A clear algorithm for solving the problem in chemistry is a great way to tune in to the final tests in this complex discipline. In 2017, significant changes were made in the structure of the exam, questions from one part of the test were removed, suggesting one answer. The wording of the questions is given so that the graduate demonstrates knowledge in various fields, for example, chemistry, and could not just put a βtickβ.
The main difficulties
The maximum difficulty for graduates is questions on the derivation of the formulas of organic compounds, they cannot make up an algorithm for solving the problem.
How to cope with such a problem? In order to cope with the proposed problem, it is important to know the algorithm for solving problems in chemistry.
The same problem is typical for other academic disciplines.
Sequencing
The most common are the tasks for determining the compound by known combustion products, therefore we propose to consider the algorithm for solving problems precisely with the example of this type of exercise.
1. The molar mass of a given substance will be determined using a known relative density for some gas (if present in the condition of the proposed problem).
2. We calculate the amount of substances formed in this process through the molar volume for the gaseous compound, through the density or mass for liquid substances.
3. We calculate the quantitative values ββof all atoms in the products of this chemical reaction, and also calculate the mass of each.
4. We summarize these values, then compare the obtained value with the given conditional mass of the organic compound.
5. If the initial mass exceeds the obtained value, we conclude about the presence of oxygen in the molecule.
6. We determine its mass, subtract for this from the given mass of the organic compound the sum of all atoms.
6. Find the number of oxygen atoms (in moles).
7. We determine the ratio of the quantities of all atoms available in the problem. We get the formula of the analyte.
8. We compose its molecular variant, molar mass.
9. If it differs from the value obtained in the first step, we increase the amount of each atom a certain number of times.
10. We compose the molecular formula of the desired substance.
11. Define the structure.
12. We write the equation of this process using the structure of organic substances.
The proposed algorithm for solving the problem is suitable for all tasks associated with the derivation of the organic compound formula. It will help high school students to cope with the exam.
Example 1
What should the solution to problems with algorithms look like?
To answer this question, we give a ready-made sample.
By burning 17.5 g of the compound, 28 L of carbon dioxide and 22.5 ml of water vapor were obtained. The vapor density of this compound is 3.125 g / L. There is information that the analyte is formed during the dehydration of tertiary saturated alcohol. Based on the proposed data:
1) make certain calculations that will be required to search for the molecular formula of this organic substance;
2) write its molecular formula;
3) make a structural view of the original compound, uniquely reflecting the connection of atoms in the proposed molecule.
Task data.
- m (starting material) - 17.5 g
- V carbon dioxide-28 l
- V water - 22.5 ml
Formulas for conducting mathematical calculations:
- β = β m * n
- β = m / Ο
If you wish, you can cope with this task in several ways.
First way
1. Determine the number of moles of all chemical reaction products using the molar volume.
nCO 2 = 1.25 mol
2. We identify the quantitative content of the first element (carbon) in the product of this process.
nC = nCO 2 =, 25 mol
3. Calculate the mass of the element.
mC = 1.25 mol * 12 g / mol = 15 g.
We determine the mass of water vapor, knowing that the density is 1g / ml.
mH 2 O is 22.5 g
We identify the amount of reaction product (water vapor).
n water = 1.25 mol
6. We calculate the quantitative content of the element (hydrogen) in the reaction product.
nH = 2n (water) = 2.5 mol
7. Determine the mass of this element.
mH = 2.5 g
8. We summarize the masses of elements to determine the presence (absence) of oxygen atoms in the molecule.
mC + mH = 1 5 g + 2.5 g = 17.5 g
This corresponds to the data of the problem, therefore, in the desired organic matter there are no oxygen atoms.
9. Find a quantitative ratio.
CH 2 is the simplest formula.
10. Calculate M of the desired substance using the density.
M substance = 70 g / mol.
n-5, the substance looks like this: C 5 H 10 .
The condition says that the substance is obtained by dehydration of alcohol, therefore, it is an alkene.
Second option
Consider another algorithm for solving the problem.
1. Knowing that this substance is obtained by dehydration of alcohols, we conclude about its possible belonging to the class of alkenes.
2. Find the value M of the desired substance using the density.
M in = 70 g / mol.
3. M (g / mol) for the compound has the form: 12n + 2n.
4. Calculate the quantitative value of carbon atoms in an ethylene hydrocarbon molecule.
14 n = 70, n = 5, so the molecular formula of the substance has the form: C 5 H 10n .
In the data of this task it is said that the substance is obtained by dehydration of tertiary alcohol, therefore, it is an alkene.
How to make an algorithm for solving the problem? The student must know how to obtain representatives of different classes of organic compounds, owns their specific chemical properties.
Example 2
Let's try to identify an algorithm for solving the problem with another example from the exam.
With the complete combustion of 22.5 grams of alpha-aminocarboxylic acid in atmospheric oxygen, 13.44 l (n.o.), carbon monoxide (4) and 3.36 l (n.o.) of nitrogen were collected. Find the formula of the proposed acid.
Data on the condition.
- m (amino acids ) - 22.5 g;
- β (carbon dioxide ) -13.44 liters;
- β (nitrogen ) -3.36 l.
Formulas
- m = M * n;
- β = β m * n.
We use the standard algorithm for solving the problem.
We find the quantitative value of the interaction products.
n (nitrogen ) = 0.15 mol.
We write the chemical equation (apply the general formula). Further, according to the reaction, owning the amount of substance, we calculate the number of moles of aminocarboxylic acid:
x - 0.3 mol.
We calculate the molar mass of the aminocarboxylic organic acid.
M (starting material ) = m / n = 22.5 g / 0.3 mol = 75 g / mol.
Calculate the molar mass of the original aminocarboxylic acid using the relative atomic masses of the elements.
M (amino acids ) = (R + 74) g / mol.
Mathematically, we determine the hydrocarbon radical.
R + 74 = 75, R = 75 - 74 = 1.
By selecting, we identify a variant of the hydrocarbon radical, write down the formula of the desired aminocarboxylic acid, formulate the answer.
Therefore, in this case there is only a hydrogen atom, therefore we have the formula CH2NH2COOH (glycine).
Answer: CH2NH2COOH.
Alternative solution
The second algorithm for solving the problem is as follows.
Calculate the quantitative expression of the reaction products using the molar volume.
n (carbon dioxide ) = 0.6 mol.
We write down the chemical process, armed with the general formula of this class of compounds. We calculate by the equation the number of moles of aminocarboxylic acid taken:
x = 0.6 * 2 / v = 1.2 / mol
Next, we calculate the molar mass of aminocarboxylic acid:
M = 75 in g / mol.
Using the relative atomic masses of the elements, we find the molar mass of aminocarboxylic acid:
M (amino acids ) = (R + 74) g / mol.
We equate the molar mass indices, then we solve the equation, determine the value of the radical:
R + 74 = 75v, R = 75v - 74 = 1 (assume in = 1).
By selection, it comes to the conclusion that there is no hydrocarbon radical, so the desired amino acid is glycine.
Therefore, R = H, we obtain the formula CH2NH2COOH (glycine).
Answer: CH2NH2COOH.
Such a solution to the problems by the algorithm method is possible only in the case when the schoolchild adequately possesses elementary mathematical skills.
Programming
What do the algorithms look like here? Examples of solving problems in computer science and computer engineering suggest a clear sequence of actions.
If the order is violated, various system errors occur that prevent the algorithm from functioning in its entirety. Developing a program using object-oriented programming consists of two stages:
- creating a graphical interface in visual mode;
- software code development.
This approach greatly simplifies the algorithm for solving programming problems.
Manually it is almost impossible to cope with this time-consuming process.
Conclusion
The standard algorithm for solving inventive problems is presented below.
This is an accurate and understandable sequence of actions. When creating it, you must own the initial data of the task, the initial state of the described object.
In order to highlight the stages of solving algorithmic problems, it is important to determine the purpose of the work, to highlight the system of commands that will be executed by the executor.
The created algorithm must have a specific set of properties:
- discreteness (division into steps);
- uniqueness (each action has one solution);
- conceptualism;
- performance.
Many algorithms are massive, that is, they can be used to solve many similar tasks.
A programming language is a special set of rules for writing data and algorithmic structures. Currently, it is used in all scientific fields. Its important aspect is speed. If the algorithm is slow, it does not guarantee a rational and quick response, it is returned for revision.
The execution time of some tasks is determined not only by the size of the input data, but also by other factors. For example, an algorithm for sorting a significant number of integers is carried out easier and faster, provided that preliminary sorting is performed.