What is the stock system name and empirical formula for this question?A Compound with 229.5g of aluminum, 357.0g of nitrogen, and 816.0g of oxygen?What i got is AlN^3O^6 for the empirical but i dont know if its right.Also could you tell me what does it mean to round to the half?Thanks in advance
Assuming that it is Al(NO2)3, the name is technically aluminum(III) nitrite, but usually is called aluminum nitrite because aluminum seldom exhibits oxidation states other than +3, and therefore, the (III) becomes optional in everyday usageIf I understand your question, you DON'T round off when the mole ratio ends up with a halfFor instance when determining the empirical formula of a compound of Mn and O, you might end up with this mole ratio Mn - 1 O - 2.5 You DON'T round off the .5, and get MnO3, you multiply both by 2 to get Mn2O5
Aluminum is perfect fit this purpose ! Easy to bend and cut and drill, good for this small repair.
