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
Chemistry
is a part of natural science that studies matter that includes the composition,
nature, and change of matter and energy that accompanies material changes.
Careful research on reagents and reaction products has spawned basic chemical
laws that show a quantitative or so-called stoichiometric relationship. The
basic chemical laws are the law of conservation of mass, the law of fixed
comparison and the law of multiple comparisons. The basic laws of chemistry are
our foundation in studying and developing the next chemistry
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
Problems example :
What is the volume of CO2 gas whose mass is 22 g (Ar: C
= 12, O = 16) when measured at 1 atm pressure?
Answer:
Mr. CO2 = 44
5.
Gas Volume in Non-Standard State
The
calculation of gas volume is not in the standard state (non-STP) used two
approaches as follows:
a.
The ideal gas equation
Assuming
that the gas to be measured is ideal, the equations that connect the number of
moles (n) of gas, pressure, temperature, and volume are:
The ideal gas law: P. V = n. R. T
Where:
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 K)
b.
With
gas conversion at the same temperature and pressure
According
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 / n2
Where:
N1 = mol of gas 1
V1 = gas volume 1
N2 = mol gas 2
V2 = gas volume 2
6.
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 = massa/mr x 1000/V
Where:
M = molarity (unit M)
Mass = in units g
Mr. = molar mass (unit g / mol)
V = volume (mL unit)
7.
Empirical Formulas and Molecular Formulas
The empirical formula is the simplest integer ratio of the number
of moles of each element in a compound. The molecular formula represents the
actual number of moles of each element in 1 mole of the compound. The molecular
formula can be identical to the empirical formula or an integer multiple of the
empirical formula. For example, phosphoric acid (H3PO4) has a molecular formula
and an identical empirical formula. Glucose has a molecular formula C6H12O6
which is a multiple of 6 times the empirical formula, CH2O
Molecular formula = (Empirical formula) n
Mr. Molecular formula = n x (Mr. Empirical Formula
N = integers
problems example :
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
What is the relationship between the molality molarity?
BalasHapusBetween the molarity and molality of different basic calculations but the formula is similar. Molarity is based on the volume of the solution (Liter), whereas the molality is based on the mass of the solvent (kg).
HapusIn molarity, the solution is measured in units of volume (volume of solute + volume of solvent). When making a solution with molar units the solute is inserted in the flask and then added the solvent until the volume reaches the limit.
1 molar = 1 mol of solute dissolved into the solvent until the solution volume becomes 1 liter
In molality, 1 mole of solute is dissolved in 1 kg of solvent so it is important to measure the solvent as much as 1 kg or 1 liter (if the solvent type is 1 g / mL)
1 molal = 1 mol of solute dissolved into 1 kg of solvent
Simulation of molar calculation
Available HCl with 36% content and density (density) of 1.2 kg / dm-3. HCl solution will be prepared with 1 molar concentration of 1 liter. How much mL HCl should be taken then dissolved in aqueous solvent to 1 liter solution? Known molar mass of HCl = 36.5 g / mol
One molar of 1 liter volume means we need 1 ml of HCl. Remember 1 M = 1 mol in 1 Liter of solution
1 mol HCl = 36.5 g HCl remember, so we need HCl as much as 36.5 g.
But the available HCl is not in grams. Therefore it needs to be converted from the available data. The question is how many mL of HCl should be taken to equal the mass of 36.5 g?
Unit value kg / dm-3 = kg / L = g / mL
Suppose 100% with a density of 1.2 g / mL then every 1 mL HCl = 1.2 g
Because the levels are only 36% then the mass of each 1 mL of the HCl
= 36% x 1.2 g / mL = 0.432 g → 1 mL = 0.432 g ↔ 0.432 g = 1 mL
Ok, if 1 g of HCl is equivalent to how 36 mL of HCl is?
1 g = → 1 g = 2,315 mL
Because the HCl needs as much as 36.5 g equivalent to (36.5 g x 2.315 mL / g) = 84.49 mL
See how much we have to take to make 1 M HCl of 1 liter, the available HCl we take as much as 84.49 mL then we dissolve in water solvent until the volume of the solution (volume of HCl + volume of water) becomes 1 liter. Practice is to prepare volumetric flask with a dose of 1 L, fill a little pure water (eg 100 mL) first, enter 84.49 mL HCl and then add water to the limit of 1 L of volumetric flask.
How to technically make a solution despite its different substances but can be seen in the video with the title of acid-base reaction from ITB below.
Well if we are going to make a 1M HCl solution only 0.5 L means we only need to take 42.245 mL (obtained from 0.5 x 84.49 mL) and then dissolved until the volume of the solution to 500 mL, use 500mL volumetric flask well, so etc.
Simulation of molal calculation (m)
How about making a 1 m HCl solution from the available HCl?
Practically the calculation of molality is equal to the calculation of molarity, the difference is in the mass of the solvent. Since 1 m equals 1 mole of solute dissolved in 1 kg of solvent, and if the solvent is water with a density of 1 g / mL then 1 kg is equivalent to 1 L. Remember 1 L of pure water.
If asked, 1 M HCl equals how many m HCl?
Molar to Molal Conversion:
1 M HCl = 1 mol, HCl = 36.5 grams dissolved in 1 liter of solution, with a density of 1.2 g / mL
Then look for mass of solution that is;
1.2 = x gram solution / 1000 ml of solution
X = 1200 grams of solution
Only solvent mass = (1200 - 36.5 g) = 1163.5 grams solvent only. (WATER)
M = 1 mol / (1163,5 / 1000) kg water = 0.859475719 molals.
Can we change the number of gas constant? Why?
BalasHapusWe can not change the gas contours because the gas constants are determined exactly. The gas constants (also called ideal, molar, universal, or universal gas constants, usually denoted by the letter R) are a physical constant that often appears in many fundamental physical equations, such as the ideal gas law and Nernst's equation. This constant is equivalent to the Boltzmann constant, but expressed in units of energy per kelvin per mole (rather than energy per kelvin per particle).
HapusThe price is:
R = 8.314472 (15) J · K-1 · mol-1
The two digits inside the brackets are the uncertainty (standard deviation) at the last two digits.
BalasHapusWhat distinguishes molarity from molality?
Molarity is the number of moles contained in each liter of the volume of the solvent
HapusM = n / V
M = molarity (molar)
N = mol
V = volume (liters)
While Molality is the number of moles contained in each kilogram of the mass of the solvent
M = n / p
M = molality (molal)
N = mol
P = mass (kilogram)
What is the molar mass of H2O?
BalasHapusIf it is known Ar H = 1; Ar O = 16
HapusWhat is the mass of 2 moles of H2O?
Answer:
Mr. H2O = (2. Ar H) + (1. Ar O)
= (2 x 1) + (1 x 16)
= 18
The molar mass of H2O = 18 g / mol
Whether at the time of adding temperature at The ideal gas equation used 273 or 273,15?
BalasHapusThe addition of temperature to the ideal gas that was from celcius to change to the temperature of kelvin is by adding 273 of the existing temperature
HapusWhat is relationship between pleasure and volume?
BalasHapusYour sense of purpose is the difference in pressure and volume. Pressure (symbol: p or P) is the unit of physics to express force (F) per unit area (A) and Volume or can also be called capacity is the calculation of how much space can be occupied in an object. The object can be a regular object or an irregular object. Regular objects such as cubes, beams, cylinders, pyramids, cones, and balls. Irregular objects such as stones found on the road. Volume is used to determine the density of an object.
HapusHow to calculate gas volume not in standard state (non-STP)?
BalasHapusAs described above, the calculation of gas volume is not in standard state (non-STP) using two approaches as follows:
HapusA. The ideal gas equation
Assuming that the measured gas is ideal, the equations connecting the number of moles (n) of gas, pressure, temperature, and volume are:
The ideal gas law: P. V = n. R. T
Where:
P = pressure (atmospheric unit, atm)
V = volume (liter, L)
N = number of moles of gas (mol unit)
R = gas constant (0.08205 L atm / mol K)
T = absolute temperature (° C + 273 K)
B. With gas conversion at the same temperature and pressure
According 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 / n2
Where:
N1 = mol of gas 1
V1 = gas volume 1
N2 = mol of gas 2
V2 = gas volume 2
For example:
In a tank with a volume of 50 dm3 there is oxygen gas at 27º C and 135 atm pressure. What is the mass of the gas?
Resolution:
R = 0.082 lt atm / molº k
P = 135 atm
V = 50 dm3
T = 300º K
N = pV / RT
= (135. 50) / (0.082 300)
= 274 mol
So:
M O2 = 16 + 16 = 32
M O2 = 32. 274
= 8768 gr
Give me an example of Law of multiple comparison
BalasHapus1081/5000
HapusExamples of Dalton's Law or Multiple Comparisons
Elements of hydrogen and oxygen can react to form water under normal circumstances, but both elements can also form hydrogen peroxide in high-energy states. In water there is 11.2% hydrogen and 88.8% oxygen, in hydrogen peroxide there is 5.93% hydrogen and 94.07% oxygen. Show that this data complies with the law of multiple comparison.
Resolution:
In water
11.2 grams of hydrogen joined with 88.8 grams of oxygen or for 1.0 grams of hydrogen joined (88.8 / 11.2) or 7.93 grams of oxygen
In hydrogen peroxide
5.93 hydrogen combines with 94.07 grams of oxygen or for 1.0 grams of hydrogen joined with (94.07 / 5.93) or 15.9 grams of oxygen.
In the case of hydrogen peroxide, the weight of oxygen per unit of hydrogen weight is 15.9, or twice the weight of oxygen per unit of hydrogen in water. In other words, for the same weight of hydrogen, the weight of oxygen in hydrogen peroxide is twice the weight of oxygen in water
If the water molecule formula is H2O, then the molecular formula of hydrogen peroxide is H2O2