Finding the simplest and true formulas of chemical compounds. Solving problems to determine the formulas of organic substances (preparation for the exam in chemistry)
Theory for task 35 from the exam in chemistry
Finding the molecular formula of a substance
Finding the chemical formula of a substance by mass fractions of elements
The mass fraction of an element is the ratio of its mass to the total mass of the substance, of which it is included:
$ W = (m (e-ta)) / (m (in-va)) $
Mass fraction of an element ($ W $) is expressed in fractions of a unit or as a percentage.
Problem 1. The elementary composition of the substance is as follows: mass fraction of iron $ 72.41% $, mass fraction of oxygen $ 27.59% $. Bring out chemical formula.
Given:
$ W (Fe) = 72.41% = 0.7241 $
$ W (O) = 27.59% = 0.2759 $
Solution:
1. For calculations, we choose the mass of the oxide $ m $ (oxide) $ = 100 $ g. Then the masses of iron and oxygen will be as follows:
$ m (Fe) = m_ (oxide) W (Fe); m (Fe) = 100 0.7241 = 72.41 $ g.
$ m (O) = m_ (oxide) W (O); m (O) = 100 0.2759 = $ 27.59 g.
2. The amounts of iron and oxygen are equal, respectively:
$ ν (Fe) = (m (Fe)) / (M (Fe)); ν (Fe) = (72.41) / (56) = 1.29. $
$ ν (O) = (m (O)) / (M (O)); ν (O) = (27.59) / (16) = 1.72. $
3. Find the ratio of the amount of iron and oxygen substances:
$ ν (Fe): ν (O) = 1.29: 1.72. $
The smaller number is taken as $ 1 (1.29 = 1) $ and we find:
$ Fe: O = 1: 1.33 $.
4. Since the formula must contain an integer number of atoms, this ratio is reduced to integers:
$ Fe: O = 1: 1.33 = 2: 2.66 = 3 3.99 = 3: 4 $.
5. Substitute the found numbers and get the oxide formula:
$ Fe: O = 3: 4 $, that is, the formula of the substance is $ Fe_3O_4 $.
Answer: $ Fe_3O_4 $.
Finding the chemical formula of a substance by mass fractions of elements, if the density or relative density of a given substance in a gaseous state is indicated
Problem 2. The mass fraction of carbon in the hydrocarbon is $ 80%. The relative density of the hydrocarbon in terms of hydrogen is $ 15 $.
Given:
Solution:
1. Let's designate the formula of substance $ C_ (x) H_ (y) $.
2. Let's find the number of moles of carbon and hydrogen atoms in $ 100 $ g of a given compound:
$ x = n (C); y = ν (H). $
$ ν (C) = (m (C)) / (M (C)) = (80) / (12) = 6.6; ν (H) = (m (H)) / (M (H)) = ( 20) / (1) = 20. $
1 way.
3. The relationship between atoms:
$ x: y = 6.6: 20 = 1: 3 $, or $ 2: 6 $.
The simplest formula substance $ CH_3 $.
4. Determine the molecular weight of the hydrocarbon by the relative density of its vapors.
$ M_r $ (substances) $ = 2D (H_2) = 32D (O_2) = 29D $ (air).
$ M_x = 2D (H_2) = 2 15 = 30 $ g / mol.
5. We calculate the relative molecular weight of the hydrocarbon using the simplest formula:
$ M_r (CH_3) = A_r (C) + 3A_r (H) = 12 + 3 = 15 $.
6. The values of $ M_x $ and $ M_r $ do not match, $ M_r = (1) / (2) M_x $, hence the hydrocarbon formula is $ C_2H_6 $.
Check: $ M_r (C_2H_6) = 2A_r (C) + 6A_r (H) = 2 12 + 6 1 = 30 $.
Answer: the molecular formula of the $ C_2H_6 $ hydrocarbon is ethane.
Method 2.
3. The relationship between atoms:
$ (x) / (y) = (6.6) / (20); (x) / (y) = (1) / (3.03); y = 3.03x. $
5. Molar mass can be represented as:
$ M_r (C_xH_y) = A_r (C) _x + A_r (H) _y; M_r (C_xH_y) = 12x + y $ or $ 30 = 12x + 1y $.
6. We solve the system of two equations with two unknowns:
$ \ (\ table \ y = 3.03x; \ 12x + y = 30; $ $ 12x + 3.03x = 30; x = 2; y = 6. $
Answer: the formula $ C_2H_6 $ is ethane.
Finding the chemical formula of a substance according to the data on the starting substance and on the products of its combustion (according to the chemical reaction equation)
Task 3. Find molecular formula of a hydrocarbon with a density of $ 1.97 $ g / L if the combustion of $ 4.4 $ g of it in oxygen forms $ 6.72 $ L of carbon monoxide (IV) (n.a.) and $ 7.2 $ g of water.
Given:
$ m (C_xH_y) = 4.4 $ g
$ ρ (C_xH_y) = 1.97 $ g / l
$ V (CO_2) = 6.72 $ l
$ m (H_2O) = 7.2 $ g
Solution:
1. Let's write the scheme of the hydrocarbon combustion equation
$ (C_xH_y) ↖ (4.4g) + O_2 → (CO_2) ↖ (6.72L) + (H_2O) ↖ (7.2g) $
2. Calculate the molar mass $ C_xH_y · M = ρ · V_m $,
$ M = 1.97 $ g / l $ 22.4 $ l / mol $ = 44 $ g / mol.
Relative molecular weight $ M_r = 44 $.
3. Determine the amount of substance:
$ ν (C_xH_y) = (m) / (M) $ or $ ν (C_xH_y) = (4.4) / (44) = 0.1 $ mol.
4. Using the value of the molar volume, we find:
$ ν (CO_2) = (m) / (M) $ or $ ν (H_2O) = (7.2) / (18) = 0.4 $ mol.
6. Therefore: $ ν (C_xH_y): ν (CO_2): νH_2O = 0.1 $ mol $: 0.3 $ mol $: 0.4 $ mol or $ 1: 3: 4 $, which should correspond to the coefficients in the equation and allows you to set the number of carbon atoms and hydrogen:
$ C_xH_y + O_2 → 3CO + 4H_2O $.
The final form of the equation:
$ C_3H_8 + 5O_2 → 3CO_2 + 4H_2O $.
Answer: the hydrocarbon formula $ C_3H_8 $ is propane.
Determination of the formula of a substance by mass fractions chemical elements(results of quantitative analysis) or according to the general formula of the substance
1. Mass fraction of an element in a substance.
The mass fraction of an element is its content in a substance as a percentage by mass. For example, a substance with the composition C2H4 contains 2 carbon atoms and 4 hydrogen atoms. If we take 1 molecule of such a substance, then its molecular weight will be equal to:
Mr (C2H4) = 2 12 + 4 1 = 28 a. eat. and it contains 2 12 a. eat. carbon.
To find the mass fraction of carbon in this substance, it is necessary to divide its mass by the mass of the whole substance:
ω (C) = 12 2/28 = 0.857 or 85.7%.
If a substance has the general formula CxHyOz, then the mass fractions of each of their atoms are also equal to the ratio of their mass to the mass of the whole substance. The mass x of the C atoms is - 12x, the mass of the H atoms is y, the mass z of the oxygen atoms is 16z.
Then
ω (C) = 12 x / (12x + y + 16z)
The formula for finding the mass fraction of an element in a substance:
ωelement =, × 100%
where Ar is the relative atomic mass of the element; n is the number of atoms of an element in a substance; Мr is the relative molecular weight of the whole substance
2. Molecular and simplest formula of a substance.
Molecular (true) formula - a formula that reflects the real number of atoms of each type included in a molecule of a substance.
For example, C6H6 is the true formula of benzene.
The simplest (empirical) formula shows the ratio of atoms in a substance. For example, for benzene the ratio C: H = 1: 1, that is, the simplest formula for benzene is CH. The molecular formula can coincide with the simplest or be a multiple of it.
3. If the problem contains only mass fractions of elements, then in the process of solving the problem it is possible to calculate only the simplest formula of the substance. To obtain the true formula in the problem, additional data are usually given - molar mass, relative or absolute density of a substance, or other data that can be used to determine the molar mass of a substance.
4. Relative density of gas X for gas Y - DpoU (X).
The relative density D is a value that shows how many times gas X is heavier than gas Y. It is calculated as the ratio of the molar masses of gases X and Y:
DpoU (X) = M (X) / M (Y)
Often used for calculations relative densities of gases by hydrogen and by air.
Relative density of gas X for hydrogen:
Dby H2 = M (gas X) / M (H2) = M (gas X) / 2
Air is a mixture of gases, so only the average molar mass can be calculated for it. Its value is taken as 29 g / mol (based on the approximate average composition). That's why:
D by air = M (gas X) / 29
5. The absolute density of the gas under normal conditions.
The absolute density of a gas is the mass of 1 liter of gas under normal conditions. It is usually measured in g / l for gases.
ρ = m (gas) / V (gas)
If we take 1 mole of gas, then: ρ = M / Vm,
and the molar mass of a gas can be found by multiplying the density by the molar volume.
Objective 1: Determine the formula of a substance if it contains 84.21% C and 15.79% H and has a relative density in air of 3.93.
1. Let the mass of the substance be equal to 100 g. Then the mass of C will be equal to 84.21 g, and the mass of H - 15.79 g.
2. Let's find the amount of matter of each atom:
ν (C) = m / M = 84.21 / 12 = 7.0175 mol,
ν (H) = 15.79 / 1 = 15.79 mol.
3. Determine the molar ratio of C and H atoms:
C: H = 7.0175: 15.79 (we divide both numbers by the smaller) = 1: 2.25 (we will multiply by 1, 2,3,4, etc. until 0 or 9 appears after the decimal point. this problem needs to be multiplied by 4) = 4: 9.
Thus, the simplest formula is C4H9.
4. Based on the relative density, we calculate the molar mass:
M = D (air) 29 = 114 g / mol.
The molar mass corresponding to the simplest formula C4H9 is 57 g / mol, which is 2 times less than the true molar mass.
Hence, the true formula is C8H18.
Task 2 : Determine the formula for alkyne with a density of 2.41 g / l under normal conditions.
General formula for alkyne СnH2n − 2
How, given the density of a gaseous alkyne, to find its molar mass? The density ρ is the mass of 1 liter of gas under normal conditions.
Since 1 mole of a substance occupies a volume of 22.4 liters, it is necessary to find out how much 22.4 liters of such a gas weigh:
M = (density ρ) (molar volume Vm) = 2.41 g / l 22.4 l / mol = 54 g / mol.
Next, let's make an equation linking molar mass and n:
14 n - 2 = 54, n = 4.
Hence, alkyne has the formula C4H6.
Problem 3 : Determine the formula of dichloroalkane containing 31.86% carbon.
The general formula of dichloroalkane: СnH2nCl2, there are 2 chlorine atoms and n carbon atoms.
Then the mass fraction of carbon is equal to:
ω (C) = (number of C atoms in a molecule) (atomic weight of C) / (molecular weight of dichloroalkane)
0.3186 = n 12 / (14n + 71)
n = 3, substance - dichloropropane. C3H6Cl2
If we know the chemical formula of a substance, then it is enough to simply calculate the relative masses of each element in it.
Apparently, it is possible to distinguish two main types of computational problems based on the forms st chemical substances. First, knowing the atomic masses of each element, you can calculate their total mass per mole of substance and determine the percentage of each element. Secondly, you can solve the inverse problem: find a chemical formula for a given percentage of elements in a substance (based on chemical analysis data)
Let's look at a few examples.
Example 1. Calculate the mass percentages of each element in phosphoric acid.
Solution. Knowing the relative atomic masses of each element, we calculate their sum for H 3 PO 4:
M r (H 3 P0 4) = 3A r (H) + A r (P) + 4A r (0) = 3. 1 + 31 + 16. 4 = 98. Then, for example, the hydrogen content is
Example 2. Iron forms three oxides with oxygen. One of them contains 77.8% iron, the other 70.0 and the third 72.4%. Determine the oxide formulas.
Solution. Let's write down the formula of iron oxide in the general case: Fe x O y. Let's find a relationship x: y and, leading to an integer ratio, we define the formulas of the oxides.
1. It was found experimentally that some substance having a molar mass of 116 g / mol contains 23 ± 2% nitrogen. The percentage of nitrogen needs to be specified.
2. Chemical analysis of a nitrogen-hydrogen compound having a relative molecular weight of 32 showed that the mass fraction of nitrogen in the compound is 66%. Prove that the analysis results are wrong.
3. Determine the formula of a substance containing 1.22 mass. parts of potassium, 1.11 mass. parts of chlorine and 2.00 mass. parts of oxygen. Are there still substances of the same qualitative composition? What can you say (in the language of formulas) about their quantitative composition?
4. Some metal chloride contains 74.7% chlorine; identify the unknown metal.
5.
A salt containing some element X has the following mass ratio of elements
X: N: N: O = 12: 5: 14: 48. What is the formula for this salt?
6. In the middle of the XIX century. uranium was assigned the following values atomic mass: 240 (Mendeleev), 180 (Armstrong), 120 (Berzelius). These values were obtained from the results of chemical analysis of uranium resin (one of the uranium oxides), which showed that it contains 84.8% uranium and 15.2% oxygen. What formula did Mendeleev, Armstrong, and Berzelius attribute to this oxide?
7. Some alum (crystalline hydrates of the composition A 1 + B 3 + (SO 4) 2. 12H 2 O) contain 51.76% oxygen and 4.53% hydrogen. Determine the alum formula.
8. The compound contains hydrogen (mass fraction - 6.33%), carbon (mass fraction - 15.19%), oxygen (mass fraction - 60.76%) and one more element, the number of atoms of which in the molecule is equal to the number of carbon atoms. Determine what kind of compound it is, what class it belongs to and how it behaves when heated.
1.23% of nitrogen is 
The composition of a substance can include only an integer number of nitrogen atoms (relative mass 14). This means that the mass of nitrogen in one mole of a substance should be a multiple of 14. Thus, 116 g of a substance should contain 14n (g) of nitrogen (14, 28, 42, 56, etc.). The closest to 26.7 is the number (multiple of 14) 28. The mass fraction of nitrogen in the substance is
2 ... If chemical analysis carried out correctly, then the molecule of this nitrogen compound with hydrogen must contain
The number of atoms in a molecule cannot be fractional, so the analysis was performed incorrectly.
3. To find the quantitative composition, we divide the mass parts of the elements by their relative atomic masses
that is, the formula of the required substance is KC1O 4 (potassium perchlorate).
The same elements are contained in potassium hypochlorite KClO, potassium chlorite KClO 2, potassium chlorate KClO 3.
| n | (Me) | Me |
| 1 | 12 | - |
| 2 | 24 | Mg |
| 3 | 36 | - |
| 4 | 48 | Ti |
| 5 | 60 | - |
Titanium or magnesium.
Solving problems to determine the formula of organic matter.
Developed by: Kust I.V. - teacher of biology and chemistry MBOU Kolyudovskaya secondary school
1. Determination of the formula of a substance by combustion products.
1.With complete combustion of the hydrocarbon, 27 g of water and 33.6 g carbon dioxide(Well.). The relative density of the hydrocarbon with respect to argon is 1.05. Establish 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 with respect to 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 burnt compound.
4. During the combustion of a secondary amine of symmetrical structure, 0.896 liters of carbon dioxide and 0.99 g of water and 0.112 liters of nitrogen were released. Establish the molecular formula for this amine.
Solution algorithm:
1.Determine the molecular weight of the hydrocarbon: M (CxHy) = M (gas) хD (gas)
2. Determine the amount of water substance: n (H2O) = t (H2O): M (H2O)
3. Determine the amount of hydrogen substance: p (H) = 2p (H2O)
4. Determine the amount of substance carbon dioxide:: n (CO2) = t (CO2): M (CO2) or
n (CO2) = V (CO2): Vm
5. Determine the amount of carbon substance: p (C) = p (CO2)
6. Define the ratio C: H = n (C): n (H) (we divide both numbers by the smallest of these numbers)
7. the simplest formula (from point 6).
8. The molecular weight of the hydrocarbon (from the first point) is divided by the molecular weight of the simplest formula (from point 7): the resulting integer means that the number of carbon and hydrogen atoms in the simplest formula must be increased by so many times.
9. Determine the molecular weight of the true formula (found in paragraph 8).
10.Write down the answer - the found formula.
Solution to problem number 1.
t (H2 O) = 27g
V (CO2) = 33.6L
D (by Ar) = 1.05
Find CxHy
Solution.
1. Determine the molecular weight of the hydrocarbon: M (CxHy) = M (gas) хD (gas)
M (CxHy) = 1.05x40g / mol = 42 g / mol
2. Determine the amount of water substance: n (H2 O) = t (H2 O): M (H2 O)
p (H2 O) = 27g: 18g / mol = 1.5 mol
3. Determine the amount of hydrogen substance: n (H) = 2n (H2 O)
n (H) = 2x1.5 mol = 3 mol
4. Determine the amount of carbon dioxide substance: n (СО2) = V (СО2): Vm
p (CO2) = 33.6 L: 22.4 L / mol = 1.5 mol
5. Determine the amount of carbon substance: n (C) = n (CO2)
p (C) = 1.5 mol
6. Ratio C: H = p (C): p (H) = 1.5 mol: 3 mol = (1.5: 1.5) :( 3: 1.5) = 1: 2
7. The simplest formula: CH2
8.42g / mol: 14 = 3
9.C3 H6 - true (M (C3 H6) = 36 + 6 = 42 g / 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 primary amine, 0.448 l of nitrogen was released. Determine the molecular formula for 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. During the interaction of 22 g of saturated monobasic acid with an excess of sodium bicarbonate solution, 5.6 liters of gas were 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 be attached to 200 ml of hydrogen.
9. Establish the molecular formula of the alkene. If it is known that 1.5 g of it is capable of attaching 600 ml of hydrogen chloride.
10. Establish the molecular formula of a cycloalkane if it is known that 3 g of it are capable of attaching 1.2 liters of hydrogen bromide.
Solution algorithm:
1. Let us determine the amount of a known substance (nitrogen, carbon dioxide, hydrogen, hydrogen chloride, hydrogen bromide): n = m: M or n = V: Vm
2.Using the equation, let's compare the amount of a substance of a known substance with the amount of a substance that needs to be determined:
3. Determine the molecular weight of the desired substance: M = m: n
4.Let's find the molecular weight of the desired substance using its general formula: (M (Cn2p) = 12p + 2p = 14p)
5. Let's equate the value of point 3 and point 4.
6. We solve the equation with one unknown, find item.
7. Let's substitute in the general formula the value of p.
8. Let's write down the answer.
Solution to problem number 5.
V (N 2) = 0.448L
t (SpN2p + 1 NH 2) = 1.8 g
Find Cn2p + 1 NH 2.
Solution.
1.Scheme of the reaction: 2 SPH2p + 1 NH 2 = N 2 (or)
2. The reaction equation: 2 SpH2 p + 1 NH 2 + (6p + 3) / 2O2 = 2nCO2 + (2p + 3) H2 O + N 2
3. Determine the amount of nitrogen substance by the formula: n = V: Vm
n (N 2) = 0.448l: 22.4l / mol = 0.02mol
4. Determine the amount of amine substance (using the equations: the coefficient in front of the amine is divided by the coefficient in front of the nitrogen)
p (CpN2 p + 1 NH 2) = 2p (N 2) = 2x0.02 mol = 0.04 mol
5. Determine the molar mass of the amine by the formula: M = m: n
M ((CpH2 n + 1 NH 2) = 1.8 g: 0.04 mol = 45 g / mol
6. Determine the molar mass of the amine according to the general formula:
M (CpN2 p + 1 NH 2) = 12p + 2p + 1 + 14 + 2 = 14p + 17g / mol
7. Equate: 14p + 17 = 45 (solve the equation)
8.Substitute in the general formula: CpN2 p + 1 NH2 = C2 H5 N H2
9.Answer: C2 H5 N H2
3. Determination of the formula of a substance using the equations of 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 the potassium salt of the saturated monobasic carboxylic acid and 3.2 g of alcohol were obtained. Establish the molecular formula of this ether.
12. An ester weighing 30 g was subjected to alkaline hydrolysis, thus obtaining 34 g sodium salt saturated monobasic acid and 16 g of alcohol. Establish the molecular formula of the ether.
1. Let's compose the hydrolysis equation.
2.According to the law of conservation of the mass of substances (the mass of the reacting substances is equal to the mass of the resulting substances): mass of ether + mass of potassium hydroxide = mass of salt + mass of alcohol.
3. Find the mass (KOH) = mass (salt) + mass (alcohol) - mass (ether)
4. Determine the amount of substance KOH: n = m (KOH): M (KOH).
5.According to the equation n (KOH) = n (ether)
6.Let's determine the molar mass of the ether: M = m: n
7.According to the equation, the amount of KOH substance = the amount of salt substance (n) = the amount of alcohol substance (n).
8. Determine the molecular weight of the salt: M = m (salt): n (salt).
9. Determine the molecular weight of the salt according to the general formula and equate the values from paragraphs 8 and 9.
10. From the molecular weight of the ester, we subtract the molecular weight of the functional group of the acid found in the previous paragraph without the weight of the metal:
11. Let's define the functional group of alcohol.
Solution to problem number 11.
Given:
t (ether) = 7.4g
t (salt) = 9.8g
t (alcohol) = 3.2g
Find the formula for ether
Solution.
1. Let's compose the hydrolysis equations ester:
SpN2p +1 COOSmN2 m + 1 + KOH = SpN2p +1 COOK + COOK2 m + 1 OH
2. Find the mass (KOH) = t (salt) + t (alcohol) -t (ether) = (9.8g + 3.2g) -7.4g = 5.6g
3.Let us define n (KOH) = t: M = 5.6g: 56g / mol = 0.1mol
4.According to the equation: p (KOH) = p (salt) = p (alcohol) = 0.1 mol
5. Determine the molar mass of the salt: M (Cn2n +1 COOK) = m: n = 9.8g: 0.1mol = 98g / mol
6. Let us determine the molar mass according to the general formula: M (CpN2p +1 COOK) = 12p + 2p + 1 + 12 + 32 + 39 = 14p + 84 (g / mol)
7. Equate: 14p + 84 = 98
CH3SOOK salt formula
8. Determine the molar mass of alcohol: M (CmH2 m + 1 OH) = 3.2 g: 0.01 mol = 32 g / mol
9.Let us define M (CmH2 m + 1 OH) = 12m + 2m + 1 + 16 + 1 = 14m + 18 (g / mol)
10. Equation: 14m + 18 = 32
Alcohol formula: CH3 OH
11.Ether formula: 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 an alkene if it is known that the same amount of it, interacting with various hydrogen halides, forms, respectively, 5.23 g of a chloro derivative, or 8.2 g of a bromo 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 is formed. Determine the alkene formula. Write its name and structural formula.
1.We write down two reaction equations (the alkene formula in general view)
2. Find the molecular weights of the products by the general formulas in the reaction equations (through n).
3. Find the amount of substance of products: n = m: M
4. We equate the found quantities of matter and solve the equations. We substitute the found n into the formula.
Solution to problem number 13.
t (CpH2p + 1 Cl) = 5.23g
t (Cn2p + 1 Br) = 8.2g
Find SPN2 p
Solution.
1. Let's compose the reaction equations:
CpH2 p + HCl = CpH2p + 1 Cl
CpN2 p + HBr = CpN2p + 1 Br
2.Let us define M (CpH2p + 1 Cl) = 12p + 2p + 1 + 35.5 = 14p + 36.5 (g / mol)
3. Let us determine n (Cn2n + 1 Cl) = m: M = 5.23g: (14n + 36.5) g / mol
4. Let us define M (Cn2p + 1 Br) = 12p + 2p + 1 + 80 = 14p + 81 (g / mol)
5.Let us define n (Cn2n + 1 Br) = m: M = 8.2g: (14n + 81) g / mol
6. Equate n (Cn2p + 1 Cl) = n (Cn2p + 1 Br)
5.23g: (14p + 36.5) g / mol = 8.2g: (14p + 81) g / mol (solve the equation)
7.Alken formula: C3 H6
5. Determination 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 was obtained. The amounts of carbon dioxide and water substances in this mixture turned out to be equal. Determine the molecular formula of the organic compound. If its relative density for oxygen is 1.81.
16. As a result of combustion of 1.32 g of an organic compound, 3.72 g of a mixture of carbon dioxide and water was obtained. The amounts of carbon dioxide and water substances in this mixture turned out to be equal. Determine the molecular formula of the organic compound. If its relative density for nitrogen is 1.5714.
Algorithm for solving the problem:
1. Determine the molar mass of organic matter: M (CxHy Oz) = D (for gas) xM (gas)
2. Determine the amount of organic matter: n (CxHy Oz) = t: M
3. Introduction of the variable x: Let X be the amount of carbon dioxide in the mixture: n (CO2) = Khmol, then the same amount of water (by condition): n (H2O) = Khmol.
4.Mass of carbon dioxide in the mixture: t (CO2) = n (CO2) xM (CO2) = 44X (g)
6.According to the condition of the problem: t (mixture) = t (H2O) + t (CO2) = 44X + 18X (We solve the equation with one unknown, we find the X-number of moles of CO2 and H2O)
11.Formula: S: N: O = n (S): n (N): n (O)
Solution to problem number 15.
1. Determine the molecular weight of organic matter: M (CxHy Oz) = 1.82x32g / mol = 58g / mol
2. Determine the amount of organic matter: n (CxHy Oz) = t: M
n (CxHy Oz) = 1.74g: 58g / mol = 0.03 mol
3.Let X-p (CO2) in the mixture, then p (H2O) -Hmol (by condition)
4..Mass of carbon dioxide in the mixture: t (CO2) = p (CO2) xM (CO2) = 44X (g)
5.Mass of water in the mixture: p (H2O) = p (H2O) xM (H2O) = 18X (g)
6. Make and solve the equation: 44X + 18X = 5.58 (the mass of the mixture according to the condition)
X = 0.09 (mol)
7. Determine the amount of carbon substance (C): n (C) = n (CO2): n (CxHy Oz)
p (C) = 0.09: 0.03 = 3 (mol) is the number of C atoms in organic matter.
8. Determine the amount of substance hydrogen (H): p (H) = 2xn (H2O): p (CxHy Oz)
p (H) = 2x0.09: 0.03 = 6 (mol) is the number of hydrogen atoms in organic matter
9.Checking the presence of oxygen in organic compound: M (O) = M (CxHy Oz) -M (C) -M (H)
M (O) = 58g / mol- (3x12) - (6x1) = 16 (g / mol)
10. Determine the amount of substance (number of atoms) oxygen: p (O) = M (O): Ar (O)
n (O) = 16: 16 = 1 (mol) is the number of oxygen atoms in organic matter.
11.Formula of the required substance: (С3Н6О).
6. Determination of the formula of a substance by the mass fraction of one of the elements included in the substance.
17. Determine the molecular formula of a dibromoalkane containing 85.11% bromine.
18. Determine the structure of an amino acid ester formed by derivatives of saturated hydrocarbons, if it is known that it contains 15.73% nitrogen.
19. Determine the molecular formula of the limiting trihydric alcohol, the mass fraction of oxygen in which is equal to 45.28%.
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 weight of the substance: W (element) = Ar (element) xn (element): Мr (substance)
Мr (substance) = Аr (element) xn (element): W (element), where n (element) is the number of atoms of a given element
2. Determine the molecular weight by the general formula
3. We equate point 1 and point 2. Solve the equation with one unknown.
4. We write down the formula, substituting the value of n into the general formula.
Solution to problem number 17.
Given:
W (Br) = 85.11%
Find SPN2 pVr 2
1. Determine the molecular weight of dibromoalkane:
Mr ( SPN 2 pVr 2) = Ar ( Vr) xn(Br): W (Br)
Mr ( SPN2 pVr 2 ) = 80x2: 0.8511 = 188
2 Determine the molecular weight by the general formula: Mr ( SPN2 pVr 2 ) = 12p + 2p + 160 = 14p + 160
3. Equate and solve the equation: 14p + 160 = 188
4.Formula: C2 H4 Br 2
7. Determination of the formula of organic matter using chemical reaction equations reflecting Chemical properties of this substance.
20.In the case of intermolecular dehydration of alcohol, 3.7 g of ether is formed. And with the intramolecular dehydration of this alcohol, 2.24 liters of ethylene hydrocarbon. Determine the formula for the alcohol.
21. During intramolecular dehydration of a certain amount of primary alcohol, 4.48 l of alkene was released, and during intermolecular dehydration, 10.2 g of simple ether is formed. What is the structure of alcohol.
Algorithm for solving the problem.
1.Writing down the equations chemical reactions, which are mentioned in the problem (be sure to equalize)
2. Determine the amount of gaseous substance by the formula: n = V: Vm
3. Determine the amount of the initial substance, then the amount of the substance of the second reaction product (according to the reaction equation and according to the condition of the problem)
4. Determine the molar mass of the second product by the formula: M = m: n
5. Determine the molar mass according to the general formula and equate (clause 4 and clause 5)
6. We solve the equation and find n - the number of carbon atoms.
7.Write down the formula.
Solution to problem number 21.
t (ether) = 10.2g
TT (CpN2 p) = 4.48L
Find SPN2p + 1 OH
Solution.
1.Write the reaction equations:
CpN2p +1 OH = CpN2p + H2 O
2СпН2п +1 ОН = СпН2 п + 1 ОСпН2п + 1 + Н2О
2. Determine the amount of alkene substance (gas): n = V: Vm
p (CpH2p) = 4.48 L: 22.4 L / mol = 0.2 mol
3. According to the first equation, the amount of alkene substance is equal to the amount of alcohol substance. According to the second equation, the amount of ether substance is 2 times less than the amount of alcohol substance i.e. p (CpH2 p + 1 OCpH2p + 1) = 0.1 mol
4. Determine the molar mass of the ether: M = m: n
M = 10.2g: 0.1mol = 102g / mol
5. Determine the molar mass according to the general formula: M (CpN2 p + 1 OSpN2p + 1) = 12p + 2p + 1 + 16 + 12p + 2p + 1 = 28p + 18
6. Equate and solve the equation: 28p + 18 = 102
7.Alcohol formula: C3 H7 OH
