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Using XAFSMass, enter W as the compound, µtd (total absorption) as 2.5, S (area of the pellet) as 0.37 (the area of a 7 mm diameter pellet expressed in cm2). Choose the energy from the drop down list to be the W L3 edge (+50 eV):

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XAFSMass reports that the edge step will be 1.57, which is almost within the ideal range of 0.5-1.5, but the mass required is very small, 3.94 mg. It is unlikely that it will be possible to successfully press a 7 mm pellet of this mass, particularly of a hard material like W. Thus, it will be necessary to dilute the W metal with cellulose. The python version of XAFSmass has a much more featureful parser than the older version, and allows us to easily express dilutions. Let's start with investigating a 50-50 mix of W metal and cellulose. Change the compound to W%50(C6H10O5) and click calculate:

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XAFSMass automatically converts our expression for dilution into a molecular formula, W0.882C6H10O5, and returns a slightly smaller edge step of 1.55 absorption units. However, the mass required is still far less than the optimal 40 mg for a 7 mm pellet. Lets try diluting the W to 10%, by entering W%10(C6H10O5):

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A simpler way of using XAFSmass is to enter only the material formula and a µtd (total absorption). XAFSmass will then return a mass and edge step. That mass of material can then be mixed with a dilutant such as cellulose to a pre-defined final mass, known to make robust pellets, such as 40 mg for 7 mm OD pellets:

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In the example above, 4.1 mg of W would be mixed with 35.9 mg of cellulose. Let us include cellulose in the calculation. 4.1 mg of W diluted with 35.9 mg of cellulose corresponds to a concentration of 10.25 wt%. This can be entered into XAFSmass via the following formula W%10.25(C6H10O5):

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If we inspect the XAFSmass window, we can see that the mass required for the total absorption we specified (2.5 abs units) is 35.2 mg. This is not equivalent to the pellet we made above, and we can confirm that by looking at the absorptance step box, where is reports that the mass of W is 3.61 mg and the edge step is 1.38. To calculate 4.1 mg of W mixed with 35.9 mg of cellulose we must manually increase the value of µtd (total absorption) until the mass is 40.0 mg AND the absorptance step box report 4.1 mg W and an edge step of 1.57:

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After changing the value a few times we find that the total absorption of the W + cellulose mix is 2.8396. This is a little higher than the 2.5 we specified in the "simple" calculation, and probably would not have too much impact on the final result.

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To demonstrate, let us compare the approach with scandium oxide, Sc2O3:

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Following the "simple" approach, we would mix 1.03 mg with 38.97 mg of cellulose to make a 7 mm pellet.

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Let's see what happens when we include cellulose in the calculation. 1.03 mg of Sc2O3 in a 40 mg pellet equates to 2.575 wt%. Enter the formula (Sc2O3)%2.575(C6H10O5) into XAFSmass, and select the Sc K-edge:

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The mass shown is 10.8 mg, which is incorrect. Our task now is to manually increase the value for µtd (total absorption) until the mass of material is 40.0 mg AND the absorptance step for Sc matches that from when we did the calculation with Sc2O3 on its own (in this case, 1.29):

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Thus, if we used the "simple" method the resulting pellet would have a µtd (total absorption) of 5.585 absorption units. This sample is far too absorbing for a transmission measurement! The cellulose contributed 4.085 absorption units that were ignored by using the "simple" method. This demonstrates that it is important to include the dilutant in calculations. The new version of XAFSmass provides a convenient way of expressing dilution, so there is no reason to use the "simple" method any longer.

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The simple method suggests 1.9 mg of Fe3O4 should be mixed with 38.1 mg cellulose:

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Include the cellulose in the calculation:

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The total absorption increase by 1.025 absorption units, and total absorption is 1.68 times higher if the cellulose is excluded.

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The energy of the Ce L3 edge is roughly halfway between Fe K and Sc K.

The simple method:

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Including cellulose:

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The total absorption has increased by 1.977 absorption units when cellulose is included in the calculation. This sample is too thick for transmission.

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Sometimes, the composition of a material makes it difficult, or impossible to make a good transmission sample that gives both a reasonable edge step and a reasonable total absorption

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This material involves many heavy elements, so the results are dominated by the high total absorption. A total thickness of 8 absorption units is required to get an edge step of only 0.1 at In L3, and even then the mass is a tiny 2.78 mg. Diluting with cellulose will only increase the total absorption further. This material is better suited to measurement in fluorescence mode.

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An example composition, in weight percent oxides:

Convert to moles of oxides. The top row is the molecular mass of the oxide:

Convert to moles of the element based on stoichiometry:

Create a formula for XAFSmass, in this case diluted to 10 wt% in cellulose:

Enter into XAFSmass:

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Your task as a user is to explore the total absorption-dilution space to find the optimum combination of dilution and total absorption that returns a reasonable edge step and a workable mass to make a pellet.

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In this example, we take advantage of the powerful parser in the python/Qt version of XAFSmass. A common type of sample measured at XAFS beamlines is a metal alloy dispersed in a matrix such as carbon. In this example, the composition is 4 wt% Pt, 26 wt. % Ni and the remainder is carbon. We can enter that into XAFSmass as the formula Pt%4Ni%26C. Helpfully, XAFSmass will immediately recalculate that for us into moles: Pt0.00352Ni0.076C. We can then take this molecular formula, and use it to explore dilution with cellulose. For example a 50 wt. % mix of the alloy+carbon material and cellulose, we can enter the following into XAFSmass: (Pt0.00352Ni0.076C)%50(C6H10O5). When you are unfamiliar with a material, it can frequently be helpful to construct a table of results of XAFSmass calculations to allow you to identify the optimum dilution. The table below shows the results of such an examination for the material Pt%4Ni%26C (7.0 mm OD pellet, diluted with cellulose):

Plotting the results:

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The intersection of grey dotted lines marks our target of 40 mg and an edge step of 1.0 absorption units. The results suggest that a total absorption of a little less than 2.0, i.e. ~1.85 and a dilution with cellulose to between 10 and 20 wt. % will give the optimum result.