The mass fraction of an element ω(E)% is the ratio of the mass of a given element m (E) in a given molecule of a substance to the molecular mass 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) = m (E) / Mr(in-va) (1)
ω% (E) = m(E) 100%/Mr(in-va)
The sum of the 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, they take a portion of a substance 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 the molecule.
So, for a substance A x B y in fractions of unity:
ω(A) = Ar(E) X / Мr(in-va) (2)
From proportion (2) we derive a 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 = 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)
Calculation formula (4) can be transformed as follows:
X: Y = ω%(A) / Ar(A) : ω%(B) / Ar(B) = X(A) : Y(B) (5)
Calculation formulas (3) and (5) are used to determine the formula of a 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(v-va) = Ar(E) X / W(A)
Examples of solving problems on 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 percentages.
Solution
1. Calculate the relative molecular weight of sulfuric acid:
Mr (H 2 SO 4) = 1 2 + 32 + 16 4 = 98
2. 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 an element with the letter ω, 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) = 64: 98 =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 percentages.
Solution
1. Calculate the relative molecular weight of aluminum oxide:
Mr(Al 2 O 3) = 27 2 + 16 3 = 102
2. 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
Mass fraction of a substance is the ratio of the mass of a given substance in a system to the mass of the entire system, i.e. ω(X) = m(X) / m,
where ω(X) is the mass fraction of substance X,
m(X) - mass of substance X,
m - mass of the entire 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) = 137+ 2 35.5 + 2 18 = 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) = m(H 2 O)/m(BaCl 2 2H 2 O) = 36 / 244 = 0.1475 = 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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We determine the amount of silver substance found in argentite: n(Ag) = m(Ag) / M(Ag) = 5.4 / 108 = 0.05 mol. From the formula Ag 2 S it follows that the amount of argentite substance is half as much as the amount of silver substance. Determine the amount of argentite substance: n(Ag 2 S) = 0.5 n(Ag) = 0.5 0.05 = 0.025 mol We calculate the mass of argentite: m(Ag 2 S) = n(Ag 2 S) M(Ag2S) = 0.025 248 = 6.2 g. Now we determine the mass fraction of argentite in a rock sample weighing 25 g. ω(Ag 2 S) = m(Ag 2 S) / m = 6.2/25 = 0.248 = 24.8%. |
A) Combustion of magnesium cool Melting of ice C) Settlement of river sand in water
D) Mixing sulfur and iron powders E) boiling water
2. The molar mass of iron is
A) 26 g/mol cool 56 g/mol C) 52 g/mol D) 112 g/mol E) 56
3. In the formula 2Na2S, the number of sodium and sulfur atoms is equal
A) 1 and 2 cool 4 and 1 C) 2 and 4 D) 4 and 2 E) 2 and 1
4. Formula of Mn(VII) oxide
1. MnO2 cool Mn2O7 C) Mn2O3 D) MnO3 E) MnO
5. In the reaction scheme P+O2? P2O5 need to put coefficients
A) 4, 5, 2 cool 2, 1, 1 C) 2, 5, 2 D 5, 4, 2 E) 2, 4, 5
6. The equation for the substitution reaction is –
A) 4Na + O2 = 2 Na2O cool CaCO3 = CaO +CO2? C) Zn + CuS = ZnS + Cu
D) 2Mg + O2 = 2MgO E) 2H2 + O2 > 2 H2O
7. An iron nail immersed in a solution of copper (II) chloride becomes covered with a red coating of copper. This is an example reaction:
A) Cool exchange Decomposition C) Substitution D) Compound E) no such reaction
8. Symbol for the chemical element manganese
A) ?е cool Mg C) О D) Mn E) Mr
9. We are talking about a chemical element, and not about a simple substance nitrogen, in the expression
A) Nitrogen is a component of air cool Nitric acid HNO3 contains nitrogen
C) Nitrogen formula N2 D) Liquid nitrogen is sometimes used to freeze food
E) nitrogen is an inert gas
10. Aluminum does not have a characteristic physical property
A) Electrical conductivity cool Thermal conductivity C) Silver-white color
D) Ability to be magnetized E) gas under normal conditions
11. A sign that allows us to call the rusting of a nail a chemical reaction is:
A) Heat release cool Gas release C) Color change
D) Odor E) sedimentation
12. Iron sulfide is a complex substance, not a mixture because
A) It can be separated by a magnet into iron and sulfur
cool It can be separated by distillation into iron and sulfur
C) Consists of atoms of different chemical elements and cannot be divided by physical methods into iron and sulfur
D) It is insoluble in water E) a gas under normal conditions
13. 3.01 * 10 23 iron atoms are
A) 2 mol cool 3 mol C) 1 mol D) 0.5 mol E) 1.5 mol
14. 69 g of sodium is
A) 3 mol cool 1 mol C) 6.3 mol D) 1.5 mol E) 0.5 mol
15. By filtering you can separate the mixture:
A) copper and iron shavings cool sugar and water C) chalk and water
D) water and acetic acid E) water and gasoline
16. The interaction of magnesium with oxygen refers to the reactions:
A) decomposition cool exchange C) connection D) substitution E) no such reaction
17. Chemical phenomena include:
A) crushing marble cool evaporation of water C) melting ice D) melting copper E) combustion of coal
19.What is the valence of aluminum?
A) 1 cool 2 C) 3 D) 4 E) 5
20.Molar mass units:
A) grams cool gram/mol C) mole D) melogram E) no unit of measurement
21. The molar mass of NaHCO3 is:
A) 156 cool 156 g/mol C) 84 g/mol D) 84 E) 84 l
22.Indicate the decomposition reaction:
A) 2H2 + O2 > 2 H2O cool 2Na + 2H2O > 2NaOH + H2
C) C + O2 > CO2 D) 2NH3 > N2 + 3H2
E) AgNO3 + HCl > AgCl +HNO3
23. The mass fraction of oxygen in sulfuric acid H2SO4 is approximately:
A) 16% cool 33% C) 65% D) 2% E) 17%
25.Which of these rows contains only metals?
A) K, Zn, Fe cool Si, Ca, Bi C)Al, C, Cr D) W, Os, B E) P, Au, Pb
26. The mass fraction of sulfur in the substance SO2 is equal to:
A) 32% cool 64% C) 50% D) 80% E) 12%
27. The mass of zinc sulfide formed by heating 10 g of sulfur with zinc is equal to:
A) 12 g cool 30.31 g C) 25.6 g D) 10.5 g E) 32.4 g
28. Symbol for the chemical element krypton
A) Ca cool Kr C) K D) Cd E) C
29. The substance is
A) Air B) copper C) Mirror D) Granite E) milk
30. The list of physical properties is redundant
A) Density cool combustion C) Thermal conductivity
D) Boiling point E) melting point
Problem 435.
How many milliliters of concentrated hydrochloric acid (p = 1.19 g/ml), containing 38% (wt.) HCI, must be taken to prepare 1 liter of 2N. solution?
Solution:
M(HCI) = M E (HCI) = 36.5 g/mol.
Let's calculate the mass of HCI in 1 liter of 2N solution: 2 .
36.5 = 72.93g.
Let's calculate the mass of a 38% solution using the formula:
Where
The volume of solution that needs to be taken to prepare 1 liter of 2N solution is calculated using the formula:
m(p-pa) = p . V,
Where p
Answer: 161.28 ml.
Problem 436.
400 ml of water were added to 100 ml of 96% (by weight) H 2 SO 4 (density 1.84 g/ml). The result was a solution with a density of 1.220 g/ml. Calculate its equivalent concentration and mass fraction of H 2 SO 4.
Solution:
We find the mass of a solution of 100 ml of a 96% solution using the formula:
m(p-pa) = p . V,
Where p is the density, and V is the volume of the solution, we get:
m(p-pa) = 1.84 . 100 = 184 g.
We find the mass of sulfuric acid in this solution using the formula:
Where
- mass fraction of dissolved substance; m (in-va) - mass of dissolved substance; m (solution) - mass of solution.
Let's calculate the mass of the solution obtained by mixing 100 ml of a 96% solution with 400 ml of water, we get:
m" (p-pa) = (100 + 400) . 1.220 = 610 g.
Let us determine the molar mass of the equivalent of H2SO)4 from the relationship:
M E (V) - molar mass of acid equivalent, g/mol; M(B) is the molar mass of the acid; Z(B) - equivalent number; Z (acid) is equal to the number of H + ions, H 2 SO 4 (((((2.
Then we find the equivalent concentration of the solution using the formula:
Where
m(B) is the mass of the dissolved substance, M E (V) is the molar mass of the equivalent of the dissolved substance, V is the volume of the solution (in l or ml).
Let's calculate the mass fraction of the resulting solution:
Answer: 7.2n; 28.96%.
m(p-pa) = p . V,
Where p is the density, and V is the volume of the solution, we get:
m(p-pa) = 1.18 . 1000 = 1180 g.
Let's calculate the mass of hydrochloric acid in solution using the formula:
Where
- mass fraction of dissolved substance; m (in-va) - mass of dissolved substance; m (solution) - mass of solution.
Let us determine the molar mass of the equivalent of HCl from the relationship:
M E (V) - molar mass of acid equivalent, g/mol; M(B) is the molar mass of the acid; Z(B) - equivalent number; Z (acid) is equal to the number of H + ions, H 2 SO 4 → 2.
Answer: 11.8n.
Problem 438.
What volume of 10% (by mass) sulfuric acid ( p= 1.07 g/ml) would be required to neutralize a solution containing 16.0 g of NaOH?
Solution:
The reaction equation for the neutralization of a NaOH solution with a H 2 SO 4 solution has the form:
H 2 SO 4 + 2NaOH ↔ Na 2 SO4 + 2H 2 O
From the reaction equation it follows that 0.5 moles of NaOH are required to neutralize 1 mole of NaOH, which means that the equivalent mass of sulfuric acid in this reaction is 49 g/mol (M/2 = 98/2 = 49).
Now let's calculate the mass of sulfuric acid required to neutralize 16 g of NaOH from the proportion:
The mass of a solution containing 19.6 g of H 2 SO 4 is calculated using the formula:
Where
- mass fraction of dissolved substance; m (in-va) - mass of dissolved substance; m (solution) - mass of solution.
The volume of the solution is calculated using the formula:
m(p-pa) = p . V,
where is the density, and V is the volume of the solution, we get:
Answer: 183.18 ml.
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