<blockquote>
  <p>The first step I did was divide each wt% by their molecular weights to get a molecular proportion:</p>
</blockquote>

<p>Actually that's not what you did. What you did is how much moles of each oxide you have assuming 100 grams of mineral. But that's less important now.</p>

<p>You are mostly correct in your steps.
Eventually you have 0.972 Fe, 0.375 Ti, and 0.972 + 2×0.375 = 1.722 O.
The (non normalised) mineral formula is then:
Fe<sub>0.972</sub>Ti<sub>0.375</sub>O<sub>1.722</sub>.</p>

<p>You can easily check that it's correct using charge balance:</p>

<p>0.972×2 + 0.375×4 - 1.722×2 = 0</p>

<p>The question now is how to normalise it. Since you were asking about 3 cations, you need to multiply everything by 3/(0.972 + 0.375), giving you:</p>

<p>Fe<sub>2.165</sub>Ti<sub>0.835</sub>O<sub>3.835</sub></p>

<p>But that's also ugly. Usually mineral formulas are normalised to round anion numbers, so let's say you multiply it all by 5/1.722 you get:</p>

<p>Fe<sub>2.822</sub>Ti<sub>1.089</sub>O<sub>5</sub></p>

<p>which is slightly prettier.</p>

<p>Without having any more information on the mineral as to what's it supposed to be, you can only guess. This is what we do in EMPA analyses - we know what mineral it is and we calculate the formula according to that. If we don't, we guess and try to figure out which one it is, and then calculate according to that.</p>

<p>I don't know much about the question, but it could be that not all iron in the mineral is Fe<sup>2+</sup>. Some of it may be Fe<sup>3+</sup>, but analysing both is not an easy task so most iron in minerals is reported as either all FeO or all Fe<sub>2</sub>O<sub>3</sub> regardless of what oxidation state it is actually in. So it's possible that you have (FeO + Fe<sub>2</sub>O<sub>3</sub>) = 70 wt%. If you play enough with that, you might end up with a better looking formula. Again, I don't know the question and I don't know if that is what you're supposed to do in the question.</p>

CC BY-SA 3.0
2015-04-20T21:16:14.093Z
Ok thank you! The question in my book literally was what I wrote, they didn't give anymore info. I just should've known that normalize to three cations meant use 3/(0.972 + 0.375) as the normalization factor.
nukenine
CC BY-SA 3.0
2015-04-20T13:17:11.537Z
2015-04-20T13:17:11.537Z
Gimelist
<p>$\ce{FeO}$ = 70 wt%</p>

<p>$\ce{TiO2}$ = 30 wt%</p>

<p>The molecular weight given of $\ce{FeO}$ = 72, and the mw of $\ce{TiO2}$ = 80. </p>

<p>The first step I did was divide each wt% by their molecular weights to get a molecular proportion:</p>

<p>$$\ce{FeO}: {70 \over 72} = 0.972 $$</p>

<p>$$\ce{TiO_2}: {30 \over 80} = 0.375$$</p>

<p>Then, I multiplied each proportion by the number of oxygens in each chemical formula</p>

<p>$$0.972 \times 1 \ \rm{oxygen} = 0.972$$
$$0.375 \times 2 \ \rm{oxygens} = 0.75$$</p>

<p>The next step is where I get really confused. I think I'm supposed to add the two numbers above together, and divide that number into a number that is supposedly the cations assumed of the mineral? (As in 5 for plagioclase, 4 for pyroxene, 3 for olivine, etc) But I do not understand how I'm supposed to gather which number to use based on only the data given. If anyone can point me in the right direction it would be greatly appreciated.</p>

<geology><geochemistry><mineralogy><homework><minerals>
How can I normalize this chemical analysis to a mineral with 3 cations?
nukenine
# How can I normalize this chemical analysis to a mineral with 3 cations?

(Asked by: username_0 on 2015-04-19)

$\ce{FeO}$ = 70 wt%

$\ce{TiO2}$ = 30 wt%

The molecular weight given of $\ce{FeO}$ = 72, and the mw of $\ce{TiO2}$ = 80.

The first step I did was divide each wt% by their molecular weights to get a molecular proportion:

$$\ce{FeO}: {70 \over 72} = 0.972 $$

$$\ce{TiO\_2}: {30 \over 80} = 0.375$$

Then, I multiplied each proportion by the number of oxygens in each chemical formula

$$0.972 \times 1 \ \rm{oxygen} = 0.972$$
$$0.375 \times 2 \ \rm{oxygens} = 0.75$$

The next step is where I get really confused. I think I'm supposed to add the two numbers above together, and divide that number into a number that is supposedly the cations assumed of the mineral? (As in 5 for plagioclase, 4 for pyroxene, 3 for olivine, etc) But I do not understand how I'm supposed to gather which number to use based on only the data given. If anyone can point me in the right direction it would be greatly appreciated.

**Tags:** geology, geochemistry, mineralogy, homework, minerals


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## 1 answers

### Answer by username_1 on 2015-04-20. Score: 5 | Accepted answer

> The first step I did was divide each wt% by their molecular weights to get a molecular proportion:

Actually that's not what you did. What you did is how much moles of each oxide you have assuming 100 grams of mineral. But that's less important now.

You are mostly correct in your steps.
Eventually you have 0.972 Fe, 0.375 Ti, and 0.972 + 2×0.375 = 1.722 O.
The (non normalised) mineral formula is then:
Fe0.972Ti0.375O1.722.

You can easily check that it's correct using charge balance:

0.972×2 + 0.375×4 - 1.722×2 = 0

The question now is how to normalise it. Since you were asking about 3 cations, you need to multiply everything by 3/(0.972 + 0.375), giving you:

Fe2.165Ti0.835O3.835

But that's also ugly. Usually mineral formulas are normalised to round anion numbers, so let's say you multiply it all by 5/1.722 you get:

Fe2.822Ti1.089O5

which is slightly prettier.

Without having any more information on the mineral as to what's it supposed to be, you can only guess. This is what we do in EMPA analyses - we know what mineral it is and we calculate the formula according to that. If we don't, we guess and try to figure out which one it is, and then calculate according to that.

I don't know much about the question, but it could be that not all iron in the mineral is Fe2+. Some of it may be Fe3+, but analysing both is not an easy task so most iron in minerals is reported as either all FeO or all Fe2O3 regardless of what oxidation state it is actually in. So it's possible that you have (FeO + Fe2O3) = 70 wt%. If you play enough with that, you might end up with a better looking formula. Again, I don't know the question and I don't know if that is what you're supposed to do in the question.

> **Comments:**
> **username_0**: Ok thank you! The question in my book literally was what I wrote, they didn't give anymore info. I just should've known that normalize to three cations meant use 3/(0.972 + 0.375) as the normalization factor.  

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