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 specific gravity substances. 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 unified system SI measurements 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.
- 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).
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 make the task easier, write down all the reactions and their coefficients.
Solution concentrations
Velikiy Novgorod
Example 1.
Solution:
ω(NaCl) = = 0.125 or 12.5%
Answer: ω(NaCl) = 0.125 or 12.5%.
Example 2.
Solution:
= 
m(FeSO 4) = = 22.8 g,
ω(FeSO 4) = = 0.076 or 7.6%
Answer: ω(FeSO 4) = 0.076 or 7.6%.
Example 3. Determine the mass fraction of hydrochloric acid if 350 l of HCl (n.s.) are dissolved in 1 liter of water.
Solution:
The mass of HCl is determined by the formula:
m(HCl) = n(HCl) m(HCl) = m(HCl) = 36.5 = 570.3 g.
Mass of solution m(solution) = m(HCl) + m(H 2 O) = m(HCl) + V(H 2 O) ρ(H 2 O)
ω(HCl) = 
= 0.363 or 36.3%
Answer: ω(HCl) = 0.363 or 36.3%.
Example 4. Determine the volume of hydrogen chloride, measured at ambient conditions, and the volume of water required to prepare 500 g of a solution with a mass fraction of HCl of 20%.
Solution:
Finding the mass of HCl:
Calculate the volume of HCl:
Calculate m(H 2 O):
V(H 2 O) = = = 400 ml
Answer
Or solvent by weight of solution
Example 5. Determine the mass of sodium nitrate and water required to prepare 800 g of solution with ω(NaNO 3) = 12%.
Solution:
Mass of dissolved salt:
m(NaNO 3) = ω(NaNO 3) m(solution) = 0.12 800 = 96 g.
m(solution) = m(NaNO 3) + m(H 2 O)
m(H 2 O) = m(solution) – m(NaNO 3) = 800 – 96 = 704 g.
Answer: m(NaNO 3) = 96 g, m(H 2 O) = 704 g.
Example 6. Determine the mass of CuSO 4 ·5H 2 O crystalline hydrate and water required to prepare 0.4 kg of solution with ω(CuSO 4) = 8%.
Solution(see example 2):
ω(CuSO 4) = = 
m(H 2 O) = m(solution) – m(CuSO 4 5H 2 O)
m(CuSO 4 5H 2 O) = n(CuSO 4 5H 2 O) M(CuSO 4 5H 2 O)
n(CuSO 4 5H 2 O) = n(CuSO 4) =
m(CuSO 4) = ω(CuSO 4) m(solution) = 0.08 400 = 32 g.
n(CuSO 4) = = 0.2 mol.
Hence m(CuSO 4 5H 2 O) = 0.2 250 = 50 g
Mass of water m(H 2 O) = 400 – 50 = 350 g
Answer: m(CuSO 4 5H 2 O) = 50 g, m(H 2 O) = 350 g.
Calculation of the mass of a solution of a certain concentration
Molar concentration
Molar concentration (molarity) is the number of moles of a substance contained in 1 liter of solution.
C(X) = , mol/l
where X is the amount of substance, mol;
V – volume of solution, l.
The volume of the solution is related to the mass of the solution as follows:
where ρ is the density of the solution, g/ml.
Molar concentration equivalent is the number of moles of equivalent substance contained in 1 liter of solution.
C(X) = , mol/l
where n( X) is the amount of equivalent substance, mol;
V – volume of solution, l.
where m(X) is the molar mass of the dissolved substance;
m(X) – mass of the dissolved substance;
m is the mass of the solution;
ω(Х) – mass fraction of solution.
The molar concentration of the equivalent is always greater than or equal to the molar concentration. This provision is used when checking the received data.
The molar concentration of the equivalent is often called normal and is denoted
1.0 n.; 0.5 n. etc.
The above calculation formulas allow you to determine the volume of the solution, the amount of substance and the amount of equivalent substance:
V= or V=
n(X) = C(X) V or n( X) = C( X) V
Literature
1. Korovin N.V. General chemistry. – M.: graduate School, 2002. – 558 p.
2. Nikolsky A. B., Suvorov A.V. Chemistry: Tutorial for universities. – St. Petersburg: Khimizdat, 2001. – 512 p.
3. Glinka N. L. Tasks and exercises general chemistry. – M.: Integral-Press, 2004. – 240 p.
4. Problems and exercises in general chemistry: Textbook / B. I. Adamson, O. N. Goncharuk, V. N. Kamyshova and others./ Ed. N.V. Korovina. – M.: Higher School, 2003. – 255 p.
5. Suvorov A.V., Nikolsky A.B. Questions and problems in general chemistry. – St. Petersburg: Khimizdat, 2002. – 304 p.
Solution concentrations
Guidelines To independent work students
Velikiy Novgorod
Concentrations of solutions: Guidelines for independent work of students / Comp. V.P. Kuzmicheva, G.N.Olisova, N.I.Ulyanova. – Veliky Novgorod: NovGU, 2006.
1. Calculation of the mass fraction of the dissolved substance………………………………………………………….4
1. 1. Calculation of the mass of a solute or solvent from the mass of a solution......5
1. 2. Calculation of the mass of a solution of a certain concentration from a given mass of a dissolved substance or solvent……………………………………………………6
1. 3. Dilution and concentration of solutions……………………………………………………….7
1. 4. Calculations related to mixing solutions………………………………………………………...9
1. 5. Tasks for independent work……………………………………………………11
2. Molar concentration……………………………………………………………………..14
2. 1. Determination of the molar concentration of a substance C(X) by the mass of the substance and the mass of the substance by a given molar concentration……………………………………………………………15
2. 2. Calculations related to dilution and concentration of solutions………………..17
2. 3. Calculations related to mixing solutions of different concentrations…………...17
2. 4. Calculations of the material balance of chemical processes: excess (deficiency)
reagents………………………………………………………………………………………...19
2. 5. Tasks for independent work……………………………………………………21
Literature…………………………………………………………………………………………………………25
1. Calculation of the mass fraction of the dissolved substance
Example 1. Calculate the mass fraction of sodium chloride in the solution if 40 g of it are dissolved in 280 ml of water.
Solution:
Mass of solution m(solution) = m(NaCl) + m(H 2 O)
m(H 2 O) = V(H 2 O) ρ(H 2 O) = 280 ml 1 g/ml = 280 g,
ω(NaCl) = = 0.125 or 12.5%
Answer: ω(NaCl) = 0.125 or 12.5%.
Example 2. 41.7 g of FeSO 4 ·7H 2 O crystalline hydrate was dissolved in 258.3 g of water. Determine the mass fraction of FeSO 4 in the resulting solution.
Solution:
First, calculate the mass of the solution:
m(solution) = m(FeSO 4 7H 2 O) + m(H 2 O) = 41.7 + 258.3 = 300 g
m(FeSO 4) = n(FeSO 4) m(FeSO 4) = n(FeSO 4 7H 2 O) m(FeSO 4) =
m(HCl) = ω(HCl) m(solution) = 0.2 500 = 100 g.
Calculate the volume of HCl:
V(HCl) = n(HCl) V M = 22.4 l/mol = 61.37 l.
Calculate m(H 2 O):
m(H 2 O) = m(solution) – m(HCl) = 500 – 100 = 400 g.
V(H 2 O) = = = 400 ml
Answer: V(HCl) = 61.37 l, V(H 2 O) = 400 ml.
Calculating the mass of a solute
Concentration calculations
dissolved substances
in solutions
Solving problems involving diluting solutions is not particularly difficult, but it requires care and some effort. Nevertheless, it is possible to simplify the solution of these problems by using the law of dilution, which is used in analytical chemistry when titrating solutions.
All chemistry problem books show solutions to problems presented as sample solutions, and all solutions use the law of dilution, the principle of which is that the amount of solute and the mass m in the original and diluted solutions remain unchanged. When we solve a problem, we keep this condition in mind, and write down the calculation in parts and gradually, step by step, approach the final result.
Let us consider the problem of solving dilution problems based on the following considerations.
Amount of solute:
= c V,
Where c– molar concentration of the dissolved substance in mol/l, V– volume of solution in l.
Solute mass m(r.v.):
m(r.v.) = m(r-ra),
Where m(solution) is the mass of the solution in g, is the mass fraction of the dissolved substance.
Let us denote the quantities in the original (or undiluted) solution c, V, m(r-ra), through With 1 ,V 1 ,
m 1 (solution), 1, and in a dilute solution - through With 2 ,V 2 ,m 2 (solution), 2 .
Let's create equations for the dilution of solutions. We will allocate the left sides of the equations for the original (undiluted) solutions, and the right sides for dilute solutions.
The constant amount of solute upon dilution will have the form:
Conservation of mass m(r.v.):
The amount of solute is related to its mass m(r.v.) with the ratio:
= m(r.v.)/ M(r.v.),
Where M(r.v.) – molar mass of the dissolved substance in g/mol.
Dilution equations (1) and (2) are related to each other as follows:
from 1 V 1 = m 2 (solution) 2 / M(r.v.),
m 1 (solution) 1 = With 2 V 2 M(r.v.).
If the volume of dissolved gas is known in the problem V(gas), then its amount of substance is related to the volume of gas (no.) by the ratio:
= V(gas)/22.4.
The dilution equations will take the following form:
V(gas)/22.4 = With 2 V 2 ,
V(gas)/22.4 = m 2 (solution) 2 / M(gas).
If the mass of a substance or the amount of substance taken to prepare a solution is known in the problem, then on the left side of the dilution equation we put m(r.v.) or, depending on the conditions of the problem.
If, according to the conditions of the problem, it is necessary to combine solutions of different concentrations of the same substance, then on the left side of the equation the masses of dissolved substances are summed up.
Quite often, problems use the density of the solution (g/ml). But since the molar concentration With is measured in mol/l, then the density should be expressed in g/l, and the volume V– in l.
Let us give examples of solving “exemplary” problems.
Task 1. What volume of 1M sulfuric acid solution must be taken to obtain 0.5 liters of 0.1M H2SO4 ?
Given:
c 1 = 1 mol/l,
V 2 = 0.5 l,
With 2 = 0.1 mol/l.
Find:
Solution
V 1 With 1 =V 2 With 2 ,
V 1 1 = 0.5 0.1; V 1 = 0.05 l, or 50 ml.
Answer.V 1 = 50 ml.
Problem 2
(,
№ 4.23). Determine the mass of the solution with mass fraction(CuSO 4) 10% and the mass of water that will be required to prepare a solution weighing 500 g with a mass fraction
(CuSO 4) 2%.
Given:
1 = 0,1,
m 2 (solution) = 500 g,
2 = 0,02.
Find:
m 1 (r-ra) = ?
m(H 2 O) = ?
Solution
m 1 (solution) 1 = m 2 (solution) 2,
m 1 (solution) 0.1 = 500 0.02.
From here m 1 (solution) = 100 g.
Let's find the mass added water:
m(H 2 O) = m 2 (size) – m 1 (solution),
m(H 2 O) = 500 – 100 = 400 g.
Answer. m 1 (solution) = 100 g, m(H 2 O) = 400 g.
Problem 3
(,
№ 4.37).What is the volume of solution with a mass fraction of sulfuric acid of 9.3%
(= 1.05 g/ml) required to prepare 0.35M solution H2SO4 40 ml volume?
Given:
1 = 0,093,
1 = 1050 g/l,
With 2 = 0.35 mol/l,
V 2 = 0.04 l,
M(H 2 SO 4) = 98 g/mol.
Find:
Solution
m 1 (solution) 1 = V 2 With 2 M(H 2 SO 4),
V 1 1 1 = V 2 With 2 M(H 2 SO 4).
We substitute the values of known quantities:
V 1 1050 0.093 = 0.04 0.35 98.
From here V 1 = 0.01405 l, or 14.05 ml.
Answer. V 1 = 14.05 ml.
Problem 4
. What volume of hydrogen chloride (NO) and water will be required to prepare 1 liter of solution (= 1.05 g/cm 3), in which the hydrogen chloride content in mass fractions is 0.1
(or 10%)?
Given:
V(solution) = 1 l,
(solution) = 1050 g/l,
= 0,1,
M(HCl) = 36.5 g/mol.
Find:
V(HCl) = ?
m(H 2 O) = ?
Solution
V(HCl)/22.4 = m(r-ra) / M(HCl),
V(HCl)/22.4 = V(r-ra) (r-ra) / M(HCl),
V(HCl)/22.4 = 1 1050 0.1/36.5.
From here V(HCl) = 64.44 l.
Let's find the mass of added water:
m(H 2 O) = m(r-ra) – m(HCl),
m(H 2 O) = V(r-ra) (r-ra) – V(HCl)/22.4 M(HCl),
m(H 2 O) = 1 1050 – 64.44/22.4 36.5 = 945 g.
Answer. 64.44 l HCl and 945 g water.
Problem 5 (, № 4.34). Determine the molar concentration of a solution with a mass fraction of sodium hydroxide of 0.2 and a density of 1.22 g/ml.
Given:
0,2,
= 1220 g/l,
M(NaOH) = 40 g/mol.
Find:
Solution
m(size) = With V M(NaOH),
m(size) = With m(r-ra) M(NaOH)/.
Let's divide both sides of the equation by m(r-ra) and substitute the numerical values of the quantities.
0,2 = c 40/1220.
From here c= 6.1 mol/l.
Answer. c= 6.1 mol/l.
Problem 6 (, № 4.30).Determine the molar concentration of the solution obtained by dissolving sodium sulfate weighing 42.6 g in water weighing 300 g, if the density of the resulting solution is 1.12 g/ml.
Given:
m(Na 2 SO 4) = 42.6 g,
m(H 2 O) = 300 g,
= 1120 g/l,
M(Na 2 SO 4) = 142 g/mol.
Find:
Solution
m(Na 2 SO 4) = With V M(Na 2 SO 4).
500 (1 – 4,5/(4,5 + 100)) = m 1 (solution) (1 – 4.1/(4.1 + 100)).
From here m 1 (solution) = 104.1/104.5 500 = 498.09 g,
m(NaF) = 500 – 498.09 = 1.91 g.
Answer. m(NaF) = 1.91 g.
LITERATURE
1.Khomchenko G.P., Khomchenko I.G. Problems in chemistry for applicants to universities. M.: New Wave, 2002.
2. Feldman F.G., Rudzitis G.E. Chemistry-9. M.: Education, 1990, p. 166.
Calculation of the mass of a solution of a certain concentration based on the mass of the solute or solvent.
Calculation of the mass of a solute or solvent from the mass of a solution and its concentration.
Calculation of the mass fraction (in percent) of the dissolved substance.
Examples typical tasks by calculating the mass fraction (in percent) of the dissolved substance.
Percentage concentration.
Mass fraction (percentage) or percentage concentration (ω) – shows the number of grams of solute contained in 100 grams of solution.
The percentage concentration or mass fraction is the ratio of the mass of the solute to the mass of the solution.
ω = msol. in-va · 100% (1),
m solution
where ω – percentage concentration (%),
m sol. in-va – mass of dissolved substance (g),
m solution – mass of solution (g).
Mass fraction is measured in fractions of a unit and is used in intermediate calculations. If the mass fraction is multiplied by 100%, the percentage concentration is obtained, which is used when the final result is given.
The mass of a solution is the sum of the mass of the solute and the mass of the solvent:
m solution = m solution + m solution. villages (2),
where m solution is the mass of the solution (g),
m r-la – mass of solvent (g),
m sol. v-va – mass of dissolved substance (g).
For example, if the mass fraction of a dissolved substance - sulfuric acid in water is 0.05, then the percentage concentration is 5%. This means that a solution of sulfuric acid weighing 100 g contains sulfuric acid weighing 5 g, and the mass of the solvent is 95 g.
EXAMPLE 1 . Calculate the percentage of crystalline hydrate and anhydrous salt if 50 g of CuSO 4 5H 2 O were dissolved in 450 g of water.
SOLUTION:
1) The total mass of the solution is 450 + 50 = 500 g.
2) We find the percentage of crystalline hydrate using formula (1):
X = 50 100 / 500 = 10%
3) Calculate the mass of anhydrous salt CuSO 4 contained in 50 g of crystalline hydrate:
4) Let's calculate molar mass CuSO 4 5H 2 O and anhydrous CuSO 4
M CuSO4 5H2O = M Cu + M s +4M o + 5M H2O = 64 + 32 + 4 16 + 5 18 = 250 g/mol
M CuSO4 = M Cu + M s + 4M o = 64 + 32 + 4 16 = 160 g/mol
5) 250 g of CuSO 4 5H 2 O contains 160 g of CuSO 4
And in 50 g CuSO 4 5H 2 O - X g CuSO 4
X = 50·160 / 250 = 32 g.
6) The percentage of anhydrous copper sulfate salt will be:
ω = 32·100 / 500 = 6.4%
ANSWER : ω СuSO4 · 5H2O = 10%, ω CuSO4 = 6.4%.
EXAMPLE 2 . How many grams of salt and water are contained in 800 g of 12% NaNO 3 solution?
SOLUTION:
1) Find the mass of the dissolved substance in 800 g of 12% NaNO 3 solution:
800 12 /100 = 96 g
2) The mass of the solvent will be: 800 –96 = 704 g.
ANSWER: Mass of HNO 3 = 96 g, mass of H 2 O = 704 g.
EXAMPLE 3 . How many grams of 3% MgSO 4 solution can be prepared from 100 g of MgSO 4 7H 2 O?
SOLUTION :
1) Calculate the molar mass of MgSO 4 7H 2 O and MgSO 4
M MgSO4 7H2O = 24 + 32 + 4 16 + 7 18 = 246 g/mol
M MgSO4 = 24 + 32 + 4 16 = 120 g/mol
2) 246 g of MgSO 4 7H 2 O contains 120 g of MgSO 4
100 g of MgSO 4 7H 2 O contains X g of MgSO 4
X = 100·120 / 246 = 48.78 g
3) According to the conditions of the problem, the mass of anhydrous salt is 3%. From here:
3% of the mass of the solution is 48.78 g
100% of the solution mass is X g
X = 100·48.78 / 3 = 1626 g
ANSWER : the mass of the prepared solution will be 1626 grams.
EXAMPLE 4. How many grams of HC1 should be dissolved in 250 g of water to obtain a 10% solution of HC1?
SOLUTION: 250 g of water constitute 100 – 10 =90% of the mass of the solution, then the mass of HC1 is 250·10 / 90 = 27.7 g of HC1.
ANSWER : The mass of HCl is 27.7 g.