Using English to Calculate Stoichiometry

Understanding Stoichiometry

" Stoichiometry is a subject in chemistry that studies the quantity of matter in a chemical reaction".
Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means element and metron means measure, so stoichiometry literally translated means the measure of elements. In order to use stoichiometry to run calculations about chemical reactions, it is important to first understand the relationships that exist between products and reactants and why they exist, which require understanding how to balanced reactions
 
THE BASIC LAW OF CHEMICAL SCIENCE

A. The Law of Conservation of Mass (Lavoisier Law)
"The mass of substances before and after the reaction is the same".
 Example:
S + O 2 → SO 2
2 gr 32 gr 64 gr

B. Law of Refine (Proust Law)"The ratio of the elemental masses in each compound is fixed"
Example:
H 2 O → mass H: mass O = 2: 16 = 1: 8

C. Double Law (Dalton's Law)
"If two elements can be formed two or more, and the mass of one element is equal, the mass of the second element is added to a simple and integer".Example:
- Element N and O can form NO and NO 2 compounds
- In the NO compound, mass N = mass O = 14: 16
- In the compound NO 2, mass N = mass O = 14: 32 - The ratio of N masses to NO and NO 2 is the same O = 16: 32 = 1: 2 mass ratio
 
D. Ideal Gas Law
For an ideal gas or a gas that is considered to be valid applies the formula:
PV = n RT





 

Information:
P = pressure (atmosphere)
V = volume (liters)
N = mol = gram / Mr
R = gas constant (lt.atm / mol.K)
T = temperature (Kelvin)

 MASS OF RELATIVE ATOM AND MASS MOLECULES RELATIVES

Having discovered highly sensitive equipment in the early twentieth century, chemists experimented on the mass of one atom. For example, experiments were performed to measure.

1. mass of one atom H = 1,66 -> 10-24 g

2. mass of one atom O = 2.70 -> 10-23 g

3. mass of one atom C = 1.99 -> 10-23 g

From the data above can be seen that the mass of one atom is very small. Experts agree to use the amount of Atomic Mass Unit (sma) or Atomic Mass Unit (amu) or commonly called also units of Dalton. In the matter of atomic structure, you have also learned that atoms are very small, therefore it is impossible to weigh the atoms using a balance sheet.

1. Relative Atomic Mass (Ar)

Experts use the C-12 carbon isotope as standard with a relative atomic mass of 12. The relative atomic mass represents the average mass ratio of one atom of an element to 1/12 of the C-12 atomic mass. Or it can be written: 1 unit of atomic mass (amu) = 1/12 mass 1 atom C-12

Example:

The average atomic mass of oxygen is 1.33 times greater than that of the carbon-12 atoms.

Then: Ar O = 1.33 -> Ar C-12

= 1.33 -> 12

= 15.96

Experts compare different atomic masses, using the relative atomic mass scale with the symbol "Ar".

The experts decided to use C-12 or 12C isotope because it has an inert core stability than other atoms. The isotope of C-12 atom has an atomic mass of 12 sma. One sma equals 1.6605655 x 10-24 g. With the use of 12C isotope as standard then can be determined the mass of atom of other element. The relative atomic mass of an element (Ar) is a declaring number

The mass ratio of one atom of that element to 1/12 mass of one C-12 atom. ArX = (average atomic mass X) / (1/2 mass of carbon atom - 12)

Problems example

If it is known that the mass of 1 oxygen atom is 2.70 x 10-23 g, what is the Ar atom O if

Atomic mass C 1,99 x 10-23 g?

Answer:

The magnitude of the Ar price is also determined by the average price of the isotope. For example, in nature there are 35Cl and 37Cl with a ratio of 75% and 25% then Ar Cl can be calculated by:

Ar Cl = (75% x 35) + (25% x 37) = 35.5

Ar is a comparative number so it has no units. Ar can be seen in the Periodic Table of the Elements (TPU) and always included in the unit of questions if necessary

2. Relative Molecular Mass (Mr)

Molecule is a combination of several elements with a certain ratio. The same elements combine to form elemental molecules, while different elements form molecules of compounds. The molecular mass of an element or compound is expressed by a molecular mass (Mr). The relative molecular mass is the ratio of the molecular mass of an element or

Compound against 1/12 x the mass of C-12 atoms.

CONCEPT MOL AND FIXED AVOGADRO

Molecular Formulas and Elemental Content in Compounds

When reacting 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).

One mole is defined as the number of substances containing the particles of the substance as much as the atoms present in 12,000 g of carbon atoms -12.

Thus, in one mole of a substance there are 6.022 x 1023 particles. The value of 6.022 x 1023 particles per mole is called the Avogadro constant, with the symbol L or N. In everyday life, the mole can be analogous to "dozen". If it's a dozen

States the number of 12 pieces, the mole states the amount of 6.022 x 10 23 particles of the substance. The word particles in NaCl, H2O, and N2 can be expressed with ions and molecules, whereas in elements like Zn, C, and Al can be expressed with atoms.

The chemical formula of a compound shows the ratio of the number of atoms present in the compound.

Table Comparison of Atom-Atoms in H2SO4

Problems example

1. Molar Mass (Mr)

The mass of one mole of substance is called the molar mass (the symbol of Mr). 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.

Molar mass = Mr or Ar substance (g / mol)

Consider the example in the following table!

Multi-Substance Molar Table

The mass of a substance is the multiplication of its molasses (g / mol) with the mol of the substance (n). Thus the mole relationship of a substance with its mass can be expressed as follows.

Mathematically, it can be stated as follows.

Molar mass = mass: mol

Mass = mol x Mr / Ar (molar mass)

2. Molar Volume (Vm)

The volume of one mole of a substance in a gas form is called the molar volume, denoted by Vm.

What is the volume of gas molar? How to calculate the volume of a certain amount of gas at a certain temperature and pressure?

Avogadro in his experiments concluded that 1 L of oxygen gas at 0 ° C and 1 atm pressure had a mass of 1.4286 g, or it could be stated that at 1 atm pressure:

Thus, under 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, a volume of 1 mole of each gas in standard conditions (0 ° C and 1 atm pressure) is applied.

Volume gas in standard state = 22.4 L

3. Gas Volume in Non-Standard State

Calculation of gas volume is not in the standard state (non-STP) used the following two approaches.

A. The ideal gas equation

Assuming the gas to be measured is ideal, the equation that links the number of moles (n) of gas, pressure, temperature, and volume

that is:

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 + 273.15 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 gas 1 V1 = gas volume 1

N2 = 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 g

Mr. = molar mass (unit g / mol)

V = volume (mL unit)

 MOLECULAR FORMULAS AND ELEMENTAL CONTENT IN COMPOUNDS

The ratio of mass and the element 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 is indicated 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.

2. Determining the formula of Chemical Hydrate (Water Crystal)

Hydrates are solid crystalline compounds containing crystalline water (H2O). The chemical formula of solid crystalline compounds is known. So basically the determination of the hydrate formula is the determination of the number of crystalline water molecules (H2O) or the value of x. In general, the hydrate formula can be written as follows.

The chemical formula of solid crystal compounds: x. H2O

For example calcium sulfate salt, has a CaSO4 chemical formula. 2H2O, meaning that in every one mole of CaSO4 there are 2 moles of H2O.

Problems example

1. A total of 5 g of copper (II) sulfate hydrate is heated until all the crystalline water evaporates. The mass of solid copper (II) sulfate formed by 3.20 g. Determine the hydrate formula! (Ar: Cu = 63,5; S = 32; O = 16; H = 1)

Answer:

Steps to determine the hydrate formula:

A. Suppose the CuSO4 hydrate formula. X H2O.

B. Write the equation of the reaction.

C. Determine the mol substances before and after the reaction.

D. Calculate the value of x, using the mole ratio of CuSO4: mol

H2O.

CuSO4. XH2O (s) -> CuSO4 (s) + xH2O

5 g 3.2 g 1.8 g

Comparison, moles CuSO4: mol H2O = 0.02: 0.10.

Comparison, moles CuSO4: mol H2O = 1: 5.

Thus, the hydrate formula of copper (II) sulfate is CuSO4. 5H2O.

3. Chemical Counts

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 picture 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?

B. Which substance and how many moles of substance is left?

Answer:

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.

B. If all F2 is reacted then S is required:

This is not possible, because S is available only 2 mol. So, the acting act as a limiting reagent is S!

The number of SF6 mols formed = x mol S.

A. Mol SF6 = 1 x 2 mol = 2 mol

B. The remaining substance is F2, as much as = 10 mol - 6 mol = 4 mol F2