Water and chalk is a way to separate. Methods for separating heterogeneous mixtures
theoretical block.
The concept of "mixture" was defined in the 17th century. English scientist Robert Boyle: "A mixture is an integral system consisting of heterogeneous components."
Comparative characteristics of a mixture and a pure substance
Signs of comparison | pure substance | Mixture |
Constant | fickle |
|
Substances | Same | Various |
Physical Properties | Permanent | Fickle |
Energy change during formation | going on | Not happening |
Separation | Through chemical reactions | Physical methods |
Mixtures differ from each other in appearance.
The classification of mixtures is shown in the table:
Here are examples of suspensions (river sand + water), emulsions (vegetable oil + water) and solutions (air in a flask, salt + water, small change: aluminum + copper or nickel + copper).
Methods for separating mixtures
In nature, substances exist in the form of mixtures. For laboratory research, industrial production, for the needs of pharmacology and medicine, pure substances are needed.
Various methods of separation of mixtures are used to purify substances.
Evaporation - the separation of solids dissolved in a liquid by converting it into vapor.
Distillation- distillation, separation of substances contained in liquid mixtures according to boiling points, followed by cooling of the vapor.
In nature, water in its pure form (without salts) does not occur. Oceanic, sea, river, well and spring water are varieties of salt solutions in water. However, often people need clean water that does not contain salts (used in car engines; in chemical production to obtain various solutions and substances; in the manufacture of photographs). Such water is called distilled, and the method of obtaining it is called distillation.
Filtration is the filtering of liquids (gases) through a filter in order to purify them from solid impurities.
These methods are based on differences in the physical properties of the components of the mixture.
Consider ways to separate heterogeneousand homogeneous mixtures.
Blend example | Separation method |
Suspension - a mixture of river sand with water | settling Separation upholding based on different densities of substances. Heavier sand settles to the bottom. You can also separate the emulsion: to separate oil or vegetable oil from water. In the laboratory, this can be done using a separating funnel. Oil or vegetable oil forms the top, lighter layer. As a result of settling, dew falls out of the fog, soot is deposited from smoke, cream is settled in milk. Separation of a mixture of water and vegetable oil by settling |
A mixture of sand and table salt in water | Filtration What is the basis for the separation of heterogeneous mixtures using filtering? On various solubility of substances in water and on various sizes of particles. Only particles of substances commensurate with them pass through the pores of the filter, while larger particles are retained on the filter. So you can separate a heterogeneous mixture of table salt and river sand. Various porous substances can be used as filters: cotton wool, coal, fired clay, pressed glass, and others. The filtering method is the basis for the operation of household appliances, such as vacuum cleaners. It is used by surgeons - gauze bandages; drillers and workers of elevators - respiratory masks. With the help of a tea strainer for filtering tea leaves, Ostap Bender, the hero of the work of Ilf and Petrov, managed to take one of the chairs from Ellochka Ogre (“The Twelve Chairs”). Separation of a mixture of starch and water by filtration |
A mixture of iron powder and sulfur | Action by magnet or water Iron powder was attracted by a magnet, but sulfur powder was not. The non-wettable sulfur powder floated to the surface of the water, while the heavy wettable iron powder settled to the bottom. Separation of a mixture of sulfur and iron using a magnet and water |
A solution of salt in water is a homogeneous mixture | Evaporation or crystallization The water evaporates and salt crystals remain in the porcelain cup. When water is evaporated from lakes Elton and Baskunchak, table salt is obtained. This separation method is based on the difference in the boiling points of the solvent and the solute. If a substance, such as sugar, decomposes when heated, then the water is not completely evaporated - the solution is evaporated, and then sugar crystals are precipitated from a saturated solution. Sometimes it is required to remove impurities from solvents with a lower boiling point, for example, water from salt. In this case, the vapors of the substance must be collected and then condensed upon cooling. This method of separating a homogeneous mixture is called distillation or distillation. In special devices - distillers, distilled water is obtained, which is used for the needs of pharmacology, laboratories, and car cooling systems. At home, you can design such a distiller: If, however, a mixture of alcohol and water is separated, then the first to be distilled off (collected in a receiving test tube) is alcohol with tboil = 78 °C, and water will remain in the test tube. Distillation is used to obtain gasoline, kerosene, gas oil from oil. Separation of homogeneous mixtures |
A special method of separating components, based on their different absorption by a certain substance, is chromatography.
Using chromatography, the Russian botanist was the first to isolate chlorophyll from the green parts of plants. In industry and laboratories, instead of filter paper for chromatography, starch, coal, limestone, and aluminum oxide are used. Are substances always required with the same degree of purification?
For different purposes, substances with different degrees of purification are needed. Cooking water is sufficiently settled to remove impurities and chlorine used to disinfect it. Drinking water must first be boiled. And in chemical laboratories for the preparation of solutions and experiments, in medicine, distilled water is needed, as purified as possible from the substances dissolved in it. Highly pure substances, the content of impurities in which does not exceed one millionth of a percent, are used in electronics, semiconductor, nuclear technology and other precision industries.
Methods for expressing the composition of mixtures.
· Mass fraction of the component in the mixture- the ratio of the mass of the component to the mass of the entire mixture. Usually the mass fraction is expressed in %, but not necessarily.
ω ["omega"] = mcomponent / mmixture
· Mole fraction of a component in a mixture- the ratio of the number of moles (amount of substance) of the component to the total number of moles of all substances in the mixture. For example, if the mixture includes substances A, B and C, then:
χ [“chi”] component A \u003d n component A / (n (A) + n (B) + n (C))
· Molar ratio of components. Sometimes in tasks for a mixture, the molar ratio of its components is indicated. For example:
ncomponent A: ncomponent B = 2: 3
· Volume fraction of the component in the mixture (only for gases)- the ratio of the volume of substance A to the total volume of the entire gas mixture.
φ ["phi"] = Vcomponent / Vmixture
Practice block.
Consider three examples of problems in which mixtures of metals react with hydrochloric acid:
Example 1When a mixture of copper and iron weighing 20 g was exposed to an excess of hydrochloric acid, 5.6 liters of gas (n.a.) were released. Determine the mass fractions of metals in the mixture.
In the first example, copper does not react with hydrochloric acid, that is, hydrogen is released when the acid reacts with iron. Thus, knowing the volume of hydrogen, we can immediately find the amount and mass of iron. And, accordingly, the mass fractions of substances in the mixture.
Example 1 solution.
n \u003d V / Vm \u003d 5.6 / 22.4 \u003d 0.25 mol.
2. According to the reaction equation:
3. The amount of iron is also 0.25 mol. You can find its mass:
mFe = 0.25 56 = 14 g.
Answer: 70% iron, 30% copper.
Example 2Under the action of an excess of hydrochloric acid on a mixture of aluminum and iron weighing 11 g, 8.96 liters of gas (n.a.) were released. Determine the mass fractions of metals in the mixture.
In the second example, the reaction is both metal. Here, hydrogen is already released from the acid in both reactions. Therefore, direct calculation cannot be used here. In such cases, it is convenient to solve using a very simple system of equations, taking for x - the number of moles of one of the metals, and for y - the amount of substance of the second.
Example 2 solution.
1. Find the amount of hydrogen:
n \u003d V / Vm \u003d 8.96 / 22.4 \u003d 0.4 mol.
2. Let the amount of aluminum be x mol, and iron y mol. Then we can express in terms of x and y the amount of hydrogen released:
2HCl = FeCl2 + |
4. We know the total amount of hydrogen: 0.4 mol. Means,
1.5x + y = 0.4 (this is the first equation in the system).
5. For a mixture of metals, you need to express masses through quantities of substances.
m = Mn
So the mass of aluminum
mAl = 27x,
mass of iron
mFe = 56y,
and the mass of the whole mixture
27x + 56y = 11 (this is the second equation in the system).
6. So, we have a system of two equations:
7. Solving such systems is much more convenient by subtracting by multiplying the first equation by 18:
27x + 18y = 7.2
and subtracting the first equation from the second:
8. (56 - 18)y \u003d 11 - 7.2
y \u003d 3.8 / 38 \u003d 0.1 mol (Fe)
x = 0.2 mol (Al)
mFe = n M = 0.1 56 = 5.6 g
mAl = 0.2 27 = 5.4 g
ωFe = mFe / mmixture = 5.6 / 11 = 0.50.91%),
respectively,
ωAl \u003d 100% - 50.91% \u003d 49.09%
Answer: 50.91% iron, 49.09% aluminum.
Example 316 g of a mixture of zinc, aluminum and copper was treated with an excess of hydrochloric acid solution. In this case, 5.6 liters of gas (n.a.) were released and 5 g of the substance did not dissolve. Determine the mass fractions of metals in the mixture.
In the third example, two metals react, but the third metal (copper) does not react. Therefore, the remainder of 5 g is the mass of copper. The quantities of the remaining two metals - zinc and aluminum (note that their total mass is 16 - 5 = 11 g) can be found using a system of equations, as in example No. 2.
Answer to Example 3: 56.25% zinc, 12.5% aluminum, 31.25% copper.
Example 4A mixture of iron, aluminum and copper was treated with an excess of cold concentrated sulfuric acid. At the same time, part of the mixture dissolved, and 5.6 liters of gas (n.a.) were released. The remaining mixture was treated with an excess of sodium hydroxide solution. 3.36 liters of gas evolved and 3 g of undissolved residue remained. Determine the mass and composition of the initial mixture of metals.
In this example, remember that cold concentrated sulfuric acid does not react with iron and aluminum (passivation), but reacts with copper. In this case, sulfur oxide (IV) is released.
With alkali reacts only aluminum- amphoteric metal (in addition to aluminum, zinc and tin also dissolve in alkalis, and beryllium can still be dissolved in hot concentrated alkali).
Example 4 solution.
1. Only copper reacts with concentrated sulfuric acid, the number of moles of gas:
nSO2 = V / Vm = 5.6 / 22.4 = 0.25 mol
2H2SO4 (conc.) = CuSO4 + |
2. (do not forget that such reactions must be equalized using an electronic balance)
3. Since the molar ratio of copper and sulfur dioxide is 1:1, then copper is also 0.25 mol. You can find the mass of copper:
mCu \u003d n M \u003d 0.25 64 \u003d 16 g.
4. Aluminum reacts with an alkali solution, and an aluminum hydroxocomplex and hydrogen are formed:
2Al + 2NaOH + 6H2O = 2Na + 3H2
Al0 − 3e = Al3+ | ||
5. Number of moles of hydrogen:
nH2 = 3.36 / 22.4 = 0.15 mol,
the molar ratio of aluminum and hydrogen is 2:3 and, therefore,
nAl = 0.15 / 1.5 = 0.1 mol.
Aluminum weight:
mAl \u003d n M \u003d 0.1 27 \u003d 2.7 g
6. The remainder is iron, weighing 3 g. You can find the mass of the mixture:
mmix \u003d 16 + 2.7 + 3 \u003d 21.7 g.
7. Mass fractions of metals:
ωCu = mCu / mmixture = 16 / 21.7 = 0.7.73%)
ωAl = 2.7 / 21.7 = 0.1.44%)
ωFe = 13.83%
Answer: 73.73% copper, 12.44% aluminum, 13.83% iron.
Example 521.1 g of a mixture of zinc and aluminum was dissolved in 565 ml of a nitric acid solution containing 20 wt. % HNO3 and having a density of 1.115 g/ml. The volume of the released gas, which is a simple substance and the only product of the reduction of nitric acid, amounted to 2.912 l (n.a.). Determine the composition of the resulting solution in mass percent. (RCTU)
The text of this problem clearly indicates the product of nitrogen reduction - "simple substance". Since nitric acid does not produce hydrogen with metals, it is nitrogen. Both metals dissolved in acid.
The problem asks not the composition of the initial mixture of metals, but the composition of the solution obtained after the reactions. This makes the task more difficult.
Example 5 solution.
1. Determine the amount of gas substance:
nN2 = V / Vm = 2.912 / 22.4 = 0.13 mol.
2. Determine the mass of the nitric acid solution, the mass and amount of the dissolved HNO3 substance:
msolution \u003d ρ V \u003d 1.115 565 \u003d 630.3 g
mHNO3 = ω msolution = 0.2 630.3 = 126.06 g
nHNO3 = m / M = 126.06 / 63 = 2 mol
Please note that since the metals have completely dissolved, it means - just enough acid(these metals do not react with water). Accordingly, it will be necessary to check Is there too much acid?, and how much of it remains after the reaction in the resulting solution.
3. Compose reaction equations ( do not forget about the electronic balance) and, for convenience of calculations, we take for 5x - the amount of zinc, and for 10y - the amount of aluminum. Then, in accordance with the coefficients in the equations, nitrogen in the first reaction will be x mol, and in the second - 3y mol:
12HNO3 = 5Zn(NO3)2 + |
Zn0 − 2e = Zn2+ | ||
36HNO3 = 10Al(NO3)3 + |
Al0 − 3e = Al3+ | ||
5. Then, given that the mass of the mixture of metals is 21.1 g, their molar masses are 65 g/mol for zinc and 27 g/mol for aluminum, we obtain the following system of equations:
6. It is convenient to solve this system by multiplying the first equation by 90 and subtracting the first equation from the second.
7. x \u003d 0.04, which means nZn \u003d 0.04 5 \u003d 0.2 mol
y \u003d 0.03, which means that nAl \u003d 0.03 10 \u003d 0.3 mol
8. Check the mass of the mixture:
0.2 65 + 0.3 27 \u003d 21.1 g.
9. Now let's move on to the composition of the solution. It will be convenient to rewrite the reactions again and write down over the reactions the amounts of all reacted and formed substances (except water):
10. The next question is: did nitric acid remain in the solution and how much is left?
According to the reaction equations, the amount of acid that reacted:
nHNO3 = 0.48 + 1.08 = 1.56 mol,
i.e. the acid was in excess and you can calculate its remainder in the solution:
nHNO3res. \u003d 2 - 1.56 \u003d 0.44 mol.
11. So, in final solution contains:
zinc nitrate in the amount of 0.2 mol:
mZn(NO3)2 = n M = 0.2 189 = 37.8 g
aluminum nitrate in the amount of 0.3 mol:
mAl(NO3)3 = n M = 0.3 213 = 63.9 g
an excess of nitric acid in an amount of 0.44 mol:
mHNO3res. = n M = 0.44 63 = 27.72 g
12. What is the mass of the final solution?
Recall that the mass of the final solution consists of those components that we mixed (solutions and substances) minus those reaction products that left the solution (precipitates and gases):
13.
Then for our task:
14. new solution \u003d mass of acid solution + mass of metal alloy - mass of nitrogen
mN2 = n M = 28 (0.03 + 0.09) = 3.36 g
new solution \u003d 630.3 + 21.1 - 3.36 \u003d 648.04 g
ωZn(NO3)2 \u003d mv-va / mr-ra \u003d 37.8 / 648.04 \u003d 0.0583
ωAl(NO3)3 \u003d mv-va / mr-ra \u003d 63.9 / 648.04 \u003d 0.0986
ωHNO3res. \u003d mv-va / mr-ra \u003d 27.72 / 648.04 \u003d 0.0428
Answer: 5.83% zinc nitrate, 9.86% aluminum nitrate, 4.28% nitric acid.
Example 6When processing 17.4 g of a mixture of copper, iron and aluminum with an excess of concentrated nitric acid, 4.48 liters of gas (n.a.) were released, and when this mixture was exposed to the same mass of excess hydrochloric acid, 8.96 l of gas (n.a.). u.). Determine the composition of the initial mixture. (RCTU)
When solving this problem, we must remember, firstly, that concentrated nitric acid with an inactive metal (copper) gives NO2, while iron and aluminum do not react with it. Hydrochloric acid, on the other hand, does not react with copper.
Answer for example 6: 36.8% copper, 32.2% iron, 31% aluminum.
Tasks for independent solution.
1. Simple problems with two mixture components.
1-1. A mixture of copper and aluminum weighing 20 g was treated with a 96% solution of nitric acid, and 8.96 liters of gas (n.a.) were released. Determine the mass fraction of aluminum in the mixture.
1-2. A mixture of copper and zinc weighing 10 g was treated with a concentrated alkali solution. In this case, 2.24 liters of gas (n. y.) were released. Calculate the mass fraction of zinc in the initial mixture.
1-3. A mixture of magnesium and magnesium oxide weighing 6.4 g was treated with a sufficient amount of dilute sulfuric acid. At the same time, 2.24 liters of gas (n.a.) were released. Find the mass fraction of magnesium in the mixture.
1-4. A mixture of zinc and zinc oxide weighing 3.08 g was dissolved in dilute sulfuric acid. Zinc sulfate weighing 6.44 g was obtained. Calculate the mass fraction of zinc in the initial mixture.
1-5. Under the action of a mixture of iron and zinc powders weighing 9.3 g on an excess of copper (II) chloride solution, 9.6 g of copper was formed. Determine the composition of the initial mixture.
1-6. What mass of a 20% hydrochloric acid solution will be required to completely dissolve 20 g of a mixture of zinc with zinc oxide, if hydrogen is released in the amount of 4.48 liters (n.a.)?
1-7. When dissolved in dilute nitric acid, 3.04 g of a mixture of iron and copper releases nitric oxide (II) with a volume of 0.896 l (n.a.). Determine the composition of the initial mixture.
1-8. When dissolving 1.11 g of a mixture of iron and aluminum filings in a 16% hydrochloric acid solution (ρ = 1.09 g / ml), 0.672 liters of hydrogen (n.a.) were released. Find the mass fractions of metals in the mixture and determine the volume of hydrochloric acid consumed.
2. Tasks are more complex.
2-1. A mixture of calcium and aluminum weighing 18.8 g was calcined without access to air with an excess of graphite powder. The reaction product was treated with dilute hydrochloric acid, and 11.2 liters of gas (n.a.) were released. Determine the mass fractions of metals in the mixture.
2-2. To dissolve 1.26 g of an alloy of magnesium with aluminum, 35 ml of a 19.6% sulfuric acid solution (ρ = 1.1 g/ml) was used. The excess acid reacted with 28.6 ml of a 1.4 mol/L potassium hydrogen carbonate solution. Determine the mass fractions of metals in the alloy and the volume of gas (n.a.) released during the dissolution of the alloy.
FROM mixture separation methods (both heterogeneous and homogeneous) are based on the fact that the substances that make up the mixture retain their individual properties. Heterogeneous mixtures may differ in composition and phase state, for example: gas + liquid; solid+liquid; two immiscible liquids, etc. The main methods for separating mixtures are shown in the diagram below. Let's consider each method separately.
Separation of heterogeneous mixtures
For separation of heterogeneous mixtures, which are solid-liquid or solid-gas systems, there are three main ways:
- filtration,
- settling (decanting,
- magnetic separation
FILTRATION
a method based on the different solubility of substances and different particle sizes of the mixture components. Filtration separates a solid from a liquid or gas.
To filter liquids, filter paper can be used, which is usually folded into fours and inserted into a glass funnel. The funnel is placed in a beaker in which filtrate is the liquid that has passed through the filter.
The pore size in the filter paper is such that it allows water molecules and solute molecules to seep through unhindered. Particles larger than 0.01 mm are retained on the filter and do notpass through it, thus forming a layer of sediment.
Remember! With the help of filtration, it is impossible to separate true solutions of substances, that is, solutions in which dissolution occurred at the level of molecules or ions.
In addition to filter paper, chemical laboratories use special filters with
different pore sizes.
Filtration of gas mixtures is not fundamentally different from filtration of liquids. The only difference is that when filtering gases from particulate matter (SPM), filters of special designs (paper, coal) and pumps are used to force the gas mixture through the filter, for example, air filtration in a car interior or an exhaust hood over a stove.
Filtering can be divided:
- cereals and water
- chalk and water
- sand and water, etc.
- dust and air (various designs of vacuum cleaners)
SETTLEMENT
The method is based on different settling rates of solid particles with different weights (densities) in a liquid or air medium. The method is used to separate two or more solid insoluble substances in water (or other solvent). A mixture of insoluble substances is placed in water, mixed thoroughly. After some time, substances with a density greater than unity settle to the bottom of the vessel, and substances with a density less than unity float. If there are several substances with different gravity in the mixture, then heavier substances will settle in the lower layer, and then lighter ones. These layers can also be separated. Previously, grains of gold were isolated from crushed gold-bearing rocks in this way. Gold-bearing sand was placed on an inclined chute, through which a stream of water was launched. The flow of water picked up and carried away the waste rock, and heavy grains of gold settled at the bottom of the gutter. In the case of gas mixtures, there is also the settling of solid particles on hard surfaces, such as dust settling on furniture or plant leaves.
Immiscible liquids can also be separated by this method. To do this, use a separating funnel.
For example, to separate gasoline and water, the mixture is placed in a separating funnel, waiting for the moment until a clear phase boundary appears. Then gently open the faucet and water flows into the glass.
Mixtures can be separated by settling:
- river sand and clay
- heavy crystalline precipitate from solution
- oil and water
- vegetable oil and water, etc.
MAGNETIC SEPARATION
The method is based on different magnetic properties of the solid components of the mixture. This method is used in the presence of ferromagnetic substances in the mixture, that is, substances with magnetic properties, such as iron.
All substances, in relation to the magnetic field, can be conditionally divided into three large groups:
- feromagnetics: attracted by magnet - Fe, Co, Ni, Gd, Dy
- paramagnets: weakly attracted-Al, Cr, Ti, V, W, Mo
- diamagnets: repelled by magnet - Cu, Ag, Au, Bi, Sn, brass
Magnetic separation can separate b:
- sulfur and iron powder
- soot and iron, etc.
Separation of homogeneous mixtures
For separation of liquid homogeneous mixtures (true solutions) use the following methods:
- evaporation (crystallization),
- distillation (distillation),
- chromatography.
EVAPORATION. CRYSTALLIZATION.
The method is based on different boiling points of solvent and solute. Used to isolate soluble solids from solutions. Evaporation is usually carried out as follows: the solution is poured into a porcelain cup and heated while constantly stirring the solution. The water gradually evaporates and a solid remains at the bottom of the cup.
DEFINITION
Crystallization- phase transition of a substance from a gaseous (vaporous), liquid or solid amorphous state to a crystalline state.
In this case, the evaporated substance (water or solvent) can be collected by condensation on a colder surface. For example, if you place a cold glass slide over an evaporating dish, water droplets form on its surface. The distillation method is based on the same principle.
DISTILLATION. DISTILLATION.
If a substance, such as sugar, decomposes when heated, then the water is not completely evaporated - the solution is evaporated, and then sugar crystals are precipitated from a saturated solution. Sometimes it is required to purify solvents from impurities, for example, water from salt. In this case, the solvent should be evaporated, and then its vapors should be collected and condensed on cooling. This method of separating a homogeneous mixture is called distillation, or distillation.
In nature, water in its pure form (without salts) does not occur. Oceanic, sea, river, well and spring water are varieties of salt solutions in water. However, often people need clean water that does not contain salts (used in car engines; in chemical production to obtain various solutions and substances; in the manufacture of photographs). This water is called distilled it is used in the laboratory for chemical experiments.
Distillation can be divided:
- water and alcohol
- oil (for various fractions)
- acetone and water, etc.
CHROMATOGRAPHY
Method for separation and analysis of mixtures of substances. Based on different rates of distribution of the test substance between two phases - stationary and mobile (eluent). The stationary phase, as a rule, is a sorbent (fine powder, such as aluminum oxide or zinc oxide or filter paper) with a developed surface, and the mobile phase is a gas or liquid flow. The flow of the mobile phase is filtered through the sorbent bed or moves along the sorbent bed, for example, on the surface of filter paper.
You can get a chromatogram yourself and see the essence of the method in practice. It is necessary to mix several inks and apply a drop of the resulting mixture on filter paper. Then, exactly in the middle of the colored spot, we will begin to pour clean water drop by drop. Each drop should be applied only after the previous one has been absorbed. Water plays the role of an eluent that transfers the test substance along the sorbent - porous paper. The substances that make up the mixture are retained by paper in different ways: some are well retained by it, while others are absorbed more slowly and continue to spread along with water for some time. Soon, a real colorful chromatogram will begin to spread across a sheet of paper: a spot of the same color in the center, surrounded by multi-colored concentric rings.
Thin-layer chromatography has become especially widespread in organic analysis. The advantage of thin layer chromatography is that it is possible to use the simplest and most sensitive detection method - visual control. Spots invisible to the eye can be developed using various reagents, as well as using ultraviolet light or autoradiography.
In the analysis of organic and inorganic substances, paper chromatography is used. Numerous methods have been developed for the separation of complex mixtures of ions, such as mixtures of rare earth elements, fission products of uranium, elements of the platinum group
MIXTURE SEPARATION METHODS USED IN INDUSTRY.
Methods for separating mixtures used in industry differ little from the laboratory methods described above.
Rectification (distillation) is most often used to separate oil. This process is described in more detail in the topic. "Oil refining".
The most common methods of purification and separation of substances in industry are settling, filtration, sorption and extraction. Filtration and settling methods are carried out similarly to the laboratory method, with the difference that settling tanks and large volume filters are used. Most often, these methods are used for wastewater treatment. Therefore, let's take a closer look at the methods extraction and sorption.
The term "extraction" can be applied to various phase equilibria (liquid-liquid, gas-liquid, liquid-solid, etc.), but more often it is applied to liquid-liquid systems, so the following definition can often be found:
DEFINITION
Extraction i - a method of separation, purification and isolation of substances, based on the process of distribution of a substance between two immiscible solvents.
One of the immiscible solvents is usually water, the other is an organic solvent, but this is not required. The extraction method is versatile; it is suitable for isolating almost all elements in various concentrations. Extraction allows you to separate complex multicomponent mixtures often more efficiently and faster than other methods. Performing an extraction separation or separation does not require complex and expensive equipment. The process can be automated, if necessary, it can be controlled remotely.
DEFINITION
Sorption- a method for the isolation and purification of substances based on the absorption by a solid body (adsorption) or liquid-sorbent (absorption) of various substances (sorbates) from gas or liquid mixtures.
Most often in industry, absorption methods are used to clean gas-air emissions from dust particles or smoke, as well as toxic gaseous substances. In the case of absorption of gaseous substances, a chemical reaction can occur between the sorbent and the solute. For example, when absorbing gaseous ammoniaNH3a solution of nitric acid HNO 3 forms ammonium nitrate NH 4 NO 3(ammonium nitrate), which can be used as a highly effective nitrogen fertilizer.
If dispersed particles are released slowly from the medium or it is necessary to pre-clarify an inhomogeneous system, methods such as flocculation, flotation, classification, coagulation, etc. are used.
Coagulation is the process of sticking together of particles in colloidal systems (emulsions or suspensions) with the formation of aggregates. Sticking occurs due to the collision of particles during Brownian motion. Coagulation refers to a spontaneous process that tends to move into a state that has a lower free energy. The coagulation threshold is the minimum concentration of an injected substance that causes coagulation. Artificial coagulation can be accelerated by adding special substances - coagulators to the colloidal system, as well as by applying an electric field to the system (electrocoagulation), mechanical action (vibration, mixing), etc.
During coagulation, coagulant chemicals are often added to the heterogeneous mixture to be separated, which destroy the solvated shells, while reducing the diffusion part of the electrical double layer located near the surface of the particles. This facilitates the agglomeration of particles and the formation of aggregates. Thus, due to the formation of larger fractions of the dispersed phase, particle settling is accelerated. Salts of iron, aluminum or salts of other polyvalent metals are used as coagulants.
Peptization is the reverse process of coagulation, which is the breakdown of aggregates into primary particles. Peptization is carried out by adding peptizing substances to the dispersion medium. This process contributes to the disaggregation of substances into primary particles. Peptizing agents can be surface-active substances (surfactants) or electrolytes such as humic acids or ferric chloride. The peptization process is used to obtain liquid dispersion systems from pastes or powders.
In turn, flocculation is a kind of coagulation. In this process, small particles that are suspended in gas or liquid media form flocculent aggregates, which are called floccules. Soluble polymers, such as polyelectrolytes, are used as flocculants. Flocculating substances can be easily removed by filtration or settling. Flocculation is used for water treatment and the separation of valuable substances from wastewater, as well as for mineral processing. In the case of water treatment, flocculants are used in low concentrations (from 0.1 to 5 mg/l).
In order to destroy aggregates in liquid systems, additives are used that induce charges on particles that prevent their convergence. This effect can also be achieved by changing the pH of the medium. This method is called deflocculation.
Flotation is the process of separating solid hydrophobic particles from a liquid continuous phase by selectively fixing them at the interface between the liquid and gaseous phases (the contact surface of liquid and gas or the surface of bubbles in the liquid phase). The resulting system of solid particles and gas inclusions is removed from the surface of the liquid phase. This process is used not only to remove particles of the dispersed phase, but also to separate different particles due to differences in their wettability. In this process, hydrophobic particles are fixed at the interface and separated from hydrophilic particles that settle to the bottom. The best flotation results occur when the particle size is between 0.1 and 0.04 mm.
There are several types of flotation: foam, oil, film, etc. The most common is froth flotation. This process allows the particles treated with reagents to be carried to the surface of the water with the help of air bubbles. This allows the formation of a foam layer, the stability of which is controlled by a foaming agent.
The classification is used in devices of variable cross section. With its help, it is possible to separate a certain amount of small particles from the main product, consisting of large particles. Classification is carried out using centrifuges and hydrocyclones due to the effect of centrifugal force.
The separation of suspensions using magnetic processing systems is a very promising method. Water that has been treated in a magnetic field retains changed properties for a long time, for example, reduced wetting ability. This process makes it possible to intensify the separation of suspensions.
While studying chemistry, I learned that there are very few pure substances in nature, technology, and everyday life. Much more common are mixtures - combinations of two or more components that are not chemically related to each other. Mixtures differ in the size of the particles of substances included in their composition, as well as the state of aggregation of the components. Chemical research requires pure substances. But how can they be obtained or isolated from the mixture? This is the question I tried to answer in my work.
In everyday life, we are surrounded by mixtures of substances. The air we breathe, the food we consume, the water we drink, and even ourselves - all these are chemically mixtures containing from 2-3 to many thousands of substances.
Mixtures are systems consisting of several components that are not chemically related to each other. Mixtures are distinguished by the size of their constituent particles of substances. Sometimes these particles are so large that they can be seen with the naked eye. Such mixtures, for example, include washing powder, culinary mixtures for baking, building mixtures. Sometimes the particles of the components in mixtures are smaller, indistinguishable to the eye. For example, flour contains grains of starch and protein that cannot be distinguished with the naked eye. Milk is also a mixture of water, which contains small droplets of fat, protein, lactose and other substances. You can see fat droplets in milk if you look at a drop of milk under a microscope. The aggregate state of substances in mixtures can be different. Toothpaste, for example, is a mixture of solid and liquid ingredients. There are mixtures, during the formation of which substances “penetrate each other” so much that they break up into tiny particles that are not distinguishable even under a microscope. No matter how we peer into the air, we will not be able to distinguish its constituent gases.
Thus, mixtures are classified:
Mixtures in which particles of substances that make up the mixture are visible to the naked eye or under a microscope are called heterogeneous or heterogeneous.
Mixtures in which even with a microscope it is impossible to see the particles of the substances that make up the mixture are called homogeneous or homogeneous.
Homogeneous mixtures according to the state of aggregation are divided into gaseous, liquid and solid. A mixture of any gases is homogeneous. For example, clean air is a homogeneous mixture of nitrogen, oxygen, carbon dioxide and noble gases. But dusty air is already a heterogeneous mixture of the same gases, only containing more dust particles. Oil is a liquid natural mixture. It contains hundreds of different components. Of course, the most common liquid mixture, or rather a solution, is the water of the seas and oceans. 1 liter of sea water contains an average of 35 grams of various salts. We encounter liquid mixtures in everyday life all the time. Shampoos and drinks, potions and household chemicals are all mixtures of substances. Even tap water cannot be considered a pure substance: it contains dissolved salts, the smallest insoluble impurities, as well as microorganisms that are disinfected by chlorination. Solid mixtures are also widespread. Rocks are a mixture of several substances. Soil, sand, clay are solid mixtures. Solid mixtures include glass, ceramics, alloys.
Chemists make mixtures by simply mixing various substances - constituents, the properties of which may be different. It is important that the properties of their constituents are preserved in mixtures. So, for example, gray paint is obtained by mixing black and white. Although we see gray, this does not mean that all particles of such gray paint are gray. Under the microscope, particles of black and white colors are sure to be found, of which black and white paint consisted.
The separation of mixtures into constituent parts (individual substances) is a more difficult task than the preparation of mixtures, but no less important. The most important ways of separating mixtures can be reflected in the scheme:
Using various methods for separating mixtures (settling, filtering, distillation, freezing, and others), oil is obtained from milk, gold from river sand, alcohol from mash, and water is purified from insoluble and soluble impurities.
Chemical laboratories and industry often require pure substances. Pure substances are substances that have constant physical properties, such as distilled water. (Practically absolutely pure substances have not been obtained.)
There are various ways to separate mixtures. Let's take a closer look at these methods.
Isolation from an inhomogeneous mixture.
1. Settling.
a) Isolation of substances of an inhomogeneous mixture formed by water-insoluble substances with different densities. For example, iron filings can be separated from wood filings by shaking this mixture with water and then settling. Iron filings sink to the bottom of the vessel, and wood filings float up, and they can be drained along with water.
b) Some substances are deposited in water at different rates. If you shake clay mixed with sand with water, the sand settles much faster. This method is used in ceramic production to separate sand from clay (production of red bricks, earthenware, etc.) c) Separation of a mixture of liquids with different densities that are slightly soluble in each other. Mixtures of gasoline with water, oil with water, vegetable oil with water quickly separate, so they can be separated using a separating funnel or column. Sometimes liquids with different densities are separated by centrifugation, such as cream from milk.
2. Filtering.
Isolation of substances from an inhomogeneous mixture formed by water-soluble substances.
To isolate table salt, its mixture with sand is shaken in water. Table salt dissolves and sand settles.
To speed up the separation of insoluble particles from the solution, the mixture is filtered. The sand remains on the filter paper, and a clear salt solution passes through the filter.
3. Action by a magnet.
Isolation from an inhomogeneous mixture of substances capable of magnetization. If there is, for example, a mixture of powders of iron and sulfur, they can be separated using a magnet.
Separation of substances from a homogeneous mixture.
4. Evaporation. Crystallization.
In order for a solute, such as table salt, to be separated from a solution, the latter is evaporated. The water evaporates, and table salt remains in the porcelain cup. Sometimes evaporation is used, i.e. partial evaporation of water. As a result, a more concentrated solution is formed, upon cooling of which the solute is released in the form of crystals. This method of purification of substances is called crystallization.
5. Distillation.
This method of separating mixtures is based on the difference in the boiling points of components soluble in each other.
Distillation (distillation) is a technique for separating homogeneous mixtures by evaporation of volatile liquids, followed by condensation of their vapors. For example, getting distilled water.
To do this, water with the substances dissolved in it is boiled in one vessel. The resulting water vapor condenses in another vessel in the form of distilled water.
6. Chromatography.
This method is based on the fact that individual substances are absorbed (bound) by the surface of another substance at different rates.
The essence of this method can be found in the following experiment.
If a strip of filter paper is hung over a vessel with red ink and only the end of the strip is immersed in them, then it can be seen that the solution will be absorbed by the paper and rise along it. However, the paint rise limit will lag behind the water rise limit. Thus, there is a separation of two substances: water and a coloring matter that gives the solution a red color.
Experimental part.
Safety rules in the home laboratory.
It is impossible to imagine chemistry without chemical experiments. Therefore, it is possible to study this science, understand its laws and, of course, fall in love with it only through an experiment. There was an opinion that a chemical experiment is complex equipment and inaccessible reagents, poisonous compounds and terrible explosions, and special conditions are necessary for practicing chemistry. However, more than 300 chemical experiments with a wide variety of substances can be performed at home. Due to the fact that there is no fume hood and other special devices in the home laboratory, it is necessary to strictly follow the safety rules:
2. Do not accumulate and store large quantities of reagents at home.
3. Chemical reagents and substances must have labels with names, concentration and production date.
4. Chemicals must not be tasted.
5. To determine the smell, you can not bring a vessel with a substance close to your face. It is necessary to make a few smooth strokes with the palm of your hand from the opening of the vessel to the nose.
6. If acid or alkali has spilled, then the substance is first neutralized or covered with sand and removed with a rag or collected in a scoop.
7. Before conducting an experiment, no matter how simple it may seem, you need to carefully read the description of the experiment and understand the properties of the substances used. For this there are textbooks, reference books and other literature.
Experience number 1. Separation of heterogeneous mixtures.
A) Prepare a heterogeneous mixture of sand and iron powder.
The purpose of the experiment: to learn how to separate heterogeneous mixtures in different ways.
Equipment: river sand, iron powder, magnet, two beakers.
Add one tablespoon of iron powder and river sand to the beaker, gently mix the mixture until the product is evenly colored. Mark its color and test its magnetic properties by holding the magnet to the outside of the glass. Determine what substances give the mixture color and magnetic properties. Separate the prepared heterogeneous mixture with a magnet. To do this, we will bring a magnet to the outer wall of the glass, and lightly tapping the magnet on the outer wall, we will collect iron powder on the inner wall of the glass. Holding the iron with a magnet on the inner wall of the glass, pour the sand into another glass. The experimental data are entered in the table.
B) Prepare a mixture of table salt, earth and shavings formed after sharpening a pencil.
Equipment: table salt, earth, shavings after sharpening a pencil, glass, water, filter, spoon, frying pan.
Experiment Method:
Prepare the mixture by mixing one teaspoon each of table salt, earth and pencil shavings. Dissolve the resulting mixture in a glass of water, remove the floating chips with a slotted spoon and lay them on a sheet of paper to dry. Make a bandage or gauze filter by folding 3-4 layers, and pull it loosely over another glass. Filter the mixture. Dry the filter with the remaining earth, then clean it off the filter. Pour the filtered liquid (filtrate) from a glass into an enameled bowl or pan and evaporate. Collect the separated salt crystals. Compare the amounts of substances before and after the experiments.
Experience number 2. Separation of homogeneous mixtures by paper chromatography.
A) Separate a homogeneous mixture of red and green dye.
Equipment: a strip of filter paper, a beaker, a cork on a beaker, red and green felt-tip pens, alcohol (70% aqueous solution).
Experiment Method:
Take a strip of filter paper, the length of which is 2-3 cm longer than the height of the beaker. In the middle of this strip, mark a point with a simple pencil, stepping back from the edge of 1.5 cm. Apply stains of dyes with a diameter of no more than 5 mm to the marked point with felt-tip pens. First, make a dot 1-2 mm in size with a red felt-tip pen, and then apply green on top of the red spot so that the green spot protrudes beyond the red border by about 1 mm. Let the stain of the mixture dry (1-2 minutes) and then carefully, so as not to damage the paper, circle it with a simple pencil along the contour.
Pour alcohol into a beaker with a layer of 0.5-1 cm. Place a paper strip with a stain of a mixture of dyes vertically into a beaker and bend the protruding part of the strip to the outer surface of the beaker. The stain of dyes should be above the liquid at a distance of 0.5 cm. Cover the glass with an inverted cork. Observe the wetting of the strip of paper and the movement of the colored spot upwards, dividing it into two spots. It will take about 20 minutes to completely separate the dye mixture. After the paper is completely saturated with alcohol, take it out and let it dry for 5-10 minutes. Mark the colors of the spot separation. Record the results of observations in the table.
B) Separate the following mixtures by chromatography on paper: an alcoholic solution of "brilliant green"; aqueous solution of black ink for drawing work.
The purpose of the experiment: to master the method of paper chromatography, to learn how to determine the difference between pure substances and mixtures.
Equipment: a chemical beaker, a strip of filter or blotting paper, an alcohol solution of "brilliant green", an aqueous solution of ink for drawing work.
Experiment Method:
A strip of filter paper must be hung over a vessel with a solution of greenery and black ink so that the paper only touches the solution.
The border of the rise of "brilliant green" and the coloring matter will lag behind the border of the rise of alcohol and water, respectively. Thus, there is a separation of two substances in the composition of homogeneous mixtures: a) alcohol and brilliant green, b) water and coloring matter.
Experience number 3. Diffusion.
The purpose of the experiment: to study in practice the process of diffusion.
Equipment: food gelatin, potassium permanganate, copper sulfate, water, saucepan, stainless steel spoon for stirring, electric or gas stove, tweezers, two transparent vials.
Experiment Method:
Dip a teaspoon of gelatin in a glass of cold water and leave for an hour or two so that the powder has time to swell. Pour the mixture into a small saucepan. Heat the mixture over low heat; make sure that it does not boil in any case! Stir the contents of the saucepan until the gelatin is completely dissolved. Pour the hot solution into two vials. When it cools down, in the middle of one of the bubbles, with a quick and careful movement, insert tweezers in which a crystal of potassium permanganate is clamped. Slightly open the tweezers and quickly remove them. In another vial, add a crystal of copper sulfate. Gelatin slows down the diffusion process, and for several hours in a row you can observe a very interesting picture: a colored ball will grow around the crystals.
Experience number 4. Separation of homogeneous mixtures by crystallization.
Grow a crystal or crystals from a saturated sodium chloride solution, copper sulphate, or potassium alum.
The purpose of the experiment: to learn how to prepare a saturated solution of table salt or other substances, grow crystals of various sizes, consolidate skills and abilities when working with substances and chemical equipment.
Equipment: a glass and a liter jar for preparing a solution, a wooden spoon or a stirring stick, salt for the experiment - table salt, blue vitriol or alum, hot water, a seed - a salt crystal suspended on a thread, a funnel and filter paper.
Experiment Method:
Prepare a saturated salt solution. To do this, first pour hot water into a jar to half its volume, then add the appropriate salt in portions, stirring constantly. Add salt until it no longer dissolves. Filter the resulting solution into a glass through a funnel with filter paper or cotton and leave the solution to cool for 2-3 hours. Introduce a seed into the cooled solution - a salt crystal suspended on a thread, carefully cover the solution with a lid and leave for a long time (2-3 days or more).
Results of the work and conclusions:
Examine your crystal and answer the questions:
How many days did you grow the crystal?
What is its shape?
What color is the crystal?
Is it transparent or not?
What are the dimensions of the crystal: height, width, thickness?
What is the mass of the crystal?
Sketch or photograph your crystal.
Experience number 5. Separation of homogeneous mixtures by distillation.
Get at home 50 ml of distilled water.
The purpose of the experiment: to learn how to separate homogeneous mixtures by distillation.
Equipment: enameled teapot, two glass jars.
Experiment Method:
Pour 1/3 of the water into an enameled teapot and place it on a gas stove so that the spout of the teapot protrudes beyond the edge of the stove. When the water boils, fasten a glass jar-refrigerator on the spout of the kettle, under which fit a second jar to collect condensate. In order for the refrigerator jar not to overheat, you can put a napkin moistened with cold water on it.
Results of the work and conclusions:
Answer the questions posed:
What is tap water?
How are homogeneous mixtures separated?
What is distilled water? Where and for what purposes is it used?
Write down your experience.
Experience number 6. Extraction of starch from potatoes.
Get a small amount of starch at home.
Equipment: 2-3 potatoes, grater, cheesecloth, small saucepan, water.
Experiment Method:
Grate the peeled potatoes on a fine grater and stir the resulting mass in water. Then filter it through cheesecloth and squeeze. Mix the rest of the mass in gauze again with water. Let the liquid settle. The starch will settle to the bottom of the dish. Drain the liquid, and stir the settled starch again. Repeat the operation several times until the starch is completely clean and white. Filter and dry the resulting starch.
What do you think, from which potato will you get more starch: from a young one (which was recently dug up) or an old one (which has been in a vegetable store all winter)?
Experience number 7. Extraction of sugar from sugar beets.
Get a small amount of sugar at home.
The purpose of the experiment: to learn how to extract substances from plant materials.
Equipment: large sugar beet, activated carbon, river sand, saucepan, two cans, cotton wool, spoon, funnel, gauze.
Experiment Method:
Cut the beets into small pieces, put them in a saucepan, pour a glass of water into it and boil for 15-20 minutes. Rub the slices of cooked beets thoroughly with a spoon or pestle. Filter this dark mass through a funnel containing cotton wool. Then filter the resulting solution through a funnel prepared in a special way. Put a piece of gauze in it, a thin layer of cotton wool on the gauze, then crushed activated carbon (4-5 tablets) and a thin layer (1 cm) of clean river sand (wash and dry the river sand in advance). The resulting solution (filtrate) is placed in a saucepan. It is necessary to evaporate part of it until transparent crystals appear. This is sugar. Taste it!
Why do you think it is necessary to filter the liquid through a layer of activated carbon?
Experience number 8. Extraction of cottage cheese from milk.
Get a few grams of cottage cheese at home.
The purpose of the experiment: to learn how to make cottage cheese at home.
Equipment: milk, vinegar, saucepan, gauze, gas stove.
Experiment Method:
There is protein in milk. If the milk boils, “runs away” over the edge, then the smell characteristic of burnt protein immediately spreads. The appearance of a characteristic smell of burnt milk indicates that the phenomenon of denaturation has occurred (folding of the protein and its transition to an insoluble form). Protein denaturation is not only due to heat.
Let's do the following experiment. Let's heat up half a glass of milk so that it becomes a little warm, and add vinegar. The milk will immediately curdle, forming large flakes. (If the milk is left in a warm place, then the protein also coagulates, but for a different reason - this is the “work” of lactic acid bacteria). The contents of the saucepan are filtered through cheesecloth, holding it by the edges. If you then connect the edges of the gauze, lift it above the glass and squeeze it out, then a thick mass will remain on it - cottage cheese.
Experience number 9. Getting butter.
Get a small amount of butter at home.
The purpose of the experiment: to learn how to extract butter from milk at home.
Equipment: milk, a glass jar, a small transparent vial with a cork or a tight-fitting lid.
Experiment Method:
Pour fresh milk into a glass jar, put it in the refrigerator. After a few hours, but better the next day, look carefully: what happened to the milk? Explain what you see.
With a small spoon, carefully scoop up the cream (the top layer of milk) and transfer it to a vial. If you need to make butter from cream, you will have to shake it for a long time and patiently for at least half an hour in a vial closed with a lid until an oil lump forms.
Experience number 10. Extraction.
Practice the extraction process.
The purpose of the experiment: to carry out the extraction process in practice.
A) Equipment: sunflower seeds, gasoline, test tube, saucer, mortar and pestle.
Experiment Method:
Grind a few sunflower seeds in a mortar. Transfer the crushed seeds into a test tube, and fill with a small amount of gasoline, shake well several times. Let the test tube stand for two hours (away from the fire), not forgetting to shake it from time to time. Drain the gasoline on a saucer and put it on the balcony. When the gasoline evaporates, there will be some oil left at the bottom that has been dissolved in the gasoline.
B) Equipment: iodine tincture, water, gasoline, test tube.
Experiment Method:
Gasoline can also extract iodine from a pharmacy iodine tincture. To do this, pour a third of water into a test tube, add about 1 ml of iodine tincture and add the same amount of gasoline to the resulting brownish solution. Shake the test tube and leave it alone. When the mixture stratifies, the upper gasoline layer will become dark brown, and the lower, aqueous layer will become almost colorless: after all, iodine does not dissolve well in water, but well in gasoline.
What is extraction? The process of separating a mixture of liquid or solid substances by means of extraction - selective dissolution in certain liquids (extractants) of one or another component of the mixture. Most often, substances are extracted from aqueous solutions with organic solvents, which are usually immiscible with water. The main requirements for extractants are: selectivity (selectivity of action), non-toxicity, possibly low volatility, chemical inertness and low cost. Extraction is used in the chemical industry, oil refining, drug production, and especially widely in non-ferrous metallurgy.
Conclusion.
Work conclusions.
In doing this work, I learned how to prepare heterogeneous and homogeneous mixtures, conducted a study of the properties of substances and found out that when simply compiling a mixture of two components, these substances do not transfer their properties to each other, but keep them to themselves. The properties of the initial components (such as: volatility, state of aggregation, ability to magnetize, solubility in water, particle size, and others) are also based on methods for their separation. When performing educational research, I mastered the following methods for separating heterogeneous mixtures: magnet action, settling, filtering and homogeneous mixtures: evaporation, crystallization, distillation, chromatography, extraction. I was able to isolate pure substances from food products: sugar from sugar beets, starch from potatoes, cottage cheese and butter from milk. I realized that chemistry is a very interesting and informative science, and that the knowledge gained in chemistry classes and after school hours will be very useful to me in life.
The results of the separation of a mixture of iron and sand.
experience #1 #1 #1 #2 #2
substance iron sand mixture part 1 part 2
color gray yellow gray-yellow gray yellow magnet attraction yes no yes yes no
sand are different sand have different properties of both iron and iron sand sand
The results of the separation of dyes on paper.
experiment No. 1 No. 2 substance mixture of dyes before separation mixture of dyes after separation color black dye No. 1 - red dye No. 2 - green conclusion this mixture is homogeneous. the mixture is divided into two starting materials; These are red and green dyes.
The material of the lesson contains information about various ways of separating mixtures and purifying substances. You will learn how to use knowledge of the differences in the properties of the components of a mixture to select the optimal method for separating a given mixture.
Topic: Initial chemical ideas
Lesson: Methods for separating mixtures and purifying substances
Let us define the difference between "methods for separating mixtures" and "methods for purifying substances." In the first case, it is important to obtain in pure form all the components that make up the mixture. When purifying a substance, obtaining impurities in a pure form is usually neglected.
SETTLEMENT
How to separate a mixture of sand and clay? This is one of the stages in ceramic production (for example, in the production of bricks). To separate such a mixture, the settling method is used. The mixture is placed in water and stirred. Clay and sand settle in water at different rates. Therefore, sand will settle much faster than clay (Fig. 1).
Rice. 1. Separation of a mixture of clay and sand by settling
The settling method is also used to separate mixtures of water-insoluble solids with different densities. For example, a mixture of iron and sawdust can be separated in this way (the sawdust will float in water, while the iron will settle).
A mixture of vegetable oil and water can also be separated by settling, because the oil does not dissolve in water and has a lower density (Fig. 2). Thus, by settling, it is possible to separate mixtures of liquids insoluble in each other with different densities.
Rice. 2. Separation of a mixture of vegetable oil and water by settling
To separate a mixture of table salt and river sand, you can use the settling method (when mixed with water, the salt will dissolve, the sand will settle), but it will be more reliable to separate the sand from the salt solution by another method - the filtration method.
Filtration of this mixture can be carried out using a paper filter and a funnel lowered into a glass. Grains of sand remain on the filter paper, and a clear solution of table salt passes through the filter. In this case, the river sand is the sediment, and the salt solution is the leachate (Fig. 3).
Rice. 3. Using filtration method to separate river sand from salt solution
Filtration can be carried out not only with filter paper, but also with other porous or loose materials. For example, bulk materials include quartz sand, and porous materials include glass wool and baked clay.
Some mixtures can be separated using the "hot filtration" method. For example, a mixture of sulfur and iron powders. Iron melts at over 1500 C and sulfur around 120 C. Molten sulfur can be separated from the iron powder using heated glass wool.
Salt can be isolated from the filtrate by evaporation, i.e. heat the mixture and the water will evaporate and the salt will remain on the porcelain cup. Evaporation, the partial evaporation of water, is sometimes used. As a result, a more concentrated solution is formed, upon cooling of which the solute is released in the form of crystals.
If a substance capable of magnetization is present in the mixture, then it is easy to isolate it in its pure form using a magnet. For example, a mixture of sulfur and iron powders can be separated in this way.
The same mixture can be separated by another method, using knowledge of the wettability of the components of the mixture with water. Iron is wetted by water, i.e. water spreads over the surface of the iron. Sulfur is not wetted by water. If you put a piece of sulfur in water, it will sink, because. The density of sulfur is greater than the density of water. But the sulfur powder will emerge, because. air bubbles stick to the grains of sulfur that are not wetted by water and push them to the surface. To separate the mixture, you need to place it in water. The sulfur powder will float and the iron will sink (Fig. 4).
Rice. 4. Separation of a mixture of sulfur and iron powders by flotation
The method of separating mixtures based on the difference in the wettability of the components is called flotation (French flotter - to float). Consider a few more methods for the separation and purification of substances.
One of the oldest methods of separating mixtures is distillation (or distillation). Using this method, it is possible to separate components that are soluble in each other and have different boiling points. This is how distilled water is obtained. Water with impurities is boiled in one vessel. The resulting water vapor condenses upon cooling in another vessel in the form of already distilled (pure) water.
Rice. 5. Obtaining distilled water
Components with similar properties can be separated using the chromatography method. This method is based on the different absorption of the substances to be separated by the surface of another substance.
For example, red ink can be separated into its components (water and dye) by chromatography.
Rice. 6. Separation of red ink by paper chromatography
In chemical laboratories, chromatography is carried out using special instruments - chromatographs, the main parts of which are a chromatographic column and a detector.
Adsorption is widely used in chemistry to purify certain substances. It is the accumulation of one substance on the surface of another substance. Adsorbents include, for example, activated carbon.
Try dropping an activated charcoal tablet into a container of colored water, stir, filter, and you will see that the filtrate has become colorless. Carbon atoms attract molecules, in this case, the dye.
Currently, adsorption is widely used for water and air purification. For example, water filters contain activated carbon as an adsorbent.
1. Collection of tasks and exercises in chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. "Chemistry, Grade 8" / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. – M.: AST: Astrel, 2006.
2. Ushakova O.V. Chemistry workbook: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. Grade 8” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M .: AST: Astrel: Profizdat, 2006. (p. 10-11)
3. Chemistry: 8th grade: textbook. for general institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005.(§4)
4. Chemistry: inorg. chemistry: textbook. for 8 cells. general institutions / G.E. Rudzitis, FuGyu Feldman. - M .: Education, JSC "Moscow textbooks", 2009. (§ 2)
5. Encyclopedia for children. Volume 17. Chemistry / Chapter. edited by V.A. Volodin, leading. scientific ed. I. Leenson. – M.: Avanta+, 2003.
Additional web resources
1. A single collection of digital educational resources ().
2. Electronic version of the journal "Chemistry and Life" ().
Homework
From the textbook P.A. Orzhekovsky and others. "Chemistry, Grade 8" With. 33 Nos. 2,4,6,T.
