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Oxides

Posted at Maret 15th, 2011 | Categorised in Calculations in Ceramic, Glaze

previously read the article glaze calculations

An oxide can be defined as a compound of two elements, one of which is oxygen.

The molecular formulae of glazes and frits are given in terms of molecular parts of the constituent oxides. The oxides are classified into three main groups:

( l ) Acidic Oxides

They are the oxides of non-metals. If soluble in water, they combine with it, forming an acid,

e.g.  SO3+ H2O = H2SO4

They combine with bases to form salts.                             .

An example, using the acidic oxide CO2, is given below:

Ca(OH)2+CO2 = CaCO3+ H2O

Silica (SiO2) and boric oxide (B203) are the two main acidic oxides used in frits and glazes. They combine with other elements to form silicate and borate glasses.

(2) Basic Oxides

They may be considered as the oxides of metals. They react with acids to form a salt and water only:

CaO+2HCl   = CaCl2= H2O

If soluble.in water they yield hydroxides:

Na2O+H2O = 2NaOH

CaO+H2O = Ca(OH)2

The main basic oxides used in the ceramic industry are listed below:

K2O Ca0         Ba0 PbO Na2O Mgo ZnO

(3) Amphoteric Oxides

An amphoteric oxide may exhibit either basic or acidic properties. Alumina, Al2O3, reacts with dilute hydrochloric acid to yield

aluminium chloride and water, and thus behaves as a basic oxide:

Al2O3+6H Cl = 2AICl3+3H2O

but with sodium hydroxide, alumina reacts as an acidic oxide and forms sodium aluminate:

Al2O3+2NaOH = 2NaAIO2+H2O

Alumina is the main amphoteric oxide in frits and glazes.

Relationship between Percentage Composition and Formula

The chemical analysis of a pure compound determines the nature and quantity of each constituent chemical substance. The result of a quantitative analysis is given in terms of percentage composition, or proportion by weight of each element present, e.g. a compound containing carbon, hydrogen, nitrogen and sulphur on analysis is found to have the following percentage composition:

C = 41·4%

H = 5·8%

N = 16·l%

S = 36.7%

If the figures for the percentage composition are divided by the atomic weights of the elements concerned, then the result of this calculation gives the simplest possible formula to the compound.

The “empirical” or “simplest possible” formula only indicates the proportions of the atoms in the molecules, but does not give their actual number.

A compound containing carbon and hydrogen only was found to have

the following percentage composition:

C = 92·31%    ‘ H = 7·69%

Calculate the empirical formula. ( Atomic weights: C = 12; H   = 1.)

The ratio of the number of carbon atoms to the number of hydrogen atoms=

(92.31: 12): (7.69:1)= 7.69: 7.69

The ratio is therefore 1 : 1 and the empirical formula = CH

Although the “simplest possible” formula has been established as CH, compounds C2H2, C3H3, C4H4, etc., would give identical results. Further evidence for this particular compound, derived from vapour density measurements, shows that the true formula is in fact C3H6 (benzene). However, the use of such data is outside the scope of this book, and it is only necessary to appreciate that the percentage composition when divided by the appropriate atomic weight gives the “simplest possible” formula and that this is adequate

for most ceramic materials, since the correct ratio between the constituent elements is maintained.

A volatile liquid has the following percentage composition:

Carbon – 10·05

Hydrogen – 0·84

Chlorine – 89.11

Calculate the empirical formula.

The relative number of atoms in the Compound is:

Carbon = 10.05 :12= 0.84

Hydrogen= 0.84 : 1 = 0.84

Chlorine = 89.11 : 35.5 = 2.51

Dividing throughout by the smallest number( 0.84)  the “simplest formula” becomes (CHCl3)

Empirical formula = CHCl3

Other chemical evidence shows that it is also its true folmula, and  the compound is in fact chloroform (CHCl3).

So far the examples have dealt with pure chemical compounds, but the same considerations apply to ceramic materials, although the formulae of some of them are expressed in terms of oxides, e.g.

potash felspar, K2O.Al2O3.6SiO2

The percentage composition of china clay is given as:

SiO2 = 46·51

Al2O3= 39·53

H2O = 13.96

Calculate its molecular formula.

SiO2  46·51 –    60 = 0·775

Al203  39 53 – 102 = 0 387

H2O  13.96 –     18 = 0·775

Dividing throughout by the smallest number (0 387) the empirical  formula = Al2O3.2SiO2 . 2H2O

The molecular formula for china clay = Al2O3. 2SiO2. 2H2O

in the above example the percentage composition is given in terms of oxides, molecular weights of which must be used in place of atomic weights. Similar principles apply in calculating the molecular formula of a glaze from it percentage composition

A simple glaze has the following percentage composition:

PbO    – 63.90

Al2O3  – 5.87

SiO2   – 30·23

Dividing the percentage composition of each oxide by its molecular weight the ratios become:

(63.90 : 223 PbO) : (5.87:102 Al2O3): (30.23: 60 SiO2)

= 0.287 PbO : 0.058 Al2O3 : 0.504 SiO2

In calculating the empirical formula, each molecular part would have been divided by the smallest number (in this case 0-058).

However, in ceramics a convention is adopted with glazes, that the sum of the basic oxides be equal to unity. This puts all glaze

FriiñiEc on a basis which enables the many ditTerent types of glaze to be compared, one with another. A knowledge of glaze formulae used in this form is of paramount importance to the glaze technologist. He can forecast such physical properties as fusibility, solubility, etc., from such information.

In the above calculation, the basic oxide is PbO and dividing throughout by 0·287 the glaze formula becomes:

1 PbO; 0.202 Al2O3; 1.756 SiO2

The oxides are usually written in the order as shown, i.e. basic, amphoteric, and acidic.

Later, the “molecular weight” of glazes is referred to. It should be noted that this weight is not the “true molecular weight”, but it really a “formula weight”. It is an arbitrary figure calculated from the formula obtained by using the above convention. The “molecular weight” figure as used in glaze calculations has no true significance, since it has been computed from this artificially devised formula.

Given the Formula, Calculate the Percentage Composition

The reverse procedure to that used in the previous problem is shown below:

Given the formula of sodium carbonate, Na2CO3, calculate the percentage composition in terms of

(a) the elements Na, C, and O ; and (b) the oxides Na2O and CO2

The molecular weight of Na2CO3 = 106

(a)                                           Sodium           (46:106)X100= 43.4%

Carbon           (12:106)X100= 11.3%

Oxygen           (48:106)X100= 45.3%

(b)                                           Na2O              (2X23)+16 = 62

(62:106)x100= 58.5%

CO2                12+(2X16) =44

(44:106)X100= 41.5%

Further example

Given the formula of a glaze as:

0.6 PbO    }                       { 1.9 SiO2

0·4 Na2O  } 0.25 Al2O3    { 0.4 B2O3

Calculate the percentage composition.

PbO                0.6  X  223   = 133.8

Na2O              0.4   X 62     =   24.8

Al2O3             0.25 ×102    =    25.5

SiO2  ·           1.9 X 60       = 114.0

B203               0.4 × 70       =    28·0

———————-

326.1

326.1 represents the “molecular weight” of the glaze.

Note. As stated previously it is not a “true molecular weight”, but, in fact, a “formula weight”

By multiplying throughout by  (100: 326.1)  the percentage composition is:

Pbo                 41.04

Na2O              7.59

Al2O3             7.82

SiO2               34.96

B203               8.59

It can be seen in the above calculation that

Molecular Weight X Molecular Parts = Parts by Weight

Using sodium oxide as an example:

Na2O molecular weight       = 62

molecular parts   = 0.4

and their product     = 24·

This relationship is used throughout glaze calculations.

This article was taken from the book “Calculations in ceramics” by R. Griffiths and C. Radford

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