How to find the mass fraction of a substance in a compound. Mass fraction of a chemical element in a complex substance
Solution A homogeneous mixture of two or more components is called.
Substances that are mixed to form a solution are called components.
The components of the solution are solute, which may be more than one, and solvent. For example, in the case of a solution of sugar in water, sugar is the solute and 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 water-containing solutions, it is traditionally customary to call water a solvent, and the second component the solute.
As a quantitative characteristic of the composition of the solution, such a concept is most often used as 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) - the mass of the substance contained in the solution (g), m (p-ra) - the 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 a unit. Due to this mass fraction can also be expressed as a percentage (%), and it is in this format that it appears in almost all tasks. The mass fraction, expressed as a percentage, is calculated using a formula similar to formula (1), with the only difference that the ratio of the mass of the solute to the mass of the entire solution is multiplied by 100%:
For a solution consisting of only two components, the solute mass fraction ω(r.v.) and the solvent mass fraction ω(solvent) can be respectively calculated.
The mass fraction of a 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 solvent is always 100%:
Obviously, in addition to the formulas written above, one should also know all those formulas that are directly mathematically derived from them. For example:
It is also necessary to remember the formula that relates 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 the mass of 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 knowing the above formulas, it is extremely important to bring the skills of their application to automaticity. This can only be achieved by solving a large number of different tasks. Tasks from real USE exams on the topic "Calculations using the concept of" mass fraction of a substance in solution "" can be solved.
Examples of tasks for 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) \u003d 5 g, and m (H 2 O) \u003d 20 g, therefore:
Example 2
What mass of water must be added to 20 g of glucose to obtain a 10% glucose solution.
Solution:
It follows from the conditions of the problem 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. Denoting the mass of water as x g, we can write the following equivalent equation based on the formula above:
Solving this equation we find x:
those. m(H 2 O) \u003d x g \u003d 180 g
Answer: m (H 2 O) \u003d 180 g
Example 3
150 g of a 15% sodium chloride solution were mixed with 100 g of a 20% solution of the same salt. What is the mass fraction of salt in the resulting solution? Give your answer to the nearest integer.
Solution:
To solve problems for the preparation of solutions, it is convenient to use the following table:
1st solution |
2nd solution |
3rd solution |
|
m r.v. |
|||
m solution |
|||
ω r.v. |
where m r.v. , m r-ra and ω r.v. are the 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%,
Inserting 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 r-ra ∙ ω r.v. / 100% , m solution = 100% ∙ m r.v. /ω r.v.
Let's start filling out the table.
If only one value is missing in a row or column, then 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 one cannot be calculated.
The first column is missing a value in only one cell. So we can calculate it:
m (1) r.v. = m (1) r-ra ∙ ω (1) r.v. /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) r-ra ∙ ω (2) r.v. /100% = 100 g ∙ 20%/100% = 20 g
Let's enter the calculated values in the table:
Now we have two values in the first line and two values in the second line. So 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 in the table, we get:
Now we have come close to calculating the desired value ω (3) r.v. . In the column where it is located, the contents of the other two cells are known, so 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
To 200 g of a 15% sodium chloride solution was added 50 ml of water. What is the mass fraction of salt in the resulting solution. Give your answer to the nearest hundredth _______%
Solution:
First of all, you should pay attention to the fact that instead of the mass of added water, we are given its volume. We 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, respectively, 0 g of sodium chloride, the problem can be solved using the same table as in the example above. Let's draw such a table and insert the values we know into it:
In the first column, two values are known, 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 \u200b\u200bare also known, so we can calculate the third:
m (3) solution = m (1) solution + m (2) solution = 200 g + 50 g = 250 g,
Enter the calculated values in 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%.
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Mass fraction of an element in complex substance
The paragraph will help you:
> find out what the mass fraction of an element in a compound is and determine its value;
> calculate the mass of the element in a certain mass of the compound, based on the mass fraction of the element;
> correctly formulate the solution of chemical problems.
Every difficult substance (chemical compound) is formed by several elements. To know the content of the elements in the compound is necessary for its effective use. For example, the best nitrogen fertilizer is considered to be the one that contains the largest amount of Nitrogen (this element is necessary for plants). Similarly, the quality of metal ore is assessed, determining how much it " rich» on a metal element.
Content element in the compound characterize its mass fraction. This value is denoted Latin letter w ("double-ve").
Let us derive a formula for calculating the mass fraction of an element in a compound from the known masses of the compound and the element. We denote the mass fraction of the element by the letter x. Taking into account that the mass of the compound is a whole, and the mass of an element is a part of the whole, we make up the proportion:
Note that the masses of the element and the compound must be taken in the same units of measurement (for example, in grams).
This is interesting
In two Sulfur compounds - SO 2 and MoS 3 - the mass fractions of the elements are the same and amount to 0.5 (or 50%) each.
The mass fraction has no dimension. It is often expressed as a percentage. In this case formula takes this form:
It is obvious that the sum of the mass fractions of all elements in the compound is 1 (or 100%).
Let us give several examples of solving computational problems. The condition of the problem and its solution are drawn up in this way. Divide a sheet of notebook or chalkboard vertical line into two unequal parts. In the left, smaller, part, the condition of the problem is abbreviated, carried out horizontal line and under it indicate what needs to be found or calculated. Mathematical formulas, explanation, calculations and answer are written on the right side.
80 g of the compound contains 32 g oxygena. Calculate the mass fraction of oxygen in the compound.
The mass fraction of an element in a compound is also calculated using the chemical formula of the compound. Since the masses of atoms and molecules are proportional to the relative atomic and molecular masses, then
where N(E) is the number of element atoms in the compound formula.
From the known mass fraction of the element, it is possible to calculate the mass of the element contained in a certain mass of the compound. From the mathematical formula for the mass fraction of an element follows:
m(E) = w(E) m(compounds).
What mass of Nitrogen is contained in ammonium nitrate (nitrogen fertilizer) weighing 1 kg, if the mass fraction of this element in the compound is 0.35?
The concept of "mass fraction" is used to characterize the quantitative composition of mixtures of substances. The corresponding mathematical formula looks like this:
conclusions
The mass fraction of an element in a compound is the ratio of the mass of the element to the corresponding mass of the compound.
The mass fraction of an element in a compound is calculated from the known masses of the element and the compound, or from its chemical formula.
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92. How to calculate the mass fraction of an element in a compound if: a) the mass of the element and the corresponding mass of the compound are known; b) the chemical formula of the compound?
93. 20 g of a substance contains 16 g of Bromine. Find the mass fraction of this element in the substance, expressing it as an ordinary fraction, decimal fraction and as a percentage.
94. Calculate (preferably orally) the mass fractions of elements in compounds with the following formulas: SO 2 , LiH, CrO 3 .
95. Comparing the formulas of substances, as well as the values of relative atomic masses, determine in which of the substances of each pair the mass fraction of the first element in the formula is greater:
a) N 2 O, NO; b) CO, CO 2 ; c) B 2 O 3, B 2 S 3.
96. Perform the necessary calculations for acetic acid CH 3 COOH and glycerol C 3 H 5 (OH) 3 and fill in the table:
| C x H y O z | M r (C x H y O z) | w(C) | W(H) | W(O) |
97. The mass fraction of Nitrogen in a certain compound is 28%. What mass of the compound contains 56 g of Nitrogen?
98. The mass fraction of Calcium in its combination with Hydrogen is 0.952. Determine the mass of hydrogen contained in 20 g of the compound.
99. Mixed 100 g of cement and 150 g of sand. What is the mass fraction of cement in the prepared mixture?
Popel P. P., Kriklya L. S., Chemistry: Pdruch. for 7 cells. zahalnosvit. navch. zakl. - K .: Exhibition Center "Academy", 2008. - 136 p.: il.
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a) In mathematics, "share" is the ratio of a part to a whole. To calculate the mass fraction of an element, multiply its relative atomic mass by the number of atoms of a given element in the formula and divide by the relative molecular mass of the substance.
b) The sum of the mass fractions of all elements that make up the substance is 1 or 100%.
2. Write down mathematical formulas for finding the mass fractions of elements if:
a) the formula of the substance is P 2 O 5, M r \u003d 2 * 31 + 5 * 16 \u003d 142
w(P) = 2*31/132 *100% = 44%
w(O) = 5*16/142*100% = 56% or w(O) = 100-44=56.
b) the formula of the substance - A x B y
w(A) = Ar(A)*x/Mr(AxBy) * 100%
w(B) = Ar(B)*y / Mr(AxBy) *100%
3. Calculate mass fractions of elements:
a) in methane (CH 4)
b) in sodium carbonate (Na 2 CO 3)
4. Compare the mass fractions of the indicated elements in substances and put a sign<, >or = :
5. In the combination of silicon with hydrogen, the mass fraction of silicon is 87.5%, hydrogen 12.5%. The relative molecular weight of the substance is 32. Determine the formula of this compound.
6. Mass fractions of elements in the compound are shown in the diagram:
Determine the formula of this substance if it is known that its relative molecular weight is 100.
7. Ethylene is a natural fruit ripening stimulant: its accumulation in fruits accelerates their ripening. The earlier the accumulation of ethylene begins, the earlier the fruits ripen. Therefore, ethylene is used to artificially accelerate fruit ripening. Derive the formula of ethylene if it is known that the mass fraction of carbon is 85.7%, the mass fraction of hydrogen is -14.3%. The relative molecular weight of this substance is 28.
8. Derive the chemical formula of the substance, if it is known that
a) w(Ca) = 36%, w(Cl) = 64%
b) w(Na) 29.1%, w(S) = 40.5%, w(O) = 30.4%.
9. Lapis has antimicrobial properties. Previously, it was used to cauterize warts. In small concentrations, it acts as an anti-inflammatory and astringent, but can cause burns. Derive the formula of lapis if it is known that it contains 63.53% silver, 8.24% nitrogen, 28.23% oxygen.
The mass fraction of the element ω (E)% is the ratio of the mass of a given element m (E) in a taken molecule of a substance to the molecular weight 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) \u003d m (E) / Mr (in-va) (1)
ω% (E) \u003d m (E) 100% / Mr (in-va)
The sum of 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, a portion of a substance is taken 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 a molecule.
So, for a substance A x B y in fractions of a unit:
ω (A) \u003d Ar (E) X / Mr (in-va) (2)
From proportion (2), we derive the calculation formula for determining 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 \u003d 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)
The calculation formula (4) can be transformed as follows:
X: Y \u003d ω% (A) / Ar (A) : ω% (B) / Ar (B) \u003d X (A) : Y (B) (5)
Calculation formulas (3) and (5) are used to determine the formula of the 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(in-va) \u003d Ar (E) X / W (A)
Examples of solving problems for 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 a percentage.
Solution
1. Calculate the relative molecular weight of sulfuric acid:
Mr (H 2 SO 4) \u003d 1 2 + 32 + 16 4 \u003d 98
2. We 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 the element with the letter ω, the 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) \u003d 64: 98 \u003d 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 a percentage.
Solution
1. Calculate the relative molecular weight of aluminum oxide:
Mr(Al 2 O 3) \u003d 27 2 + 16 3 \u003d 102
2. We 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
The mass fraction of a substance is the ratio of the mass of a given substance in the system to the mass of the entire system, i.e. ω(X) = m(X) / m,
where ω(X) - mass fraction of substance X,
m(X) - mass of substance X,
m - mass of the whole 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) \u003d 137 + 2 35.5 + 2 18 \u003d 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) \u003d m (H 2 O) / m (BaCl 2 2H 2 O) \u003d 36 / 244 \u003d 0.1475 \u003d 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.
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|
Determine the amount of silver substance in argentite: n(Ag) \u003d m (Ag) / M (Ag) \u003d 5.4 / 108 \u003d 0.05 mol. From the formula Ag 2 S it follows that the amount of argentite substance is half the amount of silver substance. Determine the amount of argentite substance: n (Ag 2 S) \u003d 0.5 n (Ag) \u003d 0.5 0.05 \u003d 0.025 mol We calculate the mass of argentite: m (Ag 2 S) \u003d n (Ag 2 S) M (Ag2S) \u003d 0.025 248 \u003d 6.2 g. Now we determine the mass fraction of argentite in a rock sample, weighing 25 g. ω (Ag 2 S) \u003d m (Ag 2 S) / m \u003d 6.2 / 25 \u003d 0.248 \u003d 24.8%. |
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 you need to calculate the mass fractions of each of the elements included in the 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%
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. It may be during training session in the laboratory, and in the course of solving a household problem, for example, 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 overall dimensions of the container (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
