Solution called a homogeneous mixture of two or more components.

The substances by mixing which produce a solution are called components.

Among the components of the solution there are solute, which may be more than one, and solvent. For example, in the case of a solution of sugar in water, the sugar is the solute and the water is the solvent.

Sometimes the concept of solvent can be applied equally to any of the components. For example, this applies to those solutions that are obtained by mixing two or more liquids that are ideally soluble in each other. So, in particular, in a solution consisting of alcohol and water, both alcohol and water can be called a solvent. However, most often in relation to aqueous solutions, the solvent is traditionally called water, and the solute is the second component.

As quantitative characteristics composition of a solution, the concept most often used is mass fraction substances in solution. The mass fraction of a substance is the ratio of the mass of this substance to the mass of the solution in which it is contained:

Where ω (in-va) – mass fraction of the substance contained in the solution (g), m(v-va) – mass of the substance contained in the solution (g), m(r-ra) – mass of the solution (g).

From formula (1) it follows that the mass fraction can take values ​​from 0 to 1, that is, it is a fraction of unity. In this regard, the mass fraction can also be expressed as a percentage (%), and it is in this format that it appears in almost all problems. The mass fraction, expressed as a percentage, is calculated using a formula similar to formula (1) with the only difference being that the ratio of the mass of the dissolved substance to the mass of the entire solution is multiplied by 100%:

For a solution consisting of only two components, the mass fraction of solute ω(s.v.) and the mass fraction of solvent ω(solvent) can be calculated accordingly.

The mass fraction of the solute is also called solution concentration.

For a two-component solution, its mass is the sum of the masses of the solute and the solvent:

Also, in the case of a two-component solution, the sum of the mass fractions of the solute and the solvent is always 100%:

It is obvious that, in addition to the formulas written above, you should also know all those formulas that are directly mathematically derived from them. For example:

It is also necessary to remember the formula connecting the mass, volume and density of a substance:

m = ρ∙V

and you also need to know that the density of water is 1 g/ml. For this reason, the volume of water in milliliters is numerically equal to mass water in grams. For example, 10 ml of water has a mass of 10 g, 200 ml - 200 g, etc.

In order to successfully solve problems, in addition to knowledge of the above formulas, it is extremely important to bring the skills of their application to automaticity. This can only be achieved by resolving large quantity various tasks. Problems from real life Unified State Examinations on the topic “Calculations using the concept of “mass fraction of a substance in solution”” can be solved.

Examples of problems involving solutions

Example 1

Calculate the mass fraction of potassium nitrate in a solution obtained by mixing 5 g of salt and 20 g of water.

Solution:

The solute in our case is potassium nitrate, and the solvent is water. Therefore, formulas (2) and (3) can be written respectively as:

From the condition m(KNO 3) = 5 g, and m(H 2 O) = 20 g, therefore:

Example 2

What mass of water must be added to 20 g of glucose to obtain a 10% glucose solution.

Solution:

From the conditions of the problem it follows that the solute is glucose and the solvent is water. Then formula (4) can be written in our case as follows:

From the condition we know the mass fraction (concentration) of glucose and the mass of glucose itself. Having designated the mass of water as x g, we can write, based on the formula above, the following equation equivalent to it:

Solving this equation we find x:

those. m(H 2 O) = x g = 180 g

Answer: m(H 2 O) = 180 g

Example 3

150 g of a 15% solution of sodium chloride was mixed with 100 g of a 20% solution of the same salt. What is the mass fraction of salt in the resulting solution? Please indicate your answer to the nearest integer.

Solution:

To solve problems for preparing solutions, it is convenient to use the following table:

where m r.v. , m solution and ω r.v. - values ​​of the mass of the dissolved substance, the mass of the solution and the mass fraction of the dissolved substance, respectively, individual for each of the solutions.

From the condition we know that:

m (1) solution = 150 g,

ω (1) r.v. = 15%,

m (2) solution = 100 g,

ω (1) r.v. = 20%,

Let's insert all these values ​​into the table, we get:

We should remember the following formulas necessary for calculations:

ω r.v. = 100% ∙ m r.v. /m solution, m r.v. = m solution ∙ ω solution /100% , m solution = 100% ∙ m solution /ω r.v.

Let's start filling out the table.

If only one value is missing from a row or column, it can be counted. The exception is the line with ω r.v., knowing the values ​​in two of its cells, the value in the third cannot be calculated.

Only one cell in the first column is missing a value. So we can calculate it:

m (1) r.v. = m (1) solution ∙ ω (1) solution /100% = 150 g ∙ 15%/100% = 22.5 g

Similarly, we know the values ​​in two cells of the second column, which means:

m (2) r.v. = m (2) solution ∙ ω (2) solution /100% = 100 g ∙ 20%/100% = 20 g

Let's enter the calculated values ​​into the table:

Now we know two values ​​in the first line and two values ​​in the second line. This means we can calculate the missing values ​​(m (3)r.v. and m (3)r-ra):

m (3)r.v. = m (1)r.v. + m (2)r.v. = 22.5 g + 20 g = 42.5 g

m (3) solution = m (1) solution + m (2) solution = 150 g + 100 g = 250 g.

Let's enter the calculated values ​​into the table and get:

Now we have come close to calculating the desired value of ω (3)r.v. . In the column where it is located, the contents of the other two cells are known, which means we can calculate it:

ω (3)r.v. = 100% ∙ m (3)r.v. /m (3) solution = 100% ∙ 42.5 g/250 g = 17%

Example 4

50 ml of water was added to 200 g of 15% sodium chloride solution. What is the mass fraction of salt in the resulting solution. Please indicate your answer to the nearest hundredth of _______%

Solution:

First of all, we should pay attention to the fact that instead of the mass of added water, we are given its volume. Let's calculate its mass, knowing that the density of water is 1 g/ml:

m ext. (H 2 O) = V ext. (H 2 O) ∙ ρ (H2O) = 50 ml ∙ 1 g/ml = 50 g

If we consider water as a 0% sodium chloride solution containing 0 g of sodium chloride, the problem can be solved using the same table as in the example above. Let's draw a table like this and insert the values ​​we know into it:

There are two known values ​​in the first column, so we can calculate the third:

m (1)r.v. = m (1)r-ra ∙ ω (1)r.v. /100% = 200 g ∙ 15%/100% = 30 g,

In the second line, two values ​​are also known, which means we can calculate the third:

m (3) solution = m (1) solution + m (2) solution = 200 g + 50 g = 250 g,

Let's enter the calculated values ​​into the appropriate cells:

Now two values ​​in the first line have become known, which means we can calculate the value of m (3)r.v. in the third cell:

m (3)r.v. = m (1)r.v. + m (2)r.v. = 30 g + 0 g = 30 g

ω (3)r.v. = 30/250 ∙ 100% = 12%.

From a chemistry course we know that the mass fraction is the content of a certain element in a substance. It would seem that such knowledge is of no use to an ordinary summer resident. But don’t rush to close the page, since the ability to calculate the mass fraction for a gardener can be very useful. However, in order not to get confused, let's talk about everything in order.

What is the essence of the concept of “mass fraction”?

The mass fraction is measured in percentages or simply in tenths. Just above we talked about classical definition, which can be found in reference books, encyclopedias or school chemistry textbooks. But it is not so easy to understand the essence of what has been said. So, suppose we have 500 g of some complex substance. Complex in this case means that it is not homogeneous in its composition. By and large, any substances we use are complex, even simple table salt, the formula of which is NaCl, that is, it consists of sodium and chlorine molecules. If we continue our reasoning using the example table salt, then we can assume that 500 grams of salt contains 400 g of sodium. Then its mass fraction will be 80% or 0.8.

Why does a summer resident need this?

I think you already know the answer to this question. Preparation of all kinds of solutions, mixtures, etc. is an integral part economic activity any gardener. Fertilizers, various nutrient mixtures, as well as other drugs, for example, growth stimulants “Epin”, “Kornevin”, etc. are used in the form of solutions. In addition, it is often necessary to mix dry substances, such as cement, sand and other components, or ordinary garden soil with a purchased substrate. Moreover, the recommended concentration of these agents and drugs in prepared solutions or mixtures in most instructions is given in mass fractions.

Thus, knowing how to calculate the mass fraction of an element in a substance will help the summer resident to correctly prepare the necessary solution of fertilizer or nutrient mixture, and this, in turn, will certainly affect the future harvest.

Calculation algorithm

So, the mass fraction of an individual component is the ratio of its mass to the total mass of the solution or substance. If the result obtained needs to be converted into a percentage, then it must be multiplied by 100. Thus, the formula for calculating the mass fraction can be written as follows:

W = Mass of substance / Mass of solution

W = (Mass of substance / Mass of solution) x 100%.

Example of determination of mass fraction

Let's assume that we have a solution for the preparation of which 5 g of NaCl was added to 100 ml of water, and now we need to calculate the concentration of table salt, that is, its mass fraction. We know the mass of the substance, and the mass of the resulting solution is the sum of two masses - salt and water and is equal to 105 g. Thus, we divide 5 g by 105 g, multiply the result by 100 and get the desired value of 4.7%. This is exactly the concentration the saline solution will have.

More practical task

In practice, a summer resident more often has to deal with problems of a different kind. For example, it is necessary to prepare an aqueous solution of some fertilizer, the concentration of which by weight should be 10%. In order to accurately observe the recommended proportions, you need to determine how much of the substance is needed and in what volume of water it will need to be dissolved.

Solving the problem begins in reverse order. First, you should divide the mass fraction expressed as a percentage by 100. As a result, we obtain W = 0.1 - this is the mass fraction of the substance in units. Now let's denote the amount of substance as x, and the final mass of the solution as M. In this case, the last value is made up of two terms - the mass of water and the mass of fertilizer. That is, M = Mv + x. So we get a simple equation:

W = x / (Mw + x)

Solving it for x, we get:

x = W x Mv / (1 – W)

Substituting the available data, we obtain the following relationship:

x = 0.1 x MV / 0.9

Thus, if we take 1 liter (that is, 1000 g) of water to prepare a solution, then to prepare a solution of the required concentration we will need approximately 111-112 g of fertilizer.

Solving dilution or addition problems

Suppose we have 10 liters (10,000 g) of ready-made aqueous solution with a concentration of a certain substance in it W1 = 30% or 0.3. How much water will need to be added to it to reduce the concentration to W2 = 15% or 0.15? In this case, the formula will help:

Мв = (W1х М1 / W2) – М1

Substituting the initial data, we find that the amount of added water should be:
Mv = (0.3 x 10,000 / 0.15) – 10,000 = 10,000 g

That is, you need to add the same 10 liters.

Now imagine the inverse problem - there are 10 liters of an aqueous solution (M1 = 10,000 g) with a concentration of W1 = 10% or 0.1. You need to get a solution with a mass fraction of fertilizer W2 = 20% or 0.2. How much will need to be added starting material? To do this you need to use the formula:

x = M1 x (W2 – W1) / (1 – W2)

Substituting the original values, we get x = 1,125 g.

Thus, knowledge of the simplest basics of school chemistry will help the gardener to correctly prepare fertilizer solutions, nutrient substrates from several elements or mixtures for construction work.

The concept of “share” is probably already familiar to you.

For example, the piece of watermelon shown in the figure is one quarter of the whole watermelon, that is, its share is 1/4 or 25%.

To better understand what a mass fraction is, imagine a kilogram of sweets (1000g) that a mother bought for her three children. From this kilogram myself youngest child they got half of all the candies (unfair of course!). The eldest - only 200g, and the middle - 300g.

This means that the mass fraction of sweets for the youngest child will be half, or 1/2, or 50%. The middle child will have 30%, and the oldest will have 20%. It should be emphasized that the mass fraction can be a dimensionless quantity (quarter, half, third, 1/5, 1/6, etc.), or can be measured as a percentage (%). When solving calculation problems, it is better to convert the mass fraction into a dimensionless quantity.

Mass fraction of substance in solution

Any solution consists of a solvent and a solute. Water is the most common inorganic solvent. Organic solvents can be alcohol, acetone, diethyl ether, etc. If the problem statement does not indicate a solvent, the solution is considered aqueous.

The mass fraction of the dissolved substance is calculated by the formula:

$\omega_\text(in-va)=\dfrac(m_\text(in-va))(m_\text(r-ra))(\cdot 100\%)$

Let's look at examples of problem solving.

How many grams of sugar and water do you need to take to prepare 150g of a 10% sugar solution?

Solution

m(solution)=150g

$\omega$(sugars)=10%=0.1

m(sugar)=?

m(sugars) = $\omega\textrm((sugars)) \cdot m(p-pa) = 0.1 \cdot 150 \textrm(g) = 15 \textrm(g)$

m(water)=m(solution) - m(sugar) = 150g - 15g=135g.

ANSWER: you need to take 15g of sugar and 135g of water.

350 ml solution. and density 1.142 g/ml contains 28 g of sodium chloride. Find the mass fraction of salt in the solution.

Solution

V(solution)=350 ml.

$\rho$(solution)=1.142 g/ml

$\omega(NaCl)$=?

m(solution) =V(solution) $\cdot \rho$(solution)=350 ml $\cdot$ 1.142 g/ml=400g

$\omega(NaCl)=\dfrac(m(NaCl))(m\textrm((solution)))=\dfrac(28\textrm(g)) (400\textrm(g)) = 0.07 $=7%

ANSWER: mass fraction of sodium chloride $\omega(NaCl)$=7%

MASS FRACTION OF AN ELEMENT IN A MOLECULE

Formula chemical substance, for example $H_2SO_4$, carries a lot of important information. It denotes either an individual molecule of a substance, which is characterized by relative atomic mass, or 1 mole of a substance, which is characterized by molar mass. The formula shows quality (consists of hydrogen, sulfur and oxygen) and quantitative composition(consists of two hydrogen atoms, a sulfur atom and four oxygen atoms). By chemical formula you can find the mass of the molecule as a whole ( molecular weight), and also calculate the ratio of the masses of elements in the molecule: m(H) : m(S) : m(O) = 2: 32: 64 = 1: 16: 32. When calculating the ratios of the masses of elements, it is necessary to take into account their atomic mass and quantity corresponding atoms: $m(H_2)=1*2=2$, $m(S)=32*1=32$, $m(O_4)=16*4=64$

The principle of calculating the mass fraction of an element is similar to the principle of calculating the mass fraction of a substance in a solution and is found using a similar formula:

$\omega_\text(elements)=\dfrac(Ar_(\text(elements))\cdot n_(\textrm(atoms)))(m_\text(molecules))(\cdot 100\%) $

Find the mass fraction of elements in sulfuric acid.

Solution

Method 1 (proportion):

Let's find the molar mass of sulfuric acid:

$M(H_2SO_4) = 1\cdot 2 + 32 + 16 \cdot 4=98\hspace(2pt)\textrm(g/mol)$

One molecule of sulfuric acid contains one sulfur atom, which means the mass of sulfur in sulfuric acid will be: $m(S) = Ar(S) \cdot n(S) = 32\textrm(g/mol) \cdot 1$= 32g/mol

Let's take the mass of the entire molecule as 100%, and the mass of sulfur as X% and make up the proportion:

$M(H_2SO_4)$=98 g/mol - 100%

m(S) = 32g/mol - X%

Where does $X=\dfrac(32\textrm(g/mol) \cdot 100\%)(98\textrm(g/mol)) =32, 65\% =32\%$

Method 2 (formula):

$\omega(S)=\dfrac(Ar_(\text(elements))\cdot n_(\textrm(atoms)))(m_\text(molecules))(\cdot 100\%)=\dfrac( Ar(S)\cdot 1)(M(H_2SO_4))(\cdot 100\%)=\dfrac(32\textrm(g/mol)\cdot 1)(98\textrm(g/mol))(\cdot 100\%) \approx32, 7\%$

Similarly, using the formula, we calculate the mass fractions of hydrogen and oxygen:

$\omega(H)=\dfrac(Ar(H)\cdot 2)(M(H_2SO_4))(\cdot 100\%)=\dfrac(1\textrm(g/mol)\cdot 2)(98\ textrm(g/mol))(\cdot 100\%)\approx2\%$

$\omega(O)=\dfrac(Ar(O)\cdot 4)(M(H_2SO_4))(\cdot 100\%)=\dfrac(16\textrm(g/mol)\cdot 4)(98\ textrm(g/mol))(\cdot 100\%)\approx65, 3\%$

The mass fraction of an element ω(E)% is the ratio of the mass of a given element m (E) in a given molecule of a substance to the molecular mass of this substance Mr (in-va).

The mass fraction of an element is expressed in fractions of a unit or as a percentage:

ω(E) = m (E) / Mr(in-va) (1)

ω% (E) = m(E) 100%/Mr(in-va)

The sum of the mass fractions of all elements of a substance is equal to 1 or 100%.

As a rule, to calculate the mass fraction of an element, they take a portion of a substance equal to the molar mass of the substance, then the mass of a given element in this portion is equal to its molar mass multiplied by the number of atoms of a given element in the molecule.

So, for a substance A x B y in fractions of unity:

ω(A) = Ar(E) X / Мr(in-va) (2)

From proportion (2) we derive calculation formula to determine the indices (x, y) in the chemical formula of a substance, if the mass fractions of both elements and the molar mass of the substance are known:

X = ω%(A) Mr(in-va) / Ar(E) 100% (3)

Dividing ω% (A) by ω% (B), i.e. transforming formula (2), we obtain:

ω(A) / ω(B) = X Ar(A) / Y Ar(B) (4)

Calculation formula (4) can be transformed as follows:

X: Y = ω%(A) / Ar(A) : ω%(B) / Ar(B) = X(A) : Y(B) (5)

Calculation formulas (3) and (5) are used to determine the formula of a substance.

If the number of atoms in a molecule of a substance for one of the elements and its mass fraction are known, the molar mass of the substance can be determined:

Mr(v-va) = Ar(E) X / W(A)

Examples of solving problems on calculating the mass fractions of chemical elements in a complex substance

Calculation of mass fractions of chemical elements in a complex substance

Example 1. Determine the mass fractions of chemical elements in sulfuric acid H 2 SO 4 and express them as percentages.

Solution

1. Calculate the relative molecular weight of sulfuric acid:

Mr (H 2 SO 4) = 1 2 + 32 + 16 4 = 98

2. Calculate the mass fractions of elements.

To do this, the numerical value of the mass of the element (taking into account the index) is divided by the molar mass of the substance:

Taking this into account and denoting the mass fraction of an element with the letter ω, calculations of mass fractions are carried out as follows:

ω(H) = 2: 98 = 0.0204, or 2.04%;

ω(S) = 32: 98 = 0.3265, or 32.65%;

ω(O) = 64: 98 =0.6531, or 65.31%

Example 2. Determine the mass fractions of chemical elements in aluminum oxide Al 2 O 3 and express them as percentages.

Solution

1. Calculate the relative molecular weight of aluminum oxide:

Mr(Al 2 O 3) = 27 2 + 16 3 = 102

2. Calculate the mass fractions of elements:

ω(Al) = 54: 102 = 0.53 = 53%

ω(O) = 48: 102 = 0.47 = 47%

How to calculate the mass fraction of a substance in a crystalline hydrate

Mass fraction of a substance is the ratio of the mass of a given substance in a system to the mass of the entire system, i.e. ω(X) = m(X) / m,

where ω(X) is the mass fraction of substance X,

m(X) - mass of substance X,

m - mass of the entire system

Mass fraction is a dimensionless quantity. It is expressed as a fraction of a unit or as a percentage.

Example 1. Determine the mass fraction of water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O.

Solution

The molar mass of BaCl 2 2H 2 O is:

M(BaCl 2 2H 2 O) = 137+ 2 35.5 + 2 18 = 244 g/mol

From the formula BaCl 2 2H 2 O it follows that 1 mol of barium chloride dihydrate contains 2 mol H 2 O. From this we can determine the mass of water contained in BaCl 2 2H 2 O:

m(H2O) = 2 18 = 36 g.

We find the mass fraction of water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O.

ω(H 2 O) = m(H 2 O)/m(BaCl 2 2H 2 O) = 36 / 244 = 0.1475 = 14.75%.

Example 2. Silver weighing 5.4 g was isolated from a rock sample weighing 25 g containing the mineral argentite Ag 2 S. Determine the mass fraction of argentite in the sample.

We determine the amount of silver substance found in argentite: n(Ag) = m(Ag) / M(Ag) = 5.4 / 108 = 0.05 mol. From the formula Ag 2 S it follows that the amount of argentite substance is half as much as the amount of silver substance. Determine the amount of argentite substance: n(Ag 2 S) = 0.5 n(Ag) = 0.5 0.05 = 0.025 mol We calculate the mass of argentite: m(Ag 2 S) = n(Ag 2 S) M(Ag2S) = 0.025 248 = 6.2 g. Now we determine the mass fraction of argentite in a rock sample weighing 25 g. ω(Ag 2 S) = m(Ag 2 S) / m = 6.2/25 = 0.248 = 24.8%.