2.91 mol Al * ( 26.982 g Al / 1 mol Al) = 78.518 grams
Empirical formula is the simplest ratio of whole numbers of components in a compound.
Assuming for 100 g of the compound
Cu As S
mass 48.41 g 19.02 g 32.57 g
number of moles 48.41 / 63.5 g/mol 19.02 / 75 g/mol 32.57 / 32 g/mol
= 0.762 mol = 0.2536 mol = 1.018 mol
divide by the least number of moles
0.762 / 0.2536 0.2536 / 0.2536 1.018 / 0.2536
= 3.00 = 1.00 = 4.01
once they are rounded off
Cu - 3
As - 1
S - 4
therefore empirical formula is Cu₃AsS₄
Answer:
Oxidized and reducing agent: manganese.
Reduced and oxidizing agent: mercury.
Explanation:
Hello!
In this case, for the reaction:

We keep in mind that the species that increase the oxidation state is the oxidized one whereas the one that decrease the oxidation state is the reduced one; therefore manganese is the oxidized one as well as the reducing agent because it goes from 0 to +2 and mercury the reduced one as well as the oxidizing agent because it goes from 2+ to 0.
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