Question

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?

Answer 1

Answer:

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)

Answer 2

Answer:

1. The number of moles of each element

Na= 0.01565mol

N= 0.015714

O= 0.047

2. The empirical formula is NaNO3

3. The name is Sodium trioxonitrate (V)

Or in short form is Sodium nitrate

Explanation:

Mass of Na= 0.360g molecular weight= 23g/mol

Mass of O= 0.752g molecular weight = 16g/mol

Mass of N= 0.220g molecular weight = 14g/mol

STEP 1: FIND THE NUMBER OF MOLES

Moles= mass÷molecular weight

For Na

0.360g÷23g/mol= 0.01565mol

For N

0.220g÷14g/mol= 0.015714mol

For O

0.752g÷16g/mol= 0.047mol

STEP 2: DIVIDE EACH MOLE WITH THE SMALLEST MOLE

The smallest mole here 0.01565mol

Therefore;

For N

0.015714mol÷0.01565mol=1.00408

For O

0.047mol÷0.01565mol= 3.003

For Na

0.01565mol÷0.01565mol=1

STEP3: TAKE THE WHOLE NUMBER OF THE FRACTIONS IN STEP2

N=1

Na=1

O=3

THE EMPIRICAL FORMULA BECOMES

NaNO3

Which is Sodium trioxonitrate (V)