Tasks of the school Olympiad in chemistry
5 - 6 grade
Test
Choose one correct answer (1 point for each answer)
1. What gas is formed during photosynthesis:
2. Atom is...
3. Is a substance:
4. To separate a mixture of water - vegetable oil, a difference in components can be used according to:
5. Chemical phenomena include:
Match: (2 points for each answer)
6.
1. simple2. complex
a) water
b) oxygen
c) nitrogen
d) carbon dioxide
e) sand
e) table salt
7.
1. pure substances2. mixtures
a) granite
b) oxygen
to the air
d) iron
e) hydrogen
f) soil
8.
1. chemical phenomena2. physical phenomena
a) iron rusting
b) metal melting
c) boiling water
d) burning food
e) leaf rot
e) dissolving sugar
9.
1. body2. substances
a) gold
b) coin
c) a chair
d) glass
e) vase
e) acetic acid
10. Distribute the ways of separating mixtures:
1. iron and sand2. water and salt
3. sand and water
a) the action of a magnet
b) filtering
c) evaporation
Tasks:
Walking through the forest in the summer, the student discovered on his way an anthill, in which a crow, spreading its wings, “took baths”, planting ants in feathers with its beak. Why did she do it? Which Chemical substance used a crow "bathing" in an anthill? (5 points)
The student decided to help his friend catch up on the missed material in chemistry, tell him about chemical phenomena: 1) heat comes from a radiator; 2) extinguishing soda with vinegar when preparing the dough; 3) melting butter in a pan; 4) adding sugar to tea; 5) juice fermentation; 6) sour milk; 7) the appearance of rust on the nails; 8) spreading the smell of perfume. Is the student right? Are all the processes listed by the student chemical? Are any of them physical? (5 points)
Cars, cars, literally everything is flooded ... What materials and substances are used to make modern cars. What phenomena (physical, chemical) are observed during the operation of the car? (7 points)
Why can't plastic birdhouses be made? (7 points)
You have been given a mixture of the following substances: iron, soot, table salt, copper. Propose a plan for the separation of these substances. What laboratory equipment would be required to separate this mixture? (7 points)
Answers to tests:
1 - b, c;2 - a, d, e, f
1 -b, d, e; 2- a, c, e
1 - a, d, e; 2 - b, c, f
1 – b, c, e; 2 - a, d, f
1-a;
2 - in;
3 - b
Answers to tasks:
2. The student is wrong. Among the listed phenomena there are also physical ones, namely: 1, 3, 4, 8.
3. Nowadays, in mechanical engineering, man-made materials are used, which are superior to metals in terms of lightness, strength, durability and other valuable properties. These are plastics, rubbers, rubber, glass, fiberglass and others. Thanks to them, modern machines can operate at high and low temperatures, deep under water, in space. The chemical energy of a fuel (usually a liquid or gaseous hydrocarbon fuel) combusted in working area, is converted into mechanical work.
4. Plastic houses are extremely dangerous for birds, because plastics, unlike wood, are not able to absorb moisture and let it out through the smallest pores. Therefore, the water vapor released during breathing is absorbed by the litter and does not leave the house. High humidity is formed in the house, which is detrimental to birds.
5. Laboratory equipment: magnet, filter paper, funnel, beaker, spirit lamp.
1) we separate the iron with a magnet;
2) we dissolve the rest of the mixture in water, the salt dissolves, soot floats on top, copper sinks to the bottom;
3) filter the mixture - soot is filtered out, copper remains at the bottom of the glass;
4) there was a salt solution. Heat a thermal glass over an alcohol lamp - the water evaporates, the salt remains.
We discussed the general algorithm for solving problem No. 35 (C5). It's time to analyze specific examples and offer you a selection of tasks for independent solution.
Example 2. Complete hydrogenation of 5.4 g of some alkyne consumes 4.48 liters of hydrogen (n.a.) Determine the molecular formula of this alkyne.
Solution. We will act in accordance with the general plan. Let the unknown alkyne molecule contain n carbon atoms. General formula of the homologous series C n H 2n-2 . Hydrogenation of alkynes proceeds in accordance with the equation:
C n H 2n-2 + 2Н 2 = C n H 2n+2.
The amount of hydrogen reacted can be found by the formula n = V/Vm. In this case, n = 4.48 / 22.4 = 0.2 mol.
The equation shows that 1 mol of alkyne adds 2 mol of hydrogen (recall that in the condition of the problem in question O complete hydrogenation), therefore, n (C n H 2n-2) = 0.1 mol.
By the mass and amount of alkyne, we find its molar mass: M (C n H 2n-2) \u003d m (mass) / n (amount) \u003d 5.4 / 0.1 \u003d 54 (g / mol).
The relative molecular weight of an alkyne is made up of n atomic masses of carbon and 2n-2 atomic masses of hydrogen. We get the equation:
12n + 2n - 2 = 54.
We decide linear equation, we get: n = 4. Alkyne formula: C 4 H 6.
Answer: C 4 H 6 .
I would like to draw attention to one significant point: the molecular formula C 4 H 6 corresponds to several isomers, including two alkynes (butyn-1 and butyn-2). Based on these problems, we will not be able to unambiguously establish the structural formula of the substance under study. However, in this case, this is not required!
Example 3. During the combustion of 112 l (n.a.) of an unknown cycloalkane in excess oxygen, 336 l of CO 2 are formed. Set the structural formula of cycloalkane.
Solution. The general formula for the homologous series of cycloalkanes is: C n H 2n. With the complete combustion of cycloalkanes, as with the combustion of any hydrocarbons, carbon dioxide and water are formed:
C n H 2n + 1.5n O 2 \u003d n CO 2 + n H 2 O.
Please note: the coefficients in the reaction equation in this case depend on n!
During the reaction, 336 / 22.4 \u003d 15 mol of carbon dioxide was formed. 112/22.4 = 5 mol of hydrocarbon entered into the reaction.
Further reasoning is obvious: if 15 moles of CO 2 are formed per 5 moles of cycloalkane, then 15 molecules of carbon dioxide are formed per 5 molecules of hydrocarbon, i.e., one molecule of cycloalkane gives 3 molecules of CO 2. Since each molecule of carbon monoxide (IV) contains one carbon atom, we can conclude that one cycloalkane molecule contains 3 carbon atoms.
Conclusion: n \u003d 3, the formula of cycloalkane is C 3 H 6.
As you can see, the solution to this problem does not "fit" into the general algorithm. We did not look for the molar mass of the compound here, did not make any equation. According to formal criteria, this example is not similar to the standard C5 problem. But above, I have already emphasized that it is important not to memorize the algorithm, but to understand the MEANING of the actions performed. If you understand the meaning, you yourself will be able to make changes to the general scheme at the exam, choose the most rational way to solve it.
In this example, there is another "strangeness": it is necessary to find not only the molecular, but also the structural formula of the compound. In the previous task, we failed to do this, but in this example - please! The fact is that the formula C 3 H 6 corresponds to only one isomer - cyclopropane.
Answer: cyclopropane.
Example 4. 116 g of some limiting aldehyde were heated long time with ammonia solution of silver oxide. During the reaction, 432 g of metallic silver was formed. Set the molecular formula of aldehyde.
Solution. The general formula for the homologous series of limiting aldehydes is: C n H 2n+1 COH. Aldehydes are easily oxidized to carboxylic acids, in particular, under the action of an ammonia solution of silver oxide:
C n H 2n + 1 COH + Ag 2 O \u003d C n H 2n + 1 COOH + 2Ag.
Note. In reality, the reaction is described by a more complex equation. When Ag 2 O is added to an aqueous solution of ammonia, a complex compound OH is formed - diammine silver hydroxide. It is this compound that acts as an oxidizing agent. During the reaction, an ammonium salt of a carboxylic acid is formed:
C n H 2n + 1 COH + 2OH \u003d C n H 2n + 1 COONH 4 + 2Ag + 3NH 3 + H 2 O.
Another important point! The oxidation of formaldehyde (HCOH) is not described by the above equation. When HCOH reacts with an ammonia solution of silver oxide, 4 mol of Ag is released per 1 mol of aldehyde:
НCOH + 2Ag 2 O \u003d CO 2 + H 2 O + 4Ag.
Be careful when solving problems related to the oxidation of carbonyl compounds!
Let's go back to our example. By the mass of released silver, you can find the amount of this metal: n(Ag) = m/M = 432/108 = 4 (mol). In accordance with the equation, 2 mol of silver is formed per 1 mol of aldehyde, therefore, n (aldehyde) \u003d 0.5n (Ag) \u003d 0.5 * 4 \u003d 2 mol.
Molar mass of aldehyde = 116/2 = 58 g/mol. Try to do the next steps yourself: you need to make an equation, solve it and draw conclusions.
Answer: C 2 H 5 COH.
Example 5. When 3.1 g of some primary amine is reacted with a sufficient amount of HBr, 11.2 g of salt is formed. Set the amine formula.
Solution. Primary amines (C n H 2n + 1 NH 2) when interacting with acids form alkylammonium salts:
C n H 2n+1 NH 2 + HBr = [C n H 2n+1 NH 3] + Br - .
Unfortunately, by the mass of the amine and the resulting salt, we will not be able to find their quantities (since the molar masses are unknown). Let's go the other way. Recall the law of conservation of mass: m(amine) + m(HBr) = m(salt), therefore, m(HBr) = m(salt) - m(amine) = 11.2 - 3.1 = 8.1.
Pay attention to this technique, which is very often used in solving C 5. Even if the mass of the reagent is not given explicitly in the condition of the problem, you can try to find it from the masses of other compounds.
So, we are back in the mainstream of the standard algorithm. By the mass of hydrogen bromide we find the amount, n(HBr) = n(amine), M(amine) = 31 g/mol.
Answer: CH 3 NH 2 .
Example 6. A certain amount of alkene X, when interacting with an excess of chlorine, forms 11.3 g of dichloride, and when reacting with an excess of bromine, 20.2 g of dibromide. Determine the molecular formula of X.
Solution. Alkenes add chlorine and bromine to form dihalogen derivatives:
C n H 2n + Cl 2 \u003d C n H 2n Cl 2,
C n H 2n + Br 2 \u003d C n H 2n Br 2.
It is pointless in this problem to try to find the amount of dichloride or dibromide (their molar masses are unknown) or the amounts of chlorine or bromine (their masses are unknown).
We use one non-standard technique. The molar mass of C n H 2n Cl 2 is 12n + 2n + 71 = 14n + 71. M (C n H 2n Br 2) = 14n + 160.
The masses of the dihalides are also known. You can find the amount of substances obtained: n (C n H 2n Cl 2) \u003d m / M \u003d 11.3 / (14n + 71). n (C n H 2n Br 2) \u003d 20.2 / (14n + 160).
By convention, the amount of dichloride is equal to the amount of dibromide. This fact gives us the opportunity to make an equation: 11.3 / (14n + 71) = 20.2 / (14n + 160).
This equation has a unique solution: n = 3.
Answer: C 3 H 6
In the final part, I offer you a selection of problems of the C5 type of varying complexity. Try to solve them yourself - it will be a great workout before passing the exam in chemistry!
In my practice, I often encounter a problem when teaching how to solve problems in chemistry. One of difficult tasks V USE assignments became task C 5.
Here are some examples:
Example 1
Determine the formula of a substance if it contains 84.21% carbon and 15.79% hydrogen, and has a relative density in air of 3.93.
Solution:
1. Let the mass of the substance be 100 g. Then the mass C will be 84.21 g, and the mass H will be 15.79 g.
2. Find the amount of substance of each atom:
n(C) \u003d m / M \u003d 84.21 / 12 \u003d 7.0175 mol,
n(H) = 15.79 / 1 = 15.79 mol.
3. Determine the molar ratio of C and H atoms:
C: H \u003d 7.0175: 15.79 (we reduce both numbers by a smaller one) \u003d 1: 2.25 (we multiply by 4) \u003d 4: 9.
Thus, the simplest formula is C 4 H 9.
4. Based on the relative density, we calculate the molar mass:
M \u003d D (air.) 29 \u003d 114 g / mol.
5. The molar mass corresponding to the simplest formula C 4 H 9 is 57 g / mol, which is 2 times less than the true molar mass.
So the true formula is C 8 H 18.
Example 2
Determine the formula of alkyne with a density of 2.41 g/l under normal conditions.
Solution:
1. General formula of alkyne С n H 2n−2
2. 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 \u003d (density ρ) (molar volume V m) \u003d 2.41 g / l 22.4 l / mol \u003d 54 g / mol.
14n − 2 = 54, n = 4.
Hence, alkyne has the formula C 4 H 6.
Answer: C 4 H 6.
Example 3
Relative vapor density organic compound nitrogen is 2. When burning 9.8 g of this compound, 15.68 liters of carbon dioxide (n.a.) and 12.6 g of water are formed. Derive the molecular formula of the organic compound.
Solution:
1. Since the substance turns into carbon dioxide and water during combustion, it means that it consists of atoms C, H and, possibly, O. Therefore, its general formula can be written as СхНуОz.
2. We can write the combustion reaction scheme (without placing the coefficients):
СхНуОz + О 2 → CO 2 + H 2 O
3. All carbon from the original substance goes into carbon dioxide, and all hydrogen goes into water.
We find the amounts of substances CO 2 and H 2 O, and determine how many moles of C and H atoms they contain:
a) n (CO 2) \u003d V / V m \u003d 15.68 / 22.4 \u003d 0.7 mol.
(There is one C atom per molecule of CO 2, which means that there are as many moles of carbon as CO 2. n (C) \u003d 0.7 mol)
b) n (H 2 O) \u003d m / M \u003d 12.6 / 18 \u003d 0.7 mol.
(One molecule of water contains two H atoms, which means the amount of hydrogen is twice as much as water. n (H) \u003d 0.7 2 \u003d 1.4 mol)
4. We check the presence of oxygen in the substance. To do this, the masses C and H must be subtracted from the mass of the entire initial substance.
m(C) = 0.7 12 = 8.4 g, m(H) = 1.4 1 = 1.4 g
The mass of the entire substance is 9.8 g.
m(O) = 9.8 − 8.4 − 1.4 = 0, i.e. There are no oxygen atoms in this substance.
5. Search for the simplest and true formulas.
C: H \u003d 0.7: 1.4 \u003d 1: 2. The simplest formula is CH 2.
6. We are looking for the true molar mass by the relative density of gas in nitrogen (do not forget that nitrogen consists of diatomic molecules N 2 and its molar mass is 28 g / mol):
M ist. \u003d D (N 2) M (N 2) \u003d 2 28 \u003d 56 g / mol.
The true formula is CH 2, its molar mass is 14. 56/14 \u003d 4. The true formula: (CH 2) 4 \u003d C 4 H 8.
Answer: C 4 H 8.
Example 4
When 25.5 g of saturated monobasic acid interacted with an excess of sodium bicarbonate solution, 5.6 l (N.O.) of gas was released. Determine the molecular formula of the acid.
Solution:
1. C n H 2n+1 COOH + NaHCO 3 à C n H 2n+1 COONa + H 2 O + CO 2
2. Find the amount of substance CO 2
n (CO 2) \u003d V / Vm \u003d 5.6 l: 22.4 l / mol \u003d 0.25 mol
3. n (CO 2) \u003d n (acids) \u003d 0.25 mol (this ratio is 1: 1 from the equation)
Then the molar mass of the acid is:
M (k-you) \u003d m / n \u003d 25.5 g: 0.25 mol \u003d 102 g / mol
4. M (k-you) \u003d 12n + 2n + 1 + 12 + 16 + 16 (from the general formula, M \u003d Ar (C) * n + Ar (H) * n + Ar (O) * n \u003d 12 * n + 1*(2n+1)+ 12+16+16+1)
M (k-you) \u003d 12n + 2n + 46 \u003d 102; n = 4; The formula of the acid is C 4 H 9 COOH.
Tasks for independent solution С5:
1. The mass fraction of oxygen in a monobasic amino acid is 42.67%. Set the molecular formula of the acid.
2. Determine the molecular formula of the tertiary amine if it is known that during its combustion 0.896 l (n.o.) of carbon dioxide, 0.99 g of water and 0.112 l (n.o.) of nitrogen were released.
3. For the complete combustion of 2 liters of gaseous hydrocarbon, 13 liters of oxygen were required, while 8 liters of carbon dioxide were formed. Find the molecular formula of hydrocarbon.
4. When burning 3 liters of gaseous hydrocarbon, 6 liters of carbon dioxide and some water were obtained. Determine the molecular formula of the hydrocarbon if it is known that 10.5 liters of oxygen were required for complete combustion.
5. The alkane dichloro derivative contains 5.31% hydrogen by mass. Determine the molecular formula of dichloroalkane. Give the structural formula of one of the possible isomers and name it
6. During the combustion of gaseous organic matter that does not contain oxygen, 4.48 liters of carbon dioxide (N.O.), 3.6 g of water and 2 g of hydrogen fluoride were released. Set the molecular formula of the compound.
