Essay on Chemistry Important Concepts

Biochemistry Fundamental Ideas

Chapter two

Overview of fundamental Concepts

Atomic excess weight

The atomic weight of element: is a mass of an atom of that element depending on a mass of accurately 12 for the carbon isotope C12.

Molecular or formula weight

It's the sum of atomic weight loads of the factors that comprise molecule from the substance i. e since given by the molecular formulation. Mwt of water (H2O) =2(1. 008) +1(16. 00) =18. 02

Mwt of hydrogen (H2) =2(1. 008) =2. 016

These weight load are all in accordance with the mass of C12 atom while 12. 1000 with (SI) system of products. Molecular weight is stated by g/mole

H. Watts

Calculate the molecular fat of: --

a) Na2SO4

b) CaSO4. 7H2O

Ans: a) 142. 04 g/mol b) 262. 25 g/mol.

The mole unit

The mole (or g. gopher or gram molecular weight) can be defined as the amount of the compound that contains volume of constitutional species (molecules) equal to that precisely 12 grams of 12C.

This quantity of molecules is known as " Avogadro s" number which is corresponding to 6. 023Г—1023molecues/mole.

Number of moles is calculated as: -

Example (1)

Calculate the quantity of moles in 500mg of sodium tungstate (Na 2WO4). Given: Na= 22. 99 W= 183. 85 O= 16

Mwt of Na2WO4= 2(22. 99) + 183. 85 + 4(16) =293. 83 g/mol.

No . of moles =

= 0. 00170 mol =1. six m mol

Example (2)

Calculate the mass of 0. two hundred fifity mmol. of ferric oxide (Fe2O3)? Fe= 55. 845 Mwt of Fe2O3 = 159. 69 mg/mmol

Mass = zero. 250 mmol Г—159. 69 mg/mmol =39. 3 magnesium

Example (3)

A) Determine the number of skin moles of K2CO3 that contains 117. 3 g. of potassium

B) Determine the mass in grams of other elements in such chemical substance? K=39. 1 O =16 C =12.

A)

g. atom of K sama dengan

Each skin mole of K2CO3 contains 2g. atom of K

ZERO of moles of k2CO3 =

Mass of Oxygen = g. atom of oxygen Г— atomic excess weight =1. 5Г—3Г—16 =72g. Mass of carbon = g. atom of carbon Г—atomic weight sama dengan 1 . five (1) (12) =18g.

Comparable weight

Equal weight: is definitely the molecular fat divided by number of reacting units

a- for stomach acids: the number of responding units is definitely the number of hydrogen ions that will furnish.

Model

HCl↔H+ +Cl-

eq. wt HCl =M. wt/1

H2SO4↔2H++SO4-2

eq. wt H2SO4 =M. wt/2

CH3COOH↔H+ +CH3COO-

eq. wt CH3COOH = Meters. wt/1

*** Note

For H3PO4: the eq. wt depends on the effect

H3PO4 ⇔ H++H2PO4-

frequency. wt: M. wt/1

H3PO4 ⇔

2H+ + HPO24-

eq. wt: M. wt/2

H3PO4 ⇔

3H++PO43-

frequency. wt: Meters. wt/3

b-

For bases: the number of responding units is definitely the hydrogen ions that

is going to react with it:

Cases

KOH + H+ в†’ K++H2O

frequency. wt sama dengan Mwt

Ba(OH) 2 & 2H+в†’Ba2++2H2O

eq. wt = M. wt

Fe2O3 & 6H+ в†’ 2Fe3+ + 3H2O

c- For sodium ex: --

eq. wt = Mwt/6

Na2CO3 & HCl в†’

NaHCO3 + NaCl

eq. wt sama dengan M. wt

Na2CO3 + 2HCl в†’

H2CO3 + 2NaCl

frequency. wt sama dengan M. wt

d- For Oxidation – reduction reaction

The number of reacting units is founded on the number of

electrons that the oxidizing or minimizing agent is going to take on or supply. It can be computed as well as the change in oxidation

number.

Case in point:

MnO4- + 8H+ & 5e- ↔

Mn2+ + 4H2O reacting = 5units

MnO4- & e- ↔

MnO42reacting = 1units

+

+

IO3 + 6H + 4e I & 3H2O

sama dengan 4 models

e- Intended for reactions regarding formation of any precipitate or perhaps

soluble complex

There is no exchange of re-acting unit, nevertheless a combination of a metal ion with the reactant. The fee of the metallic ion is usually taken as the reacting product, and the equivalent weight can be equal to the atomic excess weight divided by charge of complexing agent is its molecular pounds divided by the number of steel ions which each gopher of the agent react. Case: AgNO3 в†’Ag+ + NO3-

eq. wt = Mwt of Aktiengesellschaft NO3

HgCl2↔ Hg2+ +2Cl-

eq. wt = .5 Mwt HgCl2

Examples: Calculate the equivalent fat of the next substances: a) NH3, b) H2C2O4 (in reaction with NaOH), c) KMnO4 (Mn is reduced to Mn2+) and d) Na3 PO4 (in precipitating Ag3 PO4-).

=

a) eq. wt of NH3=

=

b) eq. wt of H2C2O4 =

&

c) MnO4- + 8H + 5e

Mn 2+ + 4H2O

No . of electrons sama dengan 5 or perhaps oxidation point out of Meters is changed from +7 to +2 Eq. wt of KMnO4 =

d) Na3PO4 + 3Ag+

sama dengan

=...