 4. H314 (severe skin burns and eye damage): goggles and gloves. H332 (harmful if inhaled) and H226 (flammable liquid and vapour): the fume cupboard and the absence of flames. 5. Rinse cautiously with water for several minutes, removing contact lenses if possible, and keep rinsing (P305+P351+P338); then seek medical advice. 6. Into the aqueous waste, after neutralisation, not down the sink. 7. CX7HX6OX3+CX4HX6OX3→CX9HX8OX4+CX2HX4OX2\ce{C7H6O3 + C4H6O3 -> C9H8O4 + C2H4O2}: C 7+4=9+27 + 4 = 9 + 2, H 6+6=8+46 + 6 = 8 + 4, O 3+3=4+23 + 3 = 4 + 2. 8. 2.00/138.0=1.449×10−2 mol2.00/138.0 = 1.449 \times 10^{-2}\,\mathrm{mol}; at most 1.449×10−2×180.0=2.61 g1.449 \times 10^{-2} \times 180.0 = 2.61\,\mathrm{g} of aspirin. 9. The crude solid contains unreacted salicylic acid, ethanoic acid and water: its mass is not that of aspirin, and its purity is unknown. 10. What stays dissolved after cooling is lost: a solubility small cold and large hot means the product dissolves when hot and comes back almost entirely when cold. 11. Every extra millilitre keeps its share of product dissolved when cold. 12. Slow growth gives larger, more regular crystals that trap less mother liquor and impurities; the ice lowers the solubility, hence the loss. 13. 0.045 L×3.3 g/L=0.15 g0.045\,\mathrm{L} \times 3.3\,\mathrm{g}/\mathrm{L} = 0.15\,\mathrm{g}. 14. 1.95/2.61=75 %1.95/2.61 = 75\,\% (74.7 %74.7\,\%). 15. 807/6=134.5 ∘C807/6 = 134.5\,{}^{\circ}\mathrm{C}. 16. Deviations −0.5-0.5, 0.50.5, −1.5-1.5, 0.50.5, −0.5-0.5, 1.51.5; sum of squares 5.5; s=5.5/5=1.05 ∘Cs = \sqrt{5.5/5} = 1.05\,{}^{\circ}\mathrm{C}. 17. uA=1.05/6=0.43 ∘Cu_A = 1.05/\sqrt6 = 0.43\,{}^{\circ}\mathrm{C}. 18. Half-width 0.5 ∘C0.5\,{}^{\circ}\mathrm{C}: 0.5/3=0.29 ∘C0.5/\sqrt3 = 0.29\,{}^{\circ}\mathrm{C}. 19. 1/3=0.58 ∘C1/\sqrt3 = 0.58\,{}^{\circ}\mathrm{C}; u=0.4282+0.2892+0.5772=0.600=0.77 ∘Cu = \sqrt{0.428^2 + 0.289^2 + 0.577^2} = \sqrt{0.600} = 0.77\,{}^{\circ}\mathrm{C}. 20. U=2×0.775=1.5 ∘CU = 2 \times 0.775 = 1.5\,{}^{\circ}\mathrm{C}: T=(134.5±1.5)T = (134.5 \pm 1.5) ∘C{}^{\circ}\mathrm{C}, k=2k = 2. 21. It melts low and over a range of 8 ∘C8\,{}^{\circ}\mathrm{C}: it is impure (salicylic acid, ethanoic acid, water). 22. Given to the unit, the value lies within ±0.5 ∘C\pm0.5\,{}^{\circ}\mathrm{C}: uref=0.5/3=0.29 ∘Cu_{\mathrm{ref}} = 0.5/\sqrt3 = 0.29\,{}^{\circ}\mathrm{C}. 23. 2.3/6.0=0.382.3/6.0 = 0.38 and 3.3/6.0=0.553.3/6.0 = 0.55. The crude product contained salicylic acid (same RfR_f as the reference) and aspirin; after recrystallisation only the aspirin spot remains: the salicylic acid has been removed, at least below what the plate detects. 24. Aspirin decomposes as it melts; heated slowly, it has time to decompose, and the mixture formed melts lower. The tabulated value refers to rapid heating, as on the bench. 25. z=∣134.5−135∣/0.600+0.083=0.5/0.827=z = |134.5 - 135|/\sqrt{0.600 + 0.083} = 0.5/0.827 = 0.60: well below 2, the measured and tabulated melting points are compatible.

Terms defined in this chapter

See all 852 terms in the glossary