Kamis, 04 Mei 2017

using english to calculate OF CHEMICAL STOIKIOMETRI



CALCULATION OF CHEMICAL STOIKIOMETRI
CALCULATION WITH CHEMICAL RESUME
AND EQUATION REACTIONS














One important aspect of chemical reactions is the quantitative relation between substances involved in chemical reactions, either as reagents or as a result of the reaction. Stoichiometry (stoi-kee-ah-met-tree) is a field in chemistry involving quantitative relationships between substances involved in chemical reactions, either as reagents or as a reaction product. 

Stoichiometry also involves the ratio of atoms between elements in a chemical formula, for example the ratio of the H atoms and the O atoms in the H2O molecule. The word stoichiometry comes from the Greek word stoicheon which means element and metron which means measure. A French chemist Jeremias BenjaminRichter (1762-1807) was the man who first laid down the basic principles of stoichiometry.

Why should we study stoichiometry? One reason, because studying chemistry can not be separated from doing experiments in the laboratory. Sometimes in our laboratory we have to react a number of grams of substance A to produce a number of grams of substance B. 
 Stoichiometry is closely related to chemical calculations.
To solve the problem of chemical calculation, the stoichiometric principles are
used, they are chemical equation and mole concept. In this lesson we will learn 
first about the principles of stoichiometry, then after that we will study
stoichiometric application on chemical calculations along with examples 
of problems and how to solve them.
The concept of a mole can be used to determine the chemical formula of 
a compound, both an empirical formula (the smallest comparison of 
atoms in a compound) and a molecular formula (number of atoms in a compound). 
1. Empirical formula, is the formula that gives the smallest comparison of 
the elemental atoms that make up each molecule of a compound2. The molecular 
formula is a formula that states exactly the atomic elements that make up 
each molecule of a compound.
 
Example: Compound Empirical formula Molecular formula
           Water        H2O           H2O
           Benzene      CH             C6H6
           Glucose      CH2O          C6H12O6
           amoniak      NH3           NH3 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
The empirical formula can be determined by knowing the mole ratio of the elements that constitute a
 compound. Whereas the molecular formula can be determined if the empirical formula and the relative 
molecular mass (Mr) of the compound are known.
The atomic ratio of the elements in a chemical formula is indicated by a round number, and not by a
 fractional number. For example, carbon monoxide (CO) has a ratio between C atoms and O atoms 
equal to 1: 1, which means the ratio of atoms to make 1 molecule of CO without any residual
 atom C or O atom we must take 1 atom C and 1 atom O according to The ratio of the atoms
 in the chemical formula of the compound.1 atom C + 1 atom O → 1 molecule CO To make 1 dozen CO molecules,
 we must take 1 dozen C atoms and 1 dozen O atoms according to the 1: 1 atomic ratio.
 1 dozen atoms C + 1 dozen atoms O → 1 dozen CO moleculesTo make 1 gross (144) CO molecule,
 the equation would be:1 gross atom C + 1 gross atom O → 1 gross molecule COSo any unit of number 
of atoms used, most important is the ratio of atoms C and the same O atom, ie 1: 1.
 According to Avogadro:"Gases of the same volume, if measured at temperatures and pressuresSame,
will have the same number of molecules ".
Therefore the ratio of the volume of hydrogen gas, oxygen gas, and water vaporIn the reaction 
of moisture formation = 2: 1: 2 then the ratio of the number of moleculesHydrogen, oxygen, 
and water vapor are also 2: 1: 2. The number of atoms of each element is notDecreases or increases 
in chemical reactions.
2. Avogadro's Mix and Fixed ConceptWhen you react one carbon atom (C)
with one molecule of oxygen (O2) it will form one molecule of CO2.
But actually what you react is not a single carbon atom with one molecule of oxygen,
but a large number of carbon atoms and a large number of oxygen molecules. 
Since the number of atoms or the number of molecules reacting is so great then to say it, 
the chemists use "mol" as the unit of the number of particles (molecules, atoms, or ions).
Gas Volume in Non-Standard StateCalculation of gas volume is not in the standard state (non-STP) 
used the following two approaches. A. The ideal gas equationAssuming the gas
to be measured is ideal, the equation that links the number of moles (n) of gas,
pressure, temperature, and volumethat is:
The ideal gas law:
P. V = n. R.
TWhere:
 P = pressure (atmospheric unit, atm)
 V = volume (liters, L)
 N = number of moles of gas (mol unit) R = gas constant (0.08205 L atm / mol K)
 T = absolute temperature (° C + 273.15 K)


With gas conversion at the same temperature and pressureAccording to Avogadro's law, 
the ratio of gases having the same number of moles has the same volume. Mathematically 
can be expressed as follows.V1 / V2 = n1 / n2Where:N1 = mol gas 1 V1 = gas volume
1N2 = mol gas 2 V2 = gas volume 2 4. Molarity (M)The amount of substances present
in a solution can be determined by using the concentration of the solution expressed 
in molarity (M). Molarity states the number of moles of substances in 1 L of solution. 
Mathematically stated as follows.Where:M = molarity (unit M)Mass = in units 
gMr. = molar mass (unit g / mol)V = volume (mL unit).
















D. Molecular Formulas and Elemental Content in Compounds

The ratio of mass and elemental content in a compound can 
be determined from its molecular formula.
1. Determination of Empirical Formulas and Molecular Formulas
The chemical formula shows the atomic type of element and the 
relative amount of each element contained in the substance.
 The number of substances contained in the substance isindicated by the index number.
The chemical formula can be an empirical formula and a molecular formula.
"Empirical formula, the formula which states the smallest comparison of atomatoms
Of the elements that make up the compound ".
"The molecular formula, the formula yamg states the number of atoms of
The elements that make up one molecule of the compound ".
Molecular formula = (Empirical formula) n
Mr. Molecular formula = n x (Mr. Empirical Formula, N = integers
The determination of the empirical formula and the molecular 
formula of a compound can be achieved by the following steps.
1. Find the mass (percentage) of each constituent compound,
2. Change to mole unit,
3. The mole ratio of each element is an empirical formula,
4. Find the molecular formula by: (Mr. empirical formula)
 n = Mr. molecular formula, n can be calculated,
5. Multiply n obtained from the count by the empirical formula.



A compound comprises 60% carbon, 5% hydrogen, and the remainder is nitrogen.
 Mr. compound it = 80 (Ar: C = 12; H = 1; N = 14). Determine 
the empirical formula and the molecular formula of the compound!
Answer:
Percentage of nitrogen = 100% - (60% + 5%) = 35%.
Suppose the mass of the compound = 100 g
Then mass C: H: N = 60: 5: 35
Comparison of mol C: mol H: mol N = 5: 5: 2,5 = 2: 2: 1
Then the empirical formula = (C2H2N) n. (Mr. empirical formula)
 n = Mr. molecular formula
(C2H2N) n = 80
(24 + 2 + 14) n = 80
40n = 80
N = 2
Thus, the molecular formula of the compound = (C2H2N) 2 = C4H4N2.
3.  Chemical Count
The determination of the amount of reactants and the reaction products involved in the
 reaction must be calculated in units of moles. That is, the units that are known must be 
converted into mol form. This method is called the mole approximation method.
4. Pereaksi Pembatas
In a chemical reaction, the mole ratio of the mixed reagents is not always the same as the 
ratio of the reaction coefficient. This means that there are reagents that will run out first.
Such reagents are called limiting reagents. How can this happen?
X + 2Y -> XY2
The above reaction shows that according to the reaction coefficient, one mole of X takes two moles of Y.
 The above figure shows that three molecules of X are reacted with four molecules 
of substance Y. After the reaction takes place, the number 
of X-substance molecules reacting only two molecules and one Molecule left.
 Meanwhile, four Y-substance molecules are reacting. Then this Y 
substance is called a limiting reagent. The limiting reagent is a 
reactant that reacts and does not remain at the end of the reaction. 
In chemical counts, the limiting reagents can be determined by dividing 
all the moles of the reactant by their coefficients, then the reactants 
having the smallest yield value being the limiting reagents.problems 
example Known reaction as follows S (s) + 3F2 (g) -> SF6 (g).If reacted 
with 2 mol of S with 10 moles of F2, determine:A. How many SF6 mols are formed?

S + 3F2 –> SF6
S + 3F2 -> SF6 
From the reaction coefficient shows that 1 mol of S requires 3 moles
of F2. The possibilities are as follows.A. If all S reacts then F2 
is required:This is possible because F2 is available 10 mol.






19 komentar:

  1. How gr mass of each? elements contained in 15.5 grams of calcium phosphate. Ca3 (PO4)2?

    BalasHapus
  2. What are the factors that affect stoichiometry?

    BalasHapus
    Balasan
    1. The factors that influence the occurrence of stiokiometri are: Amount of reagents, temperature, and volume,CONCENTRATION

      From various experiments shows that the greater the concentration of substances that react the faster the reaction takes place. The greater the concentration of the more substances that react so that the greater the likelihood of occurrence of collisions thus the greater the likelihood of reaction.
      CATALYST

      The catalyst is a substance added to a reaction in order to increase the reaction rate. The catalyst is sometimes involved in the reaction but does not undergo permanent chemical changes, in other words at the end of the catalyst reaction to be recovered in the same form and amount as before the reaction.

      Hapus
  3. Give another example of stoichiometry?

    BalasHapus
    Balasan
    1. Consider the following reaction
      H2 + CuO → Cu + H2O

      When hydrogen gas is required at 0 ° C and a pressure of 76 cmHg is 11.2 liters specify the mass of copper (Ar Cu = 63,50) generated in the above reaction:
      Discussion
      Mol hydrogen or H2 is
      Mol = 11.2 / 22.4 = 0.5 mol

      The coefficient of Cu is equal to the coefficient of H2, so its mole is also equal to 0.5 mol. The resulting Cu mass is thus:
      Gram = mol x Ar
      = 0.5 x 63,5
      = 31.74 grams

      Hapus
  4. Please explain the difference of empirical formula and molecular formula in chemical reaction?

    BalasHapus
    Balasan
    1. The empirical formula is the formula which expresses the smallest comparison of the atoms of the elements that constitute each molecule of a compound.

      The molecular formula is a formula that states exactly the number of atoms of the elements that make up each molecule of a compound.

      Compound Empirical formula Compound formula
      Water H2O H2O
      Benzene CH C6H6
      Glucose CH2O C6H12O6
      Ammonia NH3 NH3

      Theoretical results state the number or amount of products obtained from the reaction that went perfect. In practice, a reaction product can rarely achieve 100% accurate results, often resulting in less than 100%, sometimes even much lower, reaction yields. So, it is necessary to calculate the percentage of results obtained. The term commonly used to express this is the percentage of the Result which can be expressed as the equation below.

      Hapus
  5. Why in stoichiometry is there a mole approach?

    BalasHapus
    Balasan
    1. Stoichiometry is a subject in chemistry involving the linkage of reactants and products in a chemical reaction to determine the quantity of each reacting agent.Thus Stoichiometry is a chemical aspect involving the relation of various components in chemical reactions and quantitative relationships among the components.In the stoichiometry there is a mole which is the most important thing in getting the reaction

      Hapus
  6. What are the benefits of stoichiometry in everyday life?

    BalasHapus
    Balasan
    1. To solve problem solving problems, To solve the chemical calculation problems used stoichiometric principles, among others, chemical equations and the concept of moles.

      Hapus
  7. Explain stoichiometry in the gas phase!

    BalasHapus
    Balasan
    1. Gas Stoichiometry: The type of stoichiometry relates to a reaction involving a gas, in which the gas is at known temperature, pressure and volume and can be considered the ideal gas. For gas, the ideal volume ratio is equal to the ideal gas law, but the ratio of a single reaction mass 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 random, non-interacting particles that obey the ideal gas law. 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.

      Hapus

  8. One type of anaerobic respiration converts glucose, C6H12O6C_6 H_{12} O_6C​6​​H​12​​O​6​​C, start subscript, 6, end subscript, H, start subscript, 12, end subscript, O, start subscript, 6, end subscript, to ethanol C2H5OHC_2 H_5 OHC​2​​H​5​​OHC, start subscript, 2, end subscript, H, start subscript, 5, end subscript, O, H and carbon dioxide. If the molecular weight of glucose is 180180180180 grams/mol and the molar mass of ethanol is 50g/mol50 g/mol50g/mol50, g, slash, m, o, l, how many grams of carbon dioxide are produced when 1111 mol of glucose is digested via respiration?

    BalasHapus
  9. hi rini, could you explain about Chemical Count and give me example? thanks

    BalasHapus
  10. How many grams of potassium dichromate (K2Cr2O7) are needed to prepare a 250 ml solution of concentration of 2.16 M?
    Answer:
    Mol K2Cr2O7 = 250 mL solution x 2.16 mol K2Cr2O7
    1000 mL of solution
    = 0.540 mol of K2Cr2O7
    Mr K2Cr2O7 = 249.2 grams / mol, so the time required is
    Gram K2Cr2O7 = 0.540 mol K2Cr2O7 x 294.2 K2Cr2O7
    1 mol of K2Cr2O7
    = 159 grams of K2Cr2O7

    BalasHapus
  11. Please example how to calculate the volume of a solution

    BalasHapus
    Balasan
    1. The volume of HCl to be dissolved in 500 mL of water to obtain a solution having a percentage of 30% is ...
      30% = volume of HCl x 100%

      Volume HCl + 500

      30 (volume HCl + 500) = volume of HCl. 100

      30 volumes HCl + 15000 = 100. HCl volume

      15000 = (100 - 30) HCl volume

      Volume HCl = 15000 = 214.29 mL

      Hapus