Where methods of separating mixtures are used in practice. To purify substances, various methods of separating mixtures are used. Evaporation or crystallization
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In order to establish the properties of a substance, it is necessary to have it in its pure form, but substances do not occur in nature in their pure form. Each substance always contains a certain amount of impurities. A substance in which there are almost no impurities is called pure. They work with such substances in a scientific laboratory or school chemistry lab. Note that absolutely pure substances do not exist.
Mixtures include almost all natural substances, food (except salt, sugar, and some others), building materials, household chemicals, and many medicines and cosmetics.
Natural substances are mixtures, sometimes consisting of a very large number of different substances. For example, natural water always contains salts and gases dissolved in it. Sometimes a very small amount of an impurity can lead to a very strong change in some properties of the substance. For example, the content of only hundredths of iron or copper in zinc accelerates its interaction with hydrochloric acid hundreds of times. When one of the substances is in a predominant amount in a mixture, the entire mixture usually bears its name.
A component is each substance contained in a mixture.
Homogeneous mixtures.
Add a small portion of sugar to a glass of water and stir until all the sugar has dissolved. The liquid will taste sweet. Thus, the sugar did not disappear, but remained in the mixture. But we will not see its crystals, even when examining a drop of liquid through a powerful microscope.
Rice. 3. Homogeneous mixture (aqueous sugar solution)
The prepared mixture of sugar and water is homogeneous (Fig. 3); the smallest particles of these substances are evenly mixed in it.
Mixtures in which components cannot be detected with the naked eye are called homogeneous.
Water mixed with sand, chalk or clay freezes at a temperature of O 0 C and boils at 100 0 C.
Some types of heterogeneous mixtures have special names: foam (for example, polystyrene foam, soap suds), suspension (a mixture of water with a small amount of flour), emulsion (milk, well-shaken vegetable oil and water), aerosol (smoke, fog).
Rice. 5. Heterogeneous mixtures:
a - a mixture of water and sulfur;
b - a mixture of vegetable oil and water;
c - a mixture of air and water
There are different ways to separate mixtures. The choice of method for separating a mixture is influenced by the properties of the substances forming the mixture.
Let's take a closer look at each method:
Advocacy- a common method of purifying liquids from mechanical impurities insoluble in water, or liquid substances that are insoluble in each other and have different densities.
Sedimentation is used in the preparation of water for technological and domestic needs, treatment of sewage, dehydration and desalting of crude oil, and in many chemical technology processes. It is an important stage in the natural self-purification of natural and artificial reservoirs.
Filtration– separation of liquid from solid insoluble impurities; Liquid molecules pass through the pores of the filter, and large particles of impurities are retained.
Imagine that in front of you is a mixture of river sand and water. Determine the type of mixture. ( heterogeneous). Compare the physical properties of river sand and water. (These are substances that are insoluble in each other and have different densities). Suggest a method for separating this mixture ( filtering).
Action by magnet is a method of separating heterogeneous mixtures when one of the substances in the mixture is capable of being attracted by a magnet
Evaporation – this is a method of separating homogeneous mixtures, in which a solid soluble substance is released from a solution; when heated, the water evaporates, and crystals of the solid substance remain.
Distillation (Latin for "dropping") – This is a method of separating homogeneous mixtures, in which liquid mixtures are separated into fractions that differ in composition. It is carried out by partial evaporation of a liquid followed by condensation of steam. The distilled fraction (distillate) is enriched with relatively more volatile (low-boiling) substances, and the non-distilled liquid (bottoms) is enriched with relatively less volatile (high-boiling) substances.
In the laboratory, distillation is carried out using a special installation (Fig. 6). When a mixture of liquids is heated, the substance with the lowest boiling point boils first. Its vapor leaves the vessel, cools, condenses1, and the resulting liquid flows into the receiver. When this substance is no longer in the mixture, the temperature will begin to rise, and over time, another liquid component will boil. Non-volatile liquids remain in the vessel.
Rice. 6. Laboratory installation for distillation: a - conventional; b - simplified
1 - a mixture of liquids with different boiling points;
2 - thermometer;
3 - water refrigerator;
4 - receiver
Let's look at how some use methods separation of mixtures.
The filtration process underlies the operation of a respirator - a device that protects the lungs of a person working in a very dusty room. The respirator has filters that prevent dust from entering the lungs (Fig. 7). The simplest respirator is a bandage made of several layers of gauze. A vacuum cleaner also has a filter that removes dust from the air.
Rice. 7. Worker in a respirator
Conclude by what methods you can separate a mixture of soluble and insoluble substances in water.
Pure substancecontains particles only one type. Examples include silver (contains only silver atoms), sulfuric acid and carbon monoxide ( IV ) (contain only molecules of the corresponding substances). All pure substances have constant physical properties, for example, melting point (T pl) and boiling point ( ).
T baleA substance is not pure if it contains any quantity of one or more other substances –.
impurities
Contaminants lower the freezing point and raise the boiling point of a pure liquid. For example, if you add salt to water, the freezing point of the solution will decrease. Mixtures consist of two or more substances. Soil, sea water, air are all examples of different mixtures. Many mixtures can be separated into their component parts -
Components– based on the difference in their physical properties.
Traditional
methods that are used in laboratory practice to separate mixtures into individual components are:
filtration,
settling followed by decanting,
separation using a separating funnel,
centrifugation,
evaporation,
crystallization,
distillation (including fractional distillation),
chromatography, sublimation and others.Filtration. Filtration is used to separate liquids from small solid particles suspended in it.(Fig. 37), i.e. filtering liquid through finely porous materials –.
filters, which allow liquid to pass through and retain solid particles on their surface. (Fig. 37)A liquid that has passed through a filter and is freed from solid impurities in it is called filtrate
In laboratory practice it is often usedsmooth and folded paper(Fig. 38) , made from unglued filter paper.
To filter hot solutions (for example, for the purpose of recrystallization of salts), use a specialhot filter funnel. Filtration under vacuum is used to speed up filtration and more completely free the precipitate from the solution. For this purpose, a vacuum filtration device is assembled. (Fig.40) .It consists ofBunsen flask, porcelain Buchner funnel, safety bottle and vacuum pump
(usually water jet). In the case of filtering a suspension of a slightly soluble salt, the crystals of the latter can be washed with distilled water on a Buchner funnel to remove the original solution from their surface. For this purpose they usewasher .
(Fig.41). DecantationLiquids can be separated from insoluble solidsby decanting . (Fig.42)
This method can be used if the solid has a higher density than the liquid. For example, if river sand is added to a glass of water, then when it settles, it will settle to the bottom of the glass, because the density of sand is greater than water. Then the water can be separated from the sand simply by draining. This method of settling and then draining the filtrate is called decanting.Centrifugation. D To speed up the process of separating very small particles that form stable suspensions or emulsions in a liquid, the method is used centrifugation .This method can be used to separate mixtures of liquid and solid substances that differ in density. The division is carried out in .
manual or electric centrifuges (Fig.43)Separation of two immiscible liquids, having different densities and not forming stable emulsions, can be done using a separating funnel (Fig.44) 3 ) . This way you can separate, for example, a mixture of benzene and water. Benzene layer (density 3 = 0.879 g/cm
is located above a layer of water, which has a high density (= 1.0 g/cm ).
By opening the separatory funnel tap, you can carefully drain the bottom layer and separate one liquid from another. Evaporation
(Fig.45)- a method of purifying a substance based on the evaporation of a liquid when heated, followed by condensation of the resulting vapors. Purification of water from salts (or other substances, such as coloring agents) dissolved in it is called distillation. distillation, and the purified water itself is distilled.
Fractional distillation(Fig.46) used to separate mixtures of liquids with different boiling points. A liquid with a lower boiling point boils faster and passes through the fractional column(orreflux condenser).When this liquid reaches the top of the fractionation column, it entersfridge, cooled with water and throughtogethergoing toreceiver
(flask or test tube). 0 Fractional distillation can be used to separate, for example, a mixture of ethanol and water. Boiling point of ethanol 78 0 C, and water is 100
C. Ethanol evaporates more easily and is the first to pass through the refrigerator to the receiver. Sublimation –
The method is used to purify substances that, when heated, can transform from a solid state to a gaseous state, bypassing the liquid state. Next, the vapors of the substance being purified condense, and impurities that cannot sublimate are separated.
Topic: “Methods of separating mixtures” (8th grade)
Theoretical block. The definition of the concept “mixture” was given in the 17th century.: English scientist Robert Boyle
“A mixture is an integral system consisting of heterogeneous components.”
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Comparative characteristics of the mixture and pure substance |
Signs of comparison |
Pure substance |
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Mixture |
Constant |
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Fickle |
Substances |
Same |
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Various |
Physical properties |
Permanent |
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Fickle |
Energy change during formation |
Happening |
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Not happening |
Separation |
Through chemical reactions |
By physical methods
The mixtures differ from each other in appearance.
The classification of mixtures is shown in the table:
Let us give examples of suspensions (river sand + water), emulsions (vegetable oil + water) and solutions (air in a flask, table 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, and for the needs of pharmacology and medicine, pure substances are needed.
Various methods for separating mixtures are used to purify substances.
Evaporation is the separation of solids dissolved in a liquid by converting it into steam. Distillation-
In nature, water does not occur in its pure form (without salts). Ocean, sea, river, well and spring water are types of solutions of salts in water. However, people often need clean water that does not contain salts (used in car engines; in chemical production to obtain various solutions and substances; in making photographs). Such water is called distilled, and the method of obtaining it is called distillation.
Filtration - straining liquids (gases) through a filter in order to clean them from solid impurities.
These methods are based on differences in the physical properties of the components of the mixture.
Consider separation methods heterogeneous and homogeneous mixtures.
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Example of a mixture |
Separation method |
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Suspension - a mixture of river sand and water |
Advocacy Not happening defending based on different densities of substances. Heavier sand settles to the bottom. You can also separate the emulsion: separate the oil or vegetable oil from the water. In the laboratory this can be done using a separatory funnel. Petroleum or vegetable oil forms the top, lighter layer. |
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As a result of settling, dew falls out of the fog, soot settles out of the smoke, and cream settles in the milk. |
Filtration Separation of a mixture of water and vegetable oil by settling filtering A mixture of sand and table salt in water What is the basis for the separation of heterogeneous mixtures using |
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?On different solubility of substances in water and on different particle sizes. |
Only particles of substances comparable to them pass through the pores of the filter, while larger particles are retained on the filter. This way you can separate a heterogeneous mixture of table salt and river sand. Various porous substances can be used as filters: cotton wool, coal, baked clay, pressed glass and others. The filtration method is the basis for the operation of household appliances, such as vacuum cleaners. It is used by surgeons - gauze bandages; drillers and elevator workers - respiratory masks. Using a tea strainer to filter tea leaves, Ostap Bender, the hero of the work by Ilf and Petrov, managed to take one of the chairs from Ellochka the Ogress (“Twelve Chairs”). Separation of a mixture of starch and water by filtration Mixture of iron and sulfur powder
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Action by magnet or water |
Evaporation or crystallization The water evaporates, leaving salt crystals 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 boiling points of the solvent and the solute. If a substance, for example sugar, decomposes when heated, then the water is not completely evaporated - the solution is evaporated, and then sugar crystals are precipitated from the saturated solution. Sometimes it is necessary to remove impurities from solvents with a lower temperature boiling, 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 construct such a distiller: |
If you separate a mixture of alcohol and water, then the alcohol with boiling point = 78 °C will be distilled off first (collected in a receiving test tube), and water will remain in the test tube. Distillation is used to produce gasoline, kerosene, and gas oil from oil..
Separation of homogeneous mixtures
A special method for separating components, based on their different absorption by a certain substance, is
chromatography
Using chromatography, the Russian botanist M. S. Tsvet was the first to isolate chlorophyll from the green parts of plants. In industry and laboratories, starch, coal, limestone, and aluminum oxide are used instead of filter paper for chromatography. Are substances with the same degree of purification always required? For different purposes, substances with varying degrees of purification are required. Cooking water should be left to stand sufficiently to remove impurities and chlorine used to disinfect it. Water for drinking must first be boiled. And in chemical laboratories for preparing solutions and conducting experiments, in medicine, distilled water is needed, purified as much as possible from substances dissolved in it. Particularly 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 of expressing the composition of mixtures.
Mass fraction of the component in the mixture- the ratio of the number of moles (amount of substance) of a component to the total number of moles of all substances in the mixture. For example, if the mixture contains substances A, B and C, then:
χ ["chi"] component A = n component A / (n(A) + n(B) + n(C))
Molar ratio of components. Sometimes problems for a mixture indicate the molar ratio of its components.
For example:
n component A: n component 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"] = V component / V mixture
Practical block. Let's look at three examples of problems in which mixtures of metals react with salt
acid:Example 1.
When a mixture of copper and iron weighing 20 g was exposed to excess hydrochloric acid, 5.6 liters of gas (no.) 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.
Solution to example 1.
Finding the amount of hydrogen:
n = V / V m = 5.6 / 22.4 = 0.25 mol.
According to the reaction equation:
The amount of iron is also 0.25 mol. You can find its mass:
m Fe = 0.25 56 = 14 g.
Answer: 70% iron, 30% copper.Example 2.
When a mixture of aluminum and iron weighing 11 g was exposed to excess hydrochloric acid, 8.96 liters of gas (n.s.) 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 x to be the number of moles of one of the metals, and y to be the amount of substance of the second.
Let the amount of aluminum be x moles, and the amount of iron be x moles. Then we can express the amount of hydrogen released in terms of x and y:
27x + 56y = 11 (this is the second equation in the system).
So, we have a system of two equations:
It is much more convenient to solve such systems using the subtraction method, multiplying the first equation by 18:27x + 18y = 7.2
y = 3.8 / 38 = 0.1 mol (Fe)
x = 0.2 mol (Al)
Solution to example 1.
Solution to example 2.
n = V / V m = 8.96 / 22.4 = 0.4 mol.
2HCl = FeCl 2 +
We know the total amount of hydrogen: 0.4 mol.
Means, 1.5x + y = 0.4 (this is the first equation in the system). For a mixture of metals we need to express
masses
through the amount of substances.
m = Mn
So, the mass of aluminum
m Al = 27x,
mass of iron
m Fe = 56у,
and the mass of the entire mixture
m Fe = n M = 0.1 56 = 5.6 g
m Al = 0.2 27 = 5.4 g
ω Fe = m Fe / m mixture = 5.6 / 11 = 0.50909 (50.91%),
respectively,
ω Al = 100% − 50.91% = 49.09%
Answer: 50.91% iron, 49.09% aluminum.
Example 3.16 g of a mixture of zinc, aluminum and copper were treated with an excess of hydrochloric acid solution. In this case, 5.6 liters of gas (n.s.) 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 amounts 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 4.A mixture of iron, aluminum and copper was treated with an excess of cold concentrated sulfuric acid. In this case, part of the mixture dissolved, and 5.6 liters of gas (n.s.) were released. The remaining mixture was treated with an excess of sodium hydroxide solution. 3.36 liters of gas were released and 3 g of undissolved residue remained. Determine the mass and composition of the initial mixture of metals.
In this example, we must remember that cold concentrated sulfuric acid does not react with iron and aluminum (passivation), but does react with copper. This releases sulfur (IV) oxide.
With alkali reacts only aluminum- amphoteric metal (in addition to aluminum, zinc and tin also dissolve in alkalis, and beryllium can also be dissolved in hot concentrated alkali).
Solution to example 4.
(don’t forget that such reactions must be equalized using an electronic balance)
Since the molar ratio of copper and sulfur dioxide is 1:1, then copper is also 0.25 mol. You can find a mass of copper:
m Cu = n M = 0.25 64 = 16 g.Aluminum reacts with an alkali solution, resulting in the formation of a hydroxo complex of aluminum and hydrogen:
2Al + 2NaOH + 6H 2 O = 2Na + 3H 2Al 0 − 3e = Al 3+
2H + + 2e = H 2
Number of moles of hydrogen:
n H3 = 3.36 / 22.4 = 0.15 mol,
the molar ratio of aluminum and hydrogen is 2:3 and, therefore,
n Al = 0.15 / 1.5 = 0.1 mol.
Aluminum weight:
m Al = n M = 0.1 27 = 2.7 gThe remainder is iron, weighing 3 g. You can find the mass of the mixture:
m mixture = 16 + 2.7 + 3 = 21.7 g.Mass fractions of metals:
Only copper reacts with concentrated sulfuric acid, the number of moles of gas is:
n SO2 = V / Vm = 5.6 / 22.4 = 0.25 mol
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2H 2 SO 4 (conc.) = CuSO 4 + |
ω Cu = m Cu / m mixture = 16 / 21.7 = 0.7373 (73.73%)
ω Al = 2.7 / 21.7 = 0.1244 (12.44%)
ω Fe = 13.83%
Answer: 73.73% copper, 12.44% aluminum, 13.83% iron.
Example 5.21.1 g of a mixture of zinc and aluminum was dissolved in 565 ml of nitric acid solution containing 20 wt. %HNO 3 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, was 2.912 l (n.s.). Determine the composition of the resulting solution in mass percent. (RHTU)
The text of this problem clearly indicates the product of nitrogen reduction - a “simple substance”. Since nitric acid with metals does not produce hydrogen, it is nitrogen. Both metals dissolved in the acid.
The problem asks not the composition of the initial mixture of metals, but the composition of the resulting solution after the reactions. This makes the task more difficult.
Solution to example 5.
Determine the amount of gas substance:
n N2 = V / Vm = 2.912 / 22.4 = 0.13 mol.
Determine the mass of the nitric acid solution, the mass and amount of dissolved HNO3:
m solution = ρ V = 1.115 565 = 630.3 g
m HNO3 = ω m solution = 0.2 630.3 = 126.06 g
n HNO3 = m / M = 126.06 / 63 = 2 mol
Please note that since the metals have completely dissolved, it means - there was definitely 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.
We compose reaction equations ( don't forget about your electronic balance) and, for the convenience of calculations, we take 5x to be the amount of zinc, and 10y to be 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:
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12HNO 3 = 5Zn(NO 3) 2 + |
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Zn 0 − 2e = Zn 2+ |
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2N +5 + 10e = N 2 |
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36HNO3 = 10Al(NO3)3 + |
It is convenient to solve this system by multiplying the first equation by 90 and subtracting the first equation from the second.
x = 0.04, which means n Zn = 0.04 5 = 0.2 mol
y = 0.03, which means n Al = 0.03 10 = 0.3 mol
Let's check the mass of the mixture:
0.2 65 + 0.3 27 = 21.1 g.
Now let's move on to the composition of the solution. It will be convenient to rewrite the reactions again and write down above the reactions the amounts of all reacted and formed substances (except water):
The next question is: is there any nitric acid left in the solution and how much is left?
According to the reaction equations, the amount of acid that reacted:
n HNO3 = 0.48 + 1.08 = 1.56 mol,
those. the acid was in excess and you can calculate its remainder in the solution:
n HNO3 rest.
= 2 − 1.56 = 0.44 mol. So, in final solution
contains:
m Zn(NO3)2 = n M = 0.2 189 = 37.8 g
aluminum nitrate in an amount of 0.3 mol:
m Al(NO3)3 = n M = 0.3 213 = 63.9 g
excess nitric acid in an amount of 0.44 mol:
m HNO3 rest. = n M = 0.44 63 = 27.72 g
What is the mass of the final solution?
Let us remember 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):
Then for our task:
m new
solution = mass of acid solution + mass of metal alloy - mass of nitrogen
m N2 = n M = 28 (0.03 + 0.09) = 3.36 g
m new
solution = 630.3 + 21.1 − 3.36 = 648.04 g
ωZn(NO 3) 2 = m quantity / m solution = 37.8 / 648.04 = 0.0583
ωAl(NO 3) 3 = m volume / m solution = 63.9 / 648.04 = 0.0986
ω HNO3 rest. = m water / m solution = 27.72 / 648.04 = 0.0428Answer: 5.83% zinc nitrate, 9.86% aluminum nitrate, 4.28% nitric acid.
Example 6.
When 17.4 g of a mixture of copper, iron and aluminum was treated with an excess of concentrated nitric acid, 4.48 liters of gas (n.o.) were released, and when this mixture was exposed to the same mass of excess hydrochloric acid, 8.96 liters of gas (n.o.) were released. y.). Determine the composition of the initial mixture. (RHTU)
When solving this problem, we must remember, firstly, that concentrated nitric acid with an inactive metal (copper) produces NO 2, and iron and aluminum do not react with it. Hydrochloric acid, on the contrary, does not react with copper.Answer for example 6: 36.8% copper, 32.2% iron, 31% aluminum. Explanatory note. Pure substances and mixtures Methods separation mixtures. Pure substances and. Develop an understanding of pure substances and mixtures purification of substances: ... substances to various classes organic compounds. Characterize: basic
classes
organic compounds...Order of 2013 No. Work program for the academic subject "Chemistry" 8th grade (basic level 2 hours) Working programm mixtures Methods Assessing students' knowledge of opportunities and ways substances; formation of appropriate experimental skills... classification and chemical properties of basic substances
... Explanatory note, inorganic compounds, the formation of ideas about... mixtures Methods Document mixtures ways Methods. Objectives: To give the concept of pure substances and ; Consider classification mixtures Methods; Introduce students to ways... student and raises in front
class
card with the formula of an inorganic substance...
Here are the names of various chemical systems. Divide them into: mixtures; pure substances and true solutions.
Distilled water
Sea water
Oxygen
Silver
Sodium chloride solution for injection
Hydrogen
Cast iron
Carbon dioxide
Air
Oil in water emulsion
Lead
Suggest ways to separate mixtures: a) water and sand; b) wood and iron filings; c) water and ink; d) water and oil.
Pure substances and mixtures.
In everyday life, each of us encounters many mixtures of substances, dealing not only with pure, but also with contaminated substances. It is important to be able to distinguish between these concepts and be able to determine by specific characteristics what you are dealing with: a pure or contaminated substance, an individual substance or a mixture of substances. After all, a person wants to drink only water that does not contain harmful impurities. We want to breathe air that is not polluted by gases harmful to health. In medicine and the production of pharmaceuticals, the problem of obtaining and using pure substances is especially relevant.
Let's get acquainted with the basic terms of the lesson.
Pure substance- this is what is formed when two or more substances with different properties are mixed.
The substances that make up the mixture are called components. For example, air is a mixture of gases: nitrogen, oxygen, carbon dioxide and others.
If the mass of one component is tens of times less than the mass of another component of the mixture, then it is called admixture. The substance is said to be contaminated. For example, the air may be polluted with carbon monoxide, a product of incomplete combustion of organic compounds, in particular gasoline. By the way, gasoline is a mixture of organic substances - hydrocarbons.
CLASSIFICATION OF MIXTURES
The mixtures differ from each other in appearance. For example, salt water (a mixture of table salt and water) and a mixture of river sand and water. In the first case, it is impossible to see the solid-liquid interface. Such a mixture is called homogeneous (or homogeneous). Other examples of homogeneous mixtures are vinegar (a mixture of acetic acid and water), air, and sugar syrup.
A mixture of river sand and water is classified as heterogeneous (or heterogeneous) mixtures, because the composition of such a mixture is not the same at different points in the volume. Mixtures of clay and water, gasoline and water are heterogeneous.
Basically, everything that surrounds us is a mixture of substances. Moreover, there are no substances that are absolutely free of impurities.
But it is customary to talk about the relative purity of a substance, i.e. substances have different degrees of purity.
Purity of the substance
If impurities are not detected when a substance is used for technical purposes, then the substance is called technically clean. For example, the substance from which purple ink is made may contain impurities. But if these impurities do not affect the quality of the ink in any way, then it is technically pure.
If impurities are not detected by chemical reactions, then the substance is classified as chemically pure. For example, this is distilled water.
Signs of a substance’s individuality
A pure substance is sometimes called an individual substance, because it has strictly defined properties. For example, only distilled water has a melting point of 0 C, a boiling point of 100 C and is tasteless and odorless.
Do the properties of substances in a mixture change? To answer this question, let's conduct a simple experiment. Mix sulfur and iron powders. We know that iron is attracted by a magnet, but sulfur is not. Did iron retain its properties after mixing with sulfur?
CONCLUSION: The properties of the substances in the mixture do not change. Knowledge about the properties of the components of a mixture is used to separate mixtures and purify substances.
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 all the components that make up the mixture in pure form. When purifying a substance, obtaining impurities in 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 of different densities. For example, this is how you can separate a mixture of iron and wood filings (wood filings will float in water, while iron filings will settle).
A mixture of vegetable oil and water can also be separated by settling, since oil does not dissolve in water and has a lower density (Fig. 2). Thus, by settling it is possible to separate mixtures of liquids that are insoluble in each other and have different densities.
Rice. 2. Separation of a mixture of vegetable oil and water by settling
Filtration
To separate a mixture of table salt and river sand, you can use the settling method (when mixed with water, the salt will dissolve and the sand will settle), but it will be more reliable to separate the sand from the salt solution using another method - the filtration method.
Filtering this mixture can be done 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 filtrate (Fig. 3).
Rice. 3. Using the filtration method to separate river sand from the salt solution
Filtration can be carried out not only using filter paper, but also using other porous or bulk 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 temperatures above 1500 C, and sulfur at about 120 C. Molten sulfur can be separated from iron powder using heated glass wool.
1. Fill in the blanks in the text using the words “components”, “differences”, “two”, “physical”.
A mixture can be prepared by mixing at least two substances. Mixtures can be separated into their individual components using physical methods based on differences in the physical properties of the components.
2. Complete the sentences.
a) The settling method is based on The fact is that the particles of the solid substance are quite large; they quickly settle to the bottom, and the liquid can be carefully drained from the sediment.
b) The centrifugation method is based on the action of centrifugal force - heavier particles settle, and light ones end up on top.
c) The filtering method is based on passing a solution of a solid through a filter where the solid particles are retained on the filter.
3. Fill in the missing word:
a) flour and granulated sugar - a sieve; sulfur and iron filings - magnet.
b) water and sunflower oil - separating funnel; water and river sand - filter.
c) air and dust - respirator; air and poisonous gas - absorbent.
4. Make a list of necessary filtration equipment.
a) paper filter
b) a glass with a solution
c) glass funnel
d) clean glass
d) glass rod
e) tripod with foot
5. Laboratory experience. Making regular and pleated filters from filter paper or paper napkin.
Which filter do you think the solution will pass through faster - a regular one or a folded one? Why?
Through folded - the filtration contact area is larger than that of a conventional filter.
6. Suggest ways to separate the mixtures shown in Table 16.
Methods for separating some mixtures
7. Home experience. Adsorption of Pepsi-Cola colorants by activated carbon.
Reagents and equipment: carbonated drink, activated carbon; pan, funnel, filter paper, electric (gas) stove.
Progress. Pour half a cup (100 ml) of carbonated drink into the pan. Add 5 tablets of activated carbon there. Heat the pan for 10 minutes on the stove. Filter the carbon. Explain the results of the experiment.
The solution became discolored due to the absorption of colorants by activated carbon.
8. Home experience. Adsorption of odorous vapors by corn sticks.
Reagents and equipment: corn sticks, perfume or cologne; 2 identical glass jars with lids.
Progress. Place a drop of perfume into two glass jars. Place 4-5 corn sticks in one of the jars. Close both jars with lids. Shake the jar containing the corn sticks a little. For what?
To increase the rate of adsorption.
Open both jars. Explain the results of the experiment.
There is no smell in the jar where the corn sticks were, since it adsorbed the smell of perfume.