Since the 17th century chemistry has ceased to be a descriptive science. Chemical scientists began to widely use the measurement of matter. The design of balances that make it possible to determine the masses of samples has been increasingly improved. For gaseous substances, in addition to mass, volume and pressure were also measured. The use of quantitative measurements made it possible to understand the essence of chemical transformations and determine the composition of complex substances.
As you already know, a complex substance contains two or more chemical elements. It is obvious that the mass of all matter is made up of the masses of its constituent elements. This means that each element accounts for a certain portion of the mass of the substance.
The mass fraction of an element is the ratio of the mass of this element in a complex substance to the mass of the entire substance, expressed in fractions of a unit (or as a percentage):
The mass fraction of an element in a compound is indicated by a Latin small letter w(“double-ve”) and shows the share (part of the mass) attributable to a given element in the total mass of the substance. This value can be expressed in fractions of a unit or as a percentage. Of course, the mass fraction of an element in a complex substance is always less than unity (or less than 100%). After all, a part of the whole is always smaller than the whole, just as a slice of an orange is smaller than the whole orange.
For example, mercury oxide contains two elements - mercury and oxygen. When heating 50 g of this substance, 46.3 g of mercury and 3.7 g of oxygen are obtained (Fig. 57). Let's calculate the mass fraction of mercury in a complex substance:
The mass fraction of oxygen in this substance can be calculated in two ways. By definition, the mass fraction of oxygen in mercury oxide is equal to the ratio of the mass of oxygen to the mass of the oxide:
Knowing that the sum of the mass fractions of elements in a substance is equal to one (100%), the mass fraction of oxygen can be calculated from the difference:
w(O) = 1 – 0.926 = 0.074,
w(O) = 100% – 92.6% = 7.4%.
In order to find the mass fractions of elements using the proposed method, it is necessary to conduct a complex and labor-intensive chemical experiment to determine the mass of each element. If the formula of a complex substance is known, the same problem can be solved much easier.
To calculate the mass fraction of an element, you need to multiply its relative atomic mass by the number of atoms ( n) of a given element in the formula and divide by the relative molecular weight of the substance:
For example, for water (Fig. 58):
Mr(H 2 O) = 1 2 + 16 = 18,
Task 1.Calculate the mass fractions of elements in ammonia, the formula of which is NH 3 .
Given:
substance ammonia NH 3.
Find:
w(N), w(H).
Solution
1) Calculate the relative molecular weight of ammonia:
Mr(NH 3) = A r(N) + 3 A r(H) = 14 + 3 1 = 17.
2) Find the mass fraction of nitrogen in the substance:
3) Let's calculate the mass fraction of hydrogen in ammonia:
w(H) = 1 – w(N) = 1 – 0.8235 = 0.1765, or 17.65%.
Answer. w(N) = 82.35%, w(H) = 17.65%.
Task 2.Calculate the mass fractions of elements in sulfuric acid having the formula H2SO4 .
Given:
sulfuric acid H 2 SO 4.
Find:
w(H), w(S), w(O).
Solution
1) Calculate the relative molecular weight of sulfuric acid:
Mr(H2SO4) = 2 A r(H)+ A r(S)+4 A r(O) = 2 1 + 32 + 4 16 = 98.
2) Find the mass fraction of hydrogen in the substance:
3) Calculate the mass fraction of sulfur in sulfuric acid:
4. Calculate the mass fraction of oxygen in the substance:
w(O) = 1 – ( w(H)+ w(S)) = 1 – (0.0204 + 0.3265) = 0.6531, or 65.31%.
Answer. w(H) = 2.04%, w(S) = 32.65%, w(O) = 65.31%.
More often, chemists have to solve the inverse problem: using the mass fractions of elements to determine the formula of a complex substance. Let us illustrate how such problems are solved with one historical example.
Two compounds of copper with oxygen (oxides) were isolated from natural minerals - tenorite and cuprite. They differed from each other in color and mass fractions of elements. In the black oxide, the mass fraction of copper was 80%, and the mass fraction of oxygen was 20%. In red copper oxide, the mass fractions of elements were 88.9% and 11.1%, respectively. What are the formulas of these complex substances? Let's do some simple mathematical calculations.
Example 1. Calculation of the chemical formula of black copper oxide ( w(Cu) = 0.8 and w(O) = 0.2).
x, y– by the number of atoms of chemical elements in its composition: Cu x O y.
2) The ratio of the indices is equal to the ratio of the quotients of the mass fraction of the element in the compound divided by the relative atomic mass of the element:
3) The resulting relationship must be reduced to a ratio of integers: the indices in the formula showing the number of atoms cannot be fractional. To do this, divide the resulting numbers by the smaller (i.e. any) of them:
The resulting formula is CuO.
Example 2. Calculation of the formula of red copper oxide using known mass fractions w(Cu) = 88.9% and w(O) = 11.1%.
Given:
w(Cu) = 88.9%, or 0.889,
w(O) = 11.1%, or 0.111.
Find:
Solution
1) Let us denote the formula of Cu oxide x O y.
2) Find the ratio of indices x And y:
3) Let us present the ratio of indices to the ratio of integers:
Answer. The formula of the compound is Cu 2 O.
Now let's complicate the task a little.
Task 3.According to elemental analysis, the composition of calcined bitter salt, which was used by alchemists as a laxative, is as follows: mass fraction of magnesium - 20.0%, mass fraction of sulfur - 26.7%, mass fraction of oxygen - 53.3%.
Given:
w(Mg) = 20.0%, or 0.2,
w(S) = 26.7%, or 0.267,
w(O) = 53.3%, or 0.533.
Find:
Solution
1) Let us denote the formula of a substance using indices x, y, z: Mg x S y O z.
2) Let's find the ratio of indices:
3) Determine the value of the indices x, y, z:
Answer. The formula of the substance is MgSO 4.
1. What is the mass fraction of an element in a complex substance? How is this value calculated?
2.
Calculate the mass fractions of elements in the substances: a) carbon dioxide CO 2;
b) calcium sulfide CaS; c) sodium nitrate NaNO 3; d) aluminum oxide Al 2 O 3.
3. Which of the nitrogen fertilizers contains the largest mass fraction of the nutrient element nitrogen: a) ammonium chloride NH 4 Cl; b) ammonium sulfate (NH 4) 2 SO 4; c) urea (NH 2) 2 CO?
4. In the mineral pyrite, there are 8 g of sulfur per 7 g of iron. Calculate the mass fractions of each element in this substance and determine its formula.
5. The mass fraction of nitrogen in one of its oxides is 30.43%, and the mass fraction of oxygen is 69.57%. Determine the formula of the oxide.
6. In the Middle Ages, a substance called potash was isolated from the ashes of fires and was used to make soap. Mass fractions of elements in this substance: potassium - 56.6%, carbon - 8.7%, oxygen - 34.7%. Determine the formula of potash.
§ 5.1 Chemical reactions. Chemical Reaction Equations
A chemical reaction is the transformation of one substance into another. However, such a definition needs one significant addition. In a nuclear reactor or accelerator, some substances are also transformed into others, but such transformations are not called chemical. What's the matter here? Nuclear reactions occur in a nuclear reactor. They consist in the fact that the nuclei of elements, when colliding with high-energy particles (they can be neutrons, protons and nuclei of other elements), are broken into fragments, which are the nuclei of other elements. Fusion of nuclei with each other is also possible. These new nuclei then receive electrons from the environment and thus the formation of two or more new substances is completed. All these substances are some elements of the Periodic Table. Examples of nuclear reactions used to discover new elements are given in §4.4.
Unlike nuclear reactions, in chemical reactions kernels are not affected atoms. All changes occur only in the outer electron shells. Some chemical bonds are broken and others are formed.
Chemical reactions are phenomena in which some substances with a certain composition and properties are transformed into other substances - with a different composition and other properties. In this case, no changes occur in the composition of atomic nuclei.
Let's consider a typical chemical reaction: the combustion of natural gas (methane) in atmospheric oxygen. Those of you who have a gas stove at home can see this reaction in your kitchen every day. Let us write the reaction as shown in Fig. 5-1.
Rice. 5-1. Methane CH 4 and oxygen O 2 react with each other to form carbon dioxide CO 2 and water H 2 O. In this case, the bonds between C and H in the methane molecule are broken and carbon-oxygen bonds appear in their place. Hydrogen atoms that previously belonged to methane form bonds with oxygen. The figure clearly shows that for the successful implementation of the reaction to one you need to take a methane molecule two oxygen molecules.
Recording a chemical reaction using molecular drawings is not very convenient. Therefore, to record chemical reactions, abbreviated formulas of substances are used - as shown in the lower part of Fig. 5-1. This entry is called chemical reaction equation.
The number of atoms of different elements on the left and right sides of the equation is the same. On the left side one carbon atom in the methane molecule (CH 4), and on the right - same We find a carbon atom in the CO 2 molecule. We will definitely find all four hydrogen atoms from the left side of the equation on the right - in the composition of water molecules.
In the equation of a chemical reaction, to equalize the number of identical atoms in different parts of the equation, odds, which are recorded before formulas of substances. Coefficients should not be confused with indices in chemical formulas.
Let's consider another reaction - the transformation of calcium oxide CaO (quicklime) into calcium hydroxide Ca(OH) 2 (slaked lime) under the influence of water.
Rice. 5-2. Calcium oxide CaO attaches a water molecule H 2 O to form
calcium hydroxide Ca(OH) 2.
Unlike mathematical equations, chemical reaction equations cannot rearrange the left and right sides. The substances on the left side of the chemical reaction equation are called reagents, and on the right - reaction products. If you rearrange the left and right sides in the equation from Fig. 5-2, then we get the equation completely different chemical reaction:
If the reaction between CaO and H 2 O (Fig. 5-2) begins spontaneously and proceeds with the release of a large amount of heat, then strong heating is required to carry out the last reaction, where Ca (OH) 2 serves as the reagent.
Note that you can use an arrow instead of an equal sign in a chemical reaction equation. The arrow is convenient because it shows direction the course of the reaction.
Let us also add that reactants and products may not necessarily be molecules, but also atoms - if any element or elements in their pure form are involved in the reaction. For example:
H 2 + CuO = Cu + H 2 O
There are several ways to classify chemical reactions, of which we will consider two.
According to the first of them, all chemical reactions are distinguished according to the characteristic changes in the number of starting and final substances. Here you can find 4 types of chemical reactions:
Reactions CONNECTIONS,
Reactions DECOMPOSITIONS,
Reactions EXCHANGE,
Reactions SUBSTITUTIONS.
Let us give specific examples of such reactions. To do this, let’s return to the equations for producing slaked lime and the equation for producing quicklime:
CaO + H 2 O = Ca (OH) 2
Ca(OH) 2 = CaO + H 2 O
These reactions belong to different types chemical reactions. The first reaction is a typical reaction connections, since during its occurrence two substances CaO and H 2 O are combined into one: Ca (OH) 2.
The second reaction Ca(OH) 2 = CaO + H 2 O is a typical reaction decomposition: Here one substance Ca(OH) 2 decomposes to form two others.
In reactions exchange the number of reactants and products is usually the same. In such reactions, the starting substances exchange atoms and even entire components of their molecules with each other. For example, when a solution of CaBr 2 is combined with a solution of HF, a precipitate forms. In solution, calcium and hydrogen ions exchange bromine and fluorine ions with each other. The reaction occurs only in one direction because calcium and fluorine ions bind into the insoluble compound CaF 2 and after this “reverse exchange” of ions is no longer possible:
CaBr 2 + 2HF = CaF 2 ¯ + 2HBr
When merging solutions of CaCl 2 and Na 2 CO 3, a precipitate also forms, because calcium and sodium ions exchange particles of CO 3 2– and Cl– with each other to form an insoluble compound - calcium carbonate CaCO 3.
CaCl 2 + Na 2 CO 3 = CaCO 3 ¯ + 2NaCl
The arrow next to the reaction product indicates that this compound is insoluble and precipitates. Thus, the arrow can also be used to indicate the removal of a product from a chemical reaction in the form of a precipitate (¯) or gas (). For example:
Zn + 2HCl = H 2 + ZnCl 2
The last reaction belongs to another type of chemical reaction - reactions substitution. Zinc replaced hydrogen in its combination with chlorine (HCl). Hydrogen is released in the form of gas.
Substitution reactions may be externally similar to exchange reactions. The difference is that substitution reactions necessarily involve atoms of some kind simple substances that replace atoms of one of the elements in a complex substance. For example:
2NaBr + Cl 2 = 2NaCl + Br 2 - reaction substitution;
on the left side of the equation there is a simple substance - a chlorine molecule Cl 2, and on the right side there is a simple substance - a bromine molecule Br 2.
In reactions exchange both reactants and products are complex substances. For example:
CaCl 2 + Na 2 CO 3 = CaCO 3 ¯ + 2NaCl - reaction exchange;
In this equation, the reactants and products are complex substances.
The division of all chemical reactions into reactions of combination, decomposition, substitution and exchange is not the only one. There is another way of classification: based on the change (or lack of change) in the oxidation states of reactants and products. On this basis, all reactions are divided into redox reactions and all others (not redox).
The reaction between Zn and HCl is not only a substitution reaction, but also redox reaction, because the oxidation states of the reacting substances change in it:
Zn 0 + 2H +1 Cl = H 2 0 + Zn +2 Cl 2 - a substitution reaction and at the same time a redox reaction.
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.
From a chemistry course we know that the mass fraction is the content of a certain element in a substance. It would seem that such knowledge is of no use to an ordinary summer resident. But don’t rush to close the page, since the ability to calculate the mass fraction for a gardener can be very useful. However, in order not to get confused, let's talk about everything in order.What is the essence of the concept of “mass fraction”?
The mass fraction is measured in percentages or simply in tenths. Just above we talked about the classic definition, which can be found in reference books, encyclopedias or school chemistry textbooks. But it is not so easy to understand the essence of what has been said. So, suppose we have 500 g of some complex substance. Complex in this case means that it is not homogeneous in its composition. By and large, any substances we use are complex, even simple table salt, the formula of which is NaCl, that is, it consists of sodium and chlorine molecules. If we continue our reasoning using table salt as an example, we can assume that 500 grams of salt contains 400 g of sodium. Then its mass fraction will be 80% or 0.8.
Why does a summer resident need this?
I think you already know the answer to this question. The preparation of all kinds of solutions, mixtures, etc. is an integral part of the economic activity of any gardener. Fertilizers, various nutrient mixtures, as well as other drugs, for example, growth stimulants “Epin”, “Kornevin”, etc. are used in the form of solutions. In addition, it is often necessary to mix dry substances, such as cement, sand and other components, or ordinary garden soil with a purchased substrate. Moreover, the recommended concentration of these agents and drugs in prepared solutions or mixtures in most instructions is given in mass fractions.
Thus, knowing how to calculate the mass fraction of an element in a substance will help the summer resident to correctly prepare the necessary solution of fertilizer or nutrient mixture, and this, in turn, will certainly affect the future harvest.
Calculation algorithm
So, the mass fraction of an individual component is the ratio of its mass to the total mass of the solution or substance. If the result obtained needs to be converted into a percentage, then it must be multiplied by 100. Thus, the formula for calculating the mass fraction can be written as follows:
W = Mass of substance / Mass of solution
W = (Mass of substance / Mass of solution) x 100%.
Example of determination of mass fraction
Let's assume that we have a solution for the preparation of which 5 g of NaCl was added to 100 ml of water, and now we need to calculate the concentration of table salt, that is, its mass fraction. We know the mass of the substance, and the mass of the resulting solution is the sum of two masses - salt and water and is equal to 105 g. Thus, we divide 5 g by 105 g, multiply the result by 100 and get the desired value of 4.7%. This is exactly the concentration the saline solution will have.
More practical task
In practice, a summer resident more often has to deal with problems of a different kind. For example, it is necessary to prepare an aqueous solution of some fertilizer, the concentration of which by weight should be 10%. In order to accurately observe the recommended proportions, you need to determine how much of the substance is needed and in what volume of water it will need to be dissolved.
Solving the problem begins in reverse order. First, you should divide the mass fraction expressed as a percentage by 100. As a result, we obtain W = 0.1 - this is the mass fraction of the substance in units. Now let's denote the amount of substance as x, and the final mass of the solution as M. In this case, the last value is made up of two terms - the mass of water and the mass of fertilizer. That is, M = Mv + x. So we get a simple equation:
W = x / (Mw + x)
Solving it for x, we get:
x = W x Mv / (1 – W)
Substituting the available data, we obtain the following relationship:
x = 0.1 x MV / 0.9
Thus, if we take 1 liter (that is, 1000 g) of water to prepare a solution, then to prepare a solution of the required concentration we will need approximately 111-112 g of fertilizer.
Solving dilution or addition problems
Suppose we have 10 liters (10,000 g) of a ready-made aqueous solution with a concentration of a certain substance W1 = 30% or 0.3. How much water will need to be added to it to reduce the concentration to W2 = 15% or 0.15? In this case, the formula will help:
Мв = (W1х М1 / W2) – М1
Substituting the initial data, we find that the amount of added water should be:
Mv = (0.3 x 10,000 / 0.15) – 10,000 = 10,000 g
That is, you need to add the same 10 liters.
Now imagine the inverse problem - there are 10 liters of an aqueous solution (M1 = 10,000 g) with a concentration of W1 = 10% or 0.1. You need to get a solution with a mass fraction of fertilizer W2 = 20% or 0.2. How much starting material will need to be added? To do this you need to use the formula:
x = M1 x (W2 – W1) / (1 – W2)
Substituting the original values, we get x = 1,125 g.
Thus, knowledge of the simplest basics of school chemistry will help the gardener to correctly prepare fertilizer solutions, nutrient substrates from several elements or mixtures for construction work.
Chemistry is definitely an interesting science. Despite all its complexity, it allows us to better understand the nature of the world around us. And moreover, at least basic knowledge of this subject seriously helps in everyday life. For example, determining the mass fraction of a substance in a multicomponent system, that is, the ratio of the mass of any component to the total mass of the entire mixture.
Necessary:
- calculator;
— scales (if you first need to determine the masses of all components of the mixture);
— Mendeleev’s periodic table of elements.
Instructions:
- So, it became necessary for you to determine the mass fraction of the substance. Where to begin? First of all, it depends on the specific task and on the tools at hand for the work. But in any case, to determine the content of a component in a mixture, you need to know its mass and the total mass of the mixture. This can be done either on the basis of known data or on the basis of your own research. To do this, you will need to weigh the added component on a laboratory scale. After the mixture is prepared, weigh it as well.
- Write down the mass of the required substance as “ m«, total mass put the systems under the designation “ M". In this case, the formula for the mass fraction of the substance will take the following form: W=(m/M)*100. The result obtained is recorded as a percentage.
- Example: calculate the mass fraction of 15 grams of table salt dissolved in 115 g of water. Solution: the total mass of the solution is determined by the formula M=m to +m c, Where m in- mass of water, m c- mass of table salt. From simple calculations it can be determined that the total mass of the solution is 130 grams. Using the above determination formula, we find that the content of table salt in the solution will be equal to W=(15/130)*100=12%.
- A more particular situation is the need to determine mass fraction of a chemical element in a substance . It is defined in exactly the same way. The main principle of calculation will remain the same, only instead of the mass of the mixture and the specific component, you will have to deal with the molecular masses of the chemical elements.
- All the necessary information can be found in the periodic table of Mendeleev. Break down the chemical formula of a substance into its main components. Using the periodic table, determine the mass of each element. By summing them up, you get the molecular mass of your substance ( M). Similar to the previous case, the mass fraction of a substance, or, to be more precise, an element will be determined by the ratio of its mass to molecular mass. The formula will take the following form W=(m a /M)*100. Where m a- atomic mass of the element, M- molecular weight of the substance.
- Let's look at this case using a specific example. Example: determine the mass fraction of potassium in potash. Potash is potassium carbonate. Its formula K2CO3. Atomic mass of potassium - 39 , carbon - 12 , oxygen - 16 . The molecular weight of the carbonate will be determined as follows - M = 2m K +m C +2m O = 2*39+12+2*16 = 122. The potassium carbonate molecule contains two potassium atoms with an atomic mass equal to 39 . The mass fraction of potassium in the substance will be determined by the formula W = (2m K /M)*100 = (2*39/122)*100 = 63.93%.
Instructions
The mass fraction of a substance is found by the formula: w = m(in)/m(cm), where w is the mass fraction of the substance, m(in) is the mass of the substance, m(cm) is the mass of the mixture. If dissolved, then it looks like this: w = m(in)/m(solution), where m(solution) is the mass of the solution. If necessary, the mass of the solution can also be found: m(solution) = m(in) + m(solution), where m(solution) is the mass of the solvent. If desired, the mass fraction can be multiplied by 100%.
If the problem statement does not give a mass value, then it can be calculated using several formulas; the values given in the statement will help you choose the right one. The first formula for: m = V*p, where m is mass, V is volume, p is density. The following formula looks like this: m = n*M, where m is mass, n is the amount of substance, M is molar mass. The molar mass, in turn, consists of the atomic masses of the elements that make up the substance.
To better understand this material, let’s solve the problem. A mixture of copper and magnesium filings weighing 1.5 g was treated with excess . As a result of the reaction, hydrogen volume is 0.56 l (). Calculate the mass fraction of copper in the mixture.
In this problem, we write down its equation. Of the two substances with excess hydrochloric acid, only magnesium: Mg + 2HCl = MgCl2 + H2. To find the mass fraction of copper in the mixture, you need to substitute the values into the following formula: w(Cu) = m(Cu)/m(cm). The mass of the mixture is given, let’s find the mass of copper: m(Cu) = m(cm) – m(Mg). We are looking for mass: m(Mg) = n(Mg)*M(Mg). The reaction equation will help you find the amount of magnesium. We find the amount of hydrogen substance: n = V/Vm = 0.56/22.4 = 0.025 mol. The equation shows that n(H2) = n(Mg) = 0.025 mol. We calculate the mass of magnesium, knowing that the molar is 24 g/mol: m(Mg) = 0.025*24 = 0.6 g. Find the mass of copper: m(Cu) = 1.5 – 0.6 = 0.9 g. Remaining calculate the mass fraction: w(Cu) = 0.9/1.5 = 0.6 or 60%.
Video on the topic
note
The mass fraction cannot be greater than one or, if expressed as a percentage, greater than 100%.
Sources:
- "Chemistry Manual", G.P. Khomchenko, 2005.
- Calculation of the share of sales by region
The mass fraction shows, as a percentage or in fractions, the content of a substance in a solution or an element in the composition of a substance. The ability to calculate mass fraction is useful not only in chemistry lessons, but also when you want to prepare a solution or mixture, for example, for culinary purposes. Or change the percentage in your existing composition.
Instructions
For example, you need at least 15 cubic meters for the winter. meters of birch firewood.
Look for the density of birch firewood in the reference book. This is: 650 kg/m3.
Calculate the mass by substituting the values into the same specific gravity formula.
m = 650*15 = 9750 (kg)
Now, based on the load capacity and capacity of the body, you can decide on the type of vehicle and the number of trips.
Video on the topic
note
Older people are more familiar with the concept of specific gravity. The specific density of a substance is the same as specific gravity.
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%).
For example, determine the mass fraction of chemical elements in copper sulfate. Copper (copper II sulfate), has the chemical formula CuSO4. The atomic masses of the elements included in its composition are equal to Ar(Cu)=64, Ar(S)=32, Ar(O)=16, the mass numbers of these elements will be equal to M0(Cu)=64, M0(S)=32, M0(O)=16∙4=64, taking into account that the molecule contains 4 atoms. Calculate the molecular mass of the substance, it is equal to the sum of the mass numbers of the substances that make up the molecule 64+32+64=160. Determine the mass fraction of copper (Cu) in the composition of copper sulfate (ω%=(64/160)∙100%)=40%. Using the same principle, one can determine the mass fractions of all elements in this substance. Mass fraction of sulfur (S) ω%=(32/160)∙100%=20%, oxygen (O) ω%=(64/160)∙100%=40%. Please note that the sum of all mass fractions of the substance must be 100%.