readx;??? The formula for calculating a substance is the sum of the atomic weights of the atoms in the chemical formula. The steps to calculate the formula of matter www.biquge.info are as follows: (1) write the correct chemical formula of the substance, (2) multiply the atomic weight of each element by the number of atoms contained in the chemical formula to obtain the sum of the atomic weights of each element, and (3) add the atomic weight of each element to calculate the formula of the substance.

Example: Calculate the amount of alum crystal.

The chemical formula of alum crystal is kal(SO4)2?12H2O, to calculate the formula of crystal hydrate (A?NH2O), first calculate the formula amount of A, and then multiply the formula amount of water by the number of crystal water n to calculate the formula amount of crystal water contained, and add the formula amount of anhydrous salt A and the formula amount of crystal water NH2O to obtain the formula amount of crystal hydrate (A?NH2O). In the chemical formula of the crystalline hydrate A?NH2O, the small dot between A and NH2O is "?" , which is a symbol that represents addition at the time of calculation.

The chemical formula of alum is kal(SO4)2?12H2O, where the formula of anhydrous salt kal(SO4)2 is:

39x1+27x1+（32+16x4）x2=258

The formula amount of crystal water contained in alum kal(SO4)2?12H2O is:

（1x2+16）x12=216

The formula of alum crystal is:

258+216=474

Answer: The formula amount of alum crystal is 474.

Calculate the mass ratio of each element in a compound to the mass ratio of each element in a compound, which is the ratio of the sum of the atomic weights of the atoms of each element in a molecule of the compound. To calculate the mass ratio of each element in a compound, the sum of the atomic weights of each element should be calculated according to the chemical formula of the compound, and then the ratio of the sum of the atomic weights of each element should be obtained.

Example: What is the mass ratio of iron to oxygen in ferric tetroxide?

The chemical formula of ferric tetroxide is Fe3O4, in a molecule of Fe3O4, it contains 3 iron atoms and 4 oxygen atoms, and the sum of the atomic weights of the 3 iron atoms is:

3x56=168

The sum of the atomic weights of the 4 oxygen atoms is:

4x16=64

The mass ratio of iron to oxygen is 168:64, which can be simplified to 21:8.

The mass ratio of iron to oxygen in Fe3O4 is:

3x56∶4x16=21∶8

Answer: In Fe3O4, the mass ratio of Fe to O is 21 to 8.

To calculate the percentage content of each element in a compound, you must first write the chemical formula of the compound and calculate the amount of the formula, and then divide the sum of the atomic weights of each element in the chemical formula by the formula of the compound, multiply by 100%, and then calculate the percentage content of each element in the compound.

Example: Calculate the percentage content of nitrogen, hydrogen, and oxygen in ammonium nitrate (NH4NO3) in the fertilizer.

[Analysis] to calculate the percentage content of an element in a compound can be calculated using the following formula:

The percentage of nitrogen in NH4NO3 is:

The percentages of hydrogen are:

The percentage of oxygen is:

The atomic weight of hydrogen is 1, the atomic weight of nitrogen is 14, the atomic weight of oxygen is 16, and the formula of NH4NO3 is 80. The percentage of nitrogen in NH4NO3 is:

The percentages of hydrogen are:

The percentages of oxygen are:

Answer: In NH4NO3. Medium contains N35%, H5%, O60%.

Calculation of solubilityIn a saturated solution at a certain temperature, the quality of the solute, the mass of the solvent and the solubility of the substance have the following relationships:

After processing the above equation, we get:

As long as the mass of the solute in the saturated solution and the mass of the solvent are known, the solubility of the substance at that temperature can be calculated.

For example, at 20C, 25 grams of blue alum (CuSO4?5H2O) are dissolved in 71 grams of water to form a saturated copper sulfate solution, and the solubility of CuSO4 at 20C is calculated.

Blue alum is dissolved in water to form a saturated solution of copper sulfate, in which the solute is CuSO4 and the solvent is H2O. In this problem, when calculating the mass of solute CuSO4, the mass of the aqueous crystal CuSO4?5H2O should be converted to the mass of anhydrous CuSO4, and the mass of the crystal water contained in 25 grams of CuSO4?5H2O should be added when calculating the mass of the solvent.

The mass of the solute CuSO4 in this saturated copper sulphate solution is:

The quality of the solvent is:

The solubility of CuSO4 at 20C is:

A: The solubility of CuSO4 at 20C is 20 grams.

The Mass Concentration is calculated according to the definition of the Mass Percentage Concentration, which calculates the mass percentage concentration of a solution, using the following conversions:

The following formula is also applied to calculate the mass percentage concentration:

The mass of the solution = the mass of the solute + the mass of the solvent

When the density d and volume v of the solution are known, the mass of the solution and solute can be calculated, and the following formula can be applied:

Mass of solution (g) = density of solution d (g/cm3) xv (cm3)

Mass of solute (g) = d (g/cm3) x v (cm3) x mass percentage concentration

There are three basic types of calculations regarding the mass percentage concentration of a solution:

1 The mass of the solute and solvent is known, and the mass percentage concentration of the solution is calculated

Example: Dissolve 2 grams of solid NaOH in 18 grams of water and calculate the mass percentage concentration of the resulting solution.

In this NaOH solution, the mass of the solute is 2 grams, the mass of the solvent is 18 grams, and the mass of the solution is:

2 grams + 18 grams = 20 grams

The mass percentage concentration of the sodium hydroxide solution is:

[Solution]

A: The NaOH solution is 10% by mass.

2. The mass percentage concentration of the solution is known, and the mass of the solute and solvent is calculated

Example: To prepare 500 grams of a 2% Agno3 solution with a 2% mass concentration of 500 grams, how many grams of Agno3 should be weighed and how many milliliters of water should be added?

[Analysis] multiply the mass of the agno3 solution by the mass percentage concentration to obtain the mass of the solute agno3, subtract the mass of the agno3 solute from the mass of the agno3 solution to obtain the mass of the water that should be added, and divide the mass of the water by the density of the water to calculate the volume of water added.

[Solution] should be called the mass of agno3 as:

500 g x 2% = 10 g

The quality of the water that should be added is:

500 g - 10 g = 490 g

The volume of water added is:

490 g ÷ 1 g / ml = 490 ml

Answer: To prepare 500 grams of 2% Agno3 solution, 310 grams of AgNO3 should be weighed and 490 ml of water should be added.

3. Calculation of solution dilution

Example: To prepare dilute sulfuric acid with a mass percentage concentration of 20%, a solution density of 1.14 g/cm3, and a volume of 500 ml, how many milliliters of concentrated sulfuric acid with a mass percentage concentration of 98% and a density of 1.84 g/cm3 are required?

During the dilution of the solution, the volume and concentration of the solution will change, but the quality of the solute contained in the solution will always remain the same. namely

The mass of the solute contained in the original solution = the mass of the solute contained in the diluted solution

Using this rule, all types of solution dilution calculations can be solved correctly.

[Solution] Let the amount of V ml of concentrated sulfuric acid be taken, then the mass of the solute contained in the concentrated sulfuric acid is:

vx1.84x98%

The mass of the solute contained in the dilute sulfuric acid solution is:

The following equation can be listed according to the fact that the solute mass remains constant during the dilution of the solution:

vx1.84x98%=%

v=63.22 (ml)

The quality of the water required is:

=383.67 (g)

The volume of water required is:

[Note] When performing a solution dilution (or mixing) calculation, the mass of the solution and the solvent (or another solution) can be added, but their volumes cannot be added.

Answer: It is necessary to measure 63.22 ml of concentrated sulfuric acid and 383.67 ml of water.

Solubility vs. Mass Percentage Concentration Solubility is the relationship between solvent mass and solute mass in a saturated solution at a certain temperature, and mass percentage concentration is the relationship between solution mass and solute mass in any solution (saturated or unsaturated). The conversion of solubility to mass percentage concentration is that for a saturated solution at a certain temperature, the denominator of solubility is the solvent mass, and the denominator of mass percentage concentration is the solution mass.

1 Known solubility is calculated as a mass percentage concentration

Example: At 30C, the solubility of CuSO4 is 25 grams, what is the mass percentage concentration of the copper sulfate saturated solution at that temperature?

The solubility of CuSO4 at 30C is 25 grams, indicating that when 25 grams of CuSO4 is dissolved in 100 grams of water, a saturated solution of copper sulfate at this temperature can be formed. Based on the mass of the solute CuSO4 being 25 grams and the solution being 125 grams, the mass percentage concentration of the copper sulfate saturated solution can be calculated.

[Solution]

A: The mass percentage concentration of CuSO4 saturated solution at 30C is 20%.

2. The mass percentage concentration of the saturated solution is known, and the solubility of the solute at this temperature is calculated

Example: If the mass percentage concentration of a copper sulfate saturated solution is 20% at 30C, calculate how many grams of CuSO4 solubility at that temperature?

[Analysis] From the mass percentage concentration of copper sulfate saturated solution at 30C is 20%, it can be seen that when the mass of the solution is 100 grams, the mass of the solute is 20 grams, and the mass of the solvent is:

Solvent mass = solution mass - solute mass

= 100 g - 20 g = 80 g

Knowing the quality of the solute and the mass of the solvent in the saturated solution, the solubility of CuSO4 at 30C can be calculated.

The solubility of CuSO4 at 30C is:

Answer: The solubility of CuSO4 at 30C is 25 grams.

Solution crystallization calculation: correctly use the concepts of solubility and concentration to calculate the mass of solute and solvent in the original solution; according to the conditions given by the question, find out the mass relationship between the solute and solvent in the solution after crystallization; grasp that the solution after crystallization must be a saturated solution of the substance at this temperature. Grasp the above three points, for the solution crystallization calculation under various conditions, you can use a simple method to find the quality of the precipitated crystal.

1. Calculation of crystals precipitated by cooling of saturated solution

Example: The solubility of KNO3 is known to be 140 g at 72C and 30 g at 18C. There are 48 grams of 72C saturated KNo3 solution, how many grams of crystals are precipitated when cooled to 18C?

[Analysis] from the solubility of 72CKNO3 is 140 grams, it can be known that KNO3140 grams can be dissolved in 100 grams of water, according to the sum of solvent mass and solute mass is equal to the mass of the solution, it can be known that in 240 grams of potassium nitrate saturated solution contains KNO3140 grams, when cooled to 18C, KNO3 (140-30) grams can be precipitated, using the proportional relationship, you can calculate the mass of the precipitated crystals when the 48 grams of KNO3 saturated solution is cooled to 18C at 72C.

[Solution] Let the mass of the precipitated crystals of 48 grams of KNO3 saturated solution at 72C and 18 grams when cooled to 18 C.

x = 22 grams

A: Cool to 18C to precipitate crystals 22 g.

2. Calculation of crystals analyzed by evaporating water in saturated solution

[Example] If the solubility of NH4Cl is known to be 60 grams at 70C, if 320 grams of 70C NH4Cl saturated solution is heated and evaporated with 40 grams of water, and then cooled to 70C, how many grams of NH4Cl crystals will be precipitated?

At a certain temperature, the mass of the crystal precipitated by the evaporation solvent of the saturated solution is the mass of the evaporated solvent that can dissolve the solute when it reaches saturation. In this way, the mass of the crystals analyzed by the evaporated water can be calculated simply by using the relationship between the solvent mass and the solute mass.

[Solution] Let the evaporation of 40 grams of water precipitate NH4Cl crystals as x grams

x = 24 grams

Answer: Evaporation of 40 grams of water can precipitate 24 grams of NH4Cl crystals.

3. Crystallization calculation of evaporation of water and reduction of temperature of unsaturated solution

[Example] How many grams of KNO3 can be precipitated when 250 grams of 20% KNO3 solution is evaporated at 40C and 150 grams of water is evaporated, and how many grams of KNO3 can be precipitated when it is still maintained at 40C?

[Analysis] in 250 g of 20% KNo3 solution, the mass containing solute KNO3 is:

250 g x 20% = 50 g

The qualities containing the solvent are:

250 g - 50 g = 200 g

From the solubility curve, the solubility of 40CkNo3 was 64 grams, and the solubility of 10CkNuo3 was 20 grams, which showed that the 20% KnO3 solution at 40C was an unsaturated solution. For the crystallization calculation of evaporated water in an unsaturated solution, the mass of the solute contained in the original solution is subtracted from the mass of the dissolved solute in the remaining water.

[Solution] In 250 grams of 20% KNO3 solution, containing 350 grams of solute KNO, 200 grams of solvent, and after evaporating 150 grams of water, the mass of the remaining water in the solution is:

200 g - 150 g = 50 g

50 g of water dissolves KNO3 at 40 C as x g

x = 32 grams

The mass of KNO3 that can be precipitated at 40C is:

50 grams - 32 grams = 18 grams

When the temperature is continued to 10C, the mass of the precipitated KNo3 is Y grams.

y = 22 grams

The mass of the co-precipitated KNO3 is:

18 g + 22 g = 40 g

Answer: 250 grams of 20% KNO3 solution evaporated 150 grams of water and cooled to 40C to precipitate KNO 318 grams, and continued to cool down to 10C to precipitate KNO 322 grams and a total of 340 grams of KNO.

4Calculation of precipitated aqueous crystals

Example: 500 grams of copper sulfate solution with a concentration of 15%, when 300 grams of copper sulfate solution is heated and evaporated and cooled to 20 C, how many grams of blue alum crystals (CuSO4?5H2O) can be precipitated (the solubility of CuSO4 at 20C is 20 grams)

The solute in the copper sulfate solution in this question is CuSO4, and the precipitated crystal is CuSO4?5H2O. When calculating the mass of solute contained in the solution after crystallization, the mass of the precipitated aqueous crystal CuSO4?5H2O should be converted to the mass of the anhydrous crystal CuSO4, and then the total solute mass in the original solution should be subtracted from the mass of the anhydrous crystal CuSO4.

The mass of the solute CuSO4 contained in the original copper sulfate solution is:

500 g x 15% = 75 g

The quality of the solvent H2O contained is:

500 g - 75 g = 425 g

When 300 grams of water are evaporated and cooled to 20 C to precipitate CuSO4?5H2O with a mass of x grams, the mass of the solute CuSO4 in the solution after crystallization is:

The mass of water in the solution after precipitation of x grams of CuSO4?5H2O is:

According to the fact that the solution after the precipitation of x grams of CuSO4?5H2O is a saturated solution at this temperature, the following proportional relationship can be listed

x = 88 grams

Answer: CuSO4?5H2O88 g can be precipitated.

The chemical equation for calculating the mass of reactants or products according to the chemical equation can express the mass relationship between each reactant and the product in the chemical reaction, which is the basis for chemical calculation. According to the calculation of the chemical equation, the following three steps are generally followed:

The first step is to carefully review the question and write the correct chemical equation according to the meaning of the question.

The second step is to analyze the problem in depth, fully recognize the known conditions and the unknown questions, find out the connection between the known conditions and the unknown questions, and write them below the chemical formulas of the relevant substances in the chemical equations.

The third step is to list the proportional formula according to the mass relationship of each substance in the chemical equation and complete the calculation of the problem. It should be done that the solution should be based on evidence, the solution should be standardized, the writing should be neat, and the answer should be accurate.

1. Calculate the mass of the reactant or product

Example: Phosphorus pentoxide is produced by burning red phosphorus, and there are 0.62 grams of red phosphorus, how many grams of phosphorus pentoxide can be produced if the reaction is sufficient, and how many liters of oxygen is required under standard conditions? (the density of oxygen under standard conditions is 1.429 g/L)

The chemical equation [analytical] represents the mass relationship of each substance

4p+5o22p2o5

4x31 grams, 5 x 32 grams, 2 x 142 grams

According to the chemical equation of complete combustion of red phosphorus, it can be seen that 160 grams of oxygen is consumed to complete combustion of 124 grams of red phosphorus, and 284 grams of phosphorus pentoxide are generated. The problem gives that the mass of red phosphorus is 0.62 grams, and the proportional relationship can be used to calculate the mass of phosphorus pentoxide and the mass of oxygen consumed.

[Solution] Suppose 0.62 grams of red phosphorus is completely burned to produce x grams of phosphorus pentoxide, and Y grams of oxygen are consumed.

4p+5o22p2o5

x = 1.42 grams

y = 0.80 grams

The volume of 0.80 grams of oxygen in standard conditions is:

Answer: The complete combustion of 0.62 grams of red phosphorus requires 0.56 liters of oxygen under standard conditions to produce 1.42 grams of phosphorus pentoxide.

2. Calculate the purity of the substance

Example: The purity of iron powder containing 7.0 grams of rust (mainly Fe2O3) is fully reacted with a sufficient amount of dilute hydrochloric acid to produce 0.2 grams of hydrogen.

Iron powder is replaced with dilute hydrochloric acid to produce ferrous chloride and hydrogen

fe+2hcl=fecl2+h2↑

The mass of Fe can be calculated according to the mass of H2 generated, and the purity of the iron powder can be obtained by dividing the mass of Fe by the mass of impure iron powder and multiplying it by 100%.

[Solution] let the iron powder contain fe x grams

fe+2hcl=h2↑+fecl2

x = 5.6 grams

The purity of this iron powder is:

A: The purity of this iron powder is 80%.

Calculation of the mass percentage concentration of reactants or products according to a chemical equation This calculation is characterized by combining the mass percentage concentration with the calculation of the chemical equation. This type of calculation is performed by calculating the mass of the reactant or product according to the chemical equation and then dividing by the mass of the respective solution to calculate the mass percentage concentration of the reactant or product.

[Example] 3 grams of Al powder containing Al2O3, the percentage content of Al is known to be 90%, and the aluminum powder reacts completely with 100 grams of dilute sulfuric acid, what is the mass percentage concentration of dilute sulfuric acid calculated? What is the mass percentage concentration of the resulting Al2(SO4)3 solution?

[Analysis] Put Al powder containing Al2O3 into dilute sulfuric acid, and the chemical reaction that occurs is:

2al+3h2so4=al2(so4)3+3h2↑

al2o3+3h2so4=al2(so4)3+3h2o

The mass of aluminium (Al) contained in 3 grams of aluminium powder (containing Al2O3) is:

3 grams x 90% = 2.7 grams

The masses containing aluminum oxide (Al2O3) are:

3 grams x (1-90%) = 0.3 grams

The sum of the masses of H2SO4 reacting with 2.7 g of Al and H2SO4 reacting with 0.3 g of Al2O3 is the solute mass in dilute H2SO4, from which the mass percentage concentration of dilute sulfuric acid can be calculated.

The sum of the mass of Al2(SO4)3 formed by the reaction of 2.7 g of Al with H2SO4 and 0.3 g of Al2(SO4)3 formed by the reaction of Al2O3 with H2SO4 is the mass of Al2(SO4)3 solute in Al2(SO4)3 solution, from which the mass percentage concentration of the resulting Al2(SO4)3 solution can be calculated.

Let the H2SO4 reacted with 2.7 g of Al be X1 g to produce 3Y1 g of Al2 (SO4) and produce 1 g of H2Z1, and the H2SO4 reacted with 0.3 g of Al2O3 is X2 g to produce 3Y2 g of Al2 g and generate Z2 g of water.

2al+3h2so4=al2(so4)3+3h2↑

x1 = 14.7 grams

y1 = 17.1 grams

z1 = 0.3 grams

al2o3+3h2so4=al2(so4)3+3h2o

x2 = 0.86 grams

y2 = 1.01 grams

z2 = 0.16 grams

The mass percentage concentration of dilute sulfuric acid is:

The mass of the generated Al2(SO4)3 is:

The mass of Al2(SO4)3 solution is:

The mass of 18.11 g + () g + 0.16 g = 102.71 g Al2(SO4)3 solution can also be calculated by the sum of the mass of aluminum sheet containing Al2O3 and dilute sulfuric acid, and then subtracting the mass of hydrogen produced

3 g + 100 g - 0.3 g = 102.7 g

The mass percentage concentration of Al2(SO4)3 solution is:

Answer: The mass percentage concentration of dilute H2SO4 is 15.56%, and the mass percentage concentration of Al2(SO4)3 solution is 17.63%.

The mass relationship between the reactants and the products expressed by the chemical equation according to the chemical equation refers to the quantitative relationship between the pure substance, and the mass of the pure substance must be substituted into the equation when the chemical equation is used to calculate. If the question gives an impure substance, the mass of the impure substance should be converted to the mass of the pure substance before calculation, and if the purity of the substance is calculated, the mass of the pure substance should be divided by the mass of the impure substance, and then expressed as a percentage.

[Example] 15.0 grams of sodium sulfate containing insoluble impurities are dissolved in 85.8 grams of water, filtered to remove insoluble impurities, 10.0 grams of filtrate, 20.8 grams of 10% BaCl2 solution are added, and the percentage content of Na2SO4 in impure sodium sulfate is calculated.

The chemical reaction between sodium sulfate and BaCl2 solution is:

na2so4+bacl2=baso4↓+2nacl

The mass of Na2SO4 contained in 10.0 grams of filtrate can be calculated from the mass of BaCl2 consumed, and then the mass percentage concentration of Na2SO4 solution can be calculated. Then, starting from the mass percentage concentration of Na2SO4 solution, the mass of Na2SO4 contained in 15 grams of sample was calculated, and the percentage content of Na2SO4 in impure sodium sulfate was calculated.

[Solution] Let the Na2SO4 in the 10.0 g filtrate be x grams, and the Na2SO4 in the 15.0 g sample be Y grams.

na2so4+bacl2=baso4↓+2nacl

x = 1.42 grams

The mass percentage concentration of the filtrate is:

The mass of Na2SO4 contained in the sample is as follows:

y = 14.2 grams

The percentage content of Na2SO4 in impure sodium sulfate is:

Answer: Impure sodium sulfate contains %. The questioner commented: Thank you, although some of them are not very useful, thank you

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