A substance is analyzed and found to consist of 0.360 g of Na, 0.220 g of N, and 0.752 g of O. How many moles of Na, N, and O are present? What is its empirical formula? What is its name?

1. The number of mole of each element present = 1mole of Na, 1mole of N and 3moles of O

2. The empirical formula is NaNO3.

3. The name of the compound is sodium trioxonitrate (V)

Explanation:

Mass of Na = 0.360g

Mass of N = 0.220g

Mass of O = 0.752g

Number of mole present for each elements is given below:

For Na:

Molar Mass of Na = 23g/mol

Number of mole = Mass /Molar Mass

Number of mole of Na = 0.360/23 = 0.0157mol

For N:

Molar Mass of N = 14g/mol

Number of mole = Mass /Molar Mass

Number of mole of N = 0.220/14 = 0.0157mol

For O:

Molar Mass of O = 16g/mol

Number of mole = Mass /Molar Mass

Number of mole of = 0.752/16 = 0.047mol

Next we'll divide the moles of Na, N, and O obtained by the smallest mole among them i.e 0.0157

For Na:

0.0157/0.0157 = 1

For N:

0.0157/0.0157 = 1

For O:

0.047/0.0157 = 3

The number of mole of each element present = 1mole of Na, 1mole of N and 3moles of O

Therefore, the empirical formula is NaNO3.

The name of the compound is sodium trioxonitrate (V)

the answer is c. nacl