Chemistry (No. III.)

Original question.

What are the proportions respectively required of Carbonate of Potash, Carbonate of Soda, and Caustic Magnesia to neutralize a given quantity of Sulphuric Acid ?

For complete neutralization, the required masses of anhydrous potassium carbonate, anhydrous sodium carbonate and magnesium oxide are in the ratio $138 : 106 : 40 = 69 : 53 : 20$ . Here “carbonate of potash” means potassium carbonate, “carbonate of soda” means sodium carbonate, and “caustic magnesia” means magnesium oxide, rather than magnesium hydroxide or a magnesium carbonate preparation. The hydration state of the sodium carbonate matters considerably.

Each formula unit of either carbonate consumes two protons; each formula unit of magnesium oxide also consumes two protons. Thus one mole of each base neutralizes one mole of sulfuric acid when the final salts are the normal sulfates:

K_{2} C O_{3} + H_{2} S O_{4} N a_{2} C O_{3} + H_{2} S O_{4} MgO + H_{2} S O_{4} ⟶ K_{2} S O_{4} + C O_{2} + H_{2} O, ⟶ N a_{2} S O_{4} + C O_{2} + H_{2} O, ⟶ MgS O_{4} + H_{2} O .

The carbon dioxide evolved in the first two chemical reactions escapes, whereas magnesium oxide produces no gas. The stoichiometric coefficients, rather than the visibly different behaviour of the bases, determine the required amounts.

Let

m

be the mass of pure

H_{2} S O_{4}

in the given chemical solution. Using the rounded relative atomic masses

H = 1

C = 12

O = 16

Na = 23

Mg = 24

and

K = 39

gives

m_{K_{2} C O_{3}} = \frac{138}{98} m, m_{N a_{2} C O_{3}} = \frac{106}{98} m, m_{MgO} = \frac{40}{98} m .

Thus 98 mass units of pure sulfuric acid require respectively 138, 106 and 40 mass units. Present atomic weights slightly change these rounded values, without changing the one-to-one mole ratios. If

m_{s}

denotes the mass of the supplied acid and

w

its mass fraction of

H_{2} S O_{4}

, substitute

m = w m_{s}

. A volume alone is insufficient without its concentration or an equivalent density and composition.

If “carbonate of soda” is instead the crystalline decahydrate, sodium carbonate decahydrate, its rounded molar mass is

286 g mo l^{- 1}

, so the corresponding factor is

286/98

, rather than

106/98

. Likewise an impure base requires correction for its assayed content. The unspecified given quantity can therefore be handled exactly by proportional reasoning; no missing numerical measurement is needed.

2

Original question.

What formula would you assign to a substance which gave the following results to analysis :—

Constituent	Analysis
Carbon	54.35
Hydrogen	9.11
Oxygen	36.54
Total	100.00

and of which the vapour density was 3.05.

Solution

The molecular formula is $C_{4} H_{8} O_{2}$ , interpreting the vapour density as relative to air. The organic elemental analysis first determines a simpler empirical formula; the vapor-density molar mass measurement then fixes its multiple.

Take 100 mass units of the chemical substance and divide each elemental contribution by its rounded relative atomic mass:

n_{C} : n_{H} : n_{O} = \frac{54.35}{12} : \frac{9.11}{1} : \frac{36.54}{16} = 4.5292 : 9.1100 : 2.2838 ≃ 2 : 4 : 1.

The empirical formula is therefore

C_{2} H_{4} O

, with empirical formula mass approximately 44. Its calculated mass percentages are 54.55% carbon, 9.09% hydrogen and 36.36% oxygen. The small discrepancies from the given organic elemental analysis are consistent with rounded experimental data; they should not be converted into large, meaningless integer subscripts.

For equal volumes of ideal gases at equal temperature and pressure, the density ratio equals the molar mass ratio. On the air convention,

M ≃ 3.05 \times 28.97 g mo l^{- 1} = 88.36 g mo l^{- 1}, \frac{M}{44 g mo l ^{- 1}} ≃ 2.

Doubling

C_{2} H_{4} O

gives

C_{4} H_{8} O_{2}

; its rounded relative molecular mass of 88 predicts an air-relative vapor density of approximately

88/28.97 = 3.04

, close to 3.05. If the stated vapor density were instead relative to hydrogen, it would imply a relative molecular mass near 6.1, smaller than the empirical formula mass and incompatible with the analysis. That interpretation cannot describe a pure, undecomposed molecular vapor with the stated composition.

Under the older equivalent weight convention

C = 6

O = 8

H = 1

, the same composition and formula weight could be written

C_{8} H_{8} O_{4}

. This is a change of notation, not eight modern carbon chemical atoms in the molecule. The result specifies composition and molar mass; it does not distinguish among structural isomers, such as ethyl acetate and butyric acid.

3

Original question.

What observations are required for the accurate comparison of the volumes of two gases, and for what reasons ?

Solution

Observe each gas volume, temperature, pressure and moisture condition, and reduce both volumes to the same temperature, pressure and dry-gas basis. Equal scale readings alone do not establish equal amounts of gas.

Read the calibrated volume of each gas in a graduated tube or eudiometer, with the eye at the meniscus to avoid parallax. Allow the gas and the surrounding bath to reach thermal equilibrium, then read a thermometer close to the gas. Heating expands a fixed amount of gas, so a larger reading could reflect a higher temperature rather than more gas. The temperature used in a gas law must be an thermodynamic temperature.

Read the atmospheric pressure with a barometer. If a gas is confined over a liquid, also observe the difference between the inside and outside liquid levels. With

h

the internal surface height above the external surface, the total internal pressure is

p_{g} = p_{atm} - ρ_{ℓ} g h .

Here

ρ_{ℓ}

is the confining density of the liquid and

g

the gravitational acceleration; the sign reverses automatically when

h

is negative. Equalizing the levels before taking the volume reading removes this hydrostatic pressure correction. A mercury barometer can itself require temperature and calibration corrections for accurate work.

If the gas is saturated with water vapor, subtract the vapor pressure of water at its measured temperature:

p_{dry} = p_{g} - p_{H_{2} O} (T) .

The observed pressure includes both the test gas and water vapor; counting it all as test gas overestimates the amount. The saturation assumption needs time for equilibration. For a dry gas, no water vapor subtraction is made; another collecting liquid may require its own vapor pressure correction. A collecting liquid must also be chosen so that it neither appreciably dissolves nor reacts with the gas: water, for instance, is unsuitable for preserving the whole amount of freely soluble Ammonia.

For dilute gases obeying the ideal gas law, the volume

V_{0}

at a common chosen

T_{0}, p_{0}

V_{0} = V \frac{p _{dry}}{p _{0}} \frac{T _{0}}{T} .

(3)

Consequently the corrected comparison is

\frac{V _{1, 0}}{V _{2, 0}} = \frac{V _{1}}{V _{2}} \frac{p _{1, dry}}{p _{2, dry}} \frac{T _{2}}{T _{1}} .

(4)

This is also the ratio of amounts of substance in the ideal gas approximation. Check for leaks, trapped air, changing composition and incomplete transfer. At high pressure or near condensation, real gas behaviour may require compressibility factors; with

p V = Z n RT

, the inferred amount ratio contains the further factor

Z_{2} / Z_{1}

4

Original question.

Give the Chemical Formulæ of the following substances :—Hydrocyanic Acid, Formic Acid, Oxalic Acid, Calomel, Corrosive Sublimate, Arsenious Acid, Tartar-Emetic, Chlorate of Potash.

Solution

In modern notation the formulas are HCN, HCOOH, $H_{2} C_{2} O_{4}$ , $H g_{2} C l_{2}$ , $HgC l_{2}$ , $A s_{2} O_{3}$ , a hydrated potassium antimony tartrate, and $KCl O_{3}$ , respectively. Several historical names need qualification.

Hydrocyanic acid is hydrogen cyanide, $HCN$ . The pharmaceutical name normally concerns an aqueous chemical solution; its concentration is not encoded by the molecular chemical formula.
Formic acid is $HCOOH$ , equivalently $C H_{2} O_{2}$ .
Anhydrous oxalic acid is $H_{2} C_{2} O_{4}$ , or $(COOH)_{2}$ . Its common crystalline dihydrate is $H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ ; the water of crystallization must be included when calculating a weighed quantity.
Calomel is mercury(I) chloride, $H g_{2} C l_{2}$ . The formula $HgCl$ gives only the simplest composition ratio and conceals the paired mercury(I) ion.
Corrosive sublimate is mercury(II) chloride, $HgC l_{2}$ . It is chemically distinct from calomel.
The solid historically sold as arsenious acid is normally arsenic trioxide, $A s_{2} O_{3}$ , often written $As O_{3}$ in older equivalent notation. The genuinely hydrated molecular arsenous acid is $H_{3} As O_{3}$ , also written $As (OH)_{3}$ . One should not quietly assign the hydrated formula to the dry commercial oxide.
Tartar emetic, or potassium antimony tartrate, is conventionally represented as $K (SbO) C_{4} H_{4} O_{6} \cdot \frac{1}{2} H_{2} O$ . A structurally more informative doubled formula is $K_{2} S b_{2} (C_{4} H_{2} O_{6})_{2} \cdot 3 H_{2} O$ : the complete elemental totals agree. The older “antimonyl tartrate” formula is a compositional convention and does not imply isolated $Sb O^{+}$ units in the crystal.^[ref]
Chlorate of potash is potassium chlorate, $KCl O_{3}$ .

The chemical formulas above use modern atomic weights and distinguish molecular formulas, formula units and hydrates. Historical equivalent notation sometimes halves the subscripts attached to divalent chemical elements or writes water as

HO

. Such formulas must be read together with their contemporary atomic weights, rather than treated as inconsistent modern molecular formulas.

5

Original question.

Explain briefly how each of the substances named in the preceding question may be best prepared.

Solution

A suitable historical answer gives both the chemical transformation and the means of isolating a reasonably pure product. The following are preparative principles, with modern chemical formulas; they are not instructions for self-preparation of medicines. In particular, hydrogen cyanide, mercury compounds and arsenic compounds present serious toxicity hazards. Hydrogen cyanide is extremely volatile, can be absorbed by several routes, and its odour is an unreliable warning.^[ref]

For hydrocyanic acid, a customary early nineteenth-century route is the controlled distillation of potassium ferrocyanide with dilute sulfuric acid and water. Hydrogen cyanide passes into the distillate; nonvolatile salts remain behind. A generating flask, efficient laboratory condenser and chilled receiver are needed because hydrogen cyanide is readily lost. The contemporary account of hydrocyanic acid preparation in Pharmacopoeia Castrensis Ruthenica in The British and Foreign Medical Review (1842) specifically describes the ferrocyanide–sulfuric acid method and its losses on imperfect condensation. The concentration must be determined analytically before any defined dilution: a nominal starting mixture does not establish the final strength.^[ref] At the level of the released cyanide, the essential acid-base reaction is

C N^{-} + H^{+} \to HCN

. This equation alone is not a complete equation for decomposing the stable ferrocyanide coordination complex. Concentrated sulfuric acid can lead to different decomposition chemistry, so the specified dilution is meaningful.

For formic acid, gently heat oxalic acid with glycerol and some water, then separate the aqueous formic acid by distillation. The useful net transformation is

H_{2} C_{2} O_{4} ⟶ HCOOH + C O_{2} .

The glycerol participates through intermediate esters and is regenerated rather than consumed in this net equation. Marcellin Berthelot's 1856 account describes the carbon dioxide evolution, subsequent water addition and distillation, and reuse of the residual glycerol. Overheating is disadvantageous because the formic acid can decompose further to carbon monoxide and water. The method therefore separates a volatile acid from a much less volatile carrier; it is not merely the dry destructive heating of oxalic acid.^[ref]

For oxalic acid, oxidize a carbohydrate, such as sucrose or starch, with nitric acid; concentrate the resulting chemical solution appropriately and cool it to obtain crystals, followed by recrystallization. Nitrogen oxides evolve, and excessive oxidation can carry carbon onward to carbon dioxide, so an idealized single-product equation would conceal losses. The use of starch and nitric acid for oxalic acid production is explicitly documented in an 1850 account of nitric oxide recovery.^[ref] An alternative already established before 1864 is alkaline treatment of sawdust, producing an oxalate which can be purified through sparingly soluble calcium oxalate.^[ref] Decomposing the isolated calcium oxalate with sulfuric acid separates much of the calcium as calcium sulfate:

Ca C_{2} O_{4} + H_{2} S O_{4} ⟶ H_{2} C_{2} O_{4} + CaS O_{4} .

Filter off the calcium sulfate and crystallize the oxalic acid from the filtrate. The usual crystals contain two water of crystallization; a calculation for the anhydrous acid must not treat their entire mass as

H_{2} C_{2} O_{4}

For calomel, intimately combine mercury(II) sulfate, metallic Mercury and dry sodium chloride, then collect the volatile mercury(I) chloride in a cooler receiving chamber. The net change is

HgS O_{4} + Hg + 2NaCl ⟶ H g_{2} C l_{2} + N a_{2} S O_{4} .

(3)

The extra metallic Mercury lowers the average oxidation state from +2 to +1. The product is purified by appropriate sublimation and washing to remove soluble contamination, particularly mercury(II) chloride. Both the chemical composition and the fine subdivision of the product matter: condensation through a chamber can give a powder rather than an adherent crystalline crust. This physical part of calomel preparation is described in the 1843 primary account in Polytechnisches Journal volume 87 (1843); the account's historical therapeutic claims are not adopted here.^[ref]

For corrosive sublimate, heat dry mercury(II) sulfate with sodium chloride and collect the mercury(II) chloride in the cooler part of a suitable sublimation apparatus:

HgS O_{4} + 2NaCl ⟶ HgC l_{2} + N a_{2} S O_{4} .

(4)

Here no metallic Mercury is added. Nonvolatile sodium sulfate stays behind, while mercury(II) chloride is transferred and condensed. Further sublimation or recrystallization separates impurities. ^[ref] The crucial distinction from calomel preparation is whether Mercury remains in the +2 oxidation state or is converted to +1.

For the solid called arsenious acid, roast an arsenic-bearing ore in air and collect the volatile arsenic trioxide in cooler chambers; purify by resublimation. The essential oxidation is represented by

4As + 3 O_{2} ⟶ 2A s_{2} O_{3} .

(5)

Actual ores also contain other chemical elements, so their roasting equations and gaseous by-products depend on composition. A contemporary 1860 report describes roasting arseniferous ores, collecting the volatile arsenic trioxide in chambers and refining it by another sublimation. It also demonstrates that apparently good material could contain antimony(III) oxide: physical appearance alone does not establish purity.^[ref] The product intended here is dry

A s_{2} O_{3}

; dissolving it does not make the bottled solid a pure crystalline hydrate

H_{3} As O_{3}

For tartar emetic, react antimony(III) oxide with potassium hydrogen tartrate in water. A small amount of water first gives intimate contact in a paste; subsequent heating with more water completes formation of the soluble potassium antimony tartrate. Filter away unreacted insoluble matter and cool the clear filtrate to crystallize the product. This sequence is directly described in a January 1864 discussion of the newly issued British Pharmacopoeia, rather than inferred from its later editions.^[ref] In the traditional antimonyl notation the underlying reaction, before assigning crystal water of crystallization, is

S b_{2} O_{3} + 2KH C_{4} H_{4} O_{6} ⟶ 2K (SbO) C_{4} H_{4} O_{6} + H_{2} O .

(6)

This notation expresses the overall composition; tartrate coordination complexes and water of crystallization need the qualification given in 4. The historical pharmaceutical use of tartar emetic does not make it a safe contemporary emetic.^[ref]

For potassium chlorate, pass chlorine into hot concentrated potassium hydroxide. The disproportionation gives potassium chloride together with potassium chlorate:

3C l_{2} + 6KOH ⟶ KCl O_{3} + 5KCl + 3 H_{2} O .

(7)

Concentration and cooling permit fractional crystallization of the much less soluble cold-water potassium chlorate, leaving much of the potassium chloride in the mother liquor; recrystallization improves purity. Cold dilute alkali instead favours hypochlorite, so the heating condition is essential. An economical contemporary manufacturing alternative uses hot calcium hydroxide and potassium chloride; the overall accounting is

6C l_{2} + 6Ca (OH)_{2} + 2KCl ⟶ 2KCl O_{3} + 6CaC l_{2} + 6 H_{2} O .

(8)

The 1850 account in Polytechnisches Journal volume 118 (1850) describes a steam-heated stirred vessel and recovery by cooling, rather than an electrical manufacturing process introduced retrospectively.^[ref]

6

Original question.

What are the constituents of cows' milk, and how may they be separated ?

Solution

Cow's milk contains water, dispersed fat, casein, soluble milk proteins, lactose and mineral constituents, with smaller amounts of other substances. These constituents occupy different physical states, which makes their separation possible: the fat is present as an emulsion, much of the casein as colloidal casein micelles, and the lactose and many salts in chemical solution. Vitamins, enzymes and small nitrogenous constituents accompany these main fractions. The proportions vary between samples and are not fixed by the name cow's milk.^[ref]

First let fresh unhomogenized milk stand so that the less dense fat globules rise as cream; skimming gives a fat-enriched layer and skimmed milk beneath. This separates fractions rather than pure chemical substances: cream still contains water and other constituents. Churning cream coalesces the fat globules into butter; extraction of a separately prepared dried portion with a suitable organic solvent provides another way to isolate fat. A modern centrifugal separator accelerates the same density-based separation, but it is not needed to explain an 1864 bench separation.

Next add dilute acetic acid carefully to skimmed milk. Lowering its pH toward the isoelectric point of casein, about 4.6, produces a curd that can be collected by filtration and washed. Excess acid, retained fat, and shifts of calcium phosphate between curd and liquid limit purity. Rennet offers another route to a curd, but its enzyme-catalysed action and the mineral content of the resulting curd differ from simple acid precipitation; the two should not be treated as identical mechanisms.^[ref]

The liquid after removal of the curd is whey. Heat this fraction under suitable acidity to induce protein denaturation and aggregate much of its residual whey protein; collect the coagulum by filtration. Older descriptions commonly grouped this fraction under “albumin”, but modern milk chemistry distinguishes several proteins. Heating alone does not make every minor constituent fall into a single pure fraction.^[ref]

Concentrate the deproteinized whey and cool it to crystallize lactose. Filter and wash the crystals, then purify them by recrystallization; dissolved salts and residual organic matter remain substantially in the mother liquor. This provides a separation from most protein, not an automatic complete separation from all minerals.^[ref]

Finally use separate measured portions for water and mineral determinations. Controlled drying gives an operational total solids measurement; subtraction from the initial mass estimates the water lost. Ignition of a dried portion destroys most organic matter and leaves ash for mineral analysis. Ash is a transformed residue: it does not preserve all original salts unchanged, and some constituents can volatilize. Calcium, phosphorus, sodium, potassium and chloride are among the important mineral constituents. Skimming, curd formation, heat coagulation, lactose crystallization, drying and ashing separate or determine the principal fractions; none alone constitutes a complete pure-component analysis.

7

Original question.

How may Glucose be prepared and detected ?

Solution

Glucose can be obtained by hydrolysis of starch and detected as a reducing sugar by a heated alkaline copper(II) test. A positive copper test alone does not uniquely identify glucose.

Suspend starch in water, warm it to make a paste, and heat with a dilute acid, such as dilute sulfuric acid, to cause hydrolysis of the glycosidic bonds. The chemical reaction passes through shorter carbohydrates; sufficient hydrolysis ultimately gives glucose. The usual repeat-unit accounting is

(C_{6} H_{10} O_{5})_{n} + n H_{2} O ⟶ n C_{6} H_{12} O_{6} .

This idealized polymer formula omits end-group details. The acid catalyses cleavage and is not consumed in the displayed equation. Merely using amylase from malt should not be described as necessarily producing pure glucose; it can leave maltose and shorter polysaccharides.

When sulfuric acid is used, neutralization with calcium carbonate removes the acidity and much of the sulfate as calcium sulfate. Filter out the insoluble residue, then concentrate the chemical solution gently and crystallize the glucose; further recrystallization is required to distinguish a reasonably pure chemical substance from a crude syrup. Overheating and unnecessarily prolonged strong acid treatment degrade the sugar. Hydrolysis of sucrose is a less selective preparative answer because it gives fructose alongside glucose.

For the historical copper test, use freshly mixed Fehling's reagent: an aqueous copper(II) sulfate portion and an alkaline tartrate portion. The tartrate keeps copper(II) in coordination complexes instead of allowing immediate bulk precipitation of copper(II) hydroxide. Add a suitably prepared sample to a small test portion and warm in a hot water bath. A yellow-orange to brick-red precipitate of copper(I) oxide indicates reduction of the copper(II) reagent. Run a known glucose sample and a water blank alongside it. In a simple sample, a visible difference between these controls helps distinguish a true positive from a deteriorated reagent or contamination.^[ref]

The key redox accounting on the copper side is

2C u^{2 +} + 2O H^{-} + 2 e^{-} ⟶ C u_{2} O + H_{2} O .

The required electrons come from oxidation of the sugar. This half-equation deliberately omits the changing tartrate ligands and does not assign a unique organic product. The often printed claim that hot alkaline glucose simply becomes gluconate in this test is too simple: competing isomerization, oxidation and fragmentation make the organic chemistry more complicated.^[ref]

Fructose, lactose, maltose and various other reducing substances can also give a positive Fehling's test. In particular, fructose can form aldoses by tautomerization in alkaline solution; an initial ketone functional group does not make it reliably negative.^[ref] Fermentation by suitable yeast, with carbon dioxide formation, and optical rotation can supply further historical evidence, but neither is individually specific to glucose. These are anticipated diagnostic observations, not measurements supplied by the paper. In present laboratory practice an appropriately validated enzymatic glucose assay can establish glucose more selectively; the old reducing-sugar test is not by itself a diagnosis of diabetes.^[ref]

8

Original question.

Give the ratio which equal volumes of the following gases bear to the volumes of the elements of which they consist :—Hydrochloric Acid, Water, Protoxide of Nitrogen, Binoxide of Nitrogen, Ammonia.

Solution