The space around us is filled with different physical bodies, which consist of different substances with different weights. School courses chemistry and physics, introducing the concept and method of finding the mass of matter, listened to and safely forgotten by everyone who studied at school. But meanwhile theoretical knowledge, acquired once, may be needed at the most unexpected moment.
Calculation of the mass of a substance using the specific density of a substance. Example - there is a barrel of 200 liters. You need to fill the barrel with any liquid, say, light beer. How to find the mass of a filled barrel? Using the substance density formula p=m/V, where p is the specific density of the substance, m is the mass, V is the volume occupied, it is very easy to find the mass of a full barrel:- Measures of volumes - cubic centimeters, meters. That is, a barrel of 200 liters has a volume of 2 m³.
- A measure of specific gravity is found using tables and is constant value for every substance. Density is measured in kg/m³, g/cm³, t/m³. The density of light beer and other alcoholic beverages can be viewed on the website. It is 1025.0 kg/m³.
- From the density formula p \u003d m / V => m \u003d p * V: m \u003d 1025.0 kg / m³ * 2 m³ \u003d 2050 kg.
A barrel of 200 liters, completely filled with light beer, will have a mass of 2050 kg.
Finding the mass of a substance using the molar mass. M (x) \u003d m (x) / v (x) is the ratio of the mass of a substance to its quantity, where M (x) is the molar mass of X, m (x) is the mass of X, v (x) is the amount of substance X If only 1 known parameter is prescribed in the condition of the problem - the molar mass of a given substance, then finding the mass of this substance is not difficult. For example, it is necessary to find the mass of sodium iodide NaI with the amount of substance 0.6 mol.- The molar mass is calculated in the unified SI measurement system and is measured in kg / mol, g / mol. The molar mass of sodium iodide is the sum of the molar masses of each element: M (NaI)=M (Na)+M (I). The value of the molar mass of each element can be calculated from the table, or you can use the online calculator on the site: M (NaI) \u003d M (Na) + M (I) \u003d 23 + 127 \u003d 150 (g / mol).
- From the general formula M (NaI) \u003d m (NaI) / v (NaI) => m (NaI) \u003d v (NaI) * M (NaI) \u003d 0.6 mol * 150 g / mol \u003d 90 grams.
Mass of sodium iodide (NaI) with mass fraction th substance 0.6 mol is 90 grams.
- Dilution of the solution with water. The mass of the dissolved X substance does not change m (X)=m'(X). The mass of the solution increases by the mass of added water m '(p) \u003d m (p) + m (H 2 O).
- Evaporation of water from solution. The mass of the solute X does not change m (X)=m' (X). The mass of the solution is reduced by the mass of evaporated water m '(p) \u003d m (p) -m (H 2 O).
- Drainage of two solutions. The masses of solutions, as well as the masses of the solute X, add up when mixed: m '' (X) \u003d m (X) + m ' (X). m '' (p) \u003d m (p) + m '(p).
- Dropout of crystals. The masses of the dissolved substance X and the solution are reduced by the mass of the precipitated crystals: m '(X) \u003d m (X) -m (precipitate), m '(p) \u003d m (p) -m (precipitate).
Options for finding the mass of a substance - a useful course schooling, but quite practical methods. Everyone can easily find the mass of the required substance by applying the above formulas and using the proposed tables. To facilitate the task, write down all the reactions, their coefficients.
Three options for tasks: 1. The masses of the initial substance and the reaction product are given. Determine the yield of the reaction product. 2. The masses of the starting material and the yield of the reaction product are given. Determine the mass of the product. 3. Given the mass of the product and the yield of the product. Determine the mass of the starting material.
Given: m(ZnO) = 32.4 g m pr (Zn) = 24 g Find: ω out (Zn) - ? Solution: 3ZnO + 2Al = 3Zn + Al 2 O 3 3 mol substances according to the equation (mol) Under the action of aluminum on zinc oxide weighing 32.4 g, 24 g of zinc were obtained. Find the mass fraction of the yield of the reaction product.
Given: m(ZnO) = 32.4 g m pr (Zn) = 24 g Find: ω out (Zn) - ? M (ZnO) \u003d 81 g / mol Solution: 0.4 mol x 3ZnO + 2Al \u003d 3Zn + Al 2 O 3 3 mol Find the amount of ZnO substance using the formula: \u003d m / M. Let's sign its amount of substance over it in the equation. Let us sign x over Zn. Find x by composing and solving the proportion. (ZnO) \u003d 32.4 / 81 \u003d 0.4 mol 0.4 / 3 \u003d x / 3 x \u003d 0.4 mol - this is the theoretical amount of the substance found by the equation
Given: m(ZnO) = 32.4 g m pr (Zn) = 24 g Find: ω out (Zn) - ? M(ZnO)=81g/mol M(Zn)=65 g/mol Solution: 0.4 mol x 3ZnO + 2Al = 3Zn + Al 2 O 3 3 mol (Zn) = 0.4 mol × 65 g/mol = 26 g is the theoretical mass of Zn. In the problem, the practical weight of 24 g is given in the condition. Now we will find the fraction of the product yield from the theoretical possible. ω out = = = 0.92 (or 92%) m pr (Zn) m theor (Zn) 24 g 26 g Answer: ω out = 92%
Given: m (Al (OH) 3) \u003d 23.4 g ω out (Al 2 O 3) \u003d 92% Find: m pr (Al 2 O 3) -? M (Al (OH) 3) \u003d 78 g / mol Solution: 0.3 mol x 2Al (OH) 3 \u003d Al 2 O 3 + 3H 2 O 2 mol 1 mol Determine the mass of aluminum oxide, which can be obtained from 23, 4 g of aluminum hydroxide, if the yield of the reaction is 92% of the theoretically possible. M (Al 2 O 3) \u003d 102 g / mol (Al (OH) 3) \u003d 23.4 g / 78 g / mol \u003d 0.3 mol 0.3/2 \u003d x / 1 x \u003d 0.15 mol m theor. (Al 2 O 3) \u003d n M \u003d 0.15 mol 102 g / mol \u003d 15.3 g m ave. (Al 2 O 3) \u003d 15.3 g × 0.92 \u003d 14 g Answer: m ave. ( Al 2 O 3) = 14 g
Under the action of carbon monoxide (II) on iron oxide (III), iron weighing 11.2 g was obtained. Find the mass of iron oxide (III) used, given that the share of the yield of reaction products is 80% of the theoretically possible. Given: m pr (Fe) \u003d 11.2 g ω out (Fe) \u003d 80% Find: M (Fe 2 O 3) -? Solution: Fe 2 O 3 + 3CO = 2Fe + 3CO 2 1 mol 2 mol m theor = = = 14 g m pr (Fe) ω out (Fe) 11.2 g 0.8
Given: m pr (Fe) \u003d 11.2 g ω out (Fe) \u003d 80% Find: m (Fe 2 O 3) -? M (Fe) \u003d 56 g / mol M (Fe 2 O 3) \u003d 160 g / mol Solution: x 0.25 mol Fe 2 O 3 + 3CO \u003d 2Fe + 3CO 2 1 mol 2 mol We convert the found theoretical mass of iron into the amount substances according to the formula: \u003d m / M ((Fe) \u003d 14 g / 56 g / mol \u003d 0.25 mol Let's write this amount of iron over it in the equation, over the oxide we will write x. Let's solve the proportion: x / 1 \u003d 0.25 / 2, x \u003d 0.125 mol Now we translate into mass according to the formula: m \u003d × M m (Fe 2 O 3) \u003d 0.125 mol × 160 g / mol \u003d 20 g Answer: m (Fe 2 O 3) \u003d 20 g
Tasks for independent solution 1. To obtain a precipitate of barium sulfate, sulfuric acid weighing 490 g was taken. The mass fraction of the salt yield from the theoretically possible amounted to 60%. What is the mass of barium sulfate produced? 2. Calculate the yield of ammonium nitrate in % of the theoretically possible, if by passing 85 g of ammonia through a solution of nitric acid, 380 g of salt was obtained 3. As a result of the catalytic oxidation of sulfur oxide (IV) with a mass of 16 kg, sulfur oxide (VI) is formed . Calculate the mass of the reaction product if the fraction of its yield is 80% of the theoretically possible. 4. Calculate the mass of nitric acid that can be obtained from 17 g of sodium nitrate when it interacts with concentrated sulfuric acid, if the mass fraction of the acid yield is 0.96. 5. Hydrated lime, taken in the required amount, was treated with 3.15 kg of pure nitric acid. What mass of calcium nitrate was obtained if practical output in mass fractions is 0.98 or 98% compared to theoretical?
Determination of the mass or volume fraction of the yield of the reaction product from the theoretically possible
The quantitative assessment of the yield of the reaction product from the theoretically possible is expressed in fractions of a unit or in percent and is calculated by the formulas:
M practical / m theory;
M practical / m theoretical *100%,
where (etta) is the mass fraction of the yield of the reaction product from the theoretically possible;
V practical / V theory;
V practical / V theoretical * 100%,
where (phi) is the volume fraction of the yield of the reaction product from the theoretically possible.
Example 1 When copper (II) oxide weighing 96 g is reduced with hydrogen, copper weighing 56.4 g is obtained. How much will this be from the theoretically possible yield?
Decision:
1. Write down the equation chemical reaction:
CuO + H 2 \u003d Cu + H 2 O
1 mol 1 mol
2. Calculate the chemical amount of copper oxide ( II):
M (C u O) \u003d 80 g / mol,
n (CuO) \u003d 96/80 \u003d 1.2 (mol).
3. We calculate the theoretical yield of copper: based on the reaction equation, n (Cu) \u003d n (CuO) \u003d 1.2 mol,
m (C u) \u003d 1.2 64 \u003d 76.8 (g),
because M (C u) \u003d 64 g / mol
4. Calculate the mass fraction of copper yield compared to theoretically possible: = 56.4/76.8= 0.73 or 73%
Answer: 73%
Example 2 How much iodine can be obtained by the action of chlorine potassium iodide with a mass of 132.8 kg, if the loss in production is 4%?
Decision:
1. Write down the reaction equation:
2KI + Cl 2 \u003d 2KCl + I 2
2 kmol 1 kmol
2. Calculate the chemical amount of potassium iodide:
M (K I) \u003d 166 kg / kmol,
n (K I ) = 132.8/166= 0.8 (kmol).
2. We determine the theoretical yield of iodine: based on the reaction equation,
n (I 2) \u003d 1 / 2n (KI) \u003d 0.4 mol,
M (I 2) \u003d 254 kg / kmol.
From where, m (I 2) \u003d 0.4 * 254 \u003d 101.6 (kg).
3. We determine the mass fraction of the practical yield of iodine:
=(100 - 4) = 96% or 0.96
4. Determine the mass of iodine practically obtained:
m (I 2 )= 101.6 * 0.96 = 97.54 (kg).
Answer: 97.54 kg of iodine
Example 3 When burning 33.6 dm 3 of ammonia, nitrogen with a volume of 15 dm 3 was obtained. Calculate the volume fraction of nitrogen yield in % of the theoretically possible.
Decision:
1. Write down the reaction equation:
4 NH 3 + 3 O 2 \u003d 2 N 2 + 6 H 2 O
4 mol2 mol
2. Calculate the theoretical yield of nitrogen: according to the Gay-Lussac law
when burning 4 dm 3 ammonia, 2 dm 3 nitrogen is obtained, and
when burning 33.6 dm 3, dm 3 of nitrogen is obtained
x \u003d 33. 6 * 2/4 \u003d 16.8 (dm 3).
3. We calculate the volume fraction of nitrogen output from the theoretically possible:
15/16.8 =0.89 or 89%
Answer: 89%
Example 4 What mass of ammonia is needed to obtain 5 tons of nitric acid with a mass fraction of acid of 60%, assuming that the loss of ammonia in production is 2.8%?
Decision:1. We write down the equations of the reactions underlying the production of nitric acid:
4NH 3 + 5 O 2 \u003d 4NO + 6H 2 O
2NO + O 2 \u003d 2NO 2
4NO 2 + O 2 + 2H 2 O \u003d 4HNO 3
2. Based on the reaction equations, we see that from 4 moles of ammonia we get
4 mol of nitric acid. We get the scheme:
NH3HNO3
1 tmol1tmol
3. We calculate the mass and chemical amount of nitric acid, which is necessary to obtain 5 tons of a solution with a mass fraction of acid of 60%:
m (in-va) \u003d m (r-ra) * w (in-va),
m (HNO 3) \u003d 5 * 0.6 \u003d 3 (t),
4. We calculate the chemical amount of acid:
n (HNO 3 ) = 3/63 = 0.048 (tmol),
because M (HNO 3 ) \u003d 63 g / mol.
5. Based on the diagram:
n (NH 3 ) = 0.048 tmol,
and m (NH 3) \u003d 0.048 17 \u003d 0.82 (t),
because M (NH 3) \u003d 17 g / mol.
But this amount of ammonia must react, if you do not take into account the loss of ammonia in production.
6. We calculate the mass of ammonia, taking into account losses: we take the mass of ammonia involved in the reaction - 0.82 tons - for 97.2%,
Excess and deficiency of reagents
Proportional quantities and masses of reacting substances are by no means always taken. Often one of the reactants for the reaction is taken with excess, and the other with disadvantage. Obviously, if in the reaction 2H 2 + O 2 \u003d 2H 2 O to receive 2 mol H 2 O take not 1 mol O 2 and 2 mol H 2, a 2 mol H 2 and 2 mol O 2, then 1 mol O 2 will not react and will remain in excess.
The determination of the reagent taken in excess (for example, B) is carried out according to the inequalities: n A / a< n общ.В /b = (n B + n изб.В)/b , where n total- the total (taken in excess) amount of the substance, nB is the amount of substance required for the reaction, i.e. stoichiometric, and n ex.B- excess (non-reacting) amount of substance AT, and n total.V = n B + n ex.V.
Due to the fact that an excess amount of reagent AT will not react, the calculation of the resulting quantities of products must be carried out only by amount of reagent taken in short supply.
Practical yield of the product
theoretical quantity n theor. called the amount of the reaction product, which is obtained in accordance with the calculation according to the reaction equation. However, under specific reaction conditions, it may happen that less product is formed than expected from the results of the reaction equation; let's call this value practical number n pr.
Practical yield of the product is the ratio of the practical amount of the product AT(obtained in reality) to the theoretical (calculated according to the reaction equation). The practical yield of the product is denoted as ŋ B: ŋ B \u003d n pr.B / n theor.B(expressions for the mass of any product and the volume of a gaseous product have a similar form).
The practical yield of the product is represented by a fraction of unity or 100%.
In practice, most often ŋ B< 1 (100 %) because of n ex.< n теор. If under ideal conditions n pr. = n theor, then the output becomes complete, that is ŋ B = 1 (100%); it is often called theoretical output.
Mass fraction of a substance in a mixture. Degree of purity of a substance
More often, not individual substances are taken for reactions, but their mixtures, including natural ones - minerals and ores. The content of each substance in a mixture is expressed by its mass fraction.
The ratio of the mass of a substance ( m B) to the mass of the mixture ( m cm) was named mass fraction of substance B (w B) in the mixture: w B \u003d m B / m cm.
The mass fraction of a substance in a mixture is a fraction of unity or 100%.
The sum of the mass fractions of all substances in the mixture is 1 (100 %).
In a mixture, we encounter two types of substances - basic substance and impurities. main substance name the substance (B), which is in the mixture in a predominant amount; all other substances are called impurities, and the value w B considered the degree of purity of the base substance.
For example, natural calcium carbonate– limestone- may contain 82% CaCO3. In other words, 82 %
equal to the degree of purity of limestone according to CaCO3. For various impurities (sand, silicates, etc.) accounted for 
tatok in 18 %.
In minerals, ores, minerals, rocks, i.e. in natural compounds, and in industrial products, impurities are always contained.
The degree of purification of chemical reagents can be different. According to the decrease in the percentage of impurities, the following types of reagents are qualitatively distinguished: “pure”, “technical”, “chemically pure”, “pure for analysis”, “highly pure”. For example, 99, 999 % basic substance (H2SO4) contains "chemically pure" sulfuric acid . Therefore, in sulfuric acid only 0.001% impurities.
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CALCULATION OF THE YIELD OF THE REACTION PRODUCT AS A PERCENTAGE OF THE THEORETICALLY POSSIBLE IF THE MASSES OF THE STARTING SUBSTANCE AND THE REACTION PRODUCT ARE KNOWN
Task 1. Through lime water containing 3.7 g of calcium hydroxide, carbon dioxide. The precipitate that formed was filtered off, dried, and weighed. Its mass turned out to be 4.75 g. Calculate the yield of the reaction product (in percent) of the theoretically possible.
Given:I way.
Let us determine the quantities of substances given in the condition of the problem:
v=m /
M = 3.7 g /
74 g/mol = 0.05 mol;
v = 0.05 mol
v(CaCO 3
) = m(CaCO 3
) /
M(CaCO 3
) = 4.75 g /
100 g/mol = 0.0475 mol;
v(CaCO 3
) = 0.0475 mol
Let's write the chemical reaction equation:
Ca(OH)2
From the chemical reaction equation it follows that from 1 mol of Ca(OH) 2
1 mol of CaCO is formed 3
, which means that from 0.05 mol Ca(OH) 2
theoretically, the same amount should be obtained, that is, 0.05 mol of CaCO 3
. In practice, 0.0475 mol CaCO 3
, which will be:
w out.(CaCO 3
) = 0.0475 mol *
100 % /
0.05 mol = 95%
w out.(CaCO 3
) = 95 %
II way.
We take into account the mass of the starting material (calcium hydroxide) and the chemical reaction equation:
Ca(OH)2
Calculate from the reaction equation how much calcium carbonate is theoretically formed.
From 74 g Ca(OH) 2Hence x = 3.7 g* 100 g/ 74 g = 5 g, m(CaCO 3 ) = 5 g
This means that from the data on the condition of the problem of 3.7 g of calcium hydroxide theoretically (from calculations) it would be possible to obtain 5 g of calcium carbonate, but practically only 4.75 g of the reaction product was obtained. From these data, we determine the yield of calcium carbonate (in%) from the theoretically possible:
5 g CaCO 3
make up 100% yield
4.75 g CaCO 3
are x %
x = 4.75 mol*
100 %
/
5 g = 95%;
w out.
(CaCO 3
) = 95 %
Answer: the yield of calcium carbonate is 95% of the theoretically possible.
Task 2. The interaction of magnesium weighing 36 g with an excess of chlorine yielded 128.25 g of magnesium chloride. Determine the yield of the reaction product as a percentage of the theoretically possible.
Given: Let us consider two ways to solve this problem: using the quantity amount of substance and mass of matter.I way.
From the data on the condition of the problem of the values of the masses of magnesium and magnesium chloride, we calculate the values of the amount of these substances:
v(Mg) = m(Mg) /
M(Mg) = 36 g /
24 g/mol = 1.5 mol; v(Mg) = 1.5 mol
v(MgCl 2
) = m(MgCl 2
)/
M(MgCl 2
) = 128.25 g /
95 g/mol = 1.35 mol;
v(MgCl 2
) = 1.35 mol
Let's make the equation of chemical reaction:
mg
Let's use the chemical reaction equation. It follows from this equation that 1 mole of magnesium chloride can be obtained from 1 mole of magnesium, which means that from the given 1.5 moles of magnesium, the same amount can theoretically be obtained, that is, 1.5 moles of magnesium chloride. And practically received only 1.35 mol. Therefore, the yield of magnesium chloride (in%) from the theoretically possible will be:
1.5 mol MgCl 2x = 1.35 mol * 100%/ 1.5 mol = 90% i.e. w out. (MgCl 2 ) = 90%
II way.
Consider the chemical reaction equation:
mg
First of all, using the chemical reaction equation, we determine how many grams of magnesium chloride can be obtained from the data on the condition of the problem of 36 g of magnesium.
From 24 g Mg 2Hence x = 36 g * 95 g/ 24 g = 142.5 g; m(MgCl 2 ) = 142.5 g
This means that 142.5 g of magnesium chloride could be obtained from a given amount of magnesium (100% theoretical yield). And only 128.25 g of magnesium chloride were obtained (practical yield).
Consider now what percentage is the practical output of the theoretically possible:
x = 128.25 g * 100% / 142.5 g = 90%, that is w out. (MgCl 2 ) = 90%
Answer: the output of magnesium chloride is 90% of the theoretically possible.
Task 3. Potassium metal weighing 3.9 g was placed in distilled water with a volume of 50 ml. As a result of the reaction, 53.8 g of a potassium hydroxide solution with a mass fraction of the substance equal to 10% were obtained. Calculate the yield of caustic potash (in percent) of the theoretically possible.
Given:2K
Based on this chemical reaction equation, we will make calculations.
First, we determine the mass of caustic potash, which theoretically could be obtained from the mass of potassium given by the condition of the problem.
Hence: x = 3.9 g * 112 g / 78 g = 5.6 g m(KOH) = 5.6 g
From this formula we express m in-va:
m in-va \u003d m solution *
w in-va /
100%
Let's determine the mass of caustic potash, which is in 53.8 g of its 10% solution:
m(KOH) = m solution *
w(KOH) /
100% = 53.8 g *
10% /
100% = 5.38 g
m(KOH) = 5.38 g
Finally, we calculate the yield of caustic potash as a percentage of the theoretically possible:
w out. (KOH) = 5.38 g /
5.6 g *
100% = 96%
w out. (KOH) = 96%
Answer: The yield of caustic potash is 96% of the theoretically possible.
