what is the empirical formula for a compound that is 50.61% copper 11.16% nitrogen and 38.23% oxygen

CuNO3

Explanation:

In order to calculate the empirical formula in this question;

Cu = 50.61% = 50.61g

N = 11.16% = 11.16g

O = 38.23% = 38.23g

Next, we convert each gram unit to moles by dividing by their respective molar mass (Where; Cu = 63.55, N = 14, O = 16)

Cu = 50.61 ÷ 63.55 = 0.796mol

N = 11.16 ÷ 14 = 0.797mol

O = 38.23 ÷ 16 = 2.389mol

Next, we divide each mole value by the smallest (0.796)

Cu = 0.796mol ÷ 0.796 = 1

N = 0.797mol ÷ 0.796 = 1.001

O = 2.389mol ÷ 0.796 = 3.001

Approximately, the simple whole number ratio between Cu, N and O is 1:1:3, hence, the empirical formula is CuNO3.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

personal, she chose to do it.