Determine the empirical formula of an oxide of iron which has 69.9% iron and
30.1% dioxygen by mass.
| Elements | Symbol | % by mass | atomic mass | % by mass/atomic mass | simple ratio | simple whole number ratio |
| Iron | Fe | 69.9 | 55.85 | 69.9/55.85 = 1.25 | 1.25/1.25=1(divide by smaller number of privious coloumn) | 2(multiply by 2 to change in whole number ) |
| Oxygen | O | 30.1 | 16.00 | 30.1/16.00 =1.88 | 1.88/1.25=1.5 | 3 |
why are you taking atomic mass of oxygen.In question 30.1 % is mass of dioxygen so stomic mass should be 32
ReplyDeleteI too agree should take mass as 32 and calculate. Formula wil come out as Fe4O6 then,but as they want empirical formula,it is reduced to Fe2O3.
ReplyDeletewhy could you multiply the simple ratio of iron by 2 as 1 is also comes under the cateogry of whole numbers
ReplyDelete?
There is also 1.5
Deleteto make it a whole number multiply both by 2
Why from 2
DeleteCan't understand why dioxygen is taken as only a single oxygen atom
ReplyDeletePlease explain more clearly
ReplyDeleteCan we get molecular formula also if yes then show me
ReplyDeleteCan we get molecular formula also if yes then show me
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