### Alcohol Calculations

(1). To convert percentage of alcohol by volume into percentage by weight.

Multiply the volume percent age by the specific gravity of absolute alcohol (0 7936), and divide the product by the specific gravity of the liquid in question.

For let S be the specific gravity of the liquid (at 15.6°/l5.6o). Then 100 c.c. weigh 100 Sρ grams, where ρ is the sp. gr. of water at 15.6°/4°.

If V be the percentage of alcohol by volume, then 100 c.c. of the liquid contain V c.c. of absolute alcohol.

That is, 100 Sρ grams contain V c.c.;

.'. 100 grams contain V/Sρ c.c. abs. ale.

But 1 c.c. of abs. ale. weighs 0 7936 ρ gram,

.-. V/Sρ c.c. weigh V/Sρ X 0.7936 ρ gram; = V/S X 0.7936 gram.

That is, 100 grams of the liquid contain V X0.7936/S gram of alcohol, i.e., percentage by weight = Percent. by volume x 0.7936 /.Sp. gr. of the liquid

(2). To convert percentage by volume into grams per 100 c.c.

Multiply by 0.79284. For if 100 c.c. of the liquid contain V c.c. of absolute alcohol, the weight of the V c.c. is

V X 0.7936 x ρ grams; = V X 0 7936 x 0.999037 = V X 0.79284.

(3). To convert percentage of alcohol by volume into percentage of proof spirit by volume,

Multiply by 1 7535. For absolute alcohol is 75.35 "over proof " - that is, 100 volumes of absolute alcohol contain the same quantity of alcohol as do 175.35 volumes of proof spirit. Therefore 1 vol. of alcohol = 1.7535 vols. of proof spirit.

Or for summary calculations, multiply by 7/4 (= 1.75).

From the foregoing three examples it will be seen how any one denomination can be expressed in terms of any other. Thus from (3) a quantity of alcohol in terms of proof spirit can be expressed in terms of absolute alcohol by volume on dividing by 1.7535, and then either as percentage of absolute alcohol by weight, or as grams of absolute alcohol per 100 c.c, as shown in (1) and (2) respectively.

The various relations can be summarised for reference in the following "conversion equations."

Let S denote the sp. gr. of a specimen of alcohol, P the percentage of proof spirit by volume,

V the percentage of alcohol by volume, W „ „ „ weight, and G the grams per 100 c.c.

Then P = 1.7535 V,

= 2.2095 WS. V = 0.5703 P, = 1.2601 WS:

W= 1/2.2095 x P/S

= 0.7936 X V/S and G = 0.7928 F,

= 0.4521 P.

An operation frequently required is the reduction of alcohol from a higher to a lower strength.

In the laboratory, with convenient measuring vessels at hand, the operation is simple. The volumes are inversely as the strengths. We therefore take a convenient quantity of the alcohol, measured at the standard temperature, and make it up with water to such a volume, at the same temperature, that the ratio of this volume to the first shall be the ratio of the given strength to the required strength.

Example: Given strength 90 per cent., required strength 20 per cen. Dilute 20 c.c. to 90 c.c, or 222 to 100, or 1111 to 500, as may be convenient. Since the temperature of the liquid rises during the mixing, for precise work it must be adjusted before completing the volume.

It is to be carefully noted that on account of the contraction which occurs the required strength would not be given by adding 70 c.c. of water to 20 c.c. of the alcohol. More than 70 would be required.

For this reason, the calculation is less simple when, as in large operations, no suitable vessels may be available for accurately making up the diluted spirit to a required volume at a particular temperature. It is then necessary to calculate the actual quantity of water which must be added. This may be done as follows.

To find the volume of water which must be added to a given volume V1 of alcohol, in order to reduce it from a given strength S1 to a lower strength S2, the densities D1 and D2 respectively corresponding with these strengths, being known.

Let x denote the weight in grams of the water required, and V2 the resulting volume, in c.c, of the diluted spirit. Then the weight of the given volume of alcohol is V1D1 and thai of the resulting volume is V2D2.

But the latter weight = the former weight + x,

.-. V1D1 + x = V2D2,and x = V2D2 - V1D1 . (1)

Also, since the strengths are inversely as the volumes,

V2 / V1 = S1 / S2; OR V2 = V1 S1 / S2 . . . . (2)

Substituting from (2) in (1) we get: x = V1/S2 (D2S1 - D1S2) ..... (3) which gives the weight of water required, in terms of the known quantities, and expressed in grams.

For ordinary work this may be taken as the required volume of water, in c.c. The precise volume will of course depend upon the temperature of the water. At 4°, x grams = x c.c.; at 156°, x grams = x x 1.0009 c.c.

Example (I). How much water at 156° must be added to 100 c.c. of 90 per cent. alcohol in order to reduce its strength to 60 per cent. ? (Strengths by volume.)

Here V1 = 100, S1 = 90, and S2 = 60. With sufficient accuracy the values of D1 and D2 may be taken from the ordinary alcohol tables: D1 = 0.8337, and D2 = 0.9134.

.-. x = 100/60 (0.9134 x 90 - 0.8337 X 60) x 1.0009 = 53.69 c.c, or practically, 53.7 c.c.

Strictly, however, the values of D1 and D2 as taken from the ordinary alcohol tables are not the true densities (mass of unit volume), but the specific gravities at 15.6°, referred to water at that temperature as unity. Since the density of water at that temperature is 0.999037 (Despretz), and not 1, the values of D1 and D2 should be corrected accordingly. If we therefore multiply these values by 0.999037, we find the true densities D1 and D2 to be 0 8329 and 0 9125 respectively, and the corrected result is x = 53.64 c.c.

But it is to be noted that we get the same result by simply taking the values of D1 and D2 from the ordinary alcohol tables and using them in equation (3), omitting the factor 1.0009: x = 100/60 (09134 x 90 - 08337 x 60), = 53 64.

The fact is that the reciprocal of 0.999037 is 1/1.0009; and this cancels out the factor 1.0009 used in the first calculation. So that, finally, although x in equation (3) denotes the weight of water required, if we take the values of D1 and D2 as specific gravities from the ordinary alcohol tables the result expresses the required volume of water, in c.c.

This example has been elaborated a little, because the point in question is sometimes found puzzling by persons unfamiliar with alcohol calculations.

Example (2). How much water is required in order to reduce 100 gallons of spirit at 60 over proof to a strength of 20 over proof ?

From what precedes, it will be seen that equation (3) will give the answer, x, in gallons, if the specific gravities corresponding with the strengths are taken from the ordinary alcohol tables.

Here

V1 = 100, S1 = 160, S2 = 120; D1 = 0.8295, and D2 = 0.8936. Hence x = 100/120 (0.8936 x 160 - 0.8295 x 120) = 36.2 gallons.

Problems of a slightly different character are set in the next two questions.

(1). What weight of water must be added to 100 grams of an alcohol (A) of given strength (percentage by weight) in order to produce an alcohol (B) of given lower strength ?

Let a and b be the respective given strengths (percentages by weight), and let x be the weight in grams of the water required.

Then the total water present is 100 - a + x, and the weight of B produced is 100 + x.

.'. in 100 grams of B there are - 100/100+x (100 - a + x) grams of water.

But the weight of water in 100 grams of B is also 100 - b. Hence, equating,

100/100+x (100 - a + x) = 100 - b.

Solving this equation, we get x = 100/b (a - b) . . . . . . (l).

Thus if A is alcohol of 90 per cent. strength by weight, and we require to dilute it to 70 per cent., the weight of water to be added to 100 grams of A is: -

100/70 (90 - 70) = 28 4/74 grams.

(2). If in the foregoing example we have 100 c.c. of A instead of 100 grams, what is the quantity of water required ?

Let D be the density of the alcohol A. Then 100 c.c. will weigh 100 D grams. Hence the quantity of water to be added is 100 D/100 x100/b (a - b) grams == 100 (a - b) D/b - grams . . . (ii).

Here, as explained above, the value of D, if taken from the ordinary alcohol tables, is not the true density (mass of unit volume), but the specific gravity referred to water at 15.6°. Hence 100 c.c. do not weigh exactly 100 D grams if these tables are used. In this case, as in that explained above, the quantity of water given by the expression (ii) must be taken as the volume in c.c, not the weight in grams. The correction for reducing the value of D to true density, and that for converting grams of water into c.c, cancel each other out.

Multiply the volume percent age by the specific gravity of absolute alcohol (0 7936), and divide the product by the specific gravity of the liquid in question.

For let S be the specific gravity of the liquid (at 15.6°/l5.6o). Then 100 c.c. weigh 100 Sρ grams, where ρ is the sp. gr. of water at 15.6°/4°.

If V be the percentage of alcohol by volume, then 100 c.c. of the liquid contain V c.c. of absolute alcohol.

That is, 100 Sρ grams contain V c.c.;

.'. 100 grams contain V/Sρ c.c. abs. ale.

But 1 c.c. of abs. ale. weighs 0 7936 ρ gram,

.-. V/Sρ c.c. weigh V/Sρ X 0.7936 ρ gram; = V/S X 0.7936 gram.

That is, 100 grams of the liquid contain V X0.7936/S gram of alcohol, i.e., percentage by weight = Percent. by volume x 0.7936 /.Sp. gr. of the liquid

(2). To convert percentage by volume into grams per 100 c.c.

Multiply by 0.79284. For if 100 c.c. of the liquid contain V c.c. of absolute alcohol, the weight of the V c.c. is

V X 0.7936 x ρ grams; = V X 0 7936 x 0.999037 = V X 0.79284.

(3). To convert percentage of alcohol by volume into percentage of proof spirit by volume,

Multiply by 1 7535. For absolute alcohol is 75.35 "over proof " - that is, 100 volumes of absolute alcohol contain the same quantity of alcohol as do 175.35 volumes of proof spirit. Therefore 1 vol. of alcohol = 1.7535 vols. of proof spirit.

Or for summary calculations, multiply by 7/4 (= 1.75).

From the foregoing three examples it will be seen how any one denomination can be expressed in terms of any other. Thus from (3) a quantity of alcohol in terms of proof spirit can be expressed in terms of absolute alcohol by volume on dividing by 1.7535, and then either as percentage of absolute alcohol by weight, or as grams of absolute alcohol per 100 c.c, as shown in (1) and (2) respectively.

The various relations can be summarised for reference in the following "conversion equations."

Let S denote the sp. gr. of a specimen of alcohol, P the percentage of proof spirit by volume,

V the percentage of alcohol by volume, W „ „ „ weight, and G the grams per 100 c.c.

Then P = 1.7535 V,

= 2.2095 WS. V = 0.5703 P, = 1.2601 WS:

W= 1/2.2095 x P/S

= 0.7936 X V/S and G = 0.7928 F,

= 0.4521 P.

An operation frequently required is the reduction of alcohol from a higher to a lower strength.

In the laboratory, with convenient measuring vessels at hand, the operation is simple. The volumes are inversely as the strengths. We therefore take a convenient quantity of the alcohol, measured at the standard temperature, and make it up with water to such a volume, at the same temperature, that the ratio of this volume to the first shall be the ratio of the given strength to the required strength.

Example: Given strength 90 per cent., required strength 20 per cen. Dilute 20 c.c. to 90 c.c, or 222 to 100, or 1111 to 500, as may be convenient. Since the temperature of the liquid rises during the mixing, for precise work it must be adjusted before completing the volume.

It is to be carefully noted that on account of the contraction which occurs the required strength would not be given by adding 70 c.c. of water to 20 c.c. of the alcohol. More than 70 would be required.

For this reason, the calculation is less simple when, as in large operations, no suitable vessels may be available for accurately making up the diluted spirit to a required volume at a particular temperature. It is then necessary to calculate the actual quantity of water which must be added. This may be done as follows.

To find the volume of water which must be added to a given volume V1 of alcohol, in order to reduce it from a given strength S1 to a lower strength S2, the densities D1 and D2 respectively corresponding with these strengths, being known.

Let x denote the weight in grams of the water required, and V2 the resulting volume, in c.c, of the diluted spirit. Then the weight of the given volume of alcohol is V1D1 and thai of the resulting volume is V2D2.

But the latter weight = the former weight + x,

.-. V1D1 + x = V2D2,and x = V2D2 - V1D1 . (1)

Also, since the strengths are inversely as the volumes,

V2 / V1 = S1 / S2; OR V2 = V1 S1 / S2 . . . . (2)

Substituting from (2) in (1) we get: x = V1/S2 (D2S1 - D1S2) ..... (3) which gives the weight of water required, in terms of the known quantities, and expressed in grams.

For ordinary work this may be taken as the required volume of water, in c.c. The precise volume will of course depend upon the temperature of the water. At 4°, x grams = x c.c.; at 156°, x grams = x x 1.0009 c.c.

Example (I). How much water at 156° must be added to 100 c.c. of 90 per cent. alcohol in order to reduce its strength to 60 per cent. ? (Strengths by volume.)

Here V1 = 100, S1 = 90, and S2 = 60. With sufficient accuracy the values of D1 and D2 may be taken from the ordinary alcohol tables: D1 = 0.8337, and D2 = 0.9134.

.-. x = 100/60 (0.9134 x 90 - 0.8337 X 60) x 1.0009 = 53.69 c.c, or practically, 53.7 c.c.

Strictly, however, the values of D1 and D2 as taken from the ordinary alcohol tables are not the true densities (mass of unit volume), but the specific gravities at 15.6°, referred to water at that temperature as unity. Since the density of water at that temperature is 0.999037 (Despretz), and not 1, the values of D1 and D2 should be corrected accordingly. If we therefore multiply these values by 0.999037, we find the true densities D1 and D2 to be 0 8329 and 0 9125 respectively, and the corrected result is x = 53.64 c.c.

But it is to be noted that we get the same result by simply taking the values of D1 and D2 from the ordinary alcohol tables and using them in equation (3), omitting the factor 1.0009: x = 100/60 (09134 x 90 - 08337 x 60), = 53 64.

The fact is that the reciprocal of 0.999037 is 1/1.0009; and this cancels out the factor 1.0009 used in the first calculation. So that, finally, although x in equation (3) denotes the weight of water required, if we take the values of D1 and D2 as specific gravities from the ordinary alcohol tables the result expresses the required volume of water, in c.c.

This example has been elaborated a little, because the point in question is sometimes found puzzling by persons unfamiliar with alcohol calculations.

Example (2). How much water is required in order to reduce 100 gallons of spirit at 60 over proof to a strength of 20 over proof ?

From what precedes, it will be seen that equation (3) will give the answer, x, in gallons, if the specific gravities corresponding with the strengths are taken from the ordinary alcohol tables.

Here

V1 = 100, S1 = 160, S2 = 120; D1 = 0.8295, and D2 = 0.8936. Hence x = 100/120 (0.8936 x 160 - 0.8295 x 120) = 36.2 gallons.

Problems of a slightly different character are set in the next two questions.

(1). What weight of water must be added to 100 grams of an alcohol (A) of given strength (percentage by weight) in order to produce an alcohol (B) of given lower strength ?

Let a and b be the respective given strengths (percentages by weight), and let x be the weight in grams of the water required.

Then the total water present is 100 - a + x, and the weight of B produced is 100 + x.

.'. in 100 grams of B there are - 100/100+x (100 - a + x) grams of water.

But the weight of water in 100 grams of B is also 100 - b. Hence, equating,

100/100+x (100 - a + x) = 100 - b.

Solving this equation, we get x = 100/b (a - b) . . . . . . (l).

Thus if A is alcohol of 90 per cent. strength by weight, and we require to dilute it to 70 per cent., the weight of water to be added to 100 grams of A is: -

100/70 (90 - 70) = 28 4/74 grams.

(2). If in the foregoing example we have 100 c.c. of A instead of 100 grams, what is the quantity of water required ?

Let D be the density of the alcohol A. Then 100 c.c. will weigh 100 D grams. Hence the quantity of water to be added is 100 D/100 x100/b (a - b) grams == 100 (a - b) D/b - grams . . . (ii).

Here, as explained above, the value of D, if taken from the ordinary alcohol tables, is not the true density (mass of unit volume), but the specific gravity referred to water at 15.6°. Hence 100 c.c. do not weigh exactly 100 D grams if these tables are used. In this case, as in that explained above, the quantity of water given by the expression (ii) must be taken as the volume in c.c, not the weight in grams. The correction for reducing the value of D to true density, and that for converting grams of water into c.c, cancel each other out.

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