Calculating Oxide Weight Percents from Formulae and Normalizing Chemical Analyses

The atomic weights of elements tell us how many grams there are per mole.  Likewise, the atomic weight of an oxide (or any compound) tells us how many grams there are per mole of the oxide/compound.

For example, the atomic weight of quartz, SiO2, is 60.0843 (=28.0855 + 2*15.9994; make sure you know where these numbers come from).  So quartz weighs 60.0843 grams per mole.  Each mole contains 6.022 x 1023 (Avagodro’s number) of molecules.  So each SiO2 molecule weighs 60.0843÷6.022 x 1023 = a very small number!

Formulas to Weight percents (wt%)

Suppose I asked you to tell me the wt %s (weight percents) of Li, Al and Si in spodumene, LiAlSi2O6.  The calculation is as follows:
	a =	b =	c =	100 x c/d =	
ion	# atoms in spod	at wt elem	a x b	wt % elem	
Li	1	6.941	6.94	3.73	
Al	1	26.98154	26.98	14.50	
Si	2	28.0855	56.17	30.18	
O	6	15.9994	96.00	51.59	
TOTAL			d =186.09	100.00	

==>Spodumene contains 3.7 wt% Li, 14.5 wt% Al, and 30.2 wt% Si.

For reasons explained in the book, we report chemical analyses in terms of weight percents of the oxides.  So, we could redo the above calculation as follows:

		a =	b =	c =	100 x c/d = 	
ion	oxide	at wt oxide	# oxides in spod	a x b	wt % oxide in spod	
Li	Li2O	29.8814	0.5	14.94	8.03	
Al	Al2O3	101.9613	0.5	50.98	27.40	
Si	SiO2	60.0843	2	120.17	64.58	
TOTAL				d =186.09	100.00	

==>So, spodumene contains 8.0 wt % Li2O, 27. 4 wt % Al2O3 and 64.6 wt % SiO2.

(Make sure you know where all these numbers came from.)

Wt%s to Formulas

We can also do the reverse calculation.  Suppose we have a mineral analysis that is 23.9253wt% FeO, 13.4217 wt% MgO, 22.6359 wt% Al2O3 and 40.0170 wt% SiO2.  In the table below I show how to convert an analysis from oxide weight percent to a chemical formula.  (See also Box 1.5 in the text for more explanation.)
												
	A	B	C	D	E	F	G	H	I	
		at wt of	# cations	# oxygen	mineral	=E/B	cation	=F x C	= F xD	
	oxide	oxide	in oxide	in oxide	analysis			=# cations	= # oxygen	
	FeO	71.8464	1	1	23.9253	0.3330074	Fe+2	0.333	0.333	
	MgO	40.3044	1	1	13.4217	0.3330074	Mg+2	0.333	0.333	
	Al2O3	101.9613	2	3	22.6359	0.2220049	Al+2	0.444	0.666	
	SiO2	60.0843	1	2	40.0170	0.6660148	Si+4	0.666	1.332	
								Oxygen =	2.664	

Study the above table and make sure you know where all the numbers come from.  

Columns H and I allow us to write the following formula:
	Fe0.333Mg0.333Al0.444Si0.666O2.664
This is a crummy looking formula because the subscripts are not integers.  

After some trial and error, you will find an appropriate fudge factor to multiply everything by, and this formula becomes
	Fe1.5Mg1.5Al2.0Si3.0O12

If you look this up in a book, after scrounging around for a while you will find that this is the formula of a garnet that is 50 % almandine (Fe3Al2.0Si3.0O12) and 50% pyrope (Mg3Al2.0Si3.0O12).

Note that in real life, the math never comes out as exact as in the above examples (or the problems below).  Analytical error introduces some uncertainty and adds difficulties to normalizations, but the approach and principles are still the same.

Your Problems
Be warned - these can be tricky!

1.  Consider the mineral orthoclase with formula KAlSi3O8.  What are the weight percents of K2O, Al2O3 and SiO2 in orthoclase?  Show all work, don’t just look up the answers.

2.  Suppose you conducted an analysis of an unknown mineral and found it contained the following weight percents:
oxide	wt %	
Na2O	15.33079	
Al2O3	25.22062	
SiO2	59.44859	
TOTAL	100	

Normalize this analysis by using atomic weights to change it into a formula.  You will have to goof around, but if you guess the number of oxygen correctly, all the other subscripts will come out to be integers.  Name the mineral!