Acompound is found to be 38.76% calcium, 19.97% phosphorous and 41.28% oxygen. what is the empirical formula for this compound?

The empirical formula for this compound in the given case would be - Ca₃P₂O₈ or  .

Let assume the given compound is 100g than the molar quantities of each element.

=> Moles of calcium

=  

=  0.967 mol

=> Moles of phosphorus

=    

=  0.644 mol

=> Moles of oxygen

=

= 2.58 mol

Multiply integers of molecules by a number that makes them close to integers.

Element  Mass/g     Moles         Ratio       ×2       Integers       

Ca       38.76      0.967 07     1.4998       2.9995         3     

P         19.97       0.644 82     1                   2                2     

O        41.28       2.580 0       4.0011        8.0023          8

Thus, the empirical formula is Ca₃P₂O₈.  

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The empirical formula is Ca₃P₂O₈.

Assume that you have 100 g of the compound.

Then you have 38.76 g Ca, 19.97 g P, and 41.28 g O.

Now, we must convert these masses to moles and find their ratios.

If the number in the ratio are not close to integers, you multiply them by a number that makes them close to integers.

From here on, I like to summarize the calculations in a table.

Element  Mass/g     Moles         Ratio       ×2       Integers  

     Ca       38.76      0.967 07     1.4998   2.9995         3

     P         19.97       0.644 82     1             2                   2

     O        41.28       2.580 0       4.0011   8.0023          8

The empirical formula is Ca₃P₂O₈.

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b. examine and describe cells