How to find the mass fraction of an element in a substance. How to calculate the mass fraction of an element in a substance

Instructions

Determine the chemical form of the substance whose mass fractions of elements need to be found. Take the periodic table of Mendeleev and find in it the cells of elements corresponding to the atoms that make up the molecule of a given substance. In the cell, find the mass number of each such element. If the found value of the mass number element fractional, round it up to the nearest .

In the case where atoms of the same type occur several times in a molecule, multiply their atomic mass by this number. Add the masses of all the elements that make up the molecule to get the value in atomic mass units. For example, if you need to find the mass of a salt molecule, which is sulfate (Na2SO4), determines the atomic mass of sodium Ar(Na) = 23, sulfur Ar(S) = 32 and Ar(O) = 16. Since the molecule contains 2 sodium, take the value 23*2=46 for it, and 16*4=64, which has 4 atoms. Then the mass of the molecule will be sodium sulfate and will be Мr(Na2SO4)=46+32+64=142.

To calculate the mass fractions of the elements that make up the molecule of a given substance, find the ratio of the masses of the atoms included in the molecule of the substance to the mass of the molecule, and multiply the result by 100%. For example, if we consider sodium sulfate Na2SO4, calculate the mass fractions of its elements in this way: - the mass fraction of sodium will be ω(Na)= 23 2 100%/142=32.4%;
- mass fraction of sulfur will be ω(S)= 32 100%/142=22.5%;
- the mass fraction of oxygen will be ω(О)= 16 4 100%/142=45.1%.

Mass fractions show the relative elements in a given molecule of a substance. Check the correctness of the calculation by adding the mass fractions of the substance. Their sum should be 100%. In the example under consideration, 32.4%+22.5%+45.1%=100%, the calculation is made.

It is perhaps impossible to find an element as essential for life as oxygen. If a person can live without food for several weeks, without water for several days, then without oxygen - only a few minutes. This substance is widely used in various fields of industry, including the chemical industry, and also as a component of rocket fuel (oxidizer).

Instructions

Often there is a need to determine the mass of oxygen located in some closed volume, or as a result of a chemical reaction. For example: 20 grams of permanganate were subjected to thermal decomposition, the reaction was completed. How many grams of oxygen were released?

First of all, remember that potassium - aka - has the chemical formula KMnO4. When heated, it decomposes, forming potassium manganate - K2MnO4, the main one - MnO2, and O2. Having written the reaction equation and selected the coefficients, you get:

2KMnO4 = K2MnO4 + MnO2 + O2

Given that the approximate molecular weight of two molecules of potassium permanganate is 316, and the molecular weight of an oxygen molecule is, respectively, 32, by solving the proportion, calculate:

20 * 32 /316 = 2,02
That is, with the thermal decomposition of 20 grams of potassium permanganate, approximately 2.02 grams of oxygen are obtained. (Or rounded 2 grams).

Or, for example, it is necessary to determine the mass of oxygen located in a closed volume if its temperature and pressure are known. Here the universal Mendeleev-Clapeyron equation, or in other words the “equation of state of an ideal gas,” comes to the rescue. It looks like this:

PVm = MRT
P is the gas pressure,

V is its volume,

m is its molar mass,

M – mass,

R – universal gas constant,

T – temperature.

You see that the required value, that is, the mass of gas (oxygen), after bringing all the initial data into one system of units (pressure - , temperature - in degrees Kelvin, etc.), can be easily calculated using the formula:

Of course, real oxygen is not the ideal gas to describe which this equation was introduced. But at pressure and temperature values ​​close to , the deviations of the calculated values ​​from the actual ones are so insignificant that they can be safely neglected.

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What is mass fraction element? From the name itself you can understand that this is a quantity indicating the ratio of mass element, included in the composition of the substance, and the total mass of this substance. It is expressed in fractions of a unit: percent (hundredths), ppm (thousands), etc. How can you calculate the mass of something? element?

Instructions

For clarity, consider the well-known carbon, without which there would be no . If carbon is a substance (for example), then its mass share can be safely taken as one or 100%. Of course, diamond also contains impurities of other elements, but in most cases, in such small quantities that they can be neglected. But in carbon modifications such as or, the impurity content is quite high, and neglect is unacceptable.

If carbon is part of a complex substance, you must proceed as follows: write down the exact formula of the substance, then, knowing the molar masses of each element included in its composition, calculate the exact molar mass of this substance (of course, taking into account the “index” of each element). After that, determine the mass share by dividing the total molar mass element on the molar mass of the substance.

For example, you need to find the mass share carbon in acetic acid. Write the formula of acetic acid: CH3COOH. To make calculations easier, convert it to the form: C2H4O2. The molar mass of this substance is the sum of the molar masses of the elements: 24 + 4 + 32 = 60. Accordingly, the mass fraction of carbon in this substance is calculated as follows: 24/60 = 0.4.

If you need to calculate it as a percentage, respectively, 0.4 * 100 = 40%. That is, each acetic acid contains (approximately) 400 grams of carbon.

Of course, the mass fractions of all other elements can be found in a completely similar way. For example, the mass in the same acetic acid is calculated as follows: 32/60 = 0.533 or approximately 53.3%; and the mass fraction of hydrogen is 4/60 = 0.666 or approximately 6.7%.

Sources:

• mass fractions of elements

A chemical formula is a record made using generally accepted symbols that characterizes the composition of the molecule of a substance. For example, the formula of the well-known sulfuric acid is H2SO4. It can be easily seen that each sulfuric acid molecule contains two hydrogen atoms, four oxygen atoms and one atom. It must be understood that this is only an empirical formula; it characterizes the composition of the molecule, but not its “structure,” that is, the arrangement of atoms relative to each other.

You will need

• - Mendeleev table.

Instructions

First, find out the elements that make up the substance and theirs. For example: what will be the nitric oxide level? Obviously, this molecule contains two elements: nitrogen and . Both of them are gases, that is, pronounced gases. So what valence do nitrogen and oxygen have in this compound?

Remember a very important rule: non-metals have higher and lower valencies. The highest corresponds to the group number (in this case, 6 for oxygen and 5 for nitrogen), and the lowest corresponds to the difference between 8 and the group number (that is, the lowest valence for nitrogen is 3, and for oxygen is 2). The only exception to this rule is fluorine, which in all its forms exhibits one valency equal to 1.

So what valency – higher or lower – do nitrogen and oxygen have? Another rule: in compounds of two elements, the one that is located to the right and higher in the Periodic Table exhibits the lowest valence. It is quite obvious that in your case it is oxygen. Therefore, in combination with nitrogen, oxygen has a valence of 2. Accordingly, nitrogen in this compound has a higher valence of 5.

Now remember valence itself: this is the ability of an atom of any element to attach to itself a certain number of atoms of another element. Each nitrogen atom in this compound contains 5 oxygen atoms, and each oxygen atom contains 2 nitrogen atoms. What is nitrogen? That is, what indices does each element have?

Another rule will help answer this question: the sum of the valences of the elements included in the compound must be equal! What is the least common multiple of the numbers 2 and 5? Naturally, 10! By dividing it into the valence values ​​of nitrogen and oxygen, you will find the indices and the final formula compounds: N2O5.

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The mass fraction of a substance shows its content in a more complex structure, for example, in an alloy or mixture. If the total mass of a mixture or alloy is known, then knowing the mass fractions of the constituent substances, their masses can be found. You can find the mass fraction of a substance by knowing its mass and the mass of the entire mixture. This value can be expressed in fractions or percentages.

You will need

• scales;
• periodic table of chemical elements;
• calculator.

Instructions

Determine the mass fraction of the substance that is in the mixture through the masses of the mixture and the substance itself. To do this, use a scale to determine the masses that make up the mixture or. Then fold them. Take the resulting mass as 100%. To find the mass fraction of a substance in a mixture, divide its mass m by the mass of the mixture M, and multiply the result by 100% (ω%=(m/M)∙100%). For example, 20 g of table salt is dissolved in 140 g of water. To find the mass fraction of salt, add the masses of these two substances M = 140 + 20 = 160 g. Then find the mass fraction of the substance ω% = (20/160)∙100% = 12.5%.

If you need to find the mass fraction of an element in a substance with a known formula, use the periodic table of elements. Using it, find the atomic masses of the elements that are in the substance. If one is in the formula several times, multiply its atomic mass by that number and add the results. This will be the molecular weight of the substance. To find the mass fraction of any element in such a substance, divide its mass number in a given chemical formula M0 by the molecular mass of a given substance M. Multiply the result by 100% (ω%=(M0/M)∙100%).

Solution called a homogeneous mixture of two or more components.

The substances by mixing which produce a solution are called components.

Among the components of the solution there are solute, which may be more than one, and solvent. For example, in the case of a solution of sugar in water, the sugar is the solute and the water is the solvent.

Sometimes the concept of solvent can be applied equally to any of the components. For example, this applies to those solutions that are obtained by mixing two or more liquids that are ideally soluble in each other. So, in particular, in a solution consisting of alcohol and water, both alcohol and water can be called a solvent. However, most often in relation to aqueous solutions, the solvent is traditionally called water, and the solute is the second component.

As a quantitative characteristic of the composition of a solution, the concept most often used is mass fraction substances in solution. The mass fraction of a substance is the ratio of the mass of this substance to the mass of the solution in which it is contained:

Where ω (in-va) – mass fraction of the substance contained in the solution (g), m(v-va) – mass of the substance contained in the solution (g), m(r-ra) – mass of the solution (g).

From formula (1) it follows that the mass fraction can take values ​​from 0 to 1, that is, it is a fraction of unity. In this regard, the mass fraction can also be expressed as a percentage (%), and it is in this format that it appears in almost all problems. The mass fraction, expressed as a percentage, is calculated using a formula similar to formula (1) with the only difference being that the ratio of the mass of the dissolved substance to the mass of the entire solution is multiplied by 100%:

For a solution consisting of only two components, the mass fraction of solute ω(s.v.) and the mass fraction of solvent ω(solvent) can be calculated accordingly.

The mass fraction of the solute is also called solution concentration.

For a two-component solution, its mass is the sum of the masses of the solute and the solvent:

Also, in the case of a two-component solution, the sum of the mass fractions of the solute and the solvent is always 100%:

It is obvious that, in addition to the formulas written above, you should also know all those formulas that are directly mathematically derived from them. For example:

It is also necessary to remember the formula connecting the mass, volume and density of a substance:

m = ρ∙V

and you also need to know that the density of water is 1 g/ml. For this reason, the volume of water in milliliters is numerically equal to the mass of water in grams. For example, 10 ml of water has a mass of 10 g, 200 ml - 200 g, etc.

In order to successfully solve problems, in addition to knowledge of the above formulas, it is extremely important to bring the skills of their application to automaticity. This can only be achieved by solving a large number of different problems. Problems from real Unified State Examinations on the topic “Calculations using the concept of “mass fraction of a substance in solution”” can be solved.

Examples of problems involving solutions

Example 1

Calculate the mass fraction of potassium nitrate in a solution obtained by mixing 5 g of salt and 20 g of water.

Solution:

The solute in our case is potassium nitrate, and the solvent is water. Therefore, formulas (2) and (3) can be written respectively as:

From the condition m(KNO 3) = 5 g, and m(H 2 O) = 20 g, therefore:

Example 2

What mass of water must be added to 20 g of glucose to obtain a 10% glucose solution.

Solution:

From the conditions of the problem it follows that the solute is glucose and the solvent is water. Then formula (4) can be written in our case as follows:

From the condition we know the mass fraction (concentration) of glucose and the mass of glucose itself. Having designated the mass of water as x g, we can write, based on the formula above, the following equation equivalent to it:

Solving this equation we find x:

those. m(H 2 O) = x g = 180 g

Answer: m (H 2 O) \u003d 180 g

Example 3

150 g of a 15% solution of sodium chloride was mixed with 100 g of a 20% solution of the same salt. What is the mass fraction of salt in the resulting solution? Please indicate your answer to the nearest integer.

Solution:

To solve problems for preparing solutions, it is convenient to use the following table:

where m r.v. , m solution and ω r.v. - values ​​of the mass of the dissolved substance, the mass of the solution and the mass fraction of the dissolved substance, respectively, individual for each of the solutions.

From the condition, we know that:

m (1) solution = 150 g,

ω (1) r.v. = 15%,

m (2) solution = 100 g,

ω (1) r.v. = 20%,

Let's insert all these values ​​into the table, we get:

We should remember the following formulas necessary for calculations:

ω r.v. = 100% ∙ m r.v. /m solution, m r.v. = m solution ∙ ω solution /100% , m solution = 100% ∙ m solution /ω r.v.

Let's start filling out the table.

If only one value is missing from a row or column, it can be counted. The exception is the line with ω r.v., knowing the values ​​in two of its cells, the value in the third cannot be calculated.

Only one cell in the first column is missing a value. So we can calculate it:

m (1) r.v. = m (1) solution ∙ ω (1) solution /100% = 150 g ∙ 15%/100% = 22.5 g

Similarly, we know the values ​​in two cells of the second column, which means:

m (2) r.v. = m (2) r-ra ∙ ω (2) r.v. /100% = 100 g ∙ 20%/100% = 20 g

Let's enter the calculated values ​​into the table:

Now we know two values ​​in the first line and two values ​​in the second line. This means we can calculate the missing values ​​(m (3)r.v. and m (3)r-ra):

m (3)r.v. = m (1)r.v. + m (2)r.v. = 22.5 g + 20 g = 42.5 g

m (3) solution = m (1) solution + m (2) solution = 150 g + 100 g = 250 g.

Let's enter the calculated values ​​into the table and get:

Now we have come close to calculating the desired value of ω (3)r.v. . In the column where it is located, the contents of the other two cells are known, which means we can calculate it:

ω (3)r.v. = 100% ∙ m (3)r.v. / m (3) solution = 100% ∙ 42.5 g / 250 g = 17%

Example 4

50 ml of water was added to 200 g of 15% sodium chloride solution. What is the mass fraction of salt in the resulting solution. Please indicate your answer to the nearest hundredth of _______%

Solution:

First of all, we should pay attention to the fact that instead of the mass of added water, we are given its volume. Let's calculate its mass, knowing that the density of water is 1 g/ml:

m ext. (H 2 O) = V ext. (H 2 O) ∙ ρ (H2O) = 50 ml ∙ 1 g/ml = 50 g

If we consider water as a 0% sodium chloride solution containing 0 g of sodium chloride, the problem can be solved using the same table as in the example above. Let's draw a table like this and insert the values ​​we know into it:

There are two known values ​​in the first column, so we can calculate the third:

m (1)r.v. = m (1)r-ra ∙ ω (1)r.v. /100% = 200 g ∙ 15%/100% = 30 g,

In the second line, two values ​​are also known, which means we can calculate the third:

m (3) solution = m (1) solution + m (2) solution = 200 g + 50 g = 250 g,

Let's enter the calculated values ​​into the appropriate cells:

Now two values ​​in the first line have become known, which means we can calculate the value of m (3)r.v. in the third cell:

m (3) r.v. = m (1) r.v. + m (2)r.v. = 30 g + 0 g = 30 g

ω (3)r.v. = 30/250 ∙ 100% = 12%.

What is mass fraction in chemistry? Do you know the answer? How to find the mass fraction of an element in a substance? The calculation process itself is not at all that complicated. Do you still experience difficulties in such tasks? Then luck smiled on you, you found this article! Interesting? Then read quickly, now you will understand everything.

What is mass fraction?

So, first, let's find out what mass fraction is. Any chemist will answer how to find the mass fraction of an element in a substance, since they often use this term when solving problems or while in the laboratory. Of course, because calculating it is their daily task. To obtain a certain amount of a particular substance in laboratory conditions, where accurate calculations and all possible options for the outcome of reactions are very important, you need to know just a couple of simple formulas and understand the essence of the mass fraction. That's why this topic is so important.

This term is represented by the symbol “w” and is read as “omega”. It expresses the ratio of the mass of a given substance to the total mass of a mixture, solution or molecule, expressed as a fraction or percentage. Formula for calculating mass fraction:

w = m substance / m mixture.

Let's transform the formula.

We know that m=n*M, where m is mass; n is the amount of substance expressed in mole units; M is the molar mass of the substance, expressed in grams/mol. Molar mass is numerically equal to molecular mass. Only molecular weight is measured in atomic mass units or a. e. m. This unit of measurement is equal to one twelfth of the mass of the carbon nucleus 12. The value of molecular mass can be found in the periodic table.

The amount of substance n of the desired object in a given mixture is equal to the index multiplied by the coefficient for a given compound, which is very logical. For example, to calculate the number of atoms in a molecule, you need to find out how many atoms of the desired substance are in 1 molecule = index, and multiply this number by the number of molecules = coefficient.

You shouldn’t be afraid of such cumbersome definitions or formulas; they contain a certain logic, and once you understand it, you don’t even have to learn the formulas themselves. The molar mass M is equal to the sum of the atomic masses A r of a given substance. Recall that atomic mass is the mass of 1 atom of a substance. That is, the original mass fraction formula:

w = (n substance *M substance)/m mixture.

From this we can conclude that if a mixture consists of one substance, the mass fraction of which must be calculated, then w = 1, since the mass of the mixture and the mass of the substance are the same. Although a priori a mixture cannot consist of one substance.

So, we’ve sorted out the theory, but how to find the mass fraction of an element in a substance in practice? Now we will show and tell you everything.

Checking the learned material. Easy level problem

Now we will analyze two tasks: easy and medium level. Read on!

It is necessary to find out the mass fraction of iron in the iron sulfate molecule FeSO 4 * 7 H 2 O. How to solve this problem? Let's look at the solution next.

Solution:

Let's take 1 mol FeSO 4 * 7 H 2 O, then we find out the amount of iron by multiplying the iron coefficient by its index: 1 * 1 = 1. Given 1 mole of iron. Let's find out its mass in the substance: from the value in the periodic table it is clear that the atomic mass of iron is 56 a. e.m. = 56 grams/mol. In this case A r =M. Therefore, m iron = n*M = 1 mol* 56 grams/mol = 56 g.

Now we need to find the mass of the entire molecule. It is equal to the sum of the masses of the starting substances, that is, 7 mol of water and 1 mol of iron sulfate.

m= (n water * M water) + (n ferrous sulfate * M ferrous sulfate) = (7 mol*(1*2+16) gram/mol) + (1 mol* (1 mol*56 gram/mol+1 mol*32 grams/mol + 4 mol*16 grams/mol) = 126+152=278 g.

All that remains is to divide the mass of iron by the mass of the compound:

w=56g/278g=0.20143885~0.2=20%.

Answer: 20%.

Intermediate level problem

Let's solve a more complex problem. 34 g of calcium nitrate is dissolved in 500 g of water. We need to find the mass fraction of oxygen in the resulting solution.

Solution

Since when Ca(NO 3) 2 interacts with water, only the dissolution process occurs, and no reaction products are released from the solution, the mass of the mixture is equal to the sum of the masses of calcium nitrate and water.

We need to find the mass fraction of oxygen in the solution. Please note that oxygen is contained in both the solute and the solvent. Let's find the amount of the required element in water. To do this, let's calculate moles of water using the formula n=m/M.

n water =500 g/(1*2+16) gram/mol=27.7777≈28 mol

From the formula of water H 2 O we find that the amount of oxygen = the amount of water, that is, 28 mol.

Now let's find the amount of oxygen in dissolved Ca(NO 3) 2. To do this, we find out the amount of the substance itself:

n Ca(NO3)2 =34 g/(40*1+2*(14+16*3)) gram/mol≈0.2 mol.

n Ca(NO3)2 is to n O as 1 to 6, as follows from the formula of the compound. This means n O = 0.2 mol*6 = 1.2 mol. The total amount of oxygen is 1.2 mol+28 mol=29.2 mol

m O = 29.2 mol*16 grams/mol=467.2 g.

m solution = m water + m Ca(NO3) 2 = 500 g + 34 g = 534 g.

All that remains is to calculate the mass fraction of a chemical element in a substance:

w O =467.2 g /534 g≈0.87=87%.

Answer: 87%.

We hope that we have clearly explained to you how to find the mass fraction of an element in a substance. This topic is not at all difficult if you understand it well. We wish you good luck and success in your future endeavors.