How to determine the mass of a substance in chemistry. Solving calculation problems in chemistry “for the product yield from the theoretically possible” Mass fraction of the reaction product yield (ω “omega”)

Calculation of mass or volume fraction of product yield
(in percent) of the theoretically possible
Mass (molar, volume) fraction of product yield () is
ratio of mass, amount of substance or volume practically
of the resulting substance to the theoretically possible:
pract.)
m
m
(theor.)

%100


pract.)

%100

(theor.)
V

V
pract.)

100
(theor.)
%,
 =
m (ν, V) (practically) shows mass (amount of substance, volume), actually
Where
received;
m (ν, V) (theor.) shows the mass (amount of substance, volume) that
could have received if there had been no losses.
Problems on the yield of the reaction product from the theoretically possible can be
divided into three types.
1. The mass (volume) of the starting substance and the mass (volume) are known
reaction product. Determine the mass (volume) fraction of the yield
reaction product.
EXAMPLE In the laboratory, the reduction of nitrobenzene weighing 61.5 g
obtained aniline weighing 44 g. Determine the mass fraction (in%) of the yield
aniline.
x mole
Solution.
0.5 mol
C6H5NO2 + 6[H] = C6H5NH2 + 2H2O
1 mole
1. Calculate ν (C6H5NO2):
1 mole
= 0.5 (mol)
5,61
ν (C6H5NO2) = 123
2. Using the reaction equation, we determine the theoretical ν (C6H5NH2):
ν (C6H5NO2) = ν (C6H5NН2) = 0.5 mol
3. Determine the theoretical mass of aniline:
m (C6H5NH2)theor. = ν (C6H5NН2) ∙ M (C6H5NН2) =
= 0.5 ∙ 93 = 46.5 (g).
4. Determine the mass fraction of aniline yield:
pract.)
m
m
(theor.)

44
5,46
 =
= 0.946, or 94.6%.

2. The mass (volume) of the starting substance and the percentage (in %) of the yield are known
reaction product. Determine the practical mass (volume) of the product
reactions.
EXAMPLE Calculate the mass of calcium carbide formed by
the action of coal on calcium oxide weighing 16.8 g, if the mass fraction of the yield
is 80% (or 0.8).
Solution.
0.3 mol
CaO+3C
1 mole
x mole

t

CaC2 + CO
1 mole
)CaO(m
)CaO(M
8,16
56

= 0.3 (mol).
1. ν (CaO) =
2. Using the reaction equation, we determine the theoretical ν (CaС2):
ν (CaO) pract. = ν (CaС2) theor.  ν (CaС2) theor. = 0.3 (mol).
3. We calculate the practically obtained ν (CaС2):
ν (CaС2) pract. = ν (CaС2) theor. ∙  = 0.3 ∙ 0.8 = 0.24 (mol).
4. We calculate the practically obtained mass of calcium carbide:
m (CaС2) pract. = ν (CaС2) pract. ∙ M = 0.24 ∙ 64 = 15.36 (g).
3. The mass (volume) of the practically obtained substance and the proportion
the yield of this reaction product. Calculate the mass (volume) of the original
substances.
EXAMPLE Calculate what mass of sodium carbonate needs to be taken for
obtaining carbon monoxide (IV) with a volume of 28.56 l (no.) at a mass fraction
yield 85%.
Solution.
x mole
Na2CO3 + 2HC = 2NaC + H
1 mole
1. Calculate the theoretically obtained volume and amount of substance
2O + CO2
1.5 mol
1 mole
ℓ
ℓ
carbon monoxide (IV):
V (CO2)theor. =
)CO(V
2

pract.

56,28
85,0
= 33.6 (l).
)CO(V
2
6,33
4,22
V
M
= 1.5 (mol).
ν (CO2) =
2. Using the reaction equation, we determine ν (Na2CO3):
ν (Na2CO3) = ν (CO2)  ν (Na2CO3) = 1.5 (mol).
3. Determine the mass of Na2CO3:
m (Na2CO3) = ν (Na2CO3) ∙ M (Na2CO3) = 1.5 ∙ 106 = 159 (g).

Decide for yourself:
1. When magnesium weighing 1.2 g interacts with a solution of sulfuric acid
obtained salt weighing 5.5 g. Determine the mass fraction (%) of the product yield
reactions. (91.67%).
2. When sodium reacts with an amount of substance of 0.5 mol with water
obtained hydrogen with a volume of 4.2 liters. Calculate the volume fraction (%) of the gas output.
(75%.)
3. Metallic chromium is obtained by reducing its oxide Cr2O3
metallic aluminum. Calculate the mass of chromium that can be
obtain by reducing its oxide weighing 228 kg, if the mass fraction
chromium yield is 95%. (148.2 kg.)
4. When fusing sodium hydroxide weighing 60 g and silicon (IV) oxide
13 g of water vapor were formed. Determine the mass fraction (%) of the yield
water. (96.3%).
5. Determine what mass of copper will react with concentrated
sulfuric acid to obtain sulfur oxide (IV) with a volume of 3.0 l (no.), if
the volume fraction of sulfur (IV) oxide yield is 90%. (9.51 g)
6. Calculate the volume of ammonia that can be obtained by heating chloride
ammonium weighing 20 g with an excess of calcium hydroxide, if the volume fraction
ammonia yield is 98%. (8.2 l.)
7. When passing ammonia with a volume of 672 l (no.) through a solution weighing
900 g with a mass fraction of nitric acid of 40%, ammonium nitrate with a mass of
440.68 g. Determine the mass fraction (%) of the salt yield. (96%).
8. From phosphorus weighing 15.5 kg we obtained phosphoric acid weighing
41.6 kg. Calculate the mass fraction (%) of the product yield. (85%).
9. How much sulfuric acid can be obtained from elemental
sulfur weighing 192 g, if the mass fraction of the yield of the last stage is 95%.
(5.7 mol.)
10. When passing hydrogen sulfide with a volume of 2.8 l (n.o.) through an excess
solution of copper (II) sulfate, a precipitate weighing 11.4 g was formed. Determine
yield of reaction product. (95%).
11. Through a solution weighing 50 g with a mass fraction of sodium iodide 15%
missed excess chlorine. Iodine was released weighing 5.6 g. Determine the yield
reaction product. (88.2%).
12. A solution containing calcium chloride weighing 4.5 g was added
solution containing sodium phosphate weighing 4.1 g. Determine the mass
the resulting precipitate if the yield of the reaction product is 88%. (3.41 g)
13. Calculate the volume of a solution with a mass fraction of potassium hydroxide
26% ( = 1.24 g/ml) is required to react with aluminum to obtain

hydrogen with a volume of 10.64 l, if the volume fraction of hydrogen yield is
95%. (41.35 ml.)
14. Determine the amount of substance and volume (no.) of chlorine, which
required to obtain iron (III) chloride weighing 150 g at mass
the salt yield is 92.3%. (1.5 mol; 33.6 l.)
15. When passing a mixture consisting of sulfur (IV) oxide with a volume of 5 l and
oxygen with a volume of 15 liters, through a contact apparatus, the volume changed to 2 liters.
Determine the volume fraction (%) of the reaction product yield. (80%).
16. During the thermal decomposition of methane, an amount of 14 mol was obtained
acetylene, the volume of which at n. u. amounted to 120.96 l. Calculate mass
share (%) of product yield. (77%).
17. Calculate the mass of sodium acetate consumed to produce methane
weighing 80 g with a mass fraction of product yield of 70%. (586)
18. Determine the mass of acetic acid that is consumed for synthesis
ethyl acetyl ether, if the resulting mass of 70.4 g is 80% of
theoretical. (60)
19. Calculate the mass of carbon tetrachloride that can be obtained
when chlorinating methane with a volume of 11.2 l with molecular chlorine, volume
which is equal to 56 l (no.). Product yield is 70% of theoretical
possible. (53.9 g)
20. During the catalytic hydrogenation of formaldehyde, alcohol was obtained,
upon interaction with sodium metal, hydrogen was formed
volume 8.96 l (no.) The product yield at each stage of the synthesis was
80%. Determine the initial mass of formaldehyde. (37.5 g)
21. When converting equal volumes of carbon monoxide (IV) and methane, the volume
mixture increased 1.8 times. Determine your conversion rate. (90%).

In chemistry you can’t do without a lot of substances. After all, this is one of the most important parameters of a chemical element. We will tell you in this article how to find the mass of a substance in various ways.

First of all, you need to find the desired element using the periodic table, which you can download on the Internet or buy. The fractional numbers under the sign of an element are its atomic mass. It needs to be multiplied by the index. The index shows how many molecules of an element are contained in a given substance.

When you have a complex substance, you need to multiply the atomic mass of each element of the substance by its index. Now you need to add up the atomic masses you obtained. This mass is measured in units of gram/mol (g/mol). We will show how to find the molar mass of a substance using the example of calculating the molecular mass of sulfuric acid and water:
H2SO4 = (H)*2 + (S) + (O)*4 = 1*2 + 32 + 16*4 = 98g/mol;

H2O = (H)*2 + (O) = 1*2 + 16 = 18g/mol.

The molar mass of simple substances that consist of one element is calculated in the same way.
You can calculate molecular weight using an existing table of molecular weights, which can be downloaded online or purchased at a bookstore
You can calculate molar mass using formulas and equate it to molecular mass. In this case, the units of measurement must be changed from “g/mol” to “amu”.
When, for example, you know the volume, pressure, mass and temperature on the Kelvin scale (if Celsius, then you need to convert), then you can find out how to find the molecular mass of a substance using the Mendeleev-Clayperon equation:

M = (m*R*T)/(P*V),

where R is the universal gas constant; M is the molecular (molar mass), a.m.u.
You can calculate the molar mass using the formula:
where n is the amount of substance; m is the mass of a given substance. Here you need to express the amount of substance using volume (n = V/VM) or Avogadro's number (n = N/NA).
If the volume of a gas is given, then its molecular weight can be found by taking a sealed container with a known volume and pumping out the air from it. Now you need to weigh the cylinder on the scales. Next, pump gas into it and weigh it again. The difference in mass of an empty cylinder and a cylinder with gas is the mass of the gas we need.
When you need to carry out the cryoscopy process, you need to calculate the molecular weight using the formula:
M = P1*Ek*(1000/P2*Δtk),

where P1 is the mass of the dissolved substance, g; P2 is the mass of the solvent, g; Ek is the cryoscopic constant of the solvent, which can be found from the corresponding table. This constant is different for different liquids; Δtk is the temperature difference, which is measured using a thermometer.

Now you know how to find the mass of a substance, be it simple or complex, in any state of aggregation.

Sign

The word “exit” appears in the problem statement. The theoretical yield of the product is always higher than the practical one.

Concepts “theoretical mass or volume, practical mass or volume” can be used only for substances-products.

The percentage of product yield is indicated by the letter

(eta), measured as a percentage or fraction.

The quantitative output can also be used for calculations:

First type of tasks – The mass (volume) of the starting substance and the mass (volume) of the reaction product are known. It is necessary to determine the yield of the reaction product in %.

Problem 1. When magnesium weighing 1.2 g reacted with a solution of sulfuric acid, a salt weighing 5.5 g was obtained. Determine the yield of the reaction product (%).

	Given: m(Mg) = 1.2 g m practical (MgSO 4) = 5.5 g _____________________ Find:

	M (Mg) = 24 g/mol M (MgSO 4) = 24 + 32 + 4 16 = 120 g/mol
	ν( Mg ) = 1.2 g / 24 (g/mol) = 0.05 mol
5. Using CSR, we calculate the theoretical amount of substance (ν theor) and the theoretical mass (m theor) of the reaction product	m = ν M m theor (MgSO 4) = M (MgSO 4) ν theor (MgSO 4) = 120 g/mol 0.05 mol = 6 g
	(MgSO 4)=(5.5g 100%)/6g=91.7% Answer: The yield of magnesium sulfate is 91.7% compared to theoretical

Second type of tasks – The mass (volume) of the starting substance (reagent) and the yield (in%) of the reaction product are known. It is necessary to find the practical mass (volume) of the reaction product.

Problem 2. Calculate the mass of calcium carbide formed by the action of coal on calcium oxide weighing 16.8 g, if the yield is 80%.

1. Write down a brief statement of the problem	Given: m(CaO) = 16.8 g 80% or 0.8 ____________________ Find: m pract (CaC 2 ) = ?
2. Let's write down the UHR. Let's arrange the coefficients. Under the formulas (from given) we write the stoichiometric ratios displayed by the reaction equation.
3. Using PSHE, we find the molar masses of the underlined substances	M(CaO) = 40 + 16 = 56 g/mol M (CaC 2 ) = 40 + 2 12 = 64 g/mol
4. Find the amount of the reagent substance using the formulas	ν(CaO )=16.8 (g) / 56 (g/mol) = 0.3 mol
5. Using the UHR, we calculate the theoretical amount of the substance (ν theor) and the theoretical mass ( m theory ) reaction product
6. Find the mass (volume) fraction of the product yield using the formula	m practical (CaC 2 ) = 0.8 19.2 g = 15.36 g Answer: m practical (CaC 2 ) = 15.36 g

Third type of tasks– The mass (volume) of the practically obtained substance and the yield of this reaction product are known. It is necessary to calculate the mass (volume) of the starting substance.

Problem 3. Sodium carbonate reacts with hydrochloric acid. Calculate what mass of sodium carbonate must be taken to obtain carbon monoxide ( IV) with a volume of 28.56 l (n.u.). Practical product yield is 85%.

1. Write down a brief statement of the problem

Given: n. u.

V m = 22.4 l/mol

V practical (CO 2) = 28.56 l

85% or 0.85

_____________________

Find:

m(Na 2 CO 3) =?

2. Find the molar masses of substances using PSCE, if necessary

M (Na 2 CO 3) = 2 23 + 12 + 3 16 = 106 g/mol

3. We calculate the theoretically obtained volume (mass) and the amount of substance of the reaction product using the formulas:5. Determine the mass (volume) of the reagent using the formula:

m = ν M

V = ν V m

m = ν M

m (Na 2 CO 3) = 106 g/mol 1.5 mol = 159 g

SOLVE PROBLEMS

№1.

When sodium reacted with an amount of 0.5 mol of the substance with water, hydrogen with a volume of 4.2 liters was obtained. Calculate the practical gas yield (%).

Metallic chromium is obtained by reducing its oxide Cr 2 O 3 with metallic aluminum. Calculate the mass of chromium that can be obtained by reducing its oxide weighing 228 g, if the practical yield of chromium is 95%.

№3.

Determine what mass of copper will react with concentrated sulfuric acid to produce sulfur (IV) oxide with a volume of 3 liters (n.s.), if the yield of sulfur (IV) oxide is 90%.

№4.

A solution containing sodium phosphate weighing 4.1 g was added to a solution containing calcium chloride weighing 4.1 g. Determine the mass of the resulting precipitate if the yield of the reaction product is 88%.

To do this, you need to add up the masses of all the atoms in this molecule.

Example 1. In a water molecule H2O there are 2 hydrogen atoms and 1 oxygen atom. Atomic mass of hydrogen = 1, and oxygen = 16. Therefore, the molecular mass of water is 1 + 1 + 16 = 18 atomic mass units, and the molar mass of water = 18 g/mol.

Example 2. In a molecule of sulfuric acid H 2 SO 4 there are 2 hydrogen atoms, 1 sulfur atom and 4 oxygen atoms. Therefore, the molecular mass of this substance will be 1 2 + 32 + 4 16 = 98 amu, and the molar mass will be 98 g/mol.

Example 3. In the molecule of aluminum sulfate Al 2 (SO 4) 3 there are 2 aluminum atoms, 3 sulfur atoms and 12 oxygen atoms. The molecular mass of this substance is 27 · 2 + 32 · 3 + 16 · 12 = 342 amu, and the molar mass is 342 g/mol.

Mole, molar mass

Molar mass is the ratio of the mass of a substance to the amount of substance, i.e. M(x) = m(x)/n(x), (1)

where M(x) is the molar mass of substance X, m(x) is the mass of substance X, n(x) is the amount of substance X.

The SI unit for molar mass is kg/mol, but the unit commonly used is g/mol. Unit of mass - g, kg.

The SI unit for quantity of a substance is the mole.

A mole is the amount of a substance that contains 6.02·10 23 molecules of this substance.

Any problem in chemistry is solved through the quantity of a substance. You need to remember the basic formulas:

n(x) =m(x)/ M(x)

or the general formula: n(x) =m(x)/M(x) = V(x)/Vm = N/N A, (2)

where V(x) is the volume of the substance X(l), V m is the molar volume of the gas at normal conditions. (22.4 l/mol), N is the number of particles, N A is Avogadro’s constant (6.02·10 23).

Example 1. Determine the mass of sodium iodide NaI with an amount of substance of 0.6 mol.

Example 2. Determine the amount of atomic boron contained in sodium tetraborate Na 2 B 4 O 7 weighing 40.4 g.

m(Na 2 B 4 O 7) = 40.4 g.

The molar mass of sodium tetraborate is 202 g/mol.

Determine the amount of substance Na 2 B 4 O 7:

n(Na 2 B 4 O 7) = m(Na 2 B 4 O 7)/M(Na 2 B 4 O 7) = 40.4/202 = 0.2 mol.

Recall that 1 mole of sodium tetraborate molecule contains 2 moles of sodium atoms, 4 moles of boron atoms and 7 moles of oxygen atoms (see sodium tetraborate formula).

Then the amount of atomic boron substance is equal to:

n(B)= 4 n(Na 2 B 4 O 7) = 4 0.2 = 0.8 mol.

The space around us is filled with different physical bodies, which consist of different substances with different masses. School courses in chemistry and physics, introducing the concept and method of finding the mass of a substance, were 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.

Calculating the mass of a substance using the specific density of the substance. Example – there is a 200 liter barrel. You need to fill the barrel with any liquid, say, light beer. How to find the mass of a filled barrel? Using the formula for the density of a substance p=m/V, where p is the specific density of the substance, m is the mass, V is the occupied volume, it is very simple to find the mass of a full barrel:

Volume measures are cubic centimeters, meters. That is, a 200 liter barrel has a volume of 2 m³.
The measure of specific density is found using tables and is a constant value for each 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=m/V => m=p*V: m = 1025.0 kg/m³* 2 m³=2050 kg.

A 200 liter barrel completely filled with light beer will have a mass of 2050 kg.

Finding the mass of a substance using molar mass. M (x)=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 quantity of substance X If the problem statement specifies only 1 known parameter - the molar mass of a given substance, then finding the mass of this substance will not be difficult. For example, it is necessary to find the mass of sodium iodide NaI with an amount of substance of 0.6 mol.

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 using the online calculator on the website: M (NaI)=M (Na)+M (I)=23+127=150 (g/mol).
From the general formula M (NaI)=m (NaI)/v (NaI) => m (NaI)=v (NaI)*M (NaI)= 0.6 mol*150 g/mol=90 grams.

The mass of sodium iodide (NaI) with a mass fraction of 0.6 mol is 90 grams.

Finding the mass of a substance by its mass fraction in solution. The formula for the mass fraction of a substance is ω=*100%, where ω is the mass fraction of the substance, and m (substance) and m (solution) are masses, measured in grams, kilograms. The total fraction of the solution is always taken as 100%, otherwise there will be errors in the calculation. It is easy to derive the formula for the mass of a substance from the formula for the mass fraction of a substance: m (substance) = [ω*m (solution)] /100%. However, there are some features of changing the composition of the solution that need to be taken into account when solving problems on this topic:

Diluting the solution with water. The mass of the dissolved substance X does not change m (X)=m’(X). The mass of the solution increases by the mass of added water m’ (p) = m (p) + m (H 2 O).
Evaporation of water from solution. The mass of the dissolved substance X does not change m (X)=m’ (X). The mass of the solution decreases by the mass of evaporated water m’ (p) = m (p) - m (H 2 O).
Merging two solutions. The masses of solutions, as well as the masses of the dissolved substance X, when mixed, add up: m’’ (X) = m (X) + m’ (X). m’’ (p)=m (p)+m’ (p).
Loss of crystals. The masses of the dissolved substance X and the solution are reduced by the mass of the precipitated crystals: m’ (X) = m (X)-m (precipitate), m’ (p) = m (p)-m (precipitate).

An algorithm for finding the mass of a reaction product (substance) if the yield of the reaction product is known. The yield of the product is found by the formula η=*100%, where m (x practical) is the mass of product x, which is obtained as a result of the practical reaction process, m (x theoretical) is the calculated mass of substance x. Hence m (x practical)=[η*m (x theoretical)]/100% and m (x theoretical)=/η. The theoretical mass of the resulting product is always greater than the practical mass, due to the reaction error, and is 100%. If the problem does not give the mass of the product obtained in a practical reaction, then it is taken as absolute and equal to 100%.

Options for finding the mass of a substance are not a useful school course, but methods that are quite applicable in practice. Everyone can easily find the mass of the required substance by applying the above formulas and using the proposed tables. To make the task easier, write down all the reactions and their coefficients.