# Stoichiometry formula and how to use it

Stoichiometry formula and How to use it

What is stoichiometry

Stoichiometry is coined from two words: Stoicheon (atom)  and metron (measurement). It is actually deals with chemical calculation based on the elementary particles of matters such calculation highlight the relationships between the composition or chemical reactions of matters and the mole (a unit that describe a quantity of elementary particles)

Before we proceed into stoichiometry formula, stoichiometry can be group into two namely

Stoichiometry of reaction
Stoichiometry of composition.

Hope you understand the meaning stoichiometry and what is use for now let move to formulas and how to use it in questions.

The first things to note is the anion and cation Which is called electrochemical series of an element this can easily help to balance equation and also help in some aspect of solving stoichiometry questions.

Cations

Cation is a charges that metals carries, they are normally donates electron to non metal and this make them to become more positive change, their ability to transfer electron decrease as we move down the series.

For quick remember I represent each metals using mnemonic

Keep new car moving at zero free some people have cool house at ajar

K+1 ——- Keep ——— Potassium

Na+1 ——– New ———— Sodium

Ca+2 ——- Car ———— Calcium

Mg+2 ——- moving ——— magnesium

Al+3 ———At ————- Aluminium

Zn+1——–zero ————- Zinc

Fe+2  ——- free ——– iron

Sn+2 ——- some ——–zinc

Pb+2 ——- people ——- Lead

H+1 ——– have ——– Hydrogen

Cu2+ ——— cool ——- Copper

Hg+2——- house ——– Mercury

Ag+ ——– at ———– Silver

Au+—– ajar ———– Gold

That’s oxidation number, charges, valences, or cation of some common metals.

Note their electropositivity decreasing down from Potassium to Gold

Anions

Anion are charges that are carried by non-metal atoms or compounds and they .

For some nation cool breeze in and out.

F-1 ——-for ——– Fluorine

(So6)^-2 —– some —– sulphate ion

(NO)^-1 ——nation ——nitrate ion

Cl-1 ———-cool —— chlorine

Br-1 ——- breeze ——bromine

I-1 ——–In —— iodide ion

And

OH——- out ——- Hydroxide ion

### SI UNIT OF STOICHIOMETRY QUANTITY

Now let move on to some stoichiometry fomulars and how to use them in solving problems.

### Common stoichiometry formula

#### 1. Volume at STP

To calculate volume at stp

Stp means standard temperature and pressure

Which means it volume is donated as 22.4dm3

And it is

V = n x 22.4

n = no of mole

V = volume volume at dm3

22.4 = volume at stp

Given that

Calculate the volume of Na at stp when it mass is = 22g

Solution:

Mole = mass / molar mass

We need mole (n)  to calculate our volume at stp

Then

n = 22 / 22

n = 1

When volume V at stp(22.4) = ❓

V = n x 22.4

V = 22 x 22.4

V = 498.2dm3

That’s our to calculate volume at stp

#### 2. Mole

Formula to calculate mole of substance or element are

Mole = mass / molar mass

Mole = number of ion / avogadros constant

Mole = number of molecule / avogadros constant

Mole = number of particles / avogadros constant

Avogadros constant is given as = 6.02 x 10^-23

Let say

Question was given like

Calculate the mass of Na+ ion

Solution

Av = 6.02 x 10^-23

Na+ = +1

Mole =❓

Mass = ❓

Then

Mole = number ion  / 6.02 x 10^-23

Mole = 1 / 6.02x 10^-23

Mole = 1.66 x 10^22

Mass = mole x molar mass

Mass = 1.66 x 10^22   / 22

Mass = 7.5 x 10^17

#### 3. Mass concentration

Formula for calculating mass concentration of a substance is

Mass concentration = Molar concentration x Molar mass

Given that

Calculate the mass concentration of 0.2M of HCL that will neutralise 25.0cm of X of unknown base

(H= 1 Cl= 35.5)

Solution

Molar concentration of HCL = 0.2mol/dm^-3

Molar mass of HCL = 1 +35.5 = 36.5

Mass concentration = ?

If

Mass concentration (P) = Molar concentration (C) x Molar mass (M)

P = C x M

P =  0.2 x 36.5

P = 7.3gdm^-3

#### 4. Molar concentration

Formula for calculating Molar concentration is

Molar concentration (C)  = Mass (m)  / Volume (V) in dm^-3

Given that

When 0.02mole of 35.0m H2SO4 react with 2 mole of 25m of Y calculate the molar concentration of the Acid that will be required for complete neutralisation of Y

(H= 1 O= 16 S = 32)

Solution:

Mole of Acid = 0.02

Molar mass = ( 1 x 2 + 32 x 16 x 4)  = 98

Molar concentration = ?

Mass = ?

To calculate mass of an Acid

Mass (m)  = mole (n)  x molar mass (M)

m = n x M

m = 0.02 x 98

M = 1.96g

Molar concentration (C)  = Mass(m) / Volume(V) (dm^-3)

C = m / V

C = 1.96 / 35 /1000

C = 56mol/dm3

#### 5. Dilution

The formula for dilution is given as

C1V1 = C2V2

C1 = initial Concentration of a given substance

V1 = initial Volume of substance

C2 = final Concentration of the Sundance

V2 =final Volume of the substance

Thanks

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