**6. The calculation of hydro-micas (ill = illite and br = brammallite) in slatecalculation**

A positive aq value is a basis for the hydro-micas calculation (see steps of calculation 40–60 in Appendix A). The original values, as well as the results of calculation steps 9 (Al2O3 and SiO2) and 17 or 19 (in Appendix A) (MgO and FeO), are used as the starting point (Quoted from [9]):

$$\text{aq (share K}\_2\text{O)} \rightarrow \text{ill} = \text{illite} = 0.65 \text{K}\_2\text{O } 3 \text{Al}\_2 \text{O}\_3 \text{ 6SiO}\_2 \text{ 2.35H}\_2\text{O} \tag{18}$$

$$\text{lag} \left( \text{share Na}\_2\text{O} \right) \rightarrow \text{br} = \text{brammallite} = 0.65 \text{Na}\_2\text{O} \text{ 3Al}\_2\text{O}\_3 \text{ 6SiO}\_2 \text{ 2.35H}\_2\text{O} \quad \text{(19)}$$

The calculation of hydro-micas requires a recalculation of the remaining phyllosilicates, like micas (mu, pa, steps 41 and 42 in Appendix A) and chlorites (mac, mc, fac, fc, steps 47–59 in Appendix A).

If aq - contents >0, the calculation of limonite is necessary as well:

$$\text{instead of } \text{he} \to \text{lm} = \text{limonit} = \text{Fe}\_2\text{O}\_3 \text{ H}\_2\text{O} \tag{20}$$

$$\text{Residual SiO}\_2 \rightarrow \text{qz} = \text{quartz} = \text{SiO}\_2 \tag{21}$$

$$\text{Residual H}\_2\text{O} \to \text{aq} \tag{22}$$

If the ill and br values are exceptionally high, in rare cases, MgO and FeO can remain after the calculation and lead to an Al2O3-deficit. In this case, the following varieties may be calculated as hydro-micas:

$$\begin{aligned} \text{aq}(\text{share K}\_2\text{O}) &\rightarrow \text{maill} = 0.\text{65K}\_2\text{O} \, 1.05 \text{MgO} \, 1.95 \text{Al}\_2\text{O}\_3 \, 6 \text{SiO}\_2 \, 2.35\\ &\times \text{H}\_2\text{O} (\text{Mol.Wt of Mineral}: 705.21) \end{aligned} \tag{23}$$

$$\begin{aligned} \text{[aq(share K}\_2\text{O)} \rightarrow \text{fail} &= 0.65 \text{K}\_2\text{O } 1.05 \text{FeO } 1.95 \text{Al}\_2\text{O}\_3 \text{ 6SiO}\_2 \text{ 2.35} \\ &\times \text{H}\_2\text{O} (\text{Mol.Wt of mineral}: 738.32) \end{aligned} \tag{24}$$

or in combination:

$$\begin{aligned} \text{aq}(\text{share K}\_2\text{O}) &\rightarrow \text{mf} \text{fill} = 0.65 \text{K}\_2\text{O} \; 0.525 \text{MgO} \; 0.525 \text{FeO} \; 1.95 \text{Al}\_2\text{O}\_3 \; 6 \text{SiO}\_2 \\ &\times 2.35 \text{H}\_2\text{O} (\text{Mol}.\text{Wt of mineral}: 721.77) \end{aligned} \tag{25}$$

$$\begin{aligned} \text{[aq(share Na}\_2\text{O)} \rightarrow \text{mabr} = 0.65 \text{Na}\_2\text{O 1.05} \text{MgO 1.95Al}\_2\text{O}\_3 \ \text{6SiO}\_2 \ 2.35 \\ \times \text{H}\_2\text{O} (\text{Mol.Wt of mineral}: 684.27) \end{aligned} \tag{26}$$

$$\begin{aligned} \text{[aq(share Na}\_2\text{O)} \rightarrow \text{fabr} = 0.65 \text{Na}\_2\text{O } 1.05 \text{FeO } 1.95 \text{Al}\_2\text{O}\_3 \text{ 6SiO}\_2 \text{ 2.35} \\ \times \text{H}\_2\text{O} (\text{Mol.Wt of mineral}: 717.39) \end{aligned} \tag{27}$$

or in combination:

$$\begin{aligned} \text{aq}(\text{share Na}\_2\text{O}) &\rightarrow \text{mfabr} = 0.65 \text{Na}\_2\text{O} \ 0.525 \text{MgO} \ 0.525 \text{FeO} \ 1.95 \text{Al}\_2\text{O}\_3 \\ &\times 6 \text{SiO}\_2 \ 2.3 \text{SH}\_2\text{O} (\text{Mol.Wt of mineral}: 700.83) \end{aligned} \tag{28}$$

If the first calculation steps yield negative values or additional information about other minerals is available, further calculations may be considered (e.g., Alta-Quartzite-schist and others):

$$\text{CaO} \rightarrow \text{epi} = \text{epido} \\ \text{te} = 4 \text{CaO} \, \text{Fe}\_2\text{O}\_3 \, 2 \text{Al}\_2\text{O}\_3 \, 6 \text{SiO}\_2 \, \text{H}\_2\text{O} \tag{29}$$

$$\mathbf{\dot{\bf CaO}} \rightarrow \mathbf{wo} = \mathbf{w} \text{ollastonite} = \mathbf{\dot{CaO}} \,\, \mathbf{SiO\_2} \tag{30}$$

$$\text{TiO}\_2 \rightarrow \text{tit} = \text{titamine} = \text{CaO} \text{TiO}\_2 \text{SiO}\_2 \tag{31}$$

$$\text{Fe}\_2\text{O}\_3 \rightarrow \text{mt} = \text{magnetite} = \text{FeO }\text{Fe}\_2\text{O}\_3 \tag{32}$$

$$\text{MnO} \rightarrow \text{sp.} = \text{spessartime} = \text{3MnO Al}\_2\text{O}\_3 \; \text{3SiO}\_2 \tag{33}$$

As a final step the norm minerals are added up to minerals and/or mineral groups:

$$\mathbf{a}\mathbf{a} + \mathbf{a}\mathbf{b} + \mathbf{or} = \text{fields}\mathbf{s}\mathbf{a}\mathbf{s}\tag{34}$$

$$\mathbf{cc} + \mathbf{dd} + \mathbf{sid} = \mathbf{c}\mathbf{r}\mathbf{b}\mathbf{onates}\tag{35}$$

$$\mathbf{m}\mathbf{u} + \mathbf{p}\mathbf{a} = \mathbf{m}\mathbf{c}\mathbf{a} \tag{36}$$

$$\text{ill} + \text{br} = \text{hydro} - \text{micas} \tag{37}$$

$$\mathbf{m}\mathbf{a}\mathbf{c} + \mathbf{f}\mathbf{c} + \mathbf{m}\mathbf{c} + \mathbf{f}\mathbf{c} = \mathbf{c}\mathbf{h}\mathbf{l} \text{or} \mathbf{i} \tag{38}$$

pn þ pt*:* ¼ sulfides (39)

$$\mathbf{p}\mathbf{n} + \mathbf{p}\mathbf{t} + \mathbf{t}\mathbf{m} + \mathbf{i}\mathbf{m} + \mathbf{r}\mathbf{u} + \mathbf{he} \,(+\mathbf{m}\mathbf{t}) = \mathbf{ore} - \text{minerals} \tag{40}$$

$$\mathbf{q}\mathbf{z} + \mathbf{a}\mathbf{n} + \mathbf{a}\mathbf{b} + \mathbf{or}\ (+\mathbf{s}\mathbf{p}) = \text{rigid minersals}\tag{41}$$

$$\mathbf{\dot{u}} \cdot \mathbf{m} + \mathbf{p} \mathbf{a} + \mathbf{i} \mathbf{l} + \mathbf{b} \mathbf{r} + \mathbf{m} \mathbf{c} + \mathbf{m} \mathbf{a} \mathbf{c} + \mathbf{f} \mathbf{c} + \mathbf{f} \mathbf{c} = \text{elastic minernals} \tag{42}$$
