Relative mass fraction of an element. Mass fraction of element
The concept of “share” is probably already familiar to you.
For example, the piece of watermelon shown in the figure is one quarter of the whole watermelon, that is, its share is 1/4 or 25%.
To better understand what a mass fraction is, imagine a kilogram of sweets (1000g) that a mother bought for her three children. Of this kilogram, the youngest child got half of all the candies (unfair, of course!). The eldest - only 200g, and the middle - 300g.
This means that the mass fraction of sweets for the youngest child will be half, or 1/2, or 50%. The middle child will have 30%, and the oldest will have 20%. It should be emphasized that the mass fraction can be a dimensionless quantity (quarter, half, third, 1/5, 1/6, etc.), or can be measured as a percentage (%). When solving calculation problems, it is better to convert the mass fraction into a dimensionless quantity.
Mass fraction of substance in solution
Any solution consists of a solvent and a solute. Water is the most common inorganic solvent. Organic solvents can be alcohol, acetone, diethyl ether, etc. If the problem statement does not indicate a solvent, the solution is considered aqueous.
The mass fraction of the dissolved substance is calculated by the formula:
$\omega_\text(in-va)=\dfrac(m_\text(in-va))(m_\text(r-ra))(\cdot 100\%)$
Let's look at examples of problem solving.
How many grams of sugar and water do you need to take to prepare 150g of a 10% sugar solution?
Solution
m(solution)=150g
$\omega$(sugars)=10%=0.1
m(sugar)=?
m(sugars) = $\omega\textrm((sugars)) \cdot m(p-pa) = 0.1 \cdot 150 \textrm(g) = 15 \textrm(g)$
m(water)=m(solution) - m(sugar) = 150g - 15g=135g.
ANSWER: you need to take 15g of sugar and 135g of water.
350 ml solution. and density 1.142 g/ml contains 28 g of sodium chloride. Find the mass fraction of salt in the solution.
Solution
V(solution)=350 ml.
$\rho$(solution)=1.142 g/ml
$\omega(NaCl)$=?
m(solution) =V(solution) $\cdot \rho$(solution)=350 ml $\cdot$ 1.142 g/ml=400g
$\omega(NaCl)=\dfrac(m(NaCl))(m\textrm((solution)))=\dfrac(28\textrm(g)) (400\textrm(g)) = 0.07 $=7%
ANSWER: mass fraction of sodium chloride $\omega(NaCl)$=7%
MASS FRACTION OF AN ELEMENT IN A MOLECULE
The formula of a chemical substance, for example $H_2SO_4$, carries a lot of important information. It denotes either an individual molecule of a substance, which is characterized by relative atomic mass, or 1 mole of a substance, which is characterized by molar mass. The formula shows the qualitative (consists of hydrogen, sulfur and oxygen) and quantitative composition (consists of two hydrogen atoms, a sulfur atom and four oxygen atoms). Using the chemical formula, you can find the mass of the molecule as a whole (molecular mass), and also calculate the ratio of the masses of the elements in the molecule: m(H) : m(S) : m(O) = 2: 32: 64 = 1: 16: 32. When calculating the mass ratios of elements, it is necessary to take into account their atomic mass and the number of corresponding atoms: $m(H_2)=1*2=2$, $m(S)=32*1=32$, $m(O_4)=16*4 =64$
The principle of calculating the mass fraction of an element is similar to the principle of calculating the mass fraction of a substance in a solution and is found using a similar formula:
$\omega_\text(elements)=\dfrac(Ar_(\text(elements))\cdot n_(\textrm(atoms)))(m_\text(molecules))(\cdot 100\%) $
Find the mass fraction of elements in sulfuric acid.
Solution
Method 1 (proportion):
Let's find the molar mass of sulfuric acid:
$M(H_2SO_4) = 1\cdot 2 + 32 + 16 \cdot 4=98\hspace(2pt)\textrm(g/mol)$
One molecule of sulfuric acid contains one sulfur atom, which means the mass of sulfur in sulfuric acid will be: $m(S) = Ar(S) \cdot n(S) = 32\textrm(g/mol) \cdot 1$= 32g/mol
Let's take the mass of the entire molecule as 100%, and the mass of sulfur as X% and make up the proportion:
$M(H_2SO_4)$=98 g/mol - 100%
m(S) = 32g/mol - X%
Where does $X=\dfrac(32\textrm(g/mol) \cdot 100\%)(98\textrm(g/mol)) =32, 65\% =32\%$
Method 2 (formula):
$\omega(S)=\dfrac(Ar_(\text(elements))\cdot n_(\textrm(atoms)))(m_\text(molecules))(\cdot 100\%)=\dfrac( Ar(S)\cdot 1)(M(H_2SO_4))(\cdot 100\%)=\dfrac(32\textrm(g/mol)\cdot 1)(98\textrm(g/mol))(\cdot 100\%) \approx32, 7\%$
Similarly, using the formula, we calculate the mass fractions of hydrogen and oxygen:
$\omega(H)=\dfrac(Ar(H)\cdot 2)(M(H_2SO_4))(\cdot 100\%)=\dfrac(1\textrm(g/mol)\cdot 2)(98\ textrm(g/mol))(\cdot 100\%)\approx2\%$
$\omega(O)=\dfrac(Ar(O)\cdot 4)(M(H_2SO_4))(\cdot 100\%)=\dfrac(16\textrm(g/mol)\cdot 4)(98\ textrm(g/mol))(\cdot 100\%)\approx65, 3\%$
Knowing the chemical formula, you can calculate the mass fraction of chemical elements in a substance. element in substance is denoted in Greek. letter “omega” - ω E/V and is calculated using the formula:
where k is the number of atoms of this element in the molecule.
What is the mass fraction of hydrogen and oxygen in water (H 2 O)?
Solution:
M r (H 2 O) = 2*A r (H) + 1*A r (O) = 2*1 + 1* 16 = 18
2) Calculate the mass fraction of hydrogen in water:
3) Calculate the mass fraction of oxygen in water. Since water contains atoms of only two chemical elements, the mass fraction of oxygen will be equal to:
Rice. 1. Formulation of the solution to problem 1
Calculate the mass fraction of elements in the substance H 3 PO 4.
1) Calculate the relative molecular mass of the substance:
M r (H 3 PO 4) = 3*A r (N) + 1*A r (P) + 4*A r (O) = 3*1 + 1* 31 +4*16 = 98
2) Calculate the mass fraction of hydrogen in the substance:
3) Calculate the mass fraction of phosphorus in the substance:
4) Calculate the mass fraction of oxygen in the substance:
1. Collection of problems and exercises in chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry, 8th grade” / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. - M.: AST: Astrel, 2006.
2. Ushakova O.V. Chemistry workbook: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. 8th grade” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006. (p. 34-36)
3. Chemistry: 8th grade: textbook. for general education institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005.(§15)
4. Encyclopedia for children. Volume 17. Chemistry / Chapter. ed.V.A. Volodin, Ved. scientific ed. I. Leenson. - M.: Avanta+, 2003.
1. Unified collection of digital educational resources ().
2. Electronic version of the journal “Chemistry and Life” ().
4. Video lesson on the topic “Mass fraction of a chemical element in a substance” ().
Homework
1. p.78 No. 2 from the textbook “Chemistry: 8th grade” (P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005).
2. With. 34-36 No. 3.5 from the Workbook in Chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. 8th grade” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006.
LESSON TOPIC: Mass fraction of a chemical element in a compound.
THE PURPOSE OF THE LESSON: Learn to calculate the mass fraction of elements in a compound using the formula of the compound and establish the chemical formula of a complex substance using the known mass fractions of chemical elements.
Basic concepts. Mass fraction of a chemical element.
Planned learning outcomes
Subject. Be able to calculate the mass fraction of an element in a compound using its formula and establish the chemical formula of a complex substance using the known mass fractions of chemical elements.
Metasubject . Develop the ability to establish analogies and use algorithms to solve educational and cognitive problems.
Main types of student activities. Calculate the mass fraction of an element in a compound using its formula. Establish the chemical formula of a complex substance using known mass fractions of chemical elements.
Lesson structure
I. Organizational stage
II. Updating of reference knowledge
III. Learning new material
IV. Consolidation. Summing up the lesson
V. Homework
During the classes
Organizing time.
Checking homework.
Updating basic knowledge.
Give definitions: relative atomic mass, relative molecular mass.
In what units can relative atomic mass be measured?
In what units can relative molecular mass be measured?
Learning new material.
Working with the textbook. Workbook.
Guys, let's say we have a substance - sulfuric acidH 2 SO 4,
Can we find out which atoms are part of the compound?
And their number?
And in what mass ratio do they combine?
Calculation of chemical mass ratios
elements in a complex substance. (page 51)
How can you find out in what mass ratios the elements are combined in a compound whose formula isH 2 SO 4 ?
m(H): m(S): m(O)= 2*2 + 32 + 16*4= 2:32:64 = 1:16:32.
1+16+32 = 49, that is, 49 parts by mass of sulfuric acid, contains 1 part by mass of hydrogen, 16 parts by mass of sulfur, 32 parts by mass of oxygen.
Guys, what do you think, can we calculate the proportion of each element in the compound?
Today we will get acquainted with the new concept of mass fraction of an element in a compound.
W- mass fraction of the element in the compound.
n- number of atoms of the element.
Mr- relative molecular weight.
Calculation of mass fractions of chemical elements
in a complex substance. (RT)
1. Study the algorithm for calculating the mass fraction of an element in a compound.
Task No. 1 (RT)
Deriving chemical formulas if the mass fractions of chemical elements are known,
included in this substance. (RT)
2. Study the algorithm for calculating the mass fraction of an element in a compound.
Problem No. 5 (RT)
Consolidation of the studied material.
RT page 25 No. 2.
RT page 27 No. 6.
Summing up the lesson.
What new concepts did you learn in class today?
Independent work.
Homework:
study §15 pp. 51 - 53;
answer questions No. 3,4,7 pp. 53-54 (in writing).
P List of used literature.
Textbook. Chemistry 8th grade. auto G.E. Rudzitis, F.G. Feldman. Publishing house "Prosveshcheniye", 2014.
Chemistry workbook. auto Borovskikh T.A.
Mass fraction is called the ratio of the mass of a given component m(X) to the mass of the entire solution M(solution). The mass fraction is denoted by the symbol ω (omega) and expressed in fractions of a unit or as a percentage:
ω(X) = m(X)/M(solution) (in fractions of unity);
ω(X) = m(X) 100/M(solution) (in percent).
Molar concentration is the amount of dissolved substance in 1 liter of solution. It is denoted by the symbol c(X) and measured in mol/l:
c(X) = n(X)/V = m(X)/M(X) V.
In this formula, n(X) is the amount of substance X contained in the solution, M(X) is the molar mass of substance X.
Let's look at several typical tasks.
- Determine the mass of sodium bromide contained in 300 g of a 15% solution.
Solution.
The mass of sodium bromide is determined by the formula: m(NaBr) = ω M(solution)/100;
m(NaBr) = 15,300/100 = 45 g.
Answer: 45
2. The mass of potassium nitrate that needs to be dissolved in 200 g of water to obtain an 8% solution is ______ g. (Round your answer to the nearest whole number.)
Solution.
Let m(KNO 3) = x g, then M(solution) = (200 + x) g.
Mass fraction of potassium nitrate in solution:
ω(KNO 3) = x/(200 + x) = 0.08;
x = 16 + 0.08x;
0.92x = 16;
x = 17.4.
After rounding x = 17 g.
Answer: 17
3. The mass of calcium chloride that must be added to 400 g of a 5% solution of the same salt in order to double its mass fraction is equal to______ g. (Write your answer to the nearest tenth.)
Solution.
The mass of CaCl 2 in the original solution is equal to:
m(CaCl 2) = ω M(solution);
m(CaCl 2) = 0.05 400 = 20 g.
The mass fraction of CaCl 2 in the final solution is ω 1 = 0.05 2 = 0.1.
Let the mass of CaCl 2 that needs to be added to the original solution be x g.
Then the mass of the final solution M 1 (solution) = (400 + x) g.
Mass fraction of CaCl 2 in the final solution:
Solving this equation, we get x = 22.2 g.
Answer: 22.2 g.
4. The mass of alcohol that needs to be evaporated from 120 g of a 2% alcohol solution of iodine in order to increase its concentration to 5% is equal to _____________ g. (Write your answer to the nearest tenth.)
Solution.
Let us determine the mass of iodine in the original solution:
m(I 2) = ω M(solution);
m(I 2) = 0.02 120 = 2.4 g,
After evaporation, the mass of the solution became equal to:
M 1 (solution) = m(I 2)/ω 1
M 1 (solution) = 2.4/0.05 = 48 g.
Based on the difference in the masses of the solutions, we find the mass of evaporated alcohol: 120-48 = 72 g.
Answer: 72
5. The mass of water that must be added to 200 g of a 20% sodium bromide solution to obtain a 5% solution is _________ g. (Round your answer to the nearest whole number.)
Solution.
Let us determine the mass of sodium bromide in the original solution:
m(NaBr) = ω M(solution);
m(NaBr) = 0.2 200 = 40 g.
Let the mass of water that needs to be added to dilute the solution be x g, then according to the problem conditions:
From here we get x = 600 g.
Answer: 600 g.
6. The mass fraction of sodium sulfate in the solution obtained by mixing 200 g of 5% and 400 g of 10% solutions of Na 2 SO 4 is equal to _____________%. (Round your answer to the nearest tenth.)
Solution.
Let us determine the mass of sodium sulfate in the first initial solution:
m 1 (Na 2 SO 4) = 0.05 200 = 10 g.
Let us determine the mass of sodium sulfate in the second initial solution:
m 2 (Na 2 SO 4) = 0.1 400 = 40 g.
Let's determine the mass of sodium sulfate in the final solution: m(Na 2 SO 4) = 10 + 40 = 50 g.
Let us determine the mass of the final solution: M(solution) = 200 + 400 = 600 g.
Let's determine the mass fraction of Na 2 SO 4 in the final solution: 50/600 = 8.3%
Answer: 8,3%.
In addition to solving problems with solutions:
The “rule of the cross” is the diagonal scheme of the mixing rule for cases with two solutions.
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Weight of one part: 300/50 = 6 g.
Then
m1 = 6 15 = 90 g, .
m2 = 6 35 = 210 g.
You need to mix 90 g of a 60% solution and 210 g of a 10% solution.
