home |
Magnitude and its dimension |
Ratio |
|
Atomic mass of element X (relative) |
Element serial number Z=(N –) = Z=(e +) |
R |
|
Mass fraction of element E in substance X, in fractions of a unit, in %) | |
Amount of substance X, mol |
= 22.4 l/mol (n.s.) e Well. – = 101 325 Pa, T |
= 273 K |
|
Molar mass of substance X, g/mol, kg/mol |
V Mass of substance X, g, kg (X) = n (X) M |
(X) |
Amount of gas substance, mol V Molar volume of gas, l/mol, m 3 /mol |
= 22.4 l/mol at N.S. |
Amount of gas substance, mol = Amount of gas substance, mol V × (X) = |
Gas volume, m 3 |
|
Product yield |
|
Density of substance X, g/l, g/ml, kg/m 3 |
|
Density of gaseous substance X by hydrogen |
Density of gaseous substance X in air M |
(air) = 29 g/mol |
|
United Gas Law |
Mendeleev-Clapeyron equation = PV, nRT R |
= 8.314 J/mol×K |
|
Volume fraction of a gaseous substance in a mixture of gases, in fractions of a unit or in % |
|
Molar mass of a mixture of gases |
|
Mole fraction of a substance (X) in a mixture |
Amount of heat, J, kJ = (X) = n Amount of heat, J, kJ M |
Q |
Thermal effect of reactionQ =– |
H |
|
Heat of formation of substance X, J/mol, kJ/mol Speed chemical reaction |
|
(mol/lsec) Law of Mass Action |
(for a simple reaction) a A+ V B= With C + d D = u B= (for a simple reaction) k B= (A) V |
(B) |
|
Van't Hoff's rule |
|
Solubility of the substance (X) (g/100 g solvent) |
|
Mass fraction of substance X in mixture A + X, in fractions of a unit, in % |
V Weight of solution, g, kg V(rr) = V(X)+ V Weight of solution, g, kg Amount of gas substance, mol(H2O) (rr) |
(rr) |
|
Mass fraction of dissolved substance in solution, in fractions of a unit, in % |
|
Solution density |
|
Volume of solution, cm 3, l, m 3 |
|
Molar concentration, mol/l |
|
Degree of electrolyte dissociation (X), in fractions of a unit or % |
Ionic product of water K |
(H2O) = |
pH value |
pH = –lg
Main: Kuznetsova N.E. and etc
. Chemistry. 8th grade-10th grade. – M.: Ventana-Graf, 2005-2007. Kuznetsova N.E., Litvinova T.N., Levkin A.N.
Chemistry.11th grade in 2 parts, 2005-2007. Egorov A.S.
Chemistry.11th grade in 2 parts, 2005-2007. Egorov A.S. Chemistry: a modern course for preparing for the Unified State Exam. Rostov n/a: Phoenix, 2011. (2012) – 699 p.
Self-instruction manual for solving chemical problems. – Rostov-on-Don: Phoenix, 2000. – 352 p.
Chemistry/tutor manual for applicants to universities. Rostov-n/D, Phoenix, 2005– 536 p. Khomchenko G.P., Khomchenko I.G. . Problems in chemistry for applicants to universities. M.: graduate School
Additional:
Vrublevsky A.I.. Educational and training materials for preparing for centralized testing in chemistry / A.I. Vrublevsky –Mn.: Unipress LLC, 2004. – 368 p.
Vrublevsky A.I.. 1000 problems in chemistry with chains of transformations and control tests for schoolchildren and applicants. – Mn.: Unipress LLC, 2003. – 400 p.
Egorov A.S.. All types of calculation problems in chemistry for preparation for the Unified State Exam. – Rostov n/D: Phoenix, 2003. – 320 p.
Egorov A.S., Aminova G.Kh.. Typical tasks and exercises for preparing for the chemistry exam. – Rostov n/d: Phoenix, 2005. – 448 p.
Unified State Exam 2007. Chemistry. Educational and training materials for preparing students / FIPI - M.: Intellect-Center, 2007. – 272 p.
Unified State Exam 2011. Chemistry. Educational and training kit ed. A.A. Kaverina. – M.: National Education, 2011.
The only real options for tasks to prepare for the Unified State Exam. Unified State Examination 2007. Chemistry/V.Yu. Mishina, E.N. Strelnikova. M.: Federal Testing Center, 2007.–151 p.
Kaverina A.A. The optimal bank of tasks for preparing students. Unified State Exam 2012. Chemistry. Tutorial./ A.A. Kaverina, D.Yu. Dobrotin, Yu.N. Medvedev, M.G. Snastina. – M.: Intellect-Center, 2012. – 256 p.
Litvinova T.N., Vyskubova N.K., Azhipa L.T., Solovyova M.V.. Test tasks in addition to tests for students of 10-month correspondence preparatory courses (methodological instructions). Krasnodar, 2004. – P. 18 – 70.
Litvinova T.N.. Chemistry. Unified State Exam 2011. Training tests. Rostov n/d: Phoenix, 2011.– 349 p.
Litvinova T.N.. Chemistry. Tests for the Unified State Exam. Rostov n/d.: Phoenix, 2012. - 284 p.
Litvinova T.N.. Chemistry. Laws, properties of elements and their compounds. Rostov n/d.: Phoenix, 2012. - 156 p.
Litvinova T.N., Melnikova E.D., Solovyova M.V.., Azhipa L.T., Vyskubova N.K. Chemistry in tasks for applicants to universities. – M.: Onyx Publishing House LLC: Mir and Education Publishing House LLC, 2009. – 832 p.
Educational and methodological complex in chemistry for students of medical and biological classes, ed. T.N. Litvinova. – Krasnodar.: KSMU, – 2008.
Chemistry. Unified State Exam 2008. Entrance tests, teaching aid / ed. V.N. Doronkina. – Rostov n/d: Legion, 2008.– 271 p.
List of websites on chemistry:
1. Alhimik. http:// www. alhimik. ru
2. Chemistry for everyone. Electronic reference book for full course chemistry.
http:// www. informika. ru/ text/ database/ chemistry/ START. html
3. School chemistry - reference book. http:// www. schoolchemistry. by. ru
4. Chemistry tutor. http://www. chemistry.nm.ru
Internet resources
Alhimik. http:// www. alhimik. ru
Chemistry for everyone. Electronic reference book for a complete chemistry course.
http:// www. informika. ru/ text/ database/ chemistry/ START. html
School chemistry - reference book. http:// www. schoolchemistry. by. ru
http://www.classchem.narod.ru
Chemistry tutor. http://www.
chemistry.nm.ru http://www.alleng.ru/edu/chem.htm
- educational Internet resources on chemistry http://schoolchemistry.by.ru/ - school chemistry. This site has the opportunity to take On-line testing on various topics, as well as demo options
Unified State Exam http:// www. Chemistry and life—XXI century: popular science magazine.. ru
hij
E. V. Savinkina G. P. Loginova
Collection of basic formulas in chemistry
The most important chemical concepts and laws Chemical element
- this is a certain type of atom with the same nuclear charge. Relative atomic mass
(A r) shows how many times the mass of an atom of a given chemical element is greater than the mass of a carbon-12 atom (12 C). Chemical substance
– a conventional particle, the composition of which corresponds to the given chemical formula, for example:
Ar – argon substance (consists of Ar atoms),
H 2 O – the substance water (consists of H 2 O molecules),
B, A r (B): Where*T
(atom B) – mass of an atom of element B;*t and
– atomic mass unit; 1/12 *t and = T
(12 C atom) = 1.6610 24 g. Quantity of substance
B, A r (B): B, n(B), mol: N(B)
– number of particles B; N A – Avogadro's constant(N A =
6.0210 23 mol -1). Molar mass of a substance
B, A r (B): V, M(V), g/mol: t(V)
– mass B. Molar volume of gas IN, V M
B, A r (B): l/mol: V M = 22.4 l/mol (consequence from Avogadro’s law), under normal conditions (n.s. - atmospheric pressure p = 101,325 Pa (1 atm); thermodynamic temperature T = 273.15 K or Celsius temperature t =
0 °C). B for hydrogen, D
(gas B by H 2):*Density of gaseous substance IN by air, D (gas B over air): Mass fraction of element E in matterV, w(E):
The structure of the atom and the Periodic Law D.I. Mendeleev Mass number (A) – total number
Period number corresponds number energy levels, filled with electrons, and stands for the last energy level to be filled(EU).
Group number A shows And etc.
Group number B shows number of valence electrons ns And (n – 1)d.
S-elements section– the energy sublevel (ESL) is filled with electrons ns-EPU– IA- and IIA-groups, H and He.
P-elements section– filled with electrons np-EPU– IIIA-VIIIA-groups.
D-elements section– filled with electrons (P- 1) d-EPU – IB-VIIIB2-groups.
f-elements section– filled with electrons (P-2) f-EPU – lanthanides and actinides.
Volatile: SiH 4, PH 3, H 2 S, HCl.
Amphoteric: Al 2 O 3 – Al(OH) 3.
Acidic: SiO 2 – H 4 SiO 4, P 2 O 5 – H 3 PO 4, SO 3 – H 2 SO 4, Cl 2 O 7 – HClO 4.
Donor-acceptor mechanism– overlap of a free orbital of one atom with an orbital of another atom that contains a pair of electrons.
sp– linear – 180°
sp 2– triangular – 120°
sp 3– tetrahedral – 109.5°
sp 3 d– trigonal-bipyramidal – 90°; 120°
sp 3 d 2– octahedral – 90°
Solution- a homogeneous system consisting of two or more substances, the content of which can be varied within certain limits.
Solution: solvent (eg water) + solute.
True solutions contain particles smaller than 1 nanometer.
Colloidal solutions contain particles ranging in size from 1 to 100 nanometers.
Mechanical mixtures(suspensions) contain particles larger than 100 nanometers.
Suspension=> solid + liquid
Emulsion=> liquid + liquid
Foam, fog=> gas + liquid
Heterogeneous mixtures are separated settling and filtering.
Homogeneous mixtures are separated evaporation, distillation, chromatography.
Saturated solution is or may be in equilibrium with the solute (if the solute is solid, then its excess is in the precipitate).
Solubility– the content of the dissolved substance in a saturated solution at a given temperature.
Unsaturated solution less,
Supersaturated solution contains solute more, than its solubility at a given temperature.
B, A r (B): V, M(V), g/mol:– mass B,
t(r)– mass of solution.
Weight of solution, m(p), g:
ρ(p) – solution density.
Volume of solution, V(p), l:
Molar concentration, s(V), mol/l:Where n(B) is the amount of substance B;
M(B) – molar mass of substance B.
> t"(V)= t(B);
> the mass of the solution increases by the mass of added water: m"(p) = m(p) + m(H 2 O).
Evaporating water from a solution:
> the mass of the solute does not change: t"(B) = t(B).
> the mass of the solution decreases by the mass of evaporated water: m"(p) = m(p) – m(H 2 O).
Merging two solutions: The masses of solutions, as well as the masses of the dissolved substance, add up:
t"(B) = t(B) + t"(B);
t"(p) = t(p) + t"(p).
Crystal Drop: the mass of the solute and the mass of the solution are reduced by the mass of precipitated crystals:
m"(B) = m(B) – m(sediment); m"(p) = m(p) – m(sediment).
The mass of water does not change.
If during time τ in volume Amount of gas substance, mol the amount of reactant or product changed by Δ n, speed reaction:
For a monomolecular reaction A →…:
1) chemically active reagents;
2) promotion reagent concentrations;
3) increase
4) promotion temperature;
5) catalysts. Speed reaction reduce:
1) chemically inactive reagents;
2) demotion reagent concentrations;
3) decrease surfaces of solid and liquid reagents;
4) demotion temperature;
5) inhibitors.
*Temperature speed coefficient(γ) is equal to a number that shows how many times the reaction rate increases when the temperature increases by ten degrees:
*Law of mass action for chemical equilibrium: in a state of equilibrium, the ratio of the product of the molar concentrations of products in powers equal to
Their stoichiometric coefficients, to the product of the molar concentrations of the reactants in powers equal to their stoichiometric coefficients, at a constant temperature is a constant value (concentration equilibrium constant).
In a state of chemical equilibrium for a reversible reaction:
2) reducing the concentration of products;
3) increase in temperature (for an endothermic reaction);
4) decrease in temperature (for an exothermic reaction);
5) increase in pressure (for a reaction occurring with a decrease in volume);
6) decrease in pressure (for a reaction occurring with an increase in volume).
Electrolytic dissociation– the process of formation of ions (cations and anions) when certain substances are dissolved in water.
acids are formed hydrogen cations And acid anions, For example:
Dissociation degree α– the ratio of the number of dissociated particles to the number of initial particles.
At constant volume:
“Complete” ionic equation: Сu 2+ + 2Сl¯ + 2Na + + 2OH¯ = Cu(OH) 2 ↓ + 2Na + + 2Сl¯
“Short” ionic equation: Cu 2+ + 2OH¯ = Cu(OH) 2 ↓
2. Molecular equation: FeS (T) + 2HCl = FeCl 2 + H 2 S
“Complete” ionic equation: FeS + 2H + + 2Сl¯ = Fe 2+ + 2Сl¯ + H 2 S
“Short” ionic equation: FeS (T) + 2H + = Fe 2+ + H 2 S
3. Molecular equation: 3HNO 3 + K 3 PO 4 = H 3 PO 4 + 3KNO 3
“Complete” ionic equation: 3H + + 3NO 3 ¯ + 3K + + PO 4 3- = H 3 PO 4 + 3K + + 3NO 3 ¯
“Short” ionic equation: 3H + + PO 4 3- = H 3 PO 4
1. Alkali + strong acid: Ba(OH) 2 + 2HCl = BaCl 2 + 2H 2 O
Ba 2+ + 2ON¯ + 2H + + 2Сl¯ = Ba 2+ + 2Сl¯ + 2Н 2 O
H + + OH¯ = H 2 O
2. Slightly soluble base + strong acid: Cu(OH) 2(t) + 2HCl = CuCl 2 + 2H 2 O
Cu(OH) 2 + 2H + + 2Cl¯ = Cu 2+ + 2Cl¯ + 2H 2 O
Cu(OH) 2 + 2H + = Cu 2+ + 2H 2 O
*Hydrolysis– an exchange reaction between a substance and water without changing the oxidation states of atoms.
1. Irreversible hydrolysis of binary compounds:
Mg 3 N 2 + 6H 2 O = 3Mg(OH) 2 + 2NH 3
2. Reversible hydrolysis of salts:
A) Salt is formed a strong base cation and a strong acid anion:
NaCl = Na + + Сl¯
Na + + H 2 O ≠ ;
Cl¯ + H 2 O ≠
There is no hydrolysis; neutral environment, pH = 7.
B) Salt is formed a strong base cation and a weak acid anion:
Na 2 S = 2Na + + S 2-
Na + + H 2 O ≠
S 2- + H 2 O ↔ HS¯ + OH¯
Hydrolysis by anion; alkaline environment, pH >7.
B) Salt is formed a cation of a weak or slightly soluble base and an anion of a strong acid:
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several basic concepts and formulas.
All substances have different mass, density and volume. A piece of metal from one element can weigh many times more than an exactly the same size piece of another metal.
Mole(number of moles)
designation: mole, international: mol- a unit of measurement for the amount of a substance. Corresponds to the amount of substance that contains N.A. particles (molecules, atoms, ions) Therefore, a universal quantity was introduced - number of moles. A frequently encountered phrase in tasks is “received... mole of substance"
N.A.= 6.02 1023
N.A.- Avogadro's number. Also “a number by agreement.” How many atoms are there in the tip of a pencil? About a thousand. It is not convenient to operate with such quantities. Therefore, chemists and physicists around the world agreed - let’s designate 6.02 × 1023 particles (atoms, molecules, ions) as 1 mole substances.
1 mole = 6.02 1023 particles
This was the first of the basic formulas for solving problems.
6.0210 23 mol -1).
Molar mass substance is the mass of one mole of substance.
Denoted as Mr. Found according to the periodic table - it’s just a sum atomic masses substances.
For example, we are given sulfuric acid- H2SO4. Let's calculate the molar mass of a substance: atomic mass H = 1, S-32, O-16.
Mr(H2SO4)=1 2+32+16 4=98 g\mol.
The second necessary formula for solving problems is
substance mass formula:
That is, to find the mass of a substance, you need to know the number of moles (n), and we find the molar mass from the Periodic Table.
Law of conservation of mass - The mass of substances that enter into a chemical reaction is always equal to the mass of the resulting substances.
If we know the mass(es) of the substances that reacted, we can find the mass(es) of the products of that reaction. And vice versa.
The third formula for solving problems in chemistry is
volume of substance:
Sorry, this image does not meet our guidelines. To continue publishing, please delete the image or upload another one.Where did the number 22.4 come from? From Avogadro's law:
equal volumes of different gases taken at the same temperature and pressure contain the same number of molecules.
According to Avogadro's law, 1 mole of an ideal gas under normal conditions (n.s.) has the same volume Vm= 22.413 996(39) l
That is, if in the problem we are given normal conditions, then, knowing the number of moles (n), we can find the volume of the substance.
So, basic formulas for solving problems in chemistry
Avogadro's numberN.A.
6.02 1023 particles
(12 C atom) = 1.6610 24 g. n (mol)
n=V\22.4 (l\mol)
Mass of substance m (g)
Volume of substance V(l)
V=n 22.4 (l\mol)
Sorry, this image does not meet our guidelines. To continue publishing, please delete the image or upload another one.These are formulas. Often, to solve problems, you first need to write the reaction equation and (required!) arrange the coefficients - their ratio determines the ratio of moles in the process.
Modern symbols chemical elements were introduced into science in 1813 by J. Berzelius. According to his proposal, elements are designated by their initial letters Latin names. For example, oxygen (Oxygenium) is designated by the letter O, sulfur (Sulfur) by the letter S, hydrogen (Hydrogenium) by the letter H. In cases where the names of the elements begin with the same letter, one more letter is added to the first letter. Thus, carbon (Carboneum) has the symbol C, calcium (Calcium) - Ca, copper (Cuprum) - Cu.
Chemical symbols are not only abbreviated names of elements: they also express certain quantities (or masses), i.e. Each symbol represents either one atom of an element, or one mole of its atoms, or a mass of an element equal to (or proportional to) the molar mass of that element. For example, C means either one carbon atom, or one mole of carbon atoms, or 12 mass units (usually 12 g) of carbon.
Formulas of substances also indicate not only the composition of the substance, but also its quantity and mass. Each formula represents either one molecule of a substance, or one mole of a substance, or a mass of a substance equal to (or proportional to) its molar mass. For example, H2O represents either one molecule of water, or one mole of water, or 18 mass units (usually (18 g) of water.
Simple substances are also designated by formulas showing how many atoms a molecule of a simple substance consists of: for example, the formula for hydrogen H 2. If the atomic composition of a molecule of a simple substance is not precisely known or the substance consists of molecules containing a different number of atoms, and also if it has an atomic or metallic structure rather than a molecular one, the simple substance is designated by the symbol of the element. For example, the simple substance phosphorus is denoted by the formula P, since, depending on conditions, phosphorus can consist of molecules with a different number of atoms or have a polymer structure.
The formula of the substance is determined based on the results of the analysis. For example, according to analysis, glucose contains 40% (wt.) carbon, 6.72% (wt.) hydrogen and 53.28% (wt.) oxygen. Therefore, the masses of carbon, hydrogen and oxygen are in the ratio 40:6.72:53.28. Let us denote the desired formula for glucose C x H y O z, where x, y and z are the numbers of carbon, hydrogen and oxygen atoms in the molecule. The masses of the atoms of these elements are respectively equal to 12.01; 1.01 and 16.00 amu Therefore, the glucose molecule contains 12.01x amu. carbon, 1.01u amu hydrogen and 16.00zа.u.m. oxygen. The ratio of these masses is 12.01x: 1.01y: 16.00z. But we have already found this relationship based on glucose analysis data. Hence:
12.01x: 1.01y: 16.00z = 40:6.72:53.28.
According to the properties of proportion:
x: y: z = 40/12.01:6.72/1.01:53.28/16.00
or x:y:z = 3.33:6.65:3.33 = 1:2:1.
Therefore, in a glucose molecule there are two hydrogen atoms and one oxygen atom per carbon atom. This condition is satisfied by the formulas CH 2 O, C 2 H 4 O 2, C 3 H 6 O 3, etc. The first of these formulas - CH 2 O- is called the simplest or empirical formula; it has a molecular weight of 30.02. In order to find out the true or molecular formula, it is necessary to know the molecular mass of a given substance. When heated, glucose is destroyed without turning into gas. But its molecular weight can be determined by other methods: it is equal to 180. From a comparison of this molecular weight with the molecular weight corresponding to the simplest formula, it is clear that the formula C 6 H 12 O 6 corresponds to glucose.
Thus, a chemical formula is an image of the composition of a substance using symbols of chemical elements, numerical indices and some other signs. The following types of formulas are distinguished:
— simplest , which is obtained experimentally by determining the ratio of chemical elements in a molecule and using the values of their relative atomic masses (see example above);
— molecular , which can be obtained by knowing the simplest formula of a substance and its molecular weight (see example above);
— rational , displaying groups of atoms characteristic of classes of chemical elements (R-OH - alcohols, R - COOH - carboxylic acids, R - NH 2 - primary amines, etc.);
— structural (graphic) , showing mutual arrangement atoms in a molecule (can be two-dimensional (in a plane) or three-dimensional (in space));
— electronic, displaying the distribution of electrons across orbitals (written only for chemical elements, not for molecules).
Let's take a closer look at the example of the ethyl alcohol molecule:
EXAMPLE 1
Exercise | With complete combustion of an oxygen-containing organic substance weighing 13.8 g, 26.4 g were obtained carbon dioxide and 16.2 g of water. Find the molecular formula of a substance if the relative density of its vapors with respect to hydrogen is 23. |
Solution | Let’s draw up a diagram of the combustion reaction of an organic compound, designating the number of carbon, hydrogen and oxygen atoms as “x”, “y” and “z”, respectively: C x H y O z + O z →CO 2 + H 2 O. Let us determine the masses of the elements that make up this substance. Values of relative atomic masses taken from the Periodic Table of D.I. Mendeleev, round to whole numbers: Ar(C) = 12 amu, Ar(H) = 1 amu, Ar(O) = 16 amu. m(C) = n(C)×M(C) = n(CO 2)×M(C) = ×M(C); m(H) = n(H)×M(H) = 2×n(H 2 O)×M(H) = ×M(H); Let's calculate the molar masses of carbon dioxide and water. As is known, the molar mass of a molecule is equal to the sum of the relative atomic masses of the atoms that make up the molecule (M = Mr): M(CO 2) = Ar(C) + 2×Ar(O) = 12+ 2×16 = 12 + 32 = 44 g/mol; M(H 2 O) = 2×Ar(H) + Ar(O) = 2×1+ 16 = 2 + 16 = 18 g/mol. m(C) = ×12 = 7.2 g; m(H) = 2 × 16.2 / 18 × 1 = 1.8 g. m(O) = m(C x H y O z) - m(C) - m(H) = 13.8 - 7.2 - 1.8 = 4.8 g. Let's determine the chemical formula of the compound: x:y:z = m(C)/Ar(C) : m(H)/Ar(H) : m(O)/Ar(O); x:y:z = 7.2/12:1.8/1:4.8/16; x:y:z = 0.6: 1.8: 0.3 = 2: 6: 1. This means the simplest formula of the compound is C 2 H 6 O and the molar mass is 46 g/mol. Meaning molar mass An organic substance can be determined using its hydrogen density: M substance = M(H 2) × D(H 2) ; M substance = 2 × 23 = 46 g/mol. M substance / M(C 2 H 6 O) = 46 / 46 = 1. This means the formula of the organic compound will be C 2 H 6 O. |
Answer | C2H6O |
EXAMPLE 2
Exercise | The mass fraction of phosphorus in one of its oxides is 56.4%. The oxide vapor density in air is 7.59. Determine the molecular formula of the oxide. |
Solution | The mass fraction of element X in a molecule of the composition NX is calculated using the following formula: ω (X) = n × Ar (X) / M (HX) × 100%. Let's calculate mass fraction oxygen in the compound: ω(O) = 100% - ω(P) = 100% - 56.4% = 43.6%. Let us denote the number of moles of elements included in the compound as “x” (phosphorus), “y” (oxygen). Then, the molar ratio will look like this (the values of relative atomic masses taken from D.I. Mendeleev’s Periodic Table are rounded to whole numbers): x:y = ω(P)/Ar(P) : ω(O)/Ar(O); x:y = 56.4/31: 43.6/16; x:y = 1.82:2.725 = 1:1.5 = 2:3. This means that the simplest formula for combining phosphorus with oxygen will be P 2 O 3 and a molar mass of 94 g/mol. The molar mass of an organic substance can be determined using its air density: M substance = M air × D air; M substance = 29 × 7.59 = 220 g/mol. To find the true formula of an organic compound, we find the ratio of the resulting molar masses: M substance / M(P 2 O 3) = 220 / 94 = 2. This means that the indices of phosphorus and oxygen atoms should be 2 times higher, i.e. the formula of the substance will be P 4 O 6. |
Answer | P4O6 |
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