A compound contains 67.1 percent zinc and the rest is oxygen. What is the empirical formula of the compound?



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A compound contains 67.1 percent zinc and the rest is oxygen. What is the empirical formula of the compound?


The empirical formula of the compound is determined to be ZnO2.



67.1g x 1 mol Zn / 65.39g Zn = 1.026 mol Zn

32.9g x 1 mol O/ 16.00g O=2.056 mol O


1.026 / 1.026 =1 mol Znand 2.056 / 1.026 =2 mol O



For additional assistance, please see also the following data link(s):

http://en.wikipedia.org/wiki/Zinc_oxide
http://en.wikipedia.org/wiki/Zinc_dioxid...
http://chemistry.about.com/od/elementfac...
http://www.chemicalelements.com/elements/zn.html
http://chemistry.about.com/od/elementfacts/a/oxygen.htm
http://chemistry.wikia.com/wiki/Oxygen
http://chemistry.about.com/od/workedchemistryproblems/a/empirical.htm
http://chemistry.about.com/od/molecularformulas/a/Molecular-Formula-And-Empirical-Formula.htm
http://chemistry.about.com/od/workedchemistryproblems/a/How-To-Calculate-The-Empirical-And-Molecular-Formula-Of-A-Compound.htm
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm
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