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:
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)
