Kamis, 25 Mei 2017

FINAL ASSIGNMENT WULANTRI AKHIRIA (STOICHIOMETRY)



STOICHIOMETRY
1.    Understanding Stoichiometry
Stoichiometry comes from two Greek syllables Stoicheion meaning "element" and Metron which means "measurement". 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. Simply stoichiometry is a subject in chemistry that studies the quantity of matter in a chemical reaction

2.    The laws governing stoichiometry
The stoichiometry relies on law is like fixed comparative law, double comparative law and also the law of conservation of the masses
a.    The law of mass conservation
Using the laws of physics is like the law of conservation of mass, which holds that the mass of the reactant is equal to the mass of the product, Stoichiometry is used to gather information about the amount of various elements used in a chemical reaction
b.   Comparative law remains
This law states that the chemical compound (substance consisting of 2 (two) or more elements) always contains the same proportion of an element (compound with one atom type) with mass
c.    Law of multiple comparison
This law is one of the basic laws of stoichiometry, aside from the law of fixed comparison. Sometimes it is also called Dalton's law. It is said that, if 2 (two) elements constitute more than one compound between them, then the mass ratio of a second element which joins the fixed mass over the first element of both will have a ratio of a small sum of the whole

3.    Mass Molar (Mr)
The mass of one mole of substance is called the molar mass (relative molecular mass). The magnitude of the material molar mass is the relative atomic mass or the relative molecular mass of a substance expressed in units of grams per mole. The mass of a substance is the multiplication of its mole mass (g / mol) with the mol of the substance (n). Thus the mole relationship of a substance with its mass can be expressed as follows :
Molar mass = mass: mol
Mass = mol x Mr / Ar (molar mass)

4.    Molar Volume (Vm)
The volume of one mole of a substance in a gas form is called the molar volume, denoted by Vm. Avogadro in his experiments concluded that 1 L of oxygen gas at 0 ° C and 1 atm pressure has a mass of 1.4286 g Then, according to Avogadro's law it can be concluded:
1 mol of gas O2 = 22.4 L
In accordance with Avogadro's law stating that at the same temperature and pressure, the same volume of gas contains the same number of molecules or the number of moles of each gas volume the same. Under the law, the volume of 1 mole of each gas in the standard state (temperature 0 ° C and pressure 1 m) is as follows:
Volume gas in standard state = 22.4 L
5.    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 :
M = mass/mr x(1000/V)
Where:
M = molarity (unit M)
Mass = in units g
Mr. = molar mass (unit g / mol)
V = volume (mL unit)
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10 komentar:

  1. Mass conservation law
    Comparative law remains
    Double comparison law
    What distinguishes these three laws is? Include them with each example!

    BalasHapus
    Balasan
    1. 1. the law of conservation of the masses
      The law of conservation of mass: "In a closed system, the mass of matter before and after the reaction is the same". The material changes we observe in everyday life generally take place in open containers.
      Example: (Fe2O3 (s)) whose mass is equal to the mass of iron and oxygen at first.
      Fe (s) + O2 (g) -> Fe2O3 (s),
      2. Comparable Law (Proust Law)
      Based on his research on various compounds, Proust concluded that "Comparison of the mass of the elements in a single compound is certain and fixed." The same compounds, though derived from different regions or made in different ways, turn out to have the same composition
      For example, the results of the analysis of sodium chloride salts from different regions
      3. Comparative Multiplier Law (Dalton Law)
      Proust's law is further developed by scientists for unsure elements that can form more than one type of compound. One of them was John Dalton (1766 - 1844). Dalton observes the existence of an order associated with the mass ratio of the elements in a compound
      For example: if the amount of carbon reacted to each one is 1 gram, then it is observed that on carbonmonoksida formed there will be 1.33 grams of oxygen and 2.67 grams of oxygen in carbon dioxide. The oxygen mass ratio is close to 2: 1, which is a simple integer ratio

      Hapus

  2. Describe what to consider in determining Mr.?

    BalasHapus
    Balasan
    1. Relective molecule is the result of multiplication between index and coefficient. The index states the number of atoms of each element in front of it. If there is a double index (index inside brackets and indexes outside the brackets), then first multiplication of the index to get the index will be multiplied by the coefficient later.
      The coefficient denotes the total number of atomic elements present behind them. If the index and coefficients are not written then the index and the coefficient is 1.
      AXb
      Index writing and coefficients are denoted as follows
      Where,
      A = coefficient
      B = index
      X = element symbol

      Hapus
  3. why Comparative law remains "This law states that the chemical compound (substance consisting of 2 (two) or more elements) always contains the same proportion of an element (compound with one atom type) with mass" . please give me example

    BalasHapus
    Balasan
    1. Joseph Proust (1754-1826) conducted experiments, which reacted elements of hydrogen and oxygen elements. After he conducted the experiment he finally found that the element of hydrogen and oxygen elements always react to form water compounds with a constant mass ratio of 1: 8
      Hydrogen mass: oxygen mass = 1: 8
      So he also states that the chemical compound (substance consisting of 2 (two) or more elements) always contains the same proportion of an element (compound with one atom type) with mass
      Example:
      If we react 4 g of hydrogen with 40 g of oxygen, how many g of water is formed?
      Answer:
      Comparison of hydrogen mass with oxygen = 1: 8.
      Comparison of hydrogen mass with oxygen mixed = 4: 40.
      Therefore the ratio of hydrogen and oxygen = 1: 8 then 4 gr of hydrogen
      In this case the mixed oxygen does not react entirely, the oxygen remains as much (40 - 32) g = 8 g. Well, now we will calculate how much water mass is formed from 4 g of hydrogen and 32 g of oxygen.
      The answer is of course 36 g. Requiring 4 x 8 g of oxygen ie 32 g

      Hapus
  4. Why we should change the unit of mass in the example ?

    BalasHapus
    Balasan
    1. Because to balance the reaction of the coveted mole in the known mole then the unit is gram because the mole is the same gram at the same value whereas if the known unit is mmol then the mass unit must follow the units known or in question is ml or mg

      Hapus
  5. Explain about the concept mol and give an example?

    BalasHapus
    Balasan
    1. The definition of moles is the unit that states the number of particles contained in a number of substances. Under the agreement, a standard 12-carbon atom (C-12) is used with the following conditions:One mole of the substance is a number of particles contained in a substance equal to the number of atoms present in 12 grams of C-12.From the experiments performed by John Lochsmid and then justified by Avogadro through his experiments it turns out that the number of carbon atoms contained in 12 grams of C-12 is 6.02 x 10 ^23 atomic grains. This number is hereinafter called Avogadro number or avogadro constant and given the symbol L (taken from Lochsmid name)
      example:
      What is the number of carbon atoms contained in 0.05 mol of carbon?
      Resolution:
      In 1mol carbon contains 6.02 x 10 ^23 carbon atoms.
      In 0.005 mol the carbon will contain as many carbon atoms:
      (0.05mol / 1mol) x 6.02 x 10 ^23 atoms C = 3.01 x 10 ^23 atoms

      Hapus