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OpenStudy (anonymous):
What is the percent composition of a compound with the empirical formula NO3
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OpenStudy (preetha):
% of N = 14/(14+3(16)) Similarly O
OpenStudy (anonymous):
100.0gNO3(1,mole/62.0682.gNO3)(1 mole N/1mole NO3)(14.07g/1mole N)=22.6686 gN/100gNO3 or 22.66|8|%
OpenStudy (anonymous):
try to use the correct values on the periodic table to get an accurate %
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