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A. Assign oxidation numbers to each atom in the following compounds. B. Determine the...

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A. Assign oxidation numbers to each atom in the following compounds.

B. Determine the formula mass of each of the following compounds or ions.

A. HI, PBr3, GeS2, KH, As2O5, H3PO4

B. (glucose; C6H12O6), (calcium acetate; Ca(CH3COO)2), (the ammonium ion; NH+4), (the chlorate ion; ClO-3)

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The oxidation number is the number of electrons each atom donates or accepts when forming the compound.  Certain atoms such as the halogens, (F,Cl,Br, and I) always have an oxidation number of -1 (accepts 1 electron), alkalai metals (Li, Na, K, Rb, Cs) have a number of +1 (give up one electron), alkalai earths ( Be, Mg, Ca, Sr) have +2.  Some have more than one possible oxidation number.  Oxygen and sulfur generally has -2 but sometimes can have +6.  Hydrogen has +1 with non-metals and -1 with metals.

The important thing in figuring out oxidation numbers is that the sum of all the atoms in a molecule equals its charge.  In the molecules above they are all neutral so the total oxidation number is 0.  When more than one atom is present such as in H2O you have to multiply the oxidation number times the number of atoms to get the total oxidation number.  For H2O the total oxidation number = 2H + O = 2(+1) + (-2) = 0.

Lets do PBr3 above.  Br has -1.  Since it is neutral the total=0.  So P + 3Br = P + 3(-1) =0.  So the oxidation number of phosphorus here is +3 = P.  You should be able to figure out the rest.

The formula mass is the total mass that each formula would have.  What you need to do is multiply each atom in the formula by its atomic mass and add them up.  For example, the formula mass for H2O is 2*(1.01)+16.00=18.02.

In the example above the formula mass for the chlorate ion is Cl + 3O = 35.45 + 3(16.00) = 83.45.

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