How to calculate the mass score in the third year of science, and eight methods for calculating the

Common question types and basic ideas:

1) Calculation of dilution of solution with water:

According to the equality of the mass of the solute before and after dilution, it can be calculated by the following formula: m (concentrated) a% (concentrated) = [m (concentrated) + m (water)] a% (dilute).

2) Mixing with solute solutions with different solute mass fractions:

Generally, it can be calculated using the following formula:

m (concentrated) a% (concentrated) + m (dilute) a% (rare) = m (mixed) a% (mixed).

3) The problem of finding the solute mass fraction in the solution obtained after the chemical reaction:

The mass of the solute was calculated by chemical reaction, and then the total mass of the solution was obtained by analyzing the relationship between the quantities, and the mass fraction of the solute in the solution after the reaction was calculated by using the formula.

The general method for finding the mass of the solution obtained after the reaction: the mass conservation method. Solution mass = m (reactants) - m (precipitation) - m (gas).

Typical example questions. Example 1: How many grams of water should be added to the existing 100g of peracetic acid solution with a solute mass fraction of 15% to be used in a solution with a solute mass fraction?

Solution: Let the quality of water to be added be x, 100g 15%=(100g+x).

x=900g

Answer: Slightly. Example 2: How many milliliters of 98% concentrated sulfuric acid and water are needed to prepare 2000g of 2000g of dilute sulfuric acid for rust removal?

Solution: Let the mass of concentrated sulfuric acid be x

2000g×20%=x×98%

x = volume of concentrated sulfuric acid =

Volume of water = (

The solute mass is divided by the solution mass and multiplied by 100

Introduction to the high school entrance examination] The calculation of mass fraction and chemical formula in the mixture of the high school entrance examination are the most difficult points in the chemical calculation of junior high school. However, if some calculation problems are solved according to the conventional ruler method, the process is not only cumbersome, the calculation is large, and it is easy to make mistakes. If we change our thinking perspective and adopt different hypothetical strategies, we can often simplify the complex and solve the problem cleverly.

This time I will share with you the method of calculating the quality score in 8, so if you haven't got it yet, hurry up and see it!

Chapter. 1. Extreme assumptions.

Extreme assumptions are to assume the composition of a mixture as a variety of extreme cases, and to calculate and analyze each extreme case to arrive at a correct judgment.

Example 1: A certain amount of charcoal is fully burned in a closed container containing nitrogen and oxygen gas to form CO and CO2, and the mass fraction of carbon in the mixed gas of CO, CO2 and N2 obtained after the reaction is measured to be 24%, then the mass fraction of nitrogen may be ().

Analysis: This problem is easier to solve by using the extreme assumption method, and the original gas mixture is divided into two cases for extreme assumptions.

1) Suppose the gas mixture contains only N2 and CO. Let the mass fraction of CO in the gas mixture be x, then 12 28 = 24% x

x=56%, then the mass fraction of N2 in the mixed gas is: 1—56%=44%.

2) Suppose the gas mixture contains only N2 and CO2. Let the mass fraction of CO2 in the gas mixture be y, then 12 44 = 24% y

y=88%, then the mass fraction of N2 in the mixed gas is: 1-88%=12%.

Since the gas mixture is actually composed of three gases, CO, CO2 and N2, the mass fraction of N2 in the gas mixture should be between 12% and 44%, so the option that meets the topic is B.

Chapter. 2. Median hypothesis.

The median hypothesis is to assume the quantity value of a pure substance in the mixture as the intermediate value, and use the median value as a reference to analyze and reason, so as to solve the problem skillfully.

Example 2: In a mixture containing only iron oxide (Fe2O3) and ferrous oxide (FeO), the mass fraction of iron oxide in the mixture is ().

Analysis: This problem is more complicated to calculate by conventional methods. From the chemical formula, it can be seen that the mass fraction of oxygen in iron oxide is about, and the mass fraction of oxygen in ferrous oxide is about.

Assuming that they are 50% by mass in the mixture, the mass fraction of the iron element in the mixture should be: (.

The mass fraction of the iron element in the mixture is .

(1)mycl2

caco32).

Accelerates dissolution. Prevent droplets from splashing.

3）agno3

4) A because the evaporated B may contain crystalline water, that is, it contains the mass of water, not exactly nano3.

5) The ma* principle is that for every Agno3 generated, one NaCl is required. So multiply the mass of a by the NaCl quotient of agno3. This is the mass fraction of NACL.

Multiply 100% by mnacl.

The other components do not contain nitrogen, CO(NH2)2---2N

60 28x

60/x=28/x=

Conservation of nitrogen, that is to say, the mass of nitrogen in urine = the mass of nitrogen in urea, the most direct way is to use the mass fraction of nitrogen in urine Mass fraction of nitrogen in urea Below, we will explain this topic step by step.

Assuming that the total mass of urine is m, then the mass of nitrogen in urine is, and all nitrogen is provided by urea, then the mass of urea is equal to the mass fraction of nitrogen in urea, so the mass fraction of urea in the urine is m (, in fact, it is the mass fraction of nitrogen in urea.

PS: Mass fraction of nitrogen in urea CO(NH2)2 = 14 * 2 60 * 100% =

The answer to this question.

The solute is divided by the solute plus the solvent and multiplied by one hundred percent.

Relative atomic mass.

ar=the actual mass of an atom 1/12 of the actual mass of a carbon atom.

The actual mass of 1 carbon atom is minus 26 times.

The relative molecular mass such as H2O water is.

mr（h2o）=2*ar(h)+ar(o)

Relative atomic mass ar=actual mass of an atom 1/12th of the actual mass of a carbon atom.

The actual mass of 1 carbon atom is minus 26 times.

The relative molecular mass such as H2O water is mr(H2O)=2*Ar(H)+Ar(O).

If it is a chemical reaction, the law of conservation of mass should be used.

If it is just a simple substance, then the physical formula mass = density and volume should be used.

It mainly depends on the rough calculation of the relative atomic mass of the element, and before high school, you only need to memorize the relative atomic mass of the first 20 elements.

Is it relative atomic mass?