what is the empirical formula of a compound composed of 3.26g of arsenic and 1.04g of oxygen?
here is what I have done but I'm stuck after this point
3.26 + 1.04 = 4.3g
As= 3.26/4.3 x 100 = 75.8%
O= 1.04/4.3 x 100 = 24.1
As = 75.8/74.9 = 1
O = 24.1/16 = 1.5
Thank you
Empirical formula help
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because you cant have AsO1.5. Hm i dont know how to explain it exactly.. all stechiometric coeficients has to be integers (number with no numbers after decimal point, im not sure about english terminology in maths so.. hope you understand) AFAIK. So just your numbers
As = 75.8/74.9 = 1
O = 24.1/16 = 1.5
mean that in compound you have 1 atom of arsenic and 1.5 atom of oxygen. How can you have 1.5 atom of oxygen? You cant
so i multiplied the numbers so i had integrers there.
Hope I helped you..im not good at explaining things
As = 75.8/74.9 = 1
O = 24.1/16 = 1.5
mean that in compound you have 1 atom of arsenic and 1.5 atom of oxygen. How can you have 1.5 atom of oxygen? You cant
so i multiplied the numbers so i had integrers there.
Hope I helped you..im not good at explaining things
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GrahamKemp
- Distinguished Member
- Posts: 396
- Joined: Mon Mar 19, 2012 10:12 pm
It's a two step process.
That is a fairly good explanation. It's quite understandable, and also correct.
(1) Divide each elements measured mass by its atomic mass to get the stoichiometric ratio, then
(2) Multiply the results by the common factor needed to produce a ratio of lowest possible integers.
The result is As2O3: Arsenic Trioxide. also known as Arsenic Oxide.
(1) Divide each elements measured mass by its atomic mass to get the stoichiometric ratio, then
(2) Multiply the results by the common factor needed to produce a ratio of lowest possible integers.
The result is As2O3: Arsenic Trioxide. also known as Arsenic Oxide.