C - 4 Doronkin V.N. Determination of the composition of the reaction product ("acid salts")
3.1 4.48 L (N.O.) ammonia was passed through 200 g of a 4.9% solution of phosphoric acid. Name the salt formed as a result of the reaction and determine its mass.
3.2 In 240 g of 9% orthophosphoric acid, 5.68 g of phosphorus (V) oxide was dissolved and the resulting solution was boiled. What salt and in what quantity is formed if 84 g of potassium hydroxide is added to the resulting solution?
3.3 In 440 g of 8% sulfuric acid, 32 g of sulfur oxide (VI) were dissolved. What salt and in what quantity is formed if 16 g of sodium hydroxide is added to the resulting solution?
3.4 In 84 g of an 8% potassium hydroxide solution, sulfur oxide (IV) was dissolved, which was released during the firing of pyrite weighing 7.2 g. Determine mass fraction salt in the resulting solution.
3.5 The gas obtained by burning 19.2 g of sulfur reacted without residue with 682.5 ml of 5% sodium hydroxide solution (density 1.055 g/ml). Determine the composition of the resulting solution and calculate the mass fractions of substances in this solution.
3.6 Sulfur (IV) oxide, 2.24 L (n.o.) was passed through 80 g of 5% sodium hydroxide solution. Calculate the mass of salt formed as a result of the reaction.
3.7 21.6 g of silver was exposed to a 68% solution nitric acid, whose mass is 600 g. The resulting gas was passed through 300 g of a 10% cold solution of sodium hydroxide. Calculate the mass fractions of substances in the resulting solution.
3.8 In 120 g of 18% orthophosphoric acid, 5.68 g of phosphorus (V) oxide was dissolved and the resulting solution was boiled. What salt and in what quantity is formed if 60 g of sodium hydroxide is added to the resulting solution?
3.9 5.6 L (N.O.) carbon dioxide reacted without residue with 59.02 ml of 20% by mass potassium hydroxide solution (density 1.185 g/ml). Determine the mass of the substance formed as a result of the reaction.
3.10 Hydrogen sulfide with a volume of 11.2 l (n.o.) reacted without residue with 250 g of sodium hydroxide solution with a mass fraction of 8%. Determine the mass of the substance formed as a result of the reaction.
3.11 12 g of sulfur were burned in excess oxygen. The reaction product was passed through 300 g of 8% sodium hydroxide solution. Determine the mass fractions of salts in the resulting solution
3.12 The gas obtained by burning 6.4 g of sulfur reacted without residue with 138 ml of 8% sodium hydroxide solution (density 1.087 g/ml). determine the composition of the resulting solution and calculate the mass fractions of substances in this solution.
3.13 To the solution obtained by dissolving 16 g of sulfur(VI) oxide in 150 g of a 19.6% sulfuric acid solution, 16 g of sodium hydroxide was added. Determine the mass of salt formed.
3.14 To 100 g of a 24.5% solution of sulfuric acid was added 200 g of a 5% solution of sodium hydroxide. Determine the environment of the solution and the mass fraction of sodium salt in it.
3.15 44.8 liters of hydrogen sulfide were burned in excess oxygen. The combustion products were passed through 250 ml of a 25% solution caustic soda density 1.28 g/ml. Determine the mass of salt in the solution.
Calculations: mass (volume, amount of substance) of the reaction products, if one of the substances is given in excess (has impurities), if one of the substances is given in the form of a solution with a certain mass fraction of the solute (task 39 (C4)) Prepared by: teacher chemistry MOU Gymnasium 1, Zheleznogorsk Kursk region Fedorchenko S.M.
Calculations by reaction equations. 1. Copper released as a result of the reaction of 2.6 g of zinc with 160 g of a 5% copper (II) sulfate solution completely reacted with 20 ml of dilute nitric acid (ρ = 1.055 g / ml). Determine the mass fraction of copper (II) nitrate in the resulting solution. Zn + CuSO 4 \u003d ZnSO 4 + Cu (1) 3Cu + 8 HNO 3 \u003d 3Cu (NO 3) 2 + 2NO + 2H 2 O (2) n (Zn) \u003d 0.04 mol (deficiency) n (CuSO 4) \u003d 0.05 mol (excess) n (Cu) \u003d 0.04 mol m (Cu) \u003d 2.56 g. n (Cu (NO 3) 2) \u003d 0.04 mol. n(NO)=0.027 mol. m (NO) \u003d 0.81 g. m (Cu (NO 3) 2) \u003d 7.52 g. m (HNO 3) solution \u003d 21.1 g. m (solution) \u003d 22.86 g. ω (Cu (NO 3) 2) = 33%. Answer: ω (Cu (NO 3) 2) = 33%.
Calculations by reaction equations. 2. Sodium peroxide treated with excess hot water. The evolved gas was collected, and the resulting alkali solution was completely neutralized with a 10% sulfuric acid solution with a volume of 300 ml and a density of 1.08 g/ml. Determine the mass of sodium peroxide taken for the reaction and the volume collected gas. 2Na 2 O 2 + 2H 2 O \u003d 4NaOH + O 2 (1) 2NaOH + H 2 SO 4 \u003d Na 2 SO 4 + 2H 2 O (2) n (H 2 SO 4) \u003d 0.33 mol n (NaOH) \u003d 0.66 mol n (O 2) \u003d 0.165 mol V (O 2) \u003d 3.696 l 3.7 l n (Na 2 O 2) \u003d 0.33 mol m (Na 2 O 2) \u003d 25.74 g Answer : V (O 2) \u003d 3.7 l; m (Na 2 O 2) \u003d 25.74 g.
Calculations according to the reaction equations ml 34% of hydrochloric acid with a density of 1.16 g / ml was added when heated to manganese (IV) oxide weighing 2.61 g. What volume of chlorine (n.o.) will be released? How many grams of potassium carbonate can react (without heating) with the released chlorine? MnO 2 + 4HCl \u003d MnCl 2 + Cl 2 + 2H 2 O (1) 3 Cl 2 + 3K 2 CO 3 \u003d 5KCl + 3CO 2 + KClO 3 (2) n (MnO 2) \u003d 0.03 mol (deficiency) n (HCl) \u003d 0.27 mol (excess) n (Cl 2) \u003d 0.03 mol. V (Cl 2) \u003d 0.672 l. n(K 2 CO 3) \u003d 0.03 mol. m (K 2 CO 3) \u003d 4.14 g. Answer: V (Cl 2) \u003d 0.672 l; m(K 2 CO 3) = 4.14 g.
Task groups included in task 39 USE tests: 1. Calculations by reaction equations. 2. Determination of the composition of the reaction product (tasks for the "type of salt"). 3. Problems on a mixture of substances. 4. Finding the mass fraction of one of the reaction products in solution using the material balance equation. 5. Finding the mass of one of starting materials according to the material balance equation.
General principles solving computational problems in chemistry Stage 1: compose the reaction equations for those transformations that are mentioned in the condition. Step 2: Calculate quantities and masses " pure substances". Stage 3: to establish cause-and-effect relationships between the reacting substances, that is, to determine the amount of which substance needs to be found and which of the reacting substances will be used for calculation. Stage 4: make calculations according to the equation(s) of reactions, that is, calculate the amount of the desired substance, and then find its mass (or volume of gas). Step 5: Answer additional questions, formulated in the condition.
Determination of the composition of the reaction product (tasks for the "type of salt") 1. Carbon dioxide with a volume of 5.6 l (n.c.) was passed through 164 ml of a 20% sodium hydroxide solution (ρ=1.22 g/ml). Determine the composition and mass fractions of substances in the resulting solution. Given: V (CO 2) \u003d 5.6 l V (NaOH) sol. \u003d 164 ml ω (NaOH) \u003d 20% or 0.2 ρ (solution) \u003d 1.22 g / ml What is the composition of the solution? ω(substances) - ? Solution: 1. Compose the reaction equations. NaOH + CO 2 = NaHCO 3 (1) or 2NaOH + CO 2 = Na 2 CO 3 + H 2 O (2) 2. Calculate the amount of reactants ("pure" reactants). M (NaOH) \u003d 40 g / mol n (NaOH) \u003d V (solv.) (NaOH) ρ (solution) ω (NaOH) / M (NaOH) \u003d 164 1.22 0.2 / 40 \u003d 1 (mol) n (CO 2) \u003d V / Vm \u003d 5.6 / 22.4 \u003d 0.25 mol
NaOH + CO 2 = NaHCO 3 (1) or 2NaOH + CO 2 = Na 2 CO 3 + H 2 O (2) 3. We make calculations according to the reaction equations. n (CO 2) : n (NaOH) \u003d 0.25 mol: 1 mol or 1: 4, therefore, sodium hydroxide is given in excess, we calculate according to equation (2). y mol 0.25 mol x mol 2NaOH + CO 2 \u003d Na 2 CO 3 + H 2 O (2) 2 mol 1 mol 1 mol 4. It follows from the reaction equation (2) that n (NaOH) \u003d 2 n (CO 2) \u003d 0.5 mol is the amount of reacted sodium hydroxide. This means that unreacted hydroxide remained in the solution, in the amount of 0.5 mol. There are two substances in the solution, the mass fractions of which must be calculated.
NaOH + CO 2 = NaHCO 3 (1) or 2NaOH + CO 2 = Na 2 CO 3 + H 2 O (2) It follows from the reaction equation (2) that n (Na 2 CO 3) = n (CO 2) = 0.25 mol. M (Na 2 CO 3) \u003d 106 g / mol m (Na 2 CO 3) \u003d n M \u003d 0.25 106 \u003d 26.5 g. m (NaOH) \u003d n M \u003d 0.5 40 \u003d 20 g. M (CO 2) \u003d 44 g / mol. m (solution) \u003d m 1 (NaOH) + m (CO 2) \u003d V (sol.) (NaOH) ρ (solution) + n (CO 2) M (CO 2) \u003d 164 1.22 + 0 .25 44 = 211.08 (g). ω(Na 2 CO 3) \u003d 26.5 / 211.08 \u003d 0, 1255 or 12.55%. ω rest (NaOH) = 20/211.08 = 0.095 or 9.5%. Answer: ω(Na 2 CO 3) = 12.55%; ω(NaOH) = 9.5%.
Determination of the composition of the reaction product (tasks for the "type of salt") 2. The gas obtained by burning 19.2 g of sulfur reacted without residue with 682.5 ml of a 5% NaOH solution (density 1.055 g/ml). Determine the composition of the resulting solution and calculate the mass fractions of substances in this solution. Given: m (S) \u003d 19.2 g V (NaOH) sol. \u003d 682.5 ml ω (NaOH) \u003d 5% or 0.05 ρ (solution) \u003d 1.055 g / ml Solution composition -? ω(substances) - ? Solution: 1. Compose the reaction equations. S + O 2 \u003d SO 2 (1) SO 2 + NaOH \u003d NaHSO 3 (2) or SO NaOH \u003d Na 2 SO 3 + H 2 O (3) NaHSO 3 + NaOH \u003d Na 2 SO 3 + H 2 O (4 ) 2. Calculate the amount of reactants ("pure" reactants). M (S)=32 g/mol n (S)=m/M=19.2/32=0.6 (mol) M(NaOH)=40 g/mol n(NaOH)= m/M= V ( sol.) (NaOH) ρ (solution) ω (NaOH) / M (NaOH) \u003d 682.5 1.055 0.05 / 40 \u003d 0.9 (mol)
S + O 2 \u003d SO 2 (1) SO 2 + NaOH \u003d NaHSO 3 (2) or SO 2 + 2NaOH \u003d Na 2 SO 3 + H 2 O (3) NaHSO 3 + NaOH \u003d Na 2 SO 3 + H 2 O (4) 3. We make calculations according to the reaction equations. 0.6 mol x mol S + O 2 \u003d SO 2 1 mol 1 mol From the reaction equation (1) it follows that n (S) \u003d n (SO 2) \u003d 0.6 mol. According to the condition of the problem p (SO 2): n (NaOH) \u003d 0.6: 0.9 \u003d 1: 1.5. This means that the calculation should be carried out according to the reaction equation (2), because for the reaction equation (3), sodium hydroxide is not enough. 0.6 mol 0.9 mol 0.6 mol SO 2 + NaOH = NaHSO 3 (2) 1 mol 1 mol 1 mol It follows from the reaction equation (2) that sodium hydroxide will not completely react with sulfur dioxide. n (NaOH) proreact. \u003d p (SO 2) \u003d 0.6 mol n (NaOH) rest. \u003d 0.9 - 0.6 \u003d 0.3 mol The remaining sodium hydroxide is used to neutralize the acid salt - sodium hydrosulfite, the reaction equation (four).
S + O 2 \u003d SO 2 (1) SO 2 + NaOH \u003d NaHSO 3 (2) or SO 2 + 2NaOH \u003d Na 2 SO 3 + H 2 O (3) NaHSO 3 + NaOH \u003d Na 2 SO 3 + H 2 O (4) 0.6 mol 0.3 mol x mol NaHSO 3 + NaOH = Na 2 SO 3 + H 2 O (4) 1 mol 1 mol 1 mol From the reaction equation (4) it follows that n(NaHSO 3) = n(NaOH) = 0.3 mol. Sodium hydroxide is in short supply, we calculate p (Na 2 SO 3) from it. p (Na 2 SO 3) \u003d n (NaOH) \u003d 0.3 mol. After the acid salt neutralization reaction, sodium hydrosulfite remained in the solution in the amount of substance 0.3 mol. 4. After all the transformations, there are two substances in the solution: p (Na 2 SO 3) \u003d 0.3 mol. n(NaHSO 3) = 0.3 mol. m (Na 2 SO 3) \u003d n M \u003d 0.3 126 \u003d 37.8 g. n (NaHSO 3) \u003d n M \u003d 0.3104 \u003d 31.2 g. m (solution) \u003d m (NaOH ) + m (SO 2) \u003d V (sol.) (NaOH) ρ (solution) + n (SO 2) M (SO 2) \u003d 682.5 1.6 64 \u003d 758.4 (g) . ω(Na 2 SO 3)= 37.8 /758.4= 0.0498 or 4.98%. ω (NaHSO 3) \u003d 31.2 / 758.4 \u003d 0.0411 or 4.11%. Answer: ω (Na 2 SO 3) = 4.98%; ω (NaHSO 3) = 4.11%.
Determination of the composition of the reaction product (tasks for the "type of salt") 3. 32 g of sulfur oxide (VI) was dissolved in 440 g of 8% sulfuric acid. What salt and in what quantity is formed if 16 g of sodium hydroxide is added to the resulting solution? Given: m (H 2 SO 4) solution \u003d 440 g ω (H 2 SO 4) \u003d 8% or 0.08 m (SO 3) \u003d 32 g m (NaOH) \u003d 16 g n (salt) -? m(salt)-? Solution: 1. Compose the reaction equations. SO 3 + H 2 O \u003d H 2 SO 4 (1) H 2 SO 4 + NaOH \u003d NaHSO 4 + H 2 O (2) or H 2 SO 4 + 2NaOH \u003d Na 2 SO 4 + 2H 2 O (3) 2 We calculate the amounts of reacting substances (“pure” reacting substances). n (H 2 SO 4) \u003d m (H 2 SO 4) solution ω (H 2 SO 4) / M \u003d 440 0.08 / 98 \u003d 0.36 (mol) n (SO 3) \u003d m / M =32/80=0.4 mol n(NaOH)=m/M=16/40=0.4 mol
SO 3 + H 2 O \u003d H 2 SO 4 (1) H 2 SO 4 + NaOH \u003d NaHSO 4 + H 2 O (2) or H 2 SO 4 + 2NaOH \u003d Na 2 SO 4 + 2H 2 O (3) 3 From the reaction equation (1) it follows that n (H 2 SO 4) \u003d n (SO 3) \u003d 0.4 mol - the amount of acid formed. Total acids 0.36 + 0.4 = 0.76 (mol). p (H 2 SO 4): n (NaOH) \u003d 0.76: 0.4 Sodium hydroxide is in short supply - acidic salt NaHSO 4 is formed. 0.4 mol x mol H 2 SO 4 + NaOH \u003d NaHSO 4 + H 2 O 1 mol 1 mol 4. n (NaHSO 4) \u003d n (NaOH) \u003d 0.4 mol. Calculate the mass of sodium hydrosulfate. M (NaHSO 4) \u003d 120 g / mol. m (NaHSO 4) \u003d n M \u003d 0.4 120 \u003d 48 g. Answer: n (NaHSO 4) \u003d 0.4 mol; m(NaHSO 4)=48 g.
Determination of the composition of the reaction product (“type of salt” tasks) .16 g/ml). Calculate the mass fractions of the substances in the solution. Given: V(H 3 PO 4) sol. \u003d 26.25 ml ω (H 3 PO 4) \u003d 20% or 0.2 ρ (solution) \u003d 1.12 g / ml V (KOH) sol. \u003d 50 ml ω (KOH) \u003d 16% or 0.16 ρ (solution) \u003d 1.16 g / ml ω (substances) -? Solution: 1. Compose the reaction equations. H 3 PO 4 + KOH = KH 2 PO 4 + H 2 O (1) H 3 PO 4 + 2KOH = K 2 HPO 4 + 2H 2 O (2) H 3 PO 4 + 3KOH = K 3 PO 4 + 3H 2 O (3) KH 2 PO 4 + KOH \u003d K 2 HPO 4 + H 2 O (4) K 2 HPO 4 + KOH \u003d K 3 PO 4 + H 2 O (5) 2. Calculate the amount of reacting substances ("pure" reactants). n (H 3 PO 4) \u003d V ω ρ / M \u003d 26.25 0.2 1.12 / 98 \u003d 0.06 mol. n(KOH)= V ω ρ/M = 50 0.16 1.16/56= 0.16 mol. 3. Substitution of hydrogen atoms in the acid occurs in stages: H 3 PO 4 KH 2 PO 4 K 2 HPO 4 K 3 PO 4
H 3 PO 4 + KOH = KH 2 PO 4 + H 2 O (1) H 3 PO 4 + 2KOH = K 2 HPO 4 + 2H 2 O (2) H 3 PO 4 + 3KOH = K 3 PO 4 + 3H 2 O (3) KH 2 PO 4 + KOH = K 2 HPO 4 + H 2 O (4) K 2 HPO 4 + KOH = K 3 PO 4 + H 2 O (5) 0.06 mol 0.16 mol x mol H 3 PO 4 + KOH \u003d KH 2 PO 4 + H 2 O (1) 1 mol 1 mol 1 mol From the reaction equation (1) it follows that n (H 3 PO 4): n (KOH) \u003d 1: 1. According to the condition of the problem, n (H 3 PO 4) \u003d 0.06 mol, n (KOH) \u003d 0.16 mol, therefore, potassium hydroxide is in excess. The amount of reacted potassium hydroxide is 0.06 mol, the amount of remaining hydroxide is (0.16-0.06)=0.1 mol. n (KH 2 PO 4) \u003d n (H 3 PO 4) \u003d 0.6 mol. The resulting acid salt of potassium dihydroorthophosphate KH 2 PO 4 will react with the potassium hydroxide remaining in the solution - reaction equation (4). 0.06 mol 0.1 mol y mol KH 2 PO 4 + KOH = K 2 HPO 4 + H 2 O (4) 1 mol 1 mol 1 mol
H 3 PO 4 + KOH = KH 2 PO 4 + H 2 O (1) H 3 PO 4 + 2KOH = K 2 HPO 4 + 2H 2 O (2) H 3 PO 4 + 3KOH = K 3 PO 4 + 3H 2 O (3) KH 2 PO 4 + KOH \u003d K 2 HPO 4 + H 2 O (4) K 2 HPO 4 + KOH \u003d K 3 PO 4 + H 2 O (5) From the reaction equation (4) it follows that n (KH 2 PO 4): n (KOH) \u003d 1: 1, which means that KOH is in excess, it will not react completely, in the amount of 0.06 mol, and 0.04 mol will remain in solution and react with acid salt- potassium hydrogen orthophosphate - equation (5). n (K 2 HPO 4) \u003d n (KH 2 PO 4) \u003d 0.06 mol. 0.06 mol 0.04 mol z mol K 2 HPO 4 + KOH = K 3 PO 4 + H 2 O (5) 1 mol 1 mol 1 mol It follows from the reaction equation (5) that n(K 2 HPO 4) : n (KOH) \u003d 1: 1, which means that KOH is in excess, it will react completely, potassium hydrogen orthophosphate in the amount of 0.02 mol will remain in the solution. n (K 3 PO 4) \u003d n (K 2 HPO 4) \u003d 0.04 mol.
H 3 PO 4 + KOH = KH 2 PO 4 + H 2 O (1) H 3 PO 4 + 2KOH = K 2 HPO 4 + 2H 2 O (2) H 3 PO 4 + 3KOH = K 3 PO 4 + 3H 2 O (3) KH 2 PO 4 + KOH \u003d K 2 HPO 4 + H 2 O (4) K 2 HPO 4 + KOH \u003d K 3 PO 4 + H 2 O (5) 4. There are two salts in solution: n (K 2 HPO 4)=0.02 mol; n(K 3 PO 4) = 0.04 mol. m(K 2 HPO 4) \u003d n M \u003d 0.02 174 \u003d 3.48 g; m (K 3 PO 4) \u003d n M \u003d 0.04 212 \u003d 8.48 g. m (solution) \u003d m (H 3 PO 4) solution + m (KOH) solution \u003d V (H 3 PO 4) ρ(H 3 PO 4) + V (KOH) ρ(KOH) = 26.25 1, 1.16 = 87.4 (g). ω(K 2 HPO 4)=m(K 2 HPO 4)/m(solution)=3.48/87.4=0.097 or 9.7%. ω(K 3 PO 4)=m(K 3 PO 4)/m(solution)= 8.48/87.4=0.04 or 4%. Answer: ω(K 2 HPO 4)=9.7%, ω(K 3 PO 4)=4%.
Homework: 1. The gas obtained by burning 6.4 g of sulfur reacted without residue with 138 ml of 8% NaOH solution (density 1.087 g/ml). Calculate the mass fractions of substances in the resulting solution. 2. Ammonia with a volume of 4.48 l (n.o.) was passed through 200 g of a 4.9% solution of phosphoric acid. Name the salt formed as a result of the reaction, and determine its mass.
Used literature V.N. Doronkin, A.G. Berezhnaya, T.V. Sazhnev, V.A. Fevraleva, Chemistry. Preparing for the exam Legion, Rostov-on-Don, 2012 V.N. Doronkin, A.G. Berezhnaya, T.V. Sazhnev, V.A. Fevraleva, Chemistry. Thematic tests to prepare for the exam. Tasks high level complexity (C1-C5). Legion, Rostov-on-Don, 2012
Solution:
1. Write the reaction equation
according to the equation 1 mol 5 mol
by solution 0.25 mol (x) 1.25 mol
2. According to the equation, the amount of pentane substance is related to the amount of carbon dioxide substance as 1:5 (coefficients in the equation in front of these substances)
n \u003d V / V m \u003d 5.6 l / 22.4 l / mol \u003d 0.25 mol;
From here to The amount of carbon monoxide (IV) 5 * 0.25 mol \u003d 1.25 mol (proportion - find x)
4. V (CO 2) \u003d n (CO 2) * Vm \u003d 1.25 mol * 22.4 l / mol \u003d 28 l.
Although, since these substances are normal conditions gases, you can solve it even easier - the volumes of these substances are related as 1: 5, which means that the volume of carbon dioxide formed will be 5 times more volume pentane. Then the volume formed V (CO 2) \u003d 5.6 l * 5 \u003d 28 l
Task #2
Determine the mass of water that is formed during the combustion of pentane with a volume of 11.2 liters.
Solution:
1. Write the reaction equation C 5 H 12 + 8O 2 \u003d 5CO 2 + 6H 2 O
according to the equation 1 mol 6 mol
by solution 0.5 mol (x) 3 mol
2. According to the equation, the amount of pentane substance is related to the amount of water substance as 1:6 (the coefficients in the equation in front of these substances)
3. Find the amount of substance for these formulas based on the conditions of the problem (by solution)
n \u003d V / V m \u003d 11.2 l / 22.4 l / mol \u003d 0.5 mol;
From here to Amount of water substance 6 * 0.5 mol \u003d 3 mol(proportion - find x)
4. And we find a lot of water m(H 2 O) \u003d 3mol * 18g / mol \u003d 54 g
2. Finding the formula of organic matter by mass fraction of chemical elementsand relative density of gases.
1) Write down the basic formulas in your notebook:
D=Mr(1)/Mr(2)
D is the relative density of the first gas relative to the second (dimensionless value).
For example :
D(O 2) \u003d Mr (gas) / Mr (O 2) \u003d Mr (gas) / 32;
D (H 2) \u003d Mr (gas) / Mr (H 2) \u003d Mr (gas) / 2;
D(air)=Mr(gas)/Mr(air)=Mr(gas)/29.
Welement = (n * Ar (element) * 100%) / Mr (substance),
where n is the index, the number of atoms;
W is the mass fraction of the element (%).
Task #1
Find the formula of a hydrocarbon that contains 14.29% hydrogen and its nitrogen relative density is 2.
Solution:
1 .Let's find the true molar mass C x H y:
M \u003d D (N 2) ∙ 28 = 2 ∙ 28 = 56 g/mol.
2. Find the mass fraction of carbon: ω(С) = 100% - 14.29% = 85.71%.
3. Let's find the simplest formula substance and its molar mass:
x:y = 85.7 / 12: 14.29 / 1 = 7.142: 14.29 = 1: 2- CH 2
M (CH 2) \u003d 12 + 1 ∙ 2 \u003d 14 g / mol
4. Let's compare molar masses: M (C x H y) / M (CH 2) \u003d 56 / 14 \u003d 4- the true formula is C 4 H 8.
Task #2
Ethyl alcohol contains 52.18% carbon: 13.04% hydrogen: 34.78% oxygen. Alcohol vapor density by hydrogen 23. Determine the formula of ethyl alcohol.
Solution:
1. Define molecular weight the desired substance:
Mr(CxHyOz) = D(H2) Mr(H2)=23 2 =46g/mol
2. According to the formula n \u003d W element * Mr (substance) / Ar element * 100%
calculate the number of atoms C, H, O
n(C)=(52.18% 46) / 12 100% = 2
n(H)=(13.04% 46) /1 100% =6
n(O)=(34.78% 46) / 16 100% =1
We get x:y:z =2:6:1, therefore, the substance C 2 H 6 O
Check, Mr(C 2 H 6 O)= 46 g/mol
3. Finding the formula of organic matter for combustion products andrelative density of gases.
Task number 1.
When the hydrocarbon was burned, 6.6 g of carbon dioxide and 3.6 g of water were obtained. Relative density hydrocarbon by air is 1.517. Determine the formula of the substance.
1) Find m polar mass of hydrocarbon:
M in-va \u003d D air *29= 1.517 *29=44 g/mol
2) Find the amount of atomic carbon:
n (CO 2) \u003d m (CO 2) / M (CO 2) \u003d 6.6 G/44 g/mol=0.15 mol
remembering that n(CO 2) \u003d n(C) = 0.15 mol
3) Find the amount of atomic hydrogen:
n (H 2 O) \u003d m (H 2 O) / M (H 2 O) \u003d 3.6 G/18 g/mol=0.2 mol
remembering that n (H) \u003d 2 * n(H 2 O) \u003d 0.4 mol
4) We find the ratio of the quantities (C) and (H) in the formula as integers:
n (C) : n (H) \u003d 0.15: 0.4 \u003d 3: 8
5) The simplest formula of a substance: C 3 H 8
M simple \u003d 44 g / mol
M ist / M simple \u003d 44 / 44 \u003d 1
The true formula C 3 H 8 - propane
