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Naming molecular compounds and formulas

Submitted by Randi17 on Thu, 06/27/2013 - 21:21

I need help solving a few problems. I don't want the answers but how to solve it.

 

 

1. Give the correct IUPAC name for the compound with the following molecular formula: UC

For this I got Uranium Carbide, but that is wrong

 

2. Give the correct IUPAC name for the compound with the following molecular formula: Fe(CrO4)2

I am thinking this is Iron(II) Chromate, but unsure

 

3. Give the correct IUPAC name for the compound with the following molecular formula: Cu(C2H3O2)3

 

4. Give the correct IUPAC name for the compound with the following molecular formula: Zn(HCO3)4

 

5. Give the correct molecular formula for the following compound:

platinum (VI) sulfide

 

6. Give the correct molecular formula for the following compound:

manganese (V) sulfite

 

7. Give the correct molecular formula for the following compound:

nickel (I) phosphate

 

8. Give the correct molecular formula for the following compound:

gold (II) phosphide

Hint: since you cannot include subscripts in your answers, you would enter the molecular formula for sodium phosphate, Na3PO4, as Na3PO4.

Thank You! 


Uranium Carbide is an ionic compound because uranium is a metal and carbon is a non metal,,most common form of uranium are  3+ , U 4+ , UO , UO 2+

if it is forming bond with carbon which is 4- , so if it could be U4C3 ..

Iron(II) Chromate is correct Fe is showing 2+ oxidation state and Cr2O4- is an ionic compound .

Zn(HCO3)4 , \

Zn is in +4 oxidation state in this compound and trhe polyatomic ion HCO3- name is  bicarbonate, so its name should be Zinc (iV) bicarbonate

platinum (VI) sulfide,, ide subscript in sulfide indicates that it is sulfur , so it sformula should be

Pt^+4  and S^2- 

Pt2S4 in simplest form PtS2

similary try to figure out other answers also read our tutorial......

http://www.mychemistrytutor.com/tutorials/naming-ionic-compounds

 


Fe(CrO4)2 is not iron(II) chromate but iron(IV) chromate as CrO4 is really CrO42-. As there are 2 x chromates per iron, iron must be Fe4+, so iron (IV) chromate eg Fe4+(CrO42-)2

Platinum(VI) sulphide will be PtS3 where Pt is in the +6 oxidation state and each S is in the -2 OS