# Mix cross

The mixing cross (also called St. Andrew's cross ) is a clear method to calculate the proportions of two components for a mixture. It is used in chemistry to calculate concentrations and proportions in liquids or mixtures of solid components. This occurs when mixing dissolved substances (e.g. acids , salts or alkalis ) with different starting concentrations. The mixing cross is an application of the law of conservation of mass or the conservation of the amount of substance . The calculations using the mixing cross therefore only work with masses or amounts of substance. If you want to calculate with volumes, you first have to convert the individual volumes into a mass using the density . The result is a mass. This can be converted back into a volume using the density (or using a percentage calculation) (density = mass / volume in [g / ml] or [kg / l]).

The mixing cross is also used to calculate the proportions of solid substances (e.g. flour, pastries) that must be mixed into a desired mixture, or for mixed calculations in a commercial context. The analogous procedure for determining mixing temperatures is called Richmann's mixing rule .

## principle

The mixing cross is a method that can be used to calculate the mass fractions that are required to produce a solution with a specific target concentration from two stock solutions, ie solutions with known concentrations. Since the amount of substance of a dissolved substance remains constant when diluted , the product of concentration c and volume V (as a definition of the amount of substance) of a dissolved substance remains constant - provided that the concentration of the dissolved substance in the diluent is zero :

${\ displaystyle c_ {1} V_ {1} = c_ {2} V_ {2}}$

The index 1 denotes the initial state, the index 2 the final state. If the substance under consideration is present in both solutions A and B, then the following applies

${\ displaystyle c_ {A1} V_ {A1} + c_ {B1} V_ {B1} = c_ {2} V_ {2}}$

with the total volume

${\ displaystyle V_ {2} = V_ {A1} + V_ {B1}}$.

Insertion and dissolution results according to the volume ratio${\ displaystyle {\ frac {V_ {A1}} {V_ {B1}}}}$

${\ displaystyle {\ frac {V_ {A1}} {V_ {B1}}} = {\ frac {c_ {2} -c_ {B1}} {c_ {A1} -c_ {2}}}}$.

## Scheme

Mixture cross scheme using the example of a wheat mixture

To put it simply, there is a “profit type” and a “loss type” in each mix cross compared to the desired mix - they are on the left. The desired mixture is always in the middle. The aim of the calculation is to determine the mass proportions (they are calculated on the right-hand side) of the two mixing partners with which the profit and loss can be balanced out compared to the mixture. Since the mass fractions are inversely proportional to profit and loss, the “cross-calculation” is shown schematically.

Example: Two lots of wheat are to be mixed in such a way that one quintone of the mixture can be sold for 49 €. The selling price for type A is € 52 / dt, that for type B is € 45 / dt. a) In what proportions do the two types have to be mixed? b) How many dt must be taken from each variety if a total of 24 dt mixture is required? Solution a) The mixing ratio is 4: 3. The mixture is 7 parts. Solution b) 7 parts correspond to 24 German. 4 parts of type A are therefore = 13.71 German. 3 parts of type B are = 10.29 German. Both parts add up to 24 German. ${\ displaystyle {\ tfrac {24} {7}} \ cdot 4}$${\ displaystyle {\ tfrac {24} {7}} \ cdot 3}$

## Applications

### Mixing liquids

Mix cross

The known starting concentrations of the liquids are entered on the left side of the mixing cross. The desired target concentration of the mixture is written at the crossing point. Now form the difference between the known concentration at the top left and the desired target concentration in the middle and note the result at the bottom right. Then form the difference between the known concentration at the bottom left and the desired target concentration in the middle and write the result down at the top right. Negative results are noted without a sign ( amount calculation ). On the right-hand side of the mixing cross, the result is the “proportions of the total mass” (not the volume!) With which the desired target concentration can be achieved.

Example
calculation
1 (mixing with pure water): A 35% acid should be mixed with water in such a way that a target solution of 6% acid results. How much water and how much acid do you need?

The starting concentrations on the left are 35% for the acid and 0% for the water, in the middle is the desired target concentration, in this case 6%

 ${\ displaystyle {\ begin {array} {ccccc} {\ text {35}} & {} & {} & {} & \ vert 0-6 \ vert = 6 \\ {} & \ diagdown & {} & \ nearrow & {} \\ {} & {} & {\ text {6}} & {} & {} \\ {} & \ diagup & {} & \ searrow & {} \\ {\ text {0}} & {} & {} & {} & \ vert 35-6 \ vert = 29 \ end {array}}}$
• 35 - 6 make 29 parts,
• 0 - 6 result in 6 parts, (sign is omitted)

There are 35 parts in total. It therefore takes 6 parts of the 35 percent acid and 29 parts of water to produce a 6 percent acid.

If 1000 g of a 6 percent target solution are to be produced, you therefore need:

• 35 percent acidity: (1000 g / 35) * 6 = 171 g
• Water: (1000 g / 35) * 29 = 829 g

Example calculation 2 (mixing with a second acid mixture):
Instead of 0% (for the concentration of water) there could also be a value for a 15% acid on the left:

 ${\ displaystyle {\ begin {array} {ccccc} {\ text {35}} & {} & {} & {} & \ vert 15-22 \ vert = 7 \\ {} & \ diagdown & {} & \ nearrow & {} \\ {} & {} & {\ text {22}} & {} & {} \\ {} & \ diagup & {} & \ searrow & {} \\ {\ text {15}} & {} & {} & {} & \ vert 35-22 \ vert = 13 \ end {array}}}$

At a target concentration of 22%, then

• 22 - 15 = 7 parts of 35 percent acid and
• 35-22 = 13 parts 15 percent acid

be mixed. So there are a total of 20 parts.

If 1000 g of the 22 percent target solution are to be produced, you therefore need:

• 35 percent acidity: (1000 g / 20) * 7 = 350 g
• 15 percent acidity: (1000 g / 20) * 13 = 650 g

### Alloys

The mixing cross is also suitable for the approximate calculation of the mass fractions in alloys of metals, e.g. B. the proportions of zinc and copper in a brass alloy. Because of the crystal lattice structure of metals, the calculation with the mixture cross only gives approximate values. The formulas for the exact calculation can be found in the article Amount of substance .

Example calculation :

Für die Dichte einer Messinglegierung wurde durch Wägen und Volumenberechnung der
Wert 8,32 g/cm³ ermittelt.
Reines Zink besitzt nach Tabelle eine Dichte von 6,97 g/cm³ und Kupfer eine Dichte von 8,61 g/cm³.
Auf der linken Seite des Mischungskreuzes setzt man die „Ausgangskonzentrationen“
6,97 (für reines Zink) und 8,61 (für reines Kupfer) ein.
In die Mitte setzt man den Mischungswert 8,32 für Messing als Zielzahl ein.
Nun wird diagonal subtrahiert:
Subtrahiert man 8,32 von 8,61 ergibt sich 0,29 -- sind 29 Teile Zink
Subtrahiert man 8,32 von 6,97 ergibt sich 1,35 -- sind 135 Teile Kupfer.
29 Teile + 135 Teile = 164 Teile = Gesamtmasse = 100 %
29 Teile entsprechen somit 17,7 % (=Zink). 135 Teile entsprechen 82,3 % (= Kupfer)
Die vorhandene Messinglegierung besteht demnach aus ca. 18 % Zink und 82 % Kupfer.


### Mixed calculation

Mix cross

The mixing cross is also suitable for calculating mixing ratios in a commercial context.

Example calculation : Tea type 1 costs 2.60 euros per 100 g, tea type 2 costs 3.70 euros per 100 g. Calculate a mixing ratio for a tea mixture of the price 3.40 euros per 100 g.

• If you subtract 2.60 from 3.40 you get 0.80 - 8 parts of tea are 2
• If you subtract 3.40 from 3.70 you get 0.30 - 3 parts of tea are 1

8 parts + 3 parts are 11 parts. For example, 800 g of tea type 2 and 300 g of tea type 1 can be mixed into 1.1 kg of tea mixture at a price of EUR 3.40 per 100 g.

8 parts thus correspond to approx. 73% tea type 2 ; 3 parts correspond to approx. 27% tea type 1 .

## literature

• Martin Holtzhauer: Biochemical laboratory methods . 3. Edition. Springer, Berlin / Heidelberg 1997, ISBN 978-3-540-62435-6 , pp. 288 f . ( limited preview in Google Book search).
• Reiner Friebe, Karl Rauscher: Chemical tables and calculation tables for analytical practice . 11th edition. Verlag Harry Deutsch, Frankfurt am Main 2000, ISBN 978-3-8171-1621-8 .