Stoichiometry





Stoichiometry is founded on the law of conservation of mass where the total mass of the reactants equals the total mass of the products leading to the insight that the relations among quantities of reactants and products typically form a ratio of positive integers. This means that if the amounts of the separate reactants are known, then the amount of the product can be calculated. Conversely, if one reactant has a known quantity and the quantity of product can be empirically determined, then the amount of the other reactants can also be calculated.
This is illustrated in the image here, where the balanced equation is:
CH
4 + 2 O
2 → CO
2 + 2 H
2O.
Here, one molecule of methane reacts with two molecules of oxygen gas to yield one molecule of carbon dioxide and two molecules of water. Stoichiometry measures these quantitative relationships, and is used to determine the amount of products/reactants that are produced/needed in a given reaction. Describing the quantitative relationships among substances as they participate in chemical reactions is known as reaction stoichiometry. In the example above, reaction stoichiometry measures the relationship between the methane and oxygen as they react to form carbon dioxide and water.
Because of the well known relationship of moles to atomic weights, the ratios that are arrived at by stoichiometry can be used to determine quantities by weight in a reaction described by a balanced equation. This is called composition stoichiometry.
Gas stoichiometry deals with reactions involving gases, where the gases are at a known temperature, pressure, and volume and can be assumed to be ideal gases. For gases, the volume ratio is ideally the same by the ideal gas law, but the mass ratio of a single reaction has to be calculated from the molecular masses of the reactants and products. In practice, due to the existence of isotopes, molar masses are used instead when calculating the mass ratio.

Etymology
The term stoichiometry was first used by Jeremias Benjamin Richter in 1792 when the first volume of Richter's Stoichiometry or the Art of Measuring the Chemical Elements was published. The term is derived from the Greek words στοιχεῖον stoicheion "element" and μέτρον metron "measure". In patristic Greek, the word Stoichiometria was used by Nicephorus to refer to the number of line counts of the canonical New Testament and some of the Apocrypha.

A stoichiometric amount [1] or stoichiometric ratio of a reagent is the optimum amount or ratio where, assuming that the reaction proceeds to completion:
  1. All of the reagent is consumed
  2. There is no deficiency of the reagent
  3. There is no excess of the reagent.
Stoichiometry rests upon the very basic laws that help to understand it better, i.e., law of conservation of mass, the law of definite proportions (i.e., the law of constant composition), the law of multiple proportions and the law of reciprocal proportions. In general, chemical reactions combine in definite ratios of chemicals. Since chemical reactions can neither create nor destroy matter, nor transmute one element into another, the amount of each element must be the same throughout the overall reaction. For example, the number of atoms of a given element X on the reactant side must equal the number of atoms of that element on the product side, whether or not all of those atoms are actually involved in a reaction.
Chemical reactions, as macroscopic unit operations, consist of simply a very large number of elementary reactions, where a single molecule reacts with another molecule. As the reacting molecules (or moieties) consist of a definite set of atoms in an integer ratio, the ratio between reactants in a complete reaction is also in integer ratio. A reaction may consume more than one molecule, and the stoichiometric number counts this number, defined as positive for products (added) and negative for reactants (removed).[2]
Different elements have a different atomic mass, and as collections of single atoms, molecules have a definite molar mass, measured with the unit mole (6.02 × 1023 individual molecules, Avogadro's constant). By definition, carbon-12 has a molar mass of 12 g/mol. Thus, to calculate the stoichiometry by mass, the number of molecules required for each reactant is expressed in moles and multiplied by the molar mass of each to give the mass of each reactant per mole of reaction. The mass ratios can be calculated by dividing each by the total in the whole reaction.
Elements in their natural state are mixtures of isotopes of differing mass, thus atomic masses and thus molar masses are not exactly integers. For instance, instead of an exact 14:3 proportion, 17.04 kg of ammonia consists of 14.01 kg of nitrogen and 3 × 1.01 kg of hydrogen, because natural nitrogen includes a small amount of nitrogen-15, and natural hydrogen includes hydrogen-2 (deuterium).
A stoichiometric reactant is a reactant that is consumed in a reaction, as opposed to a catalytic reactant, which is not consumed in the overall reaction because it reacts in one step and is regenerated in another step.

Converting grams to moles

Stoichiometry is not only used to balance chemical equations but also used in conversions, i.e., converting from grams to moles using molar mass as the conversion factor, or from grams to milliliters using density. For example, to find the amount of NaCl (sodium chloride) in 2.00 g, one would do the following:
2.00  g NaCl 58.44  g NaCl mol − 1 = 0.034   mol {\displaystyle {\frac {2.00{\mbox{ g NaCl}}}{58.44{\mbox{ g NaCl mol}}^{-1}}}=0.034\ {\text{mol}}}
In the above example, when written out in fraction form, the units of grams form a multiplicative identity, which is equivalent to one (g/g = 1), with the resulting amount in moles (the unit that was needed), as shown in the following equation,
( 2.00  g NaCl 1 ) ( 1  mol NaCl 58.44  g NaCl ) = 0.034   mol {\displaystyle \left({\frac {2.00{\mbox{ g NaCl}}}{1}}\right)\left({\frac {1{\mbox{ mol NaCl}}}{58.44{\mbox{ g NaCl}}}}\right)=0.034\ {\text{mol}}}

Molar proportion

Stoichiometry is often used to balance chemical equations (reaction stoichiometry). For example, the two diatomic gases, hydrogen and oxygen, can combine to form a liquid, water, in an exothermic reaction, as described by the following equation:
H
2 + O
2 → 2 H
2O
Reaction stoichiometry describes the 2:1:2 ratio of hydrogen, oxygen, and water molecules in the above equation.
The molar ratio allows for conversion between moles of one substance and moles of another. For example, in the reaction
CH
3OH + 3 O
2 → 2 CO
2 + 4 H
2O
The term stoichiometry is also often used for the molar proportions of elements in stoichiometric compounds (composition stoichiometry). For example, the stoichiometry of hydrogen and oxygen in H2O is 2:1. In stoichiometric compounds, the molar proportions are whole numbers.

Determining amount of product        

Stoichiometry can also be used to find the quantity of a product yielded by a reaction. If a piece of solid copper (Cu) were added to an aqueous solution of silver nitrate (AgNO3), the silver (Ag) would be replaced in a single displacement reaction forming aqueous copper(II) nitrate (Cu(NO3)2) and solid silver. How much silver is produced if 16.00 grams of Cu is added to the solution of excess silver nitrate?
The following steps would be used:
  1. Write and balance the equation
  2. Mass to moles: Convert grams of Cu to moles of Cu
  3. Mole ratio: Convert moles of Cu to moles of Ag produced
  4. Mole to mass: Convert moles of Ag to grams of Ag produced
The complete balanced equation would be:
Cu + 2 AgNO
3Cu(NO
3)
2 + 2 Ag

Further examples

For propane (C3H8) reacting with oxygen gas (O2), the balanced chemical equation is:
C
3H
8 + 5 O
2 → 3 CO
2 + 4 H
2O
The mass of water formed if 120 g of propane (C3H8) is burned in excess oxygen is then
m H 2 O = ( 120.  g  C 3 H 8 1 ) ( 1  mol  C 3 H 8 44.09  g  C 3 H 8 ) ( 4  mol  H 2 O 1  mol  C 3 H 8 ) ( 18.02  g  H 2 O 1  mol  H 2 O ) = 196  g {\displaystyle m_{\mathrm {H_{2}O} }=\left({\frac {120.{\mbox{ g }}\mathrm {C_{3}H_{8}} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {C_{3}H_{8}} }{44.09{\mbox{ g }}\mathrm {C_{3}H_{8}} }}\right)\left({\frac {4{\mbox{ mol }}\mathrm {H_{2}O} }{1{\mbox{ mol }}\mathrm {C_{3}H_{8}} }}\right)\left({\frac {18.02{\mbox{ g }}\mathrm {H_{2}O} }{1{\mbox{ mol }}\mathrm {H_{2}O} }}\right)=196{\mbox{ g}}}

Stoichiometric ratio

Stoichiometry is also used to find the right amount of one reactant to "completely" react with the other reactant in a chemical reaction – that is, the stoichiometric amounts that would result in no leftover reactants when the reaction takes place. An example is shown below using the thermite reaction,
Fe
2O
3 + 2 AlAl
2O
3 + 2 Fe
This equation shows that 1 mole of iron(III) oxide and 2 moles of aluminum will produce 1 mole of aluminium oxide and 2 moles of iron. So, to completely react with 85.0 g of iron(III) oxide (0.532 mol), 28.7 g (1.06 mol) of aluminium are needed.
m A l = ( 85.0  g  F e 2 O 3 1 ) ( 1  mol  F e 2 O 3 159.7  g  F e 2 O 3 ) ( 2  mol Al 1  mol  F e 2 O 3 ) ( 26.98  g Al 1  mol Al ) = 28.7  g {\displaystyle m_{\mathrm {Al} }=\left({\frac {85.0{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }{159.7{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {2{\mbox{ mol Al}}}{1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {26.98{\mbox{ g Al}}}{1{\mbox{ mol Al}}}}\right)=28.7{\mbox{ g}}}

Different stoichiometries in competing reactions

Often, more than one reaction is possible given the same starting materials. The reactions may differ in their stoichiometry. For example, the methylation of benzene (C6H6), through a Friedel–Crafts reaction using AlCl3 as a catalyst, may produce singly methylated (C6H5CH3), doubly methylated (C6H4(CH3)2), or still more highly methylated (C6H6−n(CH3)n) products, as shown in the following example,
C6H6 + CH3Cl → C6H5CH3 + HCl
C6H6 + 2 CH3Cl → C6H4(CH3)2 + 2 HCl
C6H6 + n CH3Cl → C6H6−n(CH3)n + n HCl
In this example, which reaction takes place is controlled in part by the relative concentrations of the reactants.
https://en.wikipedia.org/wiki/Stoichiometry

Comments

  1. The principle that underlies stoichiometry?

    ReplyDelete
    Replies
    1. yes , ok i'll explain about your question The word stoichiometry comes from the Greek word stoicheon which means element and metron which means measure. A French chemist Jeremias Benjamin Richter (1762-1807) was the man who first laid down the basic principles of stoichiometry. He thinks stoichiometry is the science of measuring quantitative comparisons or measuring comparisons between chemical elements with each other.

      Delete
  2. The principle that underlies stoichiometry?

    ReplyDelete
  3. Can you explain why For gases, the volume ratio is ideally the same by the ideal gas law, but the mass ratio of a single reaction has to be calculated from the molecular masses of the reactants and products?

    ReplyDelete
    Replies
    1. The stoichiometric type is related to a reaction involving a gas, in which the gas is at known temperature, pressure and volume and can be an ideal gas. For gas, the ideal volume ratio is equal to the ideal gas law, the ratio of a single reaction ratio must be calculated from the molecular mass of the reactants and products, where the molecular mass is the mass of 1 molecule of the substance. The ideal gas is a theoretical gas consisting of a set of randomly moving particles, without an ideal climate-law gas interaction. The ideal gas law is the ideal gas state equation. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume and T is the absolute temperature, n is the gas mole and R is the universal gas constant.

      Delete
  4. Try to explain about:
    1. All reagents are consumed
    2. There is no shortage of reagents
    3. No excess reagents.

    ReplyDelete
    Replies
    1. i'll try to answer , and sorry i just answer who i know , no at all all the reagents are added in exactly the right mole ratio then all of each reagent is used up, so there is no excess or limiting reagent.

      In this case you can use either reagent moles to work out the amount of product formed.

      Example:

      2AgNO3+FeCl2→2AgCl(s)+Fe(NO3)2

      The mole ratio of AgNO3 : FeCl2 is 2:1

      If you add
      2 moles of AgNO3 to
      1 mole of FeCl2
      then both reagents will be fully consumed so neither is limiting.

      The reaction will produce
      2 moles of AgCl(s) and
      1 mole of Fe(NO3)2

      Delete
  5. Can you give me an example of stoichiometry other than in chemistry?

    ReplyDelete
    Replies
    1. ok , i'll explain To make a delicious cup of coffee, a recipe is required that is 9 cube sugar with 3 tablespoons of coffee.
      Stoichiometric illustration in Life
      Yes! The answer is 1 cup of coffee, with the remaining ingredients 3 cube sugar.
      How about we have 27 sugar cube and 8 spoons of coffee. How many cups of coffee can be made?
      Of course 2 glasses of coffee with the remaining 9 sugar cube and 2 coffee spoons. All absolutely must follow the recipe.

      Delete
  6. how must important stoichiometry to calculate?

    ReplyDelete
    Replies
    1. Now you're ready to use what you know about conversion factors to solve some stoichiometric problems in chemistry. Almost all stoichiometric problems can be solved in just four simple steps:

      Balance the equation.
      Convert units of a given substance to moles.
      Using the mole ratio, calculate the moles of substance yielded by the reaction.
      Convert moles of wanted substance to desired units.

      These "simple" steps probably look complicated at first glance, but relax, they will all become clear.

      Delete
  7. Could you re-explain about catalytic reactant?

    ReplyDelete
    Replies
    1. he two types of catalyst you will meet are called homogeneous and heterogeneous catalysts. Whilst they achieve the same result they work in fundamentally different ways.
      Homogeneous Catalysts

      With homogeneous catalysts the catalyst and all the reactants are in a single phase usually either liquid or an aqueous solution. Invariably the catalyst is taken up into the chemical reaction before being ‘spat out’ at the end.

      Catalyst Energy Diagram

      The following example is a real, though economically valueless, catalytic reaction. It is helpful in understanding how catalytic reactions can occur.

      Persulphate ions (S2O82-) are powerful oxidants and iodide ions (I-) are easily oxidised. However in an aqueous solution the reaction runs very slowly; it should be easy to see why:

      Equation

      If we add a small quantity of iron ions, say Fe2+ then the reaction has another route.

      Equation

      If we had started with Fe3+ ions then the reaction would simply have taken the opposite order (not the reverse reaction though).

      Equation

      The important points are that the same overall reaction is achieved and that the catalyst is returned unaltered at the end of the reaction.

      What could be preventing the reaction between the persulphate ions and the iodide ions in the uncatalysed reaction?

      Why should the introduction of Fe2+ or Fe3+ ions speed up the reaction?

      What other stage should be introduced in order to exactly duplicate the uncatalysed reaction?

      Heterogeneous Catalysts

      Most industrial catalysed reactions are of the heterogeneous type. Here the reactants and the catalyst are in different phases. Usually the catalyst is in the solid phase whilst the reactants are either liquid or gaseous. In industrial applications the catalyst is often supported on a substrate that allows the effective surface area to be increased and the catalyst to be fixed in place.

      The route by which an heterogeneous catalyst works is as follows:

      Catalyst Energy Diagram

      Firstly the reactants are adsorbed on to the surface of the catalyst. This is a chemical reaction as there is an interaction between the electrons of the reactants and the atoms on the surface of the catalyst.

      (Remember that adsorption and absorption are different. In adsorption a molecule binds to the surface of the material whilst in absorption it is taken in to the body of the material.)

      Secondly the adsorbed reactants (particularly the lighter ones such as hydrogen) are free to migrate over the surface of the catalyst.

      Thirdly, when the reactants meet they are free to react but are still bound to the surface.

      Finally the products of the reaction are desorbed from the surface allowing them to move away and freeing up catalytic surface area for further reactions.

      This points up that a good catalyst should bind moderately well to the intended reactants and products.

      Delete
  8. "please explain about law's stoychiometry "

    ReplyDelete
    Replies
    1. toichiometry rests upon the laws such as law of definite proportion and law of multiple proportions and law of conservation of mass.

      Law of conservation of mass: Using the physical laws such as law of conservation of mass, which states that the mass of the reactants equals the mass of the products, Stoichiometry is used to gather information about the amounts of various elements used in a chemical reaction, and whether they took the form of gases, solids or liquids.
      Law of definite proportion: It states that a chemical compound (substance consisting of 2 or more elements) always contains exactly the same proportion of elements (substance with one type of atom) by mass.
      Law of multiple proportions: It is one of the basic laws of stoichiometry, along side the law of definite proportions. It is sometimes called as Dalton’s law. It says that, if 2 elements form more than 1 compound between them, then the ratio of the masses of the second element which combine with the fixed mass of the first element will both ratios of small whole number.

      Delete
  9. Give me an example how to find Mr ?

    ReplyDelete
    Replies
    1. Molecular/formula mass = total of all the atomic masses of ALL the atoms in the molecule/compound.

      Watch out for brackets e.g. (OH)2 means two OH groups to add up!

      Relative molecular/formula mass calculation Example 2.1
      The diatomic molecules of the elements hydrogen H2 and chlorine Cl2
      relative atomic masses, Ar: H = 1, Cl = 35.5
      Formula masses, RMM or Mr
      relative molecular mass for hydrogen H2 = 2 x 1 = 2
      relative molecular mass for chlorine Cl2 = 2 x 35.5 = 71 respectively.

      Delete
  10. What all law of sthoiciometric can applicate in same quation??

    ReplyDelete
    Replies
    1. no , because all of law had difference using for !

      Delete
  11. 'Make an example of a problem that involves chemical calculation with its solution?'

    ReplyDelete
    Replies
    1. olution

      This sort of problem is solved by using a table to look up the change in enthalpy unless it's given to you (as it is here). The thermochemical equation tells us that ΔH ​for the decomposition of 1 mole of H2O2 is -98.2 kJ, so this relationship can be used as a conversion factor.

      Once you know the change in enthalpy, you need to know the number of moles of the relevant compound to calculate the answer. Using the Periodic Table to add up the masses of hydrogen and oxygen atoms in hydrogen peroxide, you find the molecular mass of H2O2 is 34.0 (2 x 1 for hydrogen + 2 x 16 for oxygen), which means that 1 mol H2O2 = 34.0 g H2O2.

      Using these values:

      ΔH = 1.00 g H2O2 x 1 mol H2O2 / 34.0 g H2O2 x -98.2 kJ / 1 mol H2O2

      ΔH = -2.89 kJ

      Delete
  12. This comment has been removed by a blog administrator.

    ReplyDelete
  13. This comment has been removed by a blog administrator.

    ReplyDelete
  14. Give examples Converting grams to moles

    ReplyDelete
    Replies
    1. Grams To Moles Conversion Problem

      Determine the number of moles of CO2 in 454 grams of CO2.

      Solution

      First, look up the atomic masses for carbon and oxygen from the Periodic Table. The atomic mass of C is 12.01 and the atomic mass of O is 16.00. The formula mass of CO2 is:

      12.01 + 2(16.00) = 44.01

      Thus, one mole of CO2 weighs 44.01 grams. This relation provides a conversion factor to go from grams to moles. Using the factor 1 mol/44.01 g:

      moles CO2 = 454 g x 1 mol/44.01 g = 10.3 moles

      Answer

      There are 10.3 moles CO2 in 454 grams of CO2

      Delete
  15. Replies
    1. Determine the number of moles of CO2 in 454 grams of CO2.

      Solution

      First, look up the atomic masses for carbon and oxygen from the Periodic Table. The atomic mass of C is 12.01 and the atomic mass of O is 16.00. The formula mass of CO2 is:

      12.01 + 2(16.00) = 44.01

      Thus, one mole of CO2 weighs 44.01 grams. This relation provides a conversion factor to go from grams to moles. Using the factor 1 mol/44.01 g:

      moles CO2 = 454 g x 1 mol/44.01 g = 10.3 moles

      Answer

      There are 10.3 moles CO2 in 454 grams of CO2

      Delete
  16. Why Ar is verry important in stoichiometry?

    ReplyDelete
    Replies
    1. because Dalton recognizes that it is important to determine the mass of each atom because its mass varies for each type of atom. Atoms are so small that it is impossible to determine the mass of one atom. So he focused on the relative mass values ​​and made the atomic mass table (figure 1.3) for the first time in human history. In the table, the lightest element mass, hydrogen set one as standard (H = 1). The atomic mass is a relative value, meaning a dimensionless ratio. Although some atomic masses differ from modern values, most of the proposed values ​​are in the range of compatibility with current values. This shows that his ideas and experiments are correct.

      Delete

Post a Comment

Popular posts from this blog

composition of matter in chemistry

chemical reaction in everyday life

Discussion about ionic bond and covalent bond