What is the empirical formula of a compound that breaks down into 4.12g of n and 0.88g of h? explain why.

To solve, one must calculate the number of moles per element to determine the ratio of atoms. To do this, you take the given weight (4.12g of N and 0.88g of H) and divide each by their respective atomic weight. This is found on the periodic table underneath the elemental symbol and usually has about 3 significant digits. So:

• (4.12g N)/(14.007g/mol) = .294 mol N
• (0.88g H)/(1.008g/mol) = .873 mol H

From this, you take the smallest number and assume that this number represents one atom of that element. You then divide all other numbers by this number to determine the ratio of one element in the empirical formula to another. So:

(.873 mol H)/(.294 mol N) = 2.97 atoms H to every 1.00 atom N

Because 2.97 is close to 3.00, you can assume that there are 3 hydrogens to every 1 nitrogen. To put it into empirical form, you list the main bonding atom first and the substituents second. Because hydrogen can only form one bond (because it only has one valence electron), you can assume that each of the hydrogens are bonded to the nitrogen, meaning that nitrogen is the main bonding atom. Therefore, the empirical formula is NH3 (remember to write the 3 in subscript), also known as ammonia.

4.49 gramshope that fren : )