Understanding Compound Suffixes: -ate, -ous, -ic, & More

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SUMMARY

The discussion centers on the meanings of the suffixes "-ate," "-ous," and "-ic" in organic and inorganic chemistry. The suffix "-ate" indicates a stable ion of a carboxylic acid at a specific pH, exemplified by pyruvic acid (CH3COCOOH) transitioning to its deprotonated form, pyruvate (CH3COCOO-), at pH levels above 3. In inorganic chemistry, "-ate" designates oxidation states of polyatomic ions. This distinction is crucial for understanding chemical nomenclature and behavior.

PREREQUISITES
  • Understanding of carboxylic acids and their properties
  • Familiarity with pH and its effects on chemical species
  • Knowledge of polyatomic ions and their oxidation states
  • Basic principles of organic and inorganic chemistry
NEXT STEPS
  • Research the role of pH in the behavior of carboxylic acids
  • Study the nomenclature of polyatomic ions in inorganic chemistry
  • Explore the differences between "-ate" and "-ite" suffixes in chemical compounds
  • Learn about the significance of oxidation states in chemical reactions
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Chemistry students, educators, and professionals in organic and inorganic chemistry who seek to deepen their understanding of chemical nomenclature and the implications of suffixes in compound naming.

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What does this -ate suffix with compounds mean? Can anybody tell about other suffixes like -ous -ic etc as well.

Thanks...
 
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Xishan said:
What does this -ate suffix with compounds mean? Can anybody tell about other suffixes like -ous -ic etc as well.

Thanks...

well, it depends, but in organic & biochemistry it means that you have a stable ion of a carboxylic acid at a particular pH.

for example, pyruvic acid:

CH_3COCOOH

is the carboxylic acid. but at pH > 3, most of it will be found in the deprotonated form:

CH_3COCOO^-

this is called "pyruvate". the same convention holds for other carboxylic acids.

if you are talking about general or inorganic chemistry, then "-ate" is used to designate oxidation states of polyatomic ions, as Dr. Mark's link indicates.
 
Thanks, Dr. Marks! that's all I was looking for
 

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