A molecule contains 52.20 g of manganese (Mn) and 3.80 g of carbon(C). What is its empirical formula?

              Mn₃C

Explanation:

Calculate Moles of Mn

Mole = Mass/M.Mass

Mole = 52.20 g / 54.93 g/mol

Mole = 0.95 mol

Calculate Moles of C

Mole = Mass/M.Mass

Mole = 3.80 g / 12.01 g/mol

Mole = 0.316 mol

Now,

= 0.95 / 0.316 = 3.00

This means for every one atom of C there are three atoms of Mn.

Result:

          Mn₃C

nacl is added to the aqueous of what? ? in order to get a proper answer you have to specify the exact composition of the aqueous solution.

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