In chemistry lessons, quite often it is necessary to solve problems that use mathematical methods and techniques that cause difficulties for students, and the chemistry teacher has to take on the functions of a mathematics teacher and, at the same time, problems with a chemical content, using special terms, are difficult to explain without special training for mathematics teachers. Thus, the idea was born to prepare and conduct a series of optional classes jointly by a teacher of chemistry and mathematics to solve problems on a mixture with students of grade 9.

TOPIC: SOLVING PROBLEMS USING THE CONCEPT “MASS FRACTION OF DISSOLVED SUBSTANCE. DILUTION AND CONCENTRATION OF SOLUTIONS” (INTEGRATION OF CHEMISTRY AND ALGEBRA)

GOALS :

EQUIPMENT: COMPUTER, MULTIMEDIA BOX, SCREEN, PRESENTATION.

Chemistry teacher: The quantitative composition of a solution is expressed by its concentration, which has different forms expressions. Most often, mass concentration or mass fraction of a solute is used. Recall the mathematical formula for expressing the mass fraction of a dissolved substance.

1. The mass fraction of the dissolved substance is denoted by - W r.v.
2. The mass fraction of a solute is the ratio of the mass of the solute to the mass of the solution: W (r.v.) \u003d m (r.v.) / m (p-ra) x 100%.
3. The mass of the solution is the sum of the mass of the solute and the mass of the solvent: m (r-ra) \u003d m (r.v.) + m (r-la)
4. Formula for mass fraction solute will look like this: W (r.v.) \u003d m (r.v.) / m (r.v.) + m (p-la) x 100%
5. Let's transform this formula and express the mass of the dissolved substance and the mass of the solution: m (r.v.) \u003d w (r.v.) x m (p-ra) / 100%, m (p-ra) \u003d m (r.v. )/w(r.h.) x 100%

Chemistry teacher: I propose to solve the problem using the proposed formulas.

A task. How many grams of iodine and alcohol should be taken to prepare 500 grams of 5% iodine tincture?

ANSWER: m (I2) \u003d 25 g, m (alcohol) \u003d 475 g.

Chemistry teacher: Very often in the work of chemical laboratories it is necessary to prepare solutions with a certain mass fraction of a solute by mixing two solutions or diluting a strong solution with water. Before preparing the solution, certain arithmetic calculations must be carried out.

A task. 100 grams of a solution with a mass fraction of a certain substance of 20% and 50 grams of a solution with a mass fraction of this substance of 32% are mixed. Calculate the mass fraction of the solute in the newly obtained solution.

Chemistry teacher: Let's solve this problem using the mixing rule.

Let's write the condition of the problem in the table:

Let's solve the problem using the mixing rule:

• m 1 w 1 + m 2 w 2 \u003d m 3 w 3
• m 1 w 1 + m 2 w 2 \u003d (m 1 + m 2) w 3
• m 1 w 1 + m 2 w 2 \u003d m 1 w 3 + m 2 w 3
• m 1 w 1 -m 1 w 3 \u003d m 2 w 2 -m 2 w 2
• m 1 (w 1 -w 3) \u003d m 2 (w 3 -w 2)
• m 1 /m 2 \u003d (w 3 -w 2) / (w 1 -w 3)

The ratio of the mass of the first solution to the mass of the second is equal to the ratio of the difference in the mass fractions of the mixture and the second solution to the difference in the mass fractions of the first solution and the mixture:

ANSWER: the mass fraction of the dissolved substance in the newly obtained solution is 24%.

Math teacher: This problem can be solved using algebraic transformations:

1. Find the mass of the dissolved substance in each of the solutions:

20% of 100 g 32% of 50 g

0.2x100=20(g) 0.32x50=16(g)

2. Find the mass of the dissolved substance in the mixture:

3. Find the mass of the solution:

4. Let the concentration of the resulting solution be x%, then the mass of the dissolved substance in the mixture:

0.01Xx150=1.5X

5. Let's make an equation and solve it:

ANSWER: the concentration of the resulting solution is 24%.

Chemistry teacher: In the course of chemistry there are problems that can be solved only by the method of systems of equations

Task: A 30% solution of hydrochloric acid was mixed with a 10% solution of the same acid and 600 grams of a 15% solution were obtained. How many grams of each solution was taken?

• W 1 \u003d 30% \u003d 0.3
• W 2 \u003d 10% \u003d 0.1
• W 3 \u003d 15% \u003d 0.15
• m 3 (r-ra) \u003d 600 g.

• m1(r-ra)=?
• m2(r-ra)=?

Mathematic teacher: Let's introduce the notation:

Calculate the masses of dissolved substances:

• m 1 \u003d 0.3X,
• m 2 \u003d 0.1Y,
• m 3 \u003d 600 g. x 0.15 \u003d 90 g.

Let's make a system of equations:

Solve the underlined equation:

180-0.3Y+0.1Y=90

• if Y=450 g., then X=600 g.-450 g.=150 g.

• weight of 1 solution = 150 g.
• weight 2 r-ra = 450g.

Chemistry teacher. Let's solve the same problem by the mixing method. What answer did you get? (Answers converge).

HOMEWORK.

• In what mass should 20% and 5% solutions of the same substance be mixed in order to obtain a 10% solution?

SOLUTION ALGORITHM:

• 1.Enter letter designations for mass solutions.
• 2. Calculate the masses of dissolved substances in the first, second solution and mixture.
• 3. Compose a system of equations and solve it.
• 4. Write down the answer.

1. Fill in the blanks.

a) Solution = solute+ solvent ;

b) m (solution) = m (solute)+ m (solvent).

2. Write a definition using the following words:

mass fraction, substance, mass, solution, to mass, ratio, in solution, substance, dissolved.

Answer: The mass fraction of a substance in a solution is the ratio of the mass of the solute to the mass of the solution.

3. Make formulas using the notation of velchins.

4. What is the mass fraction of the solute if it is known that 80 g of the solution contains 20 g of salt?

5. Determine the masses of salt and water that will be required to prepare 300 g of a solution with a mass fraction of salt of 20%.

6. Calculate the mass of water required to prepare 60 g of a 10% salt solution.

7. The pharmacy sells powder "Regidron", which is used for dehydration. One packet of powder contains 3.5 g of sodium chloride, 2.5 g of potassium chloride, 2.9 g of sodium citrate and 10 g of glucose. The contents of the package are dissolved in 1 liter of water. Determine the mass fractions of all components of the Regidron powder in the resulting solution.

8. 300 g of water was added to 500 g of a 20% glucose solution. Calculate the mass fraction of glucose in the new solution.

9. To 400 g of a 5% solution table salt added 50 g of salt. Calculate the mass fraction of sodium chloride in the new solution.

10. Two salt solutions were poured: 100 g of 20% and 450 g of 10%. Calculate the mass fraction of salt in the new solution.

Instruction

The mass fraction of a substance is found by the formula: w \u003d m (c) / m (cm), where w is the mass fraction of the substance, m (c) is the mass of the substance, m (cm) is the mass of the mixture. If it is dissolved, then it looks like this: w \u003d m (c) / m (p-ra), where m (p-ra) is the mass of the solution. The mass of the solution, if necessary, can also be found: m (p-ra) \u003d m (c) + m (p-la), where m (p-la) is the mass of the solvent. If desired, the mass fraction can be multiplied by 100%.

If the value of the mass is not given in the condition of the problem, then it can be calculated using several formulas, the data in the condition will help you choose the right one. The first formula for: m = V * p, where m is mass, V is volume, p is density. The following formula looks like this: m \u003d n * M, where m is the mass, n is the amount of substance, M is the molar mass. The molar mass, in turn, is made up of the atomic masses of the elements that make up the substance.

For better understanding this material let's solve the problem. A mixture of copper and magnesium filings weighing 1.5 g was treated with an excess. As a result of the reaction, hydrogen with a volume of 0.56 l (). Calculate the mass fraction of copper in the mixture.
In this problem passes, we write down its equation. Of the two substances with an excess of hydrochloric acid, only magnesium: Mg + 2HCl = MgCl2 + H2. To find the mass fraction of copper in the mixture, it is necessary to substitute the values ​​in the following formula: w(Cu) = m(Cu)/m(cm). The mass of the mixture is given, we find the mass of copper: m (Cu) \u003d m (cm) - m (Mg). We are looking for mass: m (Mg) \u003d n (Mg) * M (Mg). The reaction equation will help you find the amount of magnesium substance. We find the amount of hydrogen substance: n \u003d V / Vm \u003d 0.56 / 22.4 \u003d 0.025 mol. The equation shows that n(H2) = n(Mg) = 0.025 mol. We calculate the mass of magnesium, knowing that the molar is 24 g / mol: m (Mg) \u003d 0.025 * 24 \u003d 0.6 g. We find the mass of copper: m (Cu) \u003d 1.5 - 0.6 \u003d 0.9 g. Remaining calculate the mass fraction: w(Cu) = 0.9/1.5 = 0.6 or 60%.

Related videos

note

The mass fraction cannot be greater than one or, if expressed as a percentage, greater than 100%.

Sources:

• "Manual in Chemistry", G.P. Khomchenko, 2005.
• Calculation of the share of sales by region

Mass fraction shows as a percentage or in fractions the content of a substance in any solution or element in the composition of a substance. The ability to calculate the mass fraction is useful not only in chemistry lessons, but also when you want to prepare a solution or mixture, for example, for culinary purposes. Or change the percentage in the composition you already have.

Instruction

For example, you need at least 15 cubic meters for the winter. meters of birch firewood.
Look for the reference density of birch firewood. It is: 650 kg/m3.
Calculate the mass by substituting the values ​​into the same specific gravity formula.

m = 650*15 = 9750 (kg)

Now, based on the carrying capacity and capacity of the body, you can decide on the type vehicle and the number of trips.

Related videos

note

Older people are more familiar with the concept of specific gravity. The specific gravity of a substance is the same as specific gravity.

The mass fraction of a substance shows its content in a more complex structure, for example, in an alloy or mixture. If the total mass of a mixture or alloy is known, then knowing the mass fractions of the constituent substances, one can find their masses. To find the mass fraction of a substance, you can know its mass and the mass of the entire mixture. This value can be expressed in fractional units or percentages.

You will need

• scales;
• periodic table of chemical elements;
• calculator.

Instruction

Determine the mass fraction of the substance that is in the mixture through the masses of the mixture and the substance itself. To do this, use a balance to determine the masses that make up the mixture or . Then fold them up. Take the resulting mass as 100%. To find the mass fraction of a substance in a mixture, divide its mass m by the mass of the mixture M, and multiply the result by 100% (ω%=(m/M)∙100%). For example, 20 g of table salt are dissolved in 140 g of water. To find the mass fraction of salt, add the masses of these two substances М=140+20=160 g. Then find the mass fraction of the substance ω%=(20/160)∙100%=12.5%.

If you need to find or the mass fraction of an element in a substance with a known formula, use the periodic table of elements. From it, find the atomic masses of the elements that are in substances. If one is in the formula multiple times, multiply its atomic mass by that number and add up the results. This will be the molecular weight of the substance. To find the mass fraction of any element in such a substance, divide its mass number in the given chemical formula M0 by the molecular weight of the given substance M. Multiply the result by 100% (ω%=(M0/M)∙100%).

For example, determine the mass fraction of chemical elements in copper sulfate. Copper (copper II sulfate), has the chemical formula CuSO4. The atomic masses of the elements included in its composition are equal to Ar(Cu)=64, Ar(S)=32, Ar(O)=16, the mass numbers of these elements will be equal to M0(Cu)=64, M0(S)=32, M0(O)=16∙4=64, taking into account that the molecule contains 4 atoms. Calculate the molecular weight of a substance, it is equal to the sum of the mass numbers of the substances that make up the molecule 64+32+64=160. Determine the mass fraction of copper (Cu) in the composition blue vitriol(ω%=(64/160)∙100%)=40%. By the same principle, it is possible to determine the mass fractions of all elements in this substance. Mass fraction of sulfur (S) ω%=(32/160)∙100%=20%, oxygen (O) ω%=(64/160)∙100%=40%. Please note that the sum of all mass fractions of the substance must be 100%.

Task 3.1. Determine the mass of water in 250 g of a 10% sodium chloride solution.

Solution. From w \u003d m in-va / m solution find the mass of sodium chloride:
m in-va \u003d w m solution \u003d 0.1 250 g \u003d 25 g NaCl
Because the m r-ra = m in-va + m r-la, then we get:
m (H 2 0) \u003d m solution - m in-va \u003d 250 g - 25 g \u003d 225 g H 2 0.

Task 3.2. Determine the mass of hydrogen chloride in 400 ml of hydrochloric acid solution with a mass fraction of 0.262 and a density of 1.13 g/ml.

Solution. Because the w = m in-va / (V ρ), then we get:
m in-va \u003d w V ρ \u003d 0.262 400 ml 1.13 g / ml \u003d 118 g

Task 3.3. To 200 g of a 14% salt solution was added 80 g of water. Determine the mass fraction of salt in the resulting solution.

Solution. Find the mass of salt in the original solution:
m salt \u003d w m solution \u003d 0.14 200 g \u003d 28 g.
The same mass of salt remained in the new solution. Find the mass of the new solution:
m solution = 200 g + 80 g = 280 g.
Find the mass fraction of salt in the resulting solution:
w \u003d m salt / m solution \u003d 28 g / 280 g \u003d 0.100.

Task 3.4. What volume of a 78% sulfuric acid solution with a density of 1.70 g/ml should be taken to prepare 500 ml of a 12% sulfuric acid solution with a density of 1.08 g/ml?

Solution. For the first solution we have:
w 1 \u003d 0.78 and ρ 1 \u003d 1.70 g / ml.
For the second solution we have:
V 2 \u003d 500 ml, w 2 \u003d 0.12 and ρ 2 \u003d 1.08 g / ml.
Since the second solution is prepared from the first by adding water, the masses of the substance in both solutions are the same. Find the mass of the substance in the second solution. From w 2 \u003d m 2 / (V 2 ρ 2) we have:
m 2 \u003d w 2 V 2 ρ 2 \u003d 0.12 500 ml 1.08 g / ml \u003d 64.8 g.
m 2 \u003d 64.8 g. We find
the volume of the first solution. From w 1 = m 1 / (V 1 ρ 1) we have:
V 1 \u003d m 1 / (w 1 ρ 1) \u003d 64.8 g / (0.78 1.70 g / ml) \u003d 48.9 ml.

Task 3.5. What volume of a 4.65% sodium hydroxide solution with a density of 1.05 g/ml can be prepared from 50 ml of a 30% sodium hydroxide solution with a density of 1.33 g/ml?

Solution. For the first solution we have:
w 1 \u003d 0.0465 and ρ 1 \u003d 1.05 g / ml.
For the second solution we have:
V 2 \u003d 50 ml, w 2 \u003d 0.30 and ρ 2 \u003d 1.33 g / ml.
Since the first solution is prepared from the second by adding water, the masses of the substance in both solutions are the same. Find the mass of the substance in the second solution. From w 2 \u003d m 2 / (V 2 ρ 2) we have:
m 2 \u003d w 2 V 2 ρ 2 \u003d 0.30 50 ml 1.33 g / ml \u003d 19.95 g.
The mass of the substance in the first solution is also equal to m 2 \u003d 19.95 g.
Find the volume of the first solution. From w 1 = m 1 / (V 1 ρ 1) we have:
V 1 \u003d m 1 / (w 1 ρ 1) \u003d 19.95 g / (0.0465 1.05 g / ml) \u003d 409 ml.
Solubility coefficient (solubility) - maximum weight a substance that is soluble in 100 g of water at a given temperature. A saturated solution is a solution of a substance that is in equilibrium with the existing precipitate of that substance.

Problem 3.6. The solubility coefficient of potassium chlorate at 25 °C is 8.6 g. Determine the mass fraction of this salt in a saturated solution at 25 °C.

Solution. 8.6 g of salt dissolved in 100 g of water.
The mass of the solution is:
m solution \u003d m water + m salt \u003d 100 g + 8.6 g \u003d 108.6 g,
and the mass fraction of salt in the solution is equal to:
w \u003d m salt / m solution \u003d 8.6 g / 108.6 g \u003d 0.0792.

Problem 3.7. The mass fraction of salt in a potassium chloride solution saturated at 20 °C is 0.256. Determine the solubility of this salt in 100 g of water.

Solution. Let the solubility of the salt be X g in 100 g of water.
Then the mass of the solution is:
m solution = m water + m salt = (x + 100) g,
and the mass fraction is:
w \u003d m salt / m solution \u003d x / (100 + x) \u003d 0.256.
From here
x = 25.6 + 0.256x; 0.744x = 25.6; x = 34.4 g per 100 g of water.
Molar concentration With- the ratio of the amount of solute v (mol) to the volume of the solution V (in liters), c \u003d v (mol) / V (l), c \u003d m in-va / (M V (l)).
Molar concentration shows the number of moles of a substance in 1 liter of solution: if the solution is decimolar ( c = 0.1 M = 0.1 mol/l) means that 1 liter of the solution contains 0.1 mol of the substance.

Problem 3.8. Determine the mass of KOH required to prepare 4 liters of a 2 M solution.

Solution. For solutions with a molar concentration, we have:
c \u003d m / (M V),
where With- molar concentration,
m- the mass of the substance,
M is the molar mass of the substance,
V- the volume of the solution in liters.
From here
m \u003d c M V (l) \u003d 2 mol / l 56 g / mol 4 l \u003d 448 g KOH.

Problem 3.9. How many ml of a 98% solution of H 2 SO 4 (ρ = 1.84 g / ml) must be taken to prepare 1500 ml of a 0.25 M solution?

Solution. The task of diluting the solution. For a concentrated solution we have:
w 1 \u003d m 1 / (V 1 (ml) ρ 1).
Find the volume of this solution V 1 (ml) \u003d m 1 / (w 1 ρ 1).
Since a dilute solution is prepared from a concentrated one by mixing the latter with water, the mass of the substance in these two solutions will be the same.
For a dilute solution we have:
c 2 \u003d m 2 / (M V 2 (l)) and m 2 \u003d s 2 M V 2 (l).
We substitute the found value of the mass into the expression for the volume of the concentrated solution and carry out the necessary calculations:
V 1 (ml) \u003d m / (w 1 ρ 1) \u003d (s 2 M V 2) / (w 1 ρ 1) \u003d (0.25 mol / l 98 g / mol 1.5 l) / (0, 98 1.84 g/ml) = 20.4 ml.

Fractions of solute
ω = m1 / m,
where m1 is the mass of the solute and m is the mass of the entire solution.

If the mass fraction of the solute is needed, multiply the resulting number by 100%:
ω \u003d m1 / m x 100%

In tasks where it is necessary to calculate the mass fractions of each of the elements included in chemical, use the table D.I. Mendeleev. For example, find out the mass fractions of each of the elements that make up the hydrocarbon, which C6H12

m (C6H12) \u003d 6 x 12 + 12 x 1 \u003d 84 g / mol
ω (C) \u003d 6 m1 (C) / m (C6H12) x 100% \u003d 6 x 12 g / 84 g / mol x 100% \u003d 85%
ω (H) \u003d 12 m1 (H) / m (C6H12) x 100% \u003d 12 x 1 g / 84 g / mol x 100% \u003d 15%

Useful advice

Solve the problems of finding the mass fraction of a substance after evaporation, dilution, concentration, mixing of solutions using formulas obtained from the determination of the mass fraction. For example, the problem of evaporation can be solved using the following formula
ω 2 \u003d m1 / (m - Dm) \u003d (ω 1 m) / (m - Dm), where ω 2 is the mass fraction of the substance in one stripped off solution, Dm is the difference between the masses before and after heating.

Sources:

• how to determine the mass fraction of a substance

There are situations when it is necessary to calculate mass liquids contained in any container. This can be during a training session in the laboratory, and during the solution domestic problems e.g. when repairing or painting.

Instruction

The easiest method is to resort to weighing. First, weigh the container together with, then pour the liquid into another container that is suitable in size and weigh the empty container. And then it remains only to subtract from greater value less and you get . Of course, this method can be resorted to only when dealing with non-viscous liquids, which, after overflow, practically do not remain on the walls and bottom of the first container. That is, the quantity will remain then, but it will be so small that it can be neglected, this will hardly affect the accuracy of the calculations.

And if the liquid is viscous, for example,? How then her mass? In this case, you need to know its density (ρ) and occupied volume (V). And then everything is elementary. Mass (M) is calculated from M = ρV. Of course, before calculating it is necessary to convert the factors into a single system of units.

Density liquids can be found in a physical or chemical reference book. But it is better to use a measuring device - a density meter (densitometer). And the volume can be calculated, knowing the shape and dimensions capacity (if it has the correct geometric shape). For example, if the same glycerin is in a cylindrical barrel with a base diameter d and a height h, then the volume