How to find the simplest formula of a substance. Finding the molecular formula of a substance by mass fractions of elements
Theme 2 Tasks for the derivation of molecular formulas of substances
Subject: Determination of the molecular formula of substances by mass fractions of elements
Objectives: To know the concept of mass fraction, relative density of gases.
To be able to find mass fractions, to determine the formulas of substances by mass fractions.
Plan
Relative density of gases
Problem solving
Homework
Mass fractions of elements in a substance
The mass fraction of an element is determined by the ratio of the mass of the element to
The relative molecular weight of the substancew \u003d A r E * i / M in va
The mass fraction of an element is often expressed as a percentage, then the formula for the mass fraction of an element will look like this:
w % \u003d A r E * i / M in va * 100%
The sum of all mass fractions of the elements forming this substance is equal to one or 100%.
W 1 + W 2 + W 3 + W 4 \u003d 1 or W 1 %+ W 2 %+ W 3 %+ W 4 %=100%
If the mass fraction of one of the elements is not known, it can be defined as the difference between the unit (100%) and the sum of the known mass fractions.
W 3 = 1- ( W 1 + W 2 + W 4 ) W 3 %= 100% - ( W 1 %+ W 2 %+ W 4 %)
Algorithm for solving problems.
1. Let us denote in the formula of the substance the number of atoms by the indices x, y, z, etc. by the number of elements in the molecule.
2. If the condition does not give the mass fraction of one of the elements, we determine it by the difference of 100% minus the mass fractions of all other elements.
3. We find the ratio of indices x: y: z, which is the ratio of the quotients of the mass fraction of an element divided by its relative atomic mass. Reduce quotients from division to the ratio of integers. Determine the simplest formula of a substance.
x: y: z \u003d w 1 % /A r 1 : w 2 % / Ar 2 : w 3 % /A r 3
4. If the relative molecular mass is not given, we find it by the condition of the problem. D (N 2 ) \u003d Mg (w-va) / D (H 2 ); Mg (w-va) \u003d D (O 2 ) * Mg (O 2 );
Mg (in-va) \u003d D (air) * Mg (air); M \u003d ρ g / l * 22.4l ...
5. Compare the relative molar mass of the simplest formula of the substance with the truth found by the condition of the problem. The ratio of these masses gives the number by which you need to multiply the indices in the simplest formula.
The molar mass of the substance.It is set in the tasks: __
The ratio of the number of atoms of the elements in the molecule.
It is set: ___
1) in finished form
1) an indication of the class of substance;
2) through the density (M \u003d ρ * Vm)
2) through mass fractions of elements in a substance; ______
3) through Dr 2 (G 1 ) (M (G 1 ) \u003d Dr * M (G 2 ))
3) through the molar fractions of elements in the substance;
4) through the relation of m and V m \\ M \u003d V \\ Vm
4) through the quantity of products
the reaction in which the desired substance is involved, for example
combustion products.
Task: The oxygen density of the hydrocarbon is 1.75; the mass fraction of hydrogen in it is 14.3%. Determine the molecular formula of a hydrocarbon.
Given:Decision:
BxHy
2) w (C) \u003d 100% - 14.3% \u003d 85.7% x: y \u003d w (C) / Ar ( C): w (H) / Ar (H)
D (O 2 ) =1,75
x: y \u003d 85.7 / 12: 14.3 / 1 x: y \u003d 7.14: 14.3 x: y \u003d 1: 2
w (H) \u003d 14.3%
3) The simplest formula is CH 2 Mr (CH 2 ) =12 +1*2 =14
Find: BxHy -?
4) Mr (CxNy \u003d D (O 2 ) * Mr(ABOUT 2 ) =1,75 * 32 = 56
5) 56: 14 \u003d 4 \u003d\u003e the formula of substance C 4 N 8 is butene
6) Mr (FROM 4 N 8 ) \u003d 12 * 4 + 1 * 8 \u003d 56 The problem is solved correctly.
Answer: C 4 N 8 - butene
Example 2: The elemental composition of the substance is as follows: mass fraction of iron element 0.7241 (or 72.41%), mass fraction of oxygen 0.2759 (or 27.59%). Derive the chemical formula.
Decision:
We find the ratio of the number of atoms:
Fe: O → 72.41 / 56: 27.59 / 16 ≈ 1.29: 1.72.
We take the smallest number as a unit (divide by the smallest number in this case it is 1.29) and find the following ratio:
Fe: O ≈ 1: 1.33.
Since there must be an integer number of atoms, then this ratio is reducible to integers:
Fe: O \u003d 3: 3.99 ≈ 3: 4.
Answer: the chemical formula of this substance is Fe 3 O 4 .
Algorithm for solving problems.
1. Denote the number of atoms of the elements of the desired type (verbally): by x, y , z
2. We equate the ratio of the numbers of atoms of the elements with the relation of atomic factors: x: y: z ... \u003d a% / A 1 : b% / A 2 : c% / A 3 ... where A 1 , A 2 , A 3 - atomic masses of elements.
3. We find the simplest formula and value of the relative molecular weight.
4. Determine the relative molecular weight of the desired substance by density (M \u003d 2DH 2 ; M \u003d 29 D air or M \u003d ρ g / l * 22.4 l).
5. We find out how many times to obtain a true formula, you need to increase the number of atoms of the simplest formula.
6. We find the molecular formula of the substance.
Example 3:
Find the alkene formula if its hydrogen density is 21. Build its structural formula, name it.
Given:
M (C n H 2 n ) = D H2 * M (N 2 ) M (C n H 2 n ) =21*2 = 42
D H2 (FROM n H 2 n ) = 21
The ratio of the number of atoms of elements through the indication of the class of substance. Alkenes have the formula C n H 2 n
We express M alkene in the general form: M (C n H 2 n ) =12 n + 2 n
To findn – ?
We make equation 14n = 42 n = 3
Answer: C 3 N 6 - propene structural formula:
Scan the solution and submit to email address:bogdanowskaj@ mail. ru
The solution of computational problems to find the molecular formula of a substance.
The solution of computational problems occupies an important place in the study of the fundamentals of chemical science. Students encounter problems in deriving the chemical formula of a substance while completing a chemistry program from grades 8 to 11. This development proposes different ways to solve the problems of finding the formula of a substance based on its density, mass fraction of chemical elements in the substance and on the products of combustion.
I type of tasks.
Determination of the molecular formula of a substance based on the results of quantitative analysis (mass fraction) and relative density.
Task. Find the molecular formula of a hydrocarbon with a carbon content of 80% and a hydrogen content of 20%, the relative density for hydrogen is 15.
Given: Solution:
1st method
W (C) \u003d 80% 1. We determine Mr substances:
W (H) \u003d 20% D (H 2) \u003d Mr things / Mr (H 2)
D (H 2) \u003d 15 Mr things \u003d D (H 2) * Mr (H 2);
Mr things-va \u003d 15 * 2 \u003d 30
Molecular
Formula -? 2. Determine how much by weight is carbon:
x - 80%; x \u003d 24 (C).
3. Determine how much by weight is hydrogen:
m (C) \u003d 30 - 24 \u003d 6.
4. Determine the number of carbon and hydrogen atoms in a given substance:
n (C) \u003d 24/12 \u003d 2 atoms;
n (H) \u003d 6/1 \u003d 6 atoms.
The formula of the substance is C 2 H 6.
2nd method
1. Mr things =15*2=30.
2. We pass from mass fractions to molar fractions. For this, mass fractions must be divided into relative atomic mass.
v molar fraction \u003d w% / Ar
Find the mole fractions of carbon and hydrogen.
Denote:
x is the number of molar fractions of carbon;
y is the number of mole fractions of hydrogen.
x: y \u003d 80/12: 20/1 \u003d 6.7: 20
We take the smallest number as 1, and divide the remaining numbers by the smallest. Thus it turns out:
1: 3, then the simplest formula is CH 3. We make up the equation and determine the true formula:
12n + 3n \u003d 30, 15n \u003d 30, n \u003d 2, then the true formula is C 2 H 6.
3rd way
You can immediately determine the number of atoms of the elements that make up the substance according to the formula
n \u003d w * Mr / Ar, but Mr must be known.
2. n (C) \u003d 0.8 * 30/12 \u003d 2 atoms;
n (H) \u003d 0.2 * 30/1 \u003d 6 atoms.
So the formula is C 2 H 6.
II type of tasks.
Determination of the molecular formula of a substance based on combustion products and relative density.
Task.Upon combustion of 1.3 g of the substance, 4.4 g of carbon monoxide (IV) and 0.9 g of water are formed. The vapor density of this substance for hydrogen is 39. Determine the molecular formula of the substance.
Given: Solution:
1st method
m (v-va) \u003d 3.9 g 1. Mr (v-va) \u003d 39 * 2 \u003d 78
m (CO 2) \u003d 13.2 2. Determine the mass of carbon from carbon monoxide (IV).
m (H 2 O) \u003d 2.7 g M (CO 2) \u003d 44 g / mol, m (CO 2) \u003d 44 g.
D (H 2) \u003d 39 V 44 g (CO 2) contains 12 g (C),
And in 13.2 g (CO 2) - x g (C); x \u003d 3.6 g.
Molecular Determine the mass of hydrogen in water
Formula -? M (H 2 O) \u003d 18 g / mol, m (H 2 O) \u003d 18 g.
In 18 g (Н 2 О) - 2 g (Н),
and 2.7 g (H 2 O) y g (H); y \u003d 0.3 g (H).
3. Determine whether the substance has oxygen m (C) + m (H) \u003d 3.6 + 0.3 \u003d 3.9 g. Therefore, there is no oxygen.
4. Determine the ratio of atoms.
Let x be the number of carbon atoms, y the number of hydrogen atoms
x: y \u003d 3.6 / 12: 0.3 / 1 \u003d 0.3: 0.3 \u003d 1: 1.
The simplest formula of CH, but since Mr (substances ) \u003d 78, then we compose the equation:
12 * 1n + 1n \u003d 78
Then the true formula of the substance is C 6 H 6.
2nd method
1. Mr (in-va) \u003d 39 * 2 \u003d 78
2. The mass of carbon is determined by the mass of carbon monoxide (IV), and the mass of hydrogen is determined by the mass of water.
To determine the amount of carbon monoxide (IV) substance and the amount of water substance, and based on them
a) M (CO 2) \u003d 44 g / mol
M (H 2 O) \u003d 18 g / mol
v (C) \u003d v (CO 2) \u003d 13.2 g / 44 g / mol
v (H) \u003d 2v (H 2 O) \u003d 2 * 2.7 g / 18 g / mol;
b) determine the mass of carbon and hydrogen:
m (C) \u003d 12 * 0.3 \u003d 3.6 g
m (H) \u003d 1 * 0.3 \u003d 0.3 g
3. Determine whether there is oxygen in the substance:
m (C) + m (H) \u003d 3.6 + 0.3 \u003d 3.9 g.
So there is no oxygen.
4. We find the ratio of carbon atoms and hydrogen
v (C): v (H) \u003d 0.3: 0.3 \u003d 1: 1.
The simplest formula of the substance is CH.
5. Determine the true formula of the substance:
12 * 1 n + 1 n \u003d 78
True formula C 6 H 6.
Students choose for themselves the most appropriate way to solve such problems.
In some problems, the elemental composition of the sought substance is not obvious from the text of the condition. Most often this applies to combustion reactions of organic substances. Uncertainty in the composition is usually associated with the possibility of the presence of oxygen in a burnt substance. In the first step of solving such problems, it is necessary by calculation to reveal the elemental composition of the desired substance.
Task 2.11.
The combustion of 1.74 g of the organic compound yielded 5.58 g of a mixture of CO 2 and H 2 O. The amounts of substances CO 2 and H 2 O in this mixture were equal. Determine the molecular formula of an organic compound if the relative density of its vapors for oxygen is 1.8125.
Given:
mass of organic compound: m org v.va \u003d 1.74 g;
total mass of products of the district: m (СО 2) + m (Н 2 О) \u003d 5.58 g;
the ratio of the quantities of substances of the products of the district: n(CO 2) \u003d n(H 2 O);
relative oxygen vapor density of the starting material: D (O 2) \u003d 1.8125.
To find: the molecular formula of the burned compound.
Decision:
Step 1. The class of the burnt organic compound is not indicated, therefore, the elemental composition can only be judged by the reaction products. Carbon and hydrogen were part of the burnt substance unambiguously, since these elements are present in the combustion products, and only oxygen took part in the reaction from the air. Moreover, all carbon and all hydrogen completely transferred from the starting material to CO 2 and H 2 O. It is possible that oxygen was also part of the desired compound.
The situation with the presence or absence of oxygen can be clarified according to the data from the problem condition. We know the mass of burnt organic compounds and quantitative data,
related to products. Obviously, if the total mass of carbon from CO 2 and hydrogen from H 2 O is equal to the mass of the original organic substance, then there was no oxygen in its composition. Otherwise, if
m [(C) (in CO 2)] + m [(H) (in H 2 O)]\u003e m org. in va
oxygen was part of the starting material, and its mass is determined by the difference:
m org. in-va - m (C) (in СО 2) - m (Н) (in Н 2 О) \u003d m (O) (in the original in-ve).
We determine the mass of carbon and hydrogen in the reaction products and compare it with the mass of the starting substance.
1. The condition indicates information about the total mass of the reaction products, and therefore, first of all, we need to identify the masses of each of the products separately. For this, we denote the amount of carbon dioxide formed by the quantity " a". Then, according to the condition:
n (CO 2) \u003d n (H 2 O) \u003d a mol.
Using the value of "a" as known, we find the mass of CO 2 and H 2 O:
m (CO 2) \u003d M (CO 2). n(CO 2) \u003d (44.a) g,
m (H 2 O) \u003d M (H 2 O). n(H 2 O) \u003d (18. A) g.
We summarize the obtained expressions and equate to the value of the total mass of reaction products from the condition:
(44 . a) + (18 . a) = 5,58.
Got a mathematical equation with one unknown. Solving it, we find the value of an unknown quantity: a = 0,09.
By this value we designated the amount of substance of each of the products:
n(CO 2) \u003d n(H 2 O) \u003d 0.09 mol.
2. Find the mass of carbon in CO2 according to the algorithm:
n (CO 2) ---\u003e n (C) (in CO 2) ---\u003e m (C) (in CO 2)
n (C) (in CO 2) \u003d n (CO 2) \u003d 0.09 mol (according to the indices in the formula).
m (C) (in CO 2) \u003d n (C) (in CO 2). M (C) \u003d 0.09. 12 \u003d 1.08 g \u003d m (C) (outgoing)
3. Find the mass of hydrogen in the resulting water according to the algorithm:
n (H 2 O) ---\u003e n (H) (in H 2 O) ---\u003e m (H) (in H 2 O)
n (Н) (in Н 2 О)\u003e n (Н 2 О) 2 times (according to the indices in the formula)
n (H) (in H 2 O) \u003d 2. n (H 2 O) \u003d 2. 0.09 \u003d 0.18 mol
m (H) (in H2O) \u003d n (H) (in H2O). M (H) \u003d 0.18. 1 \u003d 0.18 g \u003d m (H) (as amended)
4. Compare the total mass of carbon and hydrogen with the mass of the starting substance:
m (C) (in CO 2) + m (H) (in H 2 O) \u003d 1.08 + 0.18 \u003d 1.26 g;
m org. v-va \u003d 1.74 g.
m (C) (in CO 2) + m (H) (in H 2 O)\u003e m org. vv-a
therefore, oxygen is part of the starting material.
m (O) (outgoing): \u003d m org. in-va - m (C) (in СО 2) - m (Н) (in Н 2 О) \u003d 1.74 -1.26 \u003d 0.48 g.
5. So, the starting material contains: carbon, hydrogen and oxygen.
Further actions are no different from examples of previously considered tasks. Denote the desired substance as C x H y O z.
Step 2 We compose a combustion reaction diagram:
C x H y O z. + O 2 ---\u003e CO 2 + H 2 O
Step 3
Define the ratio of the quantities of substance ( n) carbon, hydrogen and oxygen in the original sample of organic matter. We have already determined the amount of carbon and hydrogen in the first step.
Amount of substance ( n) oxygen we find from the data on its mass:
Step 4 We find the simplest formula:
N (C): N (H): N (O) \u003d 0.09: 0.18: 0.03
Choose the smallest value (in this case, “0.03”) and divide all three numbers into it:
Got the set of smallest integers:
N (C): N (H): N (O) \u003d 3: 6: 1
This makes it possible to write the simplest formula: C 3 H 6 O 1
Step 5
Revealing the true formula.
According to the data on the relative vapor density of the sought substance for oxygen, we determine the true molar mass:
M true. \u003d D (O 2). M (O 2) \u003d 1.8125. 32 \u003d 58 g / mol.
Define the value of the molar mass for the simplest formula:
M is simple. \u003d 3 .12 + 6. 1 +1. 16 \u003d 58 g / mol.
M is simple. \u003d M true. therefore, the simplest formula is true.
C 3 H 6 O is the molecular formula of the burnt substance.
Answer: C 3 H 6 O.
Solving problems to determine the formula of organic matter.
Developed by: Bush I.V. - teacher of biology and chemistry MBOU Kolyudovskaya secondary school
1.Determination of the formula of a substance by combustion products.
1. With the complete combustion of the hydrocarbon, 27 g of water and 33.6 g of carbon dioxide were formed (n.o.). The relative density of the hydrocarbon in argon is 1.05. Set its molecular formula.
2. During the combustion of 0.45 g of gaseous organic matter, 0.448 l of carbon dioxide, 0.63 g of water and 0.112 l of nitrogen were released. The density of the starting material for nitrogen is 1.607. Establish the molecular formula of this substance.
3. During the combustion of oxygen-free organic matter, 4.48 liters of carbon dioxide were formed. 3.6 g of water, 3.65 g of hydrogen chloride. Determine the molecular formula of the burned compound.
4. During the combustion of a secondary amine of a symmetrical structure, 0.896 l of carbon dioxide and 0.99 g of water and 0.112 l of nitrogen were released. Establish the molecular formula of this amine.
Solution Algorithm:
1.Determine the molecular weight of the hydrocarbon: M (CxHy) \u003d M (gas) xD (gas)
2. Define the amount of water substance: p (H2O) \u003d t (H2O): M (H2O)
3. Define the amount of hydrogen substance: p (H) \u003d 2p (H2O)
4. Determine the amount of carbon dioxide: p (CO2) \u003d t (CO2): M (CO2) or
p (CO2) \u003d V (CO2): Vm
5. Define the amount of carbon substance: p (C) \u003d p (CO2)
6. Define the relation C: H \u003d n (C): n (H) (we divide both numbers by the smallest of these numbers)
7. the simplest formula (from paragraph 6).
8. The molecular weight of the hydrocarbon (from the first paragraph) is divided by the molecular weight of the simplest formula (from paragraph 7): the obtained integer means - the number of carbon and hydrogen atoms in the simplest formula must be increased so many times.
9. Determine the molecular weight of the true formula (found in paragraph 8).
10. We write down the answer - the found formula.
The solution to problem number 1.
t (H2 O) \u003d 27g
V (CO2) \u003d 33.6l
D (for Ar) \u003d 1.05
Find SchNy
Decision.
1. Determine the molecular weight of the hydrocarbon: M (CxHy) \u003d M (gas) xD (gas)
M (CxHy) \u003d 1.05x40g / mol \u003d 42 g / mol
2. Determine the amount of water substance: p (H2O) \u003d t (H2O): M (H2O)
p (H2 O) \u003d 27g: 18g / mol \u003d 1.5 mol
3. Determine the amount of hydrogen substance: p (H) \u003d 2p (H2 O)
p (N) \u003d 2x1.5 mol \u003d 3 mol
4. Determine the amount of carbon dioxide: p (CO2) \u003d V (CO2): Vm
p (CO2) \u003d 33.6 L: 22.4 L / mol \u003d 1.5 mol
5. Determine the amount of carbon substance: p (C) \u003d p (CO2)
p (C) \u003d 1.5 mol
6. The ratio of C: H \u003d p (C): p (H) \u003d 1.5 mol: 3 mol \u003d (1.5: 1.5) :( 3: 1.5) \u003d 1: 2
7. The simplest formula: CH2
8.42g / mol: 14 \u003d 3
9. C3 H6 - true (M (C3 H6) \u003d 36 + 6 \u003d 42g / mol
10. Answer:. C3 H6.
2. Determination of the formula of a substance using the general formula and equations of a chemical reaction.
5. During the combustion of 1.8 g of the primary amine, 0.448 l of nitrogen was released. Determine the molecular formula of this amine.
6. During the combustion of 0.9 g of a certain limiting primary amine, 0.224 g of nitrogen was released. Determine the molecular formula of this amine.
7. In the interaction of 22 g of the limiting monobasic acid with an excess of sodium bicarbonate solution, 5.6 l of gas was released. Determine the molecular formula of the acid.
8. Establish the molecular formula of alkene, if it is known that 0.5 g of it can add 200 ml of hydrogen.
9. Establish the molecular formula of alkene. If you know that 1.5 g of it is able to attach 600 ml of hydrogen chloride.
10. Establish the molecular formula of cycloalkane, if it is known that 3g of it is able to attach 1.2 l of hydrogen bromide.
Solution Algorithm:
1. Define the amount of a known substance (nitrogen, carbon dioxide, hydrogen, hydrogen chloride, hydrogen bromide): n \u003d t: M or n \u003d V: Vm
2. By the equation, we compare the amount of a substance of a known substance with the amount of a substance that must be determined:
3. Define the molecular weight of the desired substance: M \u003d t: p
4. We will find the molecular weight of the desired substance using its general formula: (M (Spn2n) \u003d 12n + 2n \u003d 14n)
5. Equate the value of paragraph 3 and paragraph 4.
6. We solve the equation with one unknown, we find
7. We substitute the value of p.
8. Record the answer.
The solution to problem number 5.
V (N 2) \u003d 0.448l
t (SpN2p + 1 NH 2) \u003d 1.8 g
Find SpN2p + 1 NH 2.
Decision.
1. Reaction scheme: 2 SpN2p + 1 NH 2 \u003d N 2 (or)
2. Reaction equation: 2 SpN2 p + 1 NH 2 + (6p + 3) / 2О2 \u003d 2пСО2 + (2п + 3) Н2 О + N 2
3. We determine the amount of nitrogen substance by the formula: n \u003d V: Vm
p (N 2) \u003d 0.448 l: 22.4 l / mol \u003d 0.02 mol
4. Determine the amount of amine substance (using equations: the coefficient facing the amine is divided by the coefficient facing the nitrogen)
p (SpN2 p + 1 NH 2) \u003d 2p (N 2) \u003d 2x0.02 mol \u003d 0.04 mol
5. Define the molar mass of the amine according to the formula: M \u003d t: p
M ((SpN2 p + 1 NH 2) \u003d 1.8 g: 0.04 mol \u003d 45 g / mol
6. We determine the molar mass of the amine according to the general formula:
M (SpN2 p + 1 NH 2) \u003d 12p + 2p + 1 + 14 + 2 \u003d 14p + 17g / mol
7. Equate: 14p + 17 \u003d 45 (solve the equation)
8. Substitute in the general formula: SpN2 p + 1 NH 2 \u003d C2 H5 N H2
9.Answer: C2 H5 N H2
3. Determination of the formula of a substance using the equations of a chemical reaction and the law of conservation of mass of substances.
11. Some ester weighing 7.4 g was subjected to alkaline hydrolysis. In this case, 9.8 g of potassium salt of the limiting monobasic carboxylic acid and 3.2 g of alcohol were obtained. Establish the molecular formula of this ester.
12. An ester weighing 30 g was subjected to alkaline hydrolysis, while 34 g of sodium salt of limiting monobasic acid and 16 g of alcohol were obtained. Set the molecular formula for ether.
1.We compose the hydrolysis equation.
2. According to the law of conservation of mass of substances (the mass of the reacted substances is equal to the mass of the formed): mass of ether + mass of potassium hydroxide \u003d mass of salt + mass of alcohol.
3. Find mass (KOH) \u003d mass (salt) + mass (alcohol) - mass (ether)
4. Determine the amount of substance KOH: n \u003d m (KOH): M (KOH).
5. By the equation n (KOH) \u003d n (ether)
6. Define the molar mass of ether: M \u003d m: n
7. According to the equation, the amount of substance KOH \u003d the amount of salt substance (n) \u003d the amount of alcohol substance (n).
8. Define the molecular weight of the salt: M \u003d m (salt): n (salt).
9. Determine the molecular weight of the salt according to the general formula and equate the values \u200b\u200bfrom paragraphs 8 and 9.
10. From the molecular weight of the ether we take the molecular weight of the functional group of the acid found in the previous paragraph without the mass of the metal:
11. Determine the functional group of the alcohol.
The solution to problem number 11.
Given:
t (ether) \u003d 7.4g
t (salt) \u003d 9.8 g
t (alcohol) \u003d 3.2g
Find the ether formula
Decision.
1.We compose the equations for hydrolysis of the ester:
SpN2p +1 SOOSmN2 m + 1 + KOH \u003d SpN2p +1 SOOK + SmN2 m + 1 OH
2. Find the mass (KOH) \u003d t (salt) + t (alcohol) -t (ether) \u003d (9.8g + 3.2g) -7.4g \u003d 5.6g
3. Define n (KOH) \u003d t: M \u003d 5.6 g: 56 g / mol \u003d 0.1 mol
4. According to the equation: p (KOH) \u003d p (salt) \u003d p (alcohol) \u003d 0.1 mol
5. Define the molar mass of salt: M (SpN2p +1 COOK) \u003d t: p \u003d 9.8 g: 0.1 mol \u003d 98 g / mol
6. Define the molar mass according to the general formula: M (SpN2p +1 SOOK) \u003d 12p + 2p + 1 + 12 + 32 + 39 \u003d 14p + 84 (g / mol)
7. Equate: 14p + 84 \u003d 98
Formula of salt CH3COOC
8. Determine the molar mass of alcohol: M (cmH2 m + 1 OH) \u003d 3.2 g: 0.01 mol \u003d 32 g / mol
9. Define M (SmN2 m + 1 OH) \u003d 12m + 2m + 1 + 16 + 1 \u003d 14m + 18 (g / mol)
10. Equate: 14m + 18 \u003d 32
Alcohol Formula: CH3 OH
11. Formula of ether: CH3COOCH3 methyl acetate.
4. Determination of the formula of a substance using the reaction equations, written using the general formula of the class of organic compounds.
13. Determine the molecular formula of alkene, if it is known that the same amount of it, interacting with various hydrogen halides forms, respectively, either 5.23 g of chlorine derivative, or 8.2 g of bromine derivative.
14. When the same amount of alkene reacts with different halogens, 11.3 g of a dichloro derivative or 20.2 g of a dibromo derivative are formed. Define the formula of alkene. Write its name and structural formula.
1.We write down two reaction equations (alkene formula in general form)
2. We find the molecular masses of the products according to the general formulas in the reaction equations (through n).
3. Find the amount of substance products: n \u003d m: M
4. We equate the found quantities of the substance and solve the equations. Found n substitute in the formula.
The solution to problem number 13.
t (SpN2p + 1 Cl) \u003d 5.23g
t (SpN2n + 1 Br) \u003d 8.2g
Find SpN2 n
Decision.
1.We compose the reaction equations:
Spn2p + Hcl \u003d Spn2p + 1 Cl
Spn2p + HBr \u003d Spn2p + 1 Br
2. Define M (Spn2n + 1 Cl) \u003d 12n + 2n + 1 + 35.5 \u003d 14n + 36.5 (g / mol)
3.Let us define n (CnH2n + 1 Cl) \u003d t: M \u003d 5.23g: (14n + 36.5) g / mol
4.Define M (Spn2n + 1 Br) \u003d 12n + 2n + 1 + 80 \u003d 14n + 81 (g / mol)
5. Define n (CnH2n + 1 Br) \u003d t: M \u003d 8.2g: (14n + 81) g / mol
6. Equate n (SpN2n + 1 Cl) \u003d n (SpN2n + 1 Br)
5.23g: (14p + 36.5) g / mol \u003d 8.2g: (14p + 81) g / mol (solve the equation)
7.Alken formula: C3 H6
5. The definition of the formula of a substance through the introduction of the variable X.
15. As a result of burning 1.74 g of an organic compound, 5.58 g of a mixture of carbon dioxide and water are obtained. The amounts of carbon dioxide and water in this mess were equal. Determine the molecular formula of the organic compound. If the relative density of oxygen is equal to 1.81.
16. As a result of burning 1.32 g of an organic compound, 3.72 g of a mixture of carbon dioxide and water were obtained. The amounts of carbon dioxide and water in this mess were equal. Determine the molecular formula of the organic compound. If its relative density in nitrogen is 1.5714.
The algorithm for solving the problem:
1. Define the molar mass of organic matter: M (CxHy Oz) \u003d D (for gas) xM (gas)
2. Define the amount of organic matter: p (CxNy Oz) \u003d t: M
3. Introduction of the variable x: Let X be the amount of carbon dioxide in the mixture: p (CO2) \u003d Chmol, then the same amount of water (by condition): p (H2O) \u003d Chmol.
4. The mass of carbon dioxide in the mixture: t (CO2) \u003d n (CO2) xM (CO2) \u003d 44X (g)
6. By the condition of the problem: t (mixture) \u003d t (H2O) + t (CO2) \u003d 44X + 18X (We solve the equation with one unknown, we find the X-number of moles of CO2 and H2O)
11.Formula: C: H: O \u003d p (C): p (H): p (O)
The solution to problem number 15.
1. Determine the molecular weight of the organic substance: M (CxHy Oz) \u003d 1.82x32g / mol \u003d 58g / mol
2. Determine the amount of organic matter: n (CxNy Oz) \u003d t: M
n (CxNy Oz) \u003d 1.74g: 58g / mol \u003d 0.03 mol
3. Let X-p (CO2) in the mixture, then p (H2O) -Hmol (by condition)
4. The mass of carbon dioxide in the mixture: t (CO2) \u003d n (CO2) xM (CO2) \u003d 44X (g)
5. The mass of water in the mixture: p (H2O) \u003d p (H2O) xM (H2O) \u003d 18X (g)
6.We will compose and solve the equation: 44X + 18X \u003d 5.58 (mass of the mixture under the condition)
X \u003d 0.09 (mol)
7.Let us determine the amount of carbon substance (C): p (C) \u003d p (CO2): p (CxHy Oz)
p (C) \u003d 0.09: 0.03 \u003d 3 (mol) is the number of C atoms in organic matter.
8. Determine the amount of hydrogen substance (H): p (H) \u003d 2xp (H2O): p (CxHy Oz)
p (H) \u003d 2x0.09: 0.03 \u003d 6 (mol) - the number of hydrogen atoms in organic matter
9.Check the presence of oxygen in the organic compound: M (O) \u003d M (CxHy Oz) -M (C) -M (H)
M (O) \u003d 58g / mol - (3x12) - (6x1) \u003d 16 (g / mol)
10. Determine the amount of substance (number of atoms) of oxygen: p (O) \u003d M (O): Ar (O)
n (O) \u003d 16: 16 \u003d 1 (mol) is the number of oxygen atoms in organic matter.
11. The formula for the desired substance: (C3H6O).
6. Determination of the formula of a substance by the mass fraction of one of the elements included in the substance.
17. Define the molecular formula of dibromoalkane containing 85.11% bromine.
18. Determine the structure of the ester of an amino acid formed by derivatives of saturated hydrocarbons, if it is known that it contains 15.73% nitrogen.
19. Define the molecular formula of the limit trihydric alcohol, the mass fraction of oxygen in which is equal to 45.28%.
The algorithm for solving the problem.
1. Using the formula for finding the mass fraction of an element in a complex substance, we determine the molecular mass of the substance: W (element) \u003d Ar (element) xn (element): Mr (substance)
Mr (substance) \u003d Ar (element) xn (element): W (element), where n (element) is the number of atoms of this element
2. Determine the molecular weight according to the general formula
3. We equate point 1 and point 2. We solve the equation with one unknown.
4. We write the formula, substituting the value of n in the general formula.
The solution to problem No. 17.
Given:
W (Br) \u003d 85.11%
Find SpN2 pvr 2
1. Determine the molecular weight of dibromoalkane:
Mr ( Spn 2 pVr 2) = Ar ( INr) hp(Br): W (Br)
Mr (SpN2 pVr 2 ) \u003d 80x2: 0.8511 \u003d 188
2 We determine the molecular weight according to the general formula: Mr (SpN2 pVr 2 ) \u003d 12p + 2p + 160 \u003d 14p + 160
3. Equate and solve the equation: 14p + 160 \u003d 188
4.Formula: C2 H4 Br 2
7.Determination of the formula for organic matter using chemical reaction equations that reflect the chemical properties of a given substance.
20. With intermolecular dehydration of alcohol, 3.7 g of ether is formed. And with intramolecular dehydration of this alcohol 2.24 liters of ethylene hydrocarbon. Determine the formula of alcohol.
21. During intramolecular dehydration of a certain amount of primary alcohol 4.48 l of alkene was released, and with intermolecular dehydration 10.2 g of ether are formed. What structure does alcohol have?
The algorithm for solving the problem.
1.We write the equations of chemical reactions, which are mentioned in the problem (be sure to equalize)
2. Determine the amount of gaseous substance according to the formula: n \u003d V: Vm
3. We determine the amount of the starting substance, then the amount of the substance of the second reaction product (according to the reaction equation and the condition of the problem)
4. Determine the molar mass of the second product according to the formula: M \u003d t: p
5. Determine the molar mass according to the general formula and equate (p. 4 and p. 5)
6. We solve the equation and find n is the number of carbon atoms.
7.Write the formula.
The solution to problem No. 21.
t (ether) \u003d 10.2g
tT (SpN2 p) \u003d 4.48l
Find SpN2p + 1 OH
Decision.
1. We write the reaction equations:
SpN2p +1 OH \u003d SpN2p + H2 O
2SpN2p +1 OH \u003d SpN2 p + 1 OSpN2p + 1 + H2O
2. Determine the amount of alkene substance (gas): n \u003d V: Vm
p (SpN2p) \u003d 4.48 l: 22.4 l / mol \u003d 0.2 mol
3. According to the first equation, the amount of alkene is equal to the amount of alcohol. According to the second equation, the amount of ether substance is 2 times less than the amount of alcohol substance i.e. p (SpN2 p + 1 OSpN2p + 1) \u003d 0.1 mol
4. Determine the molar mass of ether: M \u003d t: p
M \u003d 10.2g: 0.1 mol \u003d 102g / mol
5. We determine the molar mass according to the general formula: M (SpN2p + 1 OSpN2p + 1) \u003d 12p + 2p + 1 + 16 + 12p + 2p + 1 \u003d 28p + 18
6. Equate and solve the equation: 28p + 18 \u003d 102
7.Alcohol formula: C3 H7 OH
I. Derivation of formulas of substances by mass fractions of elements.
1. Write the formula of the substance, denoting the indices by xyz.
2. If the mass fraction of one of the elements is unknown, then it is found by subtracting the known mass fractions from 100%.
3. Find the ratio of indices, for this, the mass fraction of each element (preferably in%) divided by its atomic mass (round to thousandths)
x: y: z \u003d ω 1 / Ar 1 : ω 2 / Ar 2 : ω 3 / Ar 3
4. The resulting numbers lead to integers. To do this, divide them by the smallest of the numbers obtained. If necessary (if it again turned out to be a fractional number), then multiply to an integer by 2, 3, 4 ... after that.
5. You get the simplest formula. For most inorganic substances, it coincides with the true, for organic, on the contrary, it does not.
Task number 1.
ω (N) \u003d 36.84% Decision:
1. We write the formula: N xO y
M.F. \u003d? 2. Find the mass fraction of oxygen:
ω (O) \u003d 100% - 36.84% \u003d 61.16%
3. Find the ratio of the indices:
x: y \u003d 36.84 / 14: 61.16 / 16 \u003d 2.631: 3.488 \u003d
2,631 / 2,631: 3,948 / 2,631 = 1: 1,5 =
1 ∙ 2: 1.5 ∙ 2 \u003d 2: 3 Þ N 2 O 3
Answer: N 2 O 3 .
II. Derivation of formulas of substances by mass fractions of elements and data for finding the true molar mass (density, mass and volume of gas or relative density).
1. Find the true molar mass:
If the density is known:
r \u003d m / V \u003d \u200b\u200bM / V m Þ M \u003d r ∙ V m \u003d r g / l ∙ 22,4 l / mol
· If the mass and volume of the gas are known, the molar mass can be found in two ways:
Through the density r \u003d m / V, M \u003d r ∙ Vm;
Through the amount of substance: n \u003d V / Vm, M \u003d m / n.
If the relative density of the first gas is known differently:
D 21 \u003d M 1 / M 2 Þ M 1 \u003d D 2 ∙ M 2
M \u003d D H2 ∙ 2 M \u003d D O2 ∙ 32
M \u003d D air. ∙ 29 M \u003d D N2 ∙ 28, etc.
2. Find the simplest formula of a substance (see previous algorithm) and its molar mass.
3. Compare the true molar mass of the substance with the simplest and increase the indices by the desired number of times.
Task number 1.
Find the formula for a hydrocarbon containing 14.29% hydrogen, and its relative density on nitrogen is 2.
ω (H) \u003d 14.29% Solution:
D ( N2 ) = 2 1. Find the true molar mass C xN at:
M \u003d D N2 ∙ 28 = 2 ∙ 28 = 56 g / mol.
M.F. \u003d? 2. Find the mass fraction of carbon:
ω (C) \u003d 100% - 14.29% \u003d 85.71%.
3. Find the simplest formula of the substance and its molar mass:
x: y \u003d 85.7 / 12: 14.29 / 1 \u003d 7.142: 14.29 \u003d 1: 2 Þ CH 2
M (CH 2 ) = 12 + 1 ∙ 2 = 14 g / mol
4. Compare the molar masses:
M (C xN at) / M (CH 2 ) \u003d 56/14 \u003d 4 Þ true formula - C 4 N 8 .
Answer: C 4 N 8 .
iii. The algorithm for solving problems on the derivation of formulas
organic matter containing oxygen.
1. Designate the formula of the substance using the indices X. Y, Z, etc. by the number of elements in the molecule. If the combustion products are СО2 and Н2О, then the substance may contain 3 elements (CxNuOZ). A special case: in addition to СО2 and Н2О, the combustion product is nitrogen (N2) for nitrogen-containing substances (Cx Well Оz Nm)
2. Make an equation for the combustion reaction without coefficients.
3. Find the amount of substance of each of the products of combustion.
5. If it is not said that the substance burned is hydrocarbon, calculate the masses of carbon and hydrogen in the combustion products. Find the mass of oxygen in a substance by the difference in mass of the starting material and m (C) + m (H). Calculate the amount of substance of oxygen atoms.
6. The ratio of the indices x: y: z is equal to the ratio of the quantities of substances v (C): v (H): v (O) reduced to the ratio of integers.
7. If necessary, according to additional data in the condition of the problem, bring the resulting empirical formula to true.
