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Posted

I have this assignment, and from this chemical equation,

 

CaCO3 + 2 CH3CO2H ----> CO2 gas + H2O + Ca + 2

CH3CO2

 

I have to name the products and I don't know what CH3CO2 is!

 

Can you please help?

 

THANKS!

Posted
when have we started actually Doing peoples homework for then?

was I away that day and mis a memo?

 

When you started actually doing people's homework.

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Posted

i think its an derivative of esters with calcium at the place of carbon hence name in iupac should be calcium methanoate or calcium ethanoate (actuall i cant remember whether carbon of coo- group should also be counted or not).

its more used name will be calcium acetate:eyebrow:

Posted

Hello,

 

Whether you had calcium acetate (calcium ethanoate, Ca(OAc)2) or the free ions Ca++ and 2-OAc will depend upon the pH of the observed solution. I would like to suggest looking up and learning about pH (and pKa, leave out activity, for now:eyebrow: ).

 

Once learned, this can help you to predict what-reacts-with-what (and at what pH it might occur) and what products you will have (at the observed pH).

 

Cheers,

 

O3

Posted

What's the world coming to? Don't they teach kids how to use Google anymore?

 

 

""Naming of the organic acids

Organic acids are derived from alkanes, the IUPAC names of organic acids are also derived from the IUPAC names of alkanes. The carboxylic group is given a suffix “oic” which is then attached to the IUPAC name of the alkane. A word acid is added in the end. The suffix “oic” shows that there is a presence of a carboxylic group.

 

For naming of the organic acids we will use the rules laid down by IUPAC :

Find the alkane from which the alcohol is made. Name it similar to the nomenclature for straight chain hydrocarbons. This is called the parent hydrocarbon. Add an “oic” suffix to the name of the alkane by removing the “e” of the alkane name.

 

Find where the carboxylic group is attached : in the branches or in the side chains. The –COOH group is generally attached at the end of a branch.

 

 

Examples below will clarify the way the IUPAC nomenclature is used.

 

Example 1 : Let us see how formic acid HCOOH is named by IUPAC method.

 

 

 

The compound contains one carbon atom, so its parent hydrocarbon is methane. The carboxyl group –COOH indicates that it is an acid. Hence the suffix “oic” is attached to methane by replacing the last “e” of methane. Hence the IUPAC name of HCOOH is methan-oic acid or methanoic acid. Its common name is formic acid.

 

Example 2 : Let us see how acetic acid CH3COOH is named by IUPAC method.

 

 

 

The compound has 2 carbon atoms. So its parent hydrocarbon is an ethane. The presence of -COOH shows that the compound is an organic acid. Hence attach the suffix “oic” and thus the compound CH3COOH is named as something acid by IUPAC. Its common name is acetic acid.

 

Example 3 : Let us see how C2H5COOH is named by IUPAC method.

 

 

 

The compound has 3 carbon atoms. So its parent hydrocarbon is a propane. The presence of -COOH shows that the compound is an organic acid. Hence attach the suffix “oic” and thus the compound C3H5COOH is named as what ?acid by IUPAC. What acid can also be written as CH3CH2COOH, as this molecular formula clearly indicates its structure.

 

Example 4 : Let us see how C3H7COOH is named by IUPAC method.

 

 

 

The compound has 4 carbon atoms in a straight chain. So its parent hydrocarbon is a butane. The presence of -COOH shows that the compound is an organic acid. Hence attach the suffix “oic” and thus the compound C4H7COOH is named as hello acid by IUPAC. Hello acid can also be written as CH3CH2CH2COOH, as this molecular formula clearly indicates its structure.""

 

PS: for the deprotonated form (anion), like CH3COO, skip the "-oic acid" part and use "aote" instead...eg, propanoic acid becomes propanoate

This is a good site for learning naming conventions.

http://home.att.net/~cat6a/org_chem-IX.htm

Posted

Several monts on I think it's fair to say the OP has handed in his assignment by now. The fact that he forgot to put the charges on the ions in his equations probably doesn't matter any more.

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