USING ENGLISH FOR CALCULATE



Stoichiometry


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:
1.  Balance the equation.
2.  Convert units of a given substance to moles.
3.  Using the mole ratio, calculate the moles of substance yielded by the reaction.
4.  Convert moles of wanted substance to desired units.
These "simple" steps probably look complicated at first glance, but relax,
they will all become clear.

Let's begin our tour of stoichiometry by looking at the equation for how iron rusts:
Fe + O2→Fe2O3
   

Step 1. Balancing the Equation

The constituent parts of a chemical equation are never destroyed or lost: the yield of a reaction must exactly correspond to the original reagents. This fact holds not just for the type of elements in the yield, but also the number. Given our unbalanced equation:
Fe + O2→Fe2O3
   

This equation states that 1 iron (Fe) atom will react with two oxygen (O) atoms to yield 2 iron atoms and 3 oxygen atoms. (The subscript number, such as the two in O2 describe how many atoms of an element are in a molecule.) This unbalanced reaction can't possibly represent a real reaction because it describes a reaction in which one Fe atom magically becomes two Fe atoms.
Therefore, we must balance the equation by placing coefficients before the various molecules and atoms to ensure that the number of atoms on the left side of the arrow corresponds exactly to the number of elements on the right.
4Fe +3O2→2Fe2O3
   

Let's count up the atoms in this new, balanced version of the reaction. On the left of the arrow we have 4 atoms of iron and 6 atoms of oxygen (since 3×2 = 6). On the right we also have 4 iron (since 2×2 = 4 ) and 6 oxygen ( 2×3 = 6 ). The atoms on both sides of the equation match.
The process of balancing an equation is basically trial and error. It gets easier and easier with practice. You will likely start to balance equations almost automatically in your mind.


Step 2. Converting Given Units of a Substance to Moles

The process of converting given units into moles involves conversion factors. Below we will provide the most common and important conversion factors to convert between moles and grams, moles and volumes of gases, moles and molecules, and moles and solutions. Note also that though these conversion factors focus on converting from some other unit to moles, they can also be turned around, allowing you to convert from moles to some other unit.

1.      Converting from Grams to Moles

The gram formula mass of a compound (or element) can be defined as the mass of one mole of the compound. As the definition suggests, it is measured in grams/mole and is found by summing the atomic weights of every atom in the compound. Atomic weights on the periodic table are given in terms of amu (atomic mass units), but, by design, amu correspond to the gram formula mass. In other words, a mole of a 12 amu carbon atom will weigh 12 grams.

The gram formula mass can be used as a conversion factor in stoichiometric calculations through the following equation:
Moles = (grams) : (gram formula mass)

Gram formula mass is also known as GFM. You may also see the term gram molecular mass, abbreviated GMM. This term is often used instead of GFM when the substance is molecular and not ionic. However, only the terminology is different, GMM is used in the same way as GFM. Therefore, I will use the catch- all term GFM in this study guide.

2.      Converting between Volume of a Gas and Moles

The Ideal Gas law, discussed at length in the Sparknote on Gases, provides a handy means of converting between moles and a gas, provided you know certain qualities of that gas. The Ideal Gas Law isPV = nRT , with n representing the number of moles. If we rearrange the equation to solve for n , we get:
n = PV / RT

with P representing pressure in atm, V representing volume in liters, Trepresenting temperature in Kelvins, and R the gas constant, which equals .0821 L-atm/mol-K. Given P , V , and T , you can calculate the number of moles of substance in a gas.
In those instances when a problem specifies that the calculations are to be made at STP (Standard Temperature and Pressure; P = 1 atm, T= 273 K)), the problem becomes even simpler. At STP, a mole of gas will always occupy 22.4 L of volume. If you are given a volume of a gas at STP, you can calculate the moles in that gas by calculating the volume you are given as a fraction of 22.4 L. At STP, 11.2 L of a gas will be .5 moles; 89.6 L of gas will be 4 moles.

3.      Converting between Individual Particles and Moles
Avogadro's Number provides the conversion factor for moving from number of particles to moles. There are 6.02×1023formula units of particles in every mole of substance, with formula unit describing the substance we are looking at, whether it is a compound, molecule, atom, or ion. A formula unit is the smallest unit of a substance that still retains that substance's properties and is the simplest way to write the formula of the substance without coefficients. Some representative formula units are listed below.
·                     Compounds: Cu2S , NaCl
·                     Molecules: N2 , H2
·                     Atoms: Fe, Na
·                     Ions: Na+(aq) , Cl-(aq)
Since 1 mole = 6.02×1023 formula units, the conversion from formula units to moles is simple:
Moles = (formula units) : (6.02 x 1023
)

4.      Converting between Solutions and Moles
It is possible, and fairly easy to convert between the measures of solution (molarity and molality) and moles.
Molarity is defined as the number of moles of solute divided by the number of liters of solvent. Rearranging the equation to solve for moles yields:
Moles = molarity × liters of solution
Molality is defined as the number of moles of solute divided by the number of kilograms of solvent. Rearranging the equation to solve for moles yields:
Moles = molality × kilograms of solution


Step 3. Using the Mole Ratio to Calculate Yield

Before demonstrating how to calculate how much yield a reaction will produce, we must first explain what the mole ratio is.

The Mole Ratio
Let's look once again at our balanced demonstration reaction:
4Fe +3O2→2Fe2O3
   

The coefficients in front of iron, oxygen, and iron (III) oxide are ratios that govern the reaction; in other words, these numbers do not demand that the reaction can only take place with the presence of exactly 4 moles of iron and 3 moles of oxygen, producing 2 moles of iron (III) oxide. Instead, these numbers state the ratio of the reaction: the amount of iron and oxygen reaction together will follow a ratio of 4 to 3. The mole ratio describes exactly what its name suggests, the molar ratio at which a reaction will proceed. For example, 2 moles of Fe will react with 1.5 moles of O2 to yield 1 mole of Fe2O3 . Alternatively, 20 moles of Fe will react with 15 moles of O2 to yield 10 moles of Fe2O3 . Each of these examples of the reaction follow the 4:3:2 ratio described by the coefficients.
Now, with a balanced equation, the given units converted to moles, and our understanding of the mole ration, which will allow us to see the ratio of reactants to each other and to their product, we can calculate the yield of a reaction in moles.


Step 4. Convert moles of wanted substance to desired units

Demands that we be able to convert from moles to back to the units requested in a specific problem, but that only involves turning backwards the specific converstion factors described above.


Sample Problems

Problem 1: Given the following equation at STP:
N2(g) + H2(g)→NH3(g)
   
Determine what volume of H2(g) is needed to produce 224 L of NH3(g).
Solution:
Step 1: Balance the equation.
N2(g) + 3H2(g)→2NH3(g)
   
Step 2: Convert the given quantity to moles. Note in this step, 22.4 L is on the denominator of the conversion factors since we want to convert from liters to moles. Remember your conversion factors must always be arranged so that the units cancel.
 = 10 moles of NH3(g)
   
Step 3: Mole ratio.


= 15 moles H2(g)
   
Step 4: Convert to desired units
 = 336 L H2(g)
   


Problem 2: 
Given the following reaction:
2H2S(g) + O2(g)→SO2(g) + 2H2O(s)
   
How many atoms of oxygen do I need in order to get 18 g of ice?
Solution:
Step 1: The equation is partially balanced already, but let's finish the job.
2H2S(g) +3O2(g)→2SO2(g) + 2H2O(s)
   
Step 2: Convert to moles: 
1 formula unit of H2O has 2 atoms of H and 1 atom of O 
The atomic mass of H is 1 gram/mole
Atomic mass of O = 16 grams/mole
GFM of H2O(s) :
 +  =  18 grams / mole

   
×1 mole = 1 mole of H2O(s)
   

Step 3: Mole ratio.
×3 moles O2(g) = 1.5 moles O2(g)
   
Step 4: Convert to desired units.
 = 9.03×1023 molecules O2(g)

   

Is this the answer? No. The question asks for ATOMS of oxygen. There are two atoms of oxygen in each molecule of O2(g).

×2 atoms O = 1.806×1024 atoms

Komentar

  1. Why does lavoiser make stoichiometry an aspect that handles quantitative chemical reactions into basic chemical methodologies ?

    BalasHapus
    Balasan
    1. Lavoisier establishes the law of conservation of mass, and provides the basic concepts equivalent to its accurate and creative experiments. Thus, stoichiometry which is quantitatively responsible for chemical reactions becomes a basic chemical methodology. All the fundamental laws, from the law of conservation of the masses, the law of the law. These fundamental laws form the basis of atomic theory, and However, interesting to note, the equivalent concept was used before the atomic theory was introduced.

      Hapus
  2. Could you explain the meaning of stoichiometry as simply?

    BalasHapus
    Balasan
    1. Basic chemical calculations that represent quantitative relation of chemical formulas and chemical equations

      Hapus
  3. In the reaction between magnesium metal as much as 10 grams with 6 grams of oxygen according to the equation of the reaction:
    2Mg (s) + O2 (g) 2MgO (S)
    Apparently from the experiment generated 15 grams of magnesium as much as 1 gram. This fact is according to the law ..... (Ar Mg = 24, O = 16). Whether this is included in Dalton's law, lavoiser, Boyle, proust, or Gay lussac.

    BalasHapus
  4. Give me an example of applying the formula in Convert between Gas Volume and Moles

    BalasHapus
    Balasan
    1. The amount of gas molar volume can be determined by the ideal gas equation: PV = nRT

      P = pressure = 1 atm

      N = mol = 1 mole of gas

      T = temperature in Kelvin = 273 K

      R = gas constant = 0.082 liter atm / mol K

      Then:

         P V = nRT

      V = 1 x 0.082 x 273

      V = 22,389

      V = 22.4 liters

      Thus, the standard volume = VSTP = 22.4 Liter / mol.

      Can be formulated: V = n x Vm

      N = number of moles

      Vm = VSTP = molar volume

      Hapus
  5. how to Converting Given Units of a Substance to Moles

    BalasHapus
    Balasan
    1. Relation of Moles to Mass

      Before discussing the relationship of moles with the masses, you must first remember about Relative Atomic Mass (Ar) and Relative Molecular Mass (Mr). Still remember? Then we check your memories by doing the following.

      Calculate Mr. H2SO4 (Ar H = 1, S = 32, and O = 16)!
      Given the relative atomic mass (Ar) of several elements as follows.
          Ca = 40

          O = 16

          H = 1

      Determine the relative molecular mass (Mr) of Ca (OH) 2!

      Do you remember? Then we go directly to the next matter that is about the molar mass.

      The molar mass represents the mass possessed by 1 mol of substance, equal to the Ar or Mr.

      For element:

      1 mol element = Ar gram, then it can be formulated:

      Mass 1 mol of substance = Ar substance expressed in grams

                                          or

      The molar mass of the substance = large Ar of gram / mol

       

      For compounds:

      1 mol of compound = Mr gram, then it can be formulated:

      The mass of 1mol substances = Mr substances expressed in grams

                                          or

      The molar mass of the substance = large Mr. gram / mol substance

      Thus the difference between the molar mass and the relative molecular mass is in the units. The molar mass has a gram / mol unit while the relative molecular mass has no units.

      The relationship between mol and mass is:

      Quantity (in mol) = Mass of compound or element (gram) / Molar mass of compound or element (gram / mol)

      Hapus
  6. What is the relationship of moles and avogadro number?

    BalasHapus
    Balasan
    1. When in daily life we know the number of units in the form of dozens (12 pieces), kodi (20 pieces), and rim (500 pieces) then in chemistry we know moles as units of quantities. One mole is defined as the number of substances containing particles by the number of atoms in 12 grams of carbon-12. Since the mass of 1 C-12 atom is 1,992 × 10-23 gram, then the number of atoms in 1 mol (12 gram) C-12 can be calculated in the following way:
      Number of atoms
      This 6.02 x 1023 number is called the Avogadro number and is given the symbol L or NA. Thus, 1 mole of the substance is the amount of the substance containing 6.02 x 1023 particles. If n denotes the number of moles and N states the number of atoms or molecules, then the relation n and N can be written in the following formula.
      Mol

      Hapus
  7. I want to ask, what calculations are included in stoichiometry?

    BalasHapus
    Balasan
    1. A. MOL CONCEPT
      1. Mol = gram
                       Ar or Mr.
      2. Mol = number of particles
                            6.02 x 1023
          Number of particles = molecule / ion / atom
      3. Mol = volume
                            22.4 L (in STP state)
          When not in STP state (00C, 1 atm)
                       Mol 1 = volume 1
                       Mol 2 volume 2
      4. Mol = M x V
          M = Mol
                           V
          M = concentration (M)
          V = volume (L)
      Calculation steps:
      To connect one formula to another the formula is always changed in MOL

      B. CHEMICAL FORM
      1. Empirical Formulas
      The empirical formula is the formula which expresses the smallest ratio of the atoms that make up a compound
      Calculation steps:
      Comparison of the constituent atoms is equal to the ratio of mass (gr) or percent of mass divided by their respective Ar
      2. Molecular Formulas
      The molecular formula is a formula that states the number of actual atoms that make up a compound
      (Mr. empirical formula) x = Mr. molecular formula
      X = multiples

      C. COMPARATIVE LAW STATUTE (PROUST)
      Comparative law remains in AxBy:
      Mass A = x.Ar A
      Mass B y.Ar B
        Mass A = x.Ar A
      Mass AxBy Mr. AxBy
      % A = x.Ar A x 100%
                       Mr. AxBy
      % A = mass A x 100%
                      Mass AxBy

      D. EQUATION REACTIONS
      The reaction coefficient represents the mole ratio of the substances in the reaction
      If the moles of the reacted substance are not comparable, the mole of the substance used is a mole which, when divided by the coefficient, is the least yield

      The combustion reaction of organic compounds:
      CxHy + (x + ¼ y) O2 -> x CO2 + ½ y H2O
      Mol O2: mol CO2: mol H2O = (x + ¼ y): x: ½ y

      Warm up seny. Hydrate (containing water)
      Compounds.xH2O -> Compounds + xH2O
      X = mol H2O
                Mol Compounds

      E. GAS LAW
      Formula: PV = nRT
      P = pressure (atm)
      V = volume (L)
      N = mol substance
      R = 0.0082
      T = temperature (0K), x0C = (x + 273) 0K

      Reaction coefficient = volume ratio

      Hapus

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