A2.241-g sample of nickel reacts with oxygen to form 2.852 g of the metal oxide. calculate the empirical formula of the oxide?

NiO

Explanation:

Firstly, we need to know the mass of oxygen used up. This is the difference between the mass of the nickel and the mass of the oxide. This is : 2.852 - 2.241 = 0.611g

Now we convert these masses to mole by dividing by their atomic masses. The atomic mass of oxygen is 16 while that of nickel is 59

O = 0.611/16 = 0.0381875

Ni = 2.241/59 = 0.037983050847458

The next thing is to divide each by the smaller number of moles, which is that of the nickel.

O = 0381875 /0.037983050847458= 1

Ni= 0.037983050847458/0.037983050847458= 1

The empirical formula is thus NiO

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"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."