Using Alligation Alternate to Solve Composition Problems
Anatol Mancot!
Queensborough Community College of the City University of New York Boyside, New York 11364
Problems involving the composition of mixtures may be solved by utilizing the relatively obscure method of "alligation alternate" in lieu of the standard algebraic procedure with no loss in accuracy. The definition and mathematical validity of the alligation alternate method presented by Bradley, et al.' is as follows. Alligation alternate is a method by which we may calculate the number of parts of two or more components of a given strength when they are to be mixed to prepare a mixture of desired strength. A final proportion permits us to translate relative parts to any specific denomination. If x parts of a higher value component (A) are to be mixed with y parts of a lower value component (B) to produce an intermediate value of x y parts (C), the ratio of x to y is derived by alligation alternate in the following manner.
+
Higher value component
Proportional Given Values Desired Value Parts Reauired A
.
z - = -C - B y A-C
resulting in the identical answer as that of the method of alligation alternate. Let us examine the following composition problem, solved first by the standard algebraic procedure and then by the method of alligation alternate. Problem I
There are only two naturally occurring lithium isotopes. These are 'Li with a nuclidic mass of 7.0160 and %i with a nuclidic mass of 6.0151. Find the percent distributions of these two isotopes which account for lithium having an atomic weight of 6.9398. Solution by Standard Algebraic Method
Let X = percent distribution of =Li; then (100 - X) is the percent distribution of 'Li 6.0151X
+ 7.0160(100 - X)= 6,9398 100
Lower value component
B
Set up the given and desired values as shown in the above diagram. Then proceed as follows: A (1) minus C gives y; B (2) from C gives x; the ratio of A to B is x parts of A as shown by arrow (3) to y parts of B as shown by arrow (4). Therefore -B 2=C y A-C
Solution by Alligation Alternate
Nuclidic Masses Desired Value Ratio of Atomic Weight 'Li to 'Li Given Values Higher nuclidic 7.0160 (3)-0.9247 msss ('Li)
The algebraic solution is as follows BRADLEY, W. T., GUSTAFSON, C. B., AND STOKLOSA, M. J., "Phmmeceutical C~lculations"(4th ed.), Lea & Febiger, Philadelphia, Pa., 1963, pp. 165-6.
2.936
Lower nuclidic mass ('Li)
6.0151
1.0009 (5)
Volume 49, Number I , January 1972
/
ST
Set up the given and desired values as shown in the diagram. Then proceed as follows: 7.0160 (1) minus 6.9398 gives 0.0762; 6.0151 (2) from 6.9398 gives 0.9247; the ratio of 'Li to OLi is 0.9247 parts of 'Li as shown by arrow (3) to 0.0762 parts of 6Lias shown by arrow (4); the total value of 1.0009 (5) represents 100% of the components making up the atomic weight of 6.9398; therefore, the percent distribution of ?Li and sLi are
The following problems will be solved only by alligation alternate. It is optional as to whether the higher or lower component is presented first, as problem I1 will show.
Solution by Alligation Alternate (Simplified Format)
2.017
& : 2,000
\
3.864
,4.500A 8.364
A 2 . 4 8 3
'nv
X 2,000 = 1,217.6 rnl of the 2.017 M HCI
- 1,217.6 = 782.4 ml of the 8.364 M KC1
Problem IV
The density of mercury = 13.6 g/ml, of silver = 10.5 g/ml. How many grams of mercury and silver would be needed to produce an alloy weighing 120 g and having a density of 12.4 g/ml? Solution by Alligotion Alternate (Simplified Format)
Problem II
Boron is composed of two naturally occurring isotopes, (l0B and llB) with nuclidic masses 10.0129 and 11.0093. Find the ratio of I% to llB that would account for boron having an atomic weight of 10.811.
The densities as shown above must be converted to grams
Solution by Alligotion Alternate
Y
Nuclidic Masses Desired Value Given Values Atomic Weight Lmuer nuclidic mass (IOB)
10 0129
Ratio of ' 0 6 to "B
\/ (3)-0.1983
11.0093
0.9964 (5)
Set up the given and desired values as shown in the diagram. Then proceed as follows: 10.0129 (1) from 10.811 gives 0.7981; 11.0093 (2) minus 10.811 gives 0.1983; the ratio of l0B to "B is 0.1983 parts of O 'B as shown by arrow (3) to 0.7981 parts of "B as shown by arrow (4). Therefore
Problem Ill
Two solutions of HCI have been assayed and found to be 2.017 M and 8.364 M. How many ml of each solution would be mixed to prepare 2,000 ml of a 4.500 M solution of HCI?
58
/
Journal of Chemical Education
Y
=
38.4 g of alloy, therefore
E 8X 120 = 80.6 g of mercury in the 120-g dloy 38.4 120 - 80.6 = 39.4 g of silver in the 120-g aUoy Problem V
10.811
Higher uuclidic mass ("B)
12.4 = 3.1
The ionization potentials of the elements lithium and rubidium are respectively 124 and 96 kcal/mole. A mixture of lithium and rubidium containing 6.02 X lozaatoms required 102 kcal for complete ionization. How many atoms of lithium were present in the mixture? Solution by Alligdion Alternote (Simplified Format)
6 28
X 6.02 X 10" = 1.29 X 10' atoms of Li
As has been shown, the solution to composition problems involves the simplest of arithmetical operations to arrive at the correct answer when using the method of alligation alternate.
