Question

4.19-g sample of nitrous oxide (an anesthetic, sometimes called laughing gas) contains 5.73 × 1022 N2O molecules. How many nitrogen atoms?

Answer 1

Answer:

1.14×10²³ atoms of N are contained in 4.19 g of N₂O

Explanation:

Nitrous oxide → N₂O

You must begin with this rule.

1 mol of compound has x moles of each element, for example:

1 mol of water has 2 moles of H and 1 mol of O (H₂O)

1 mol of ammonia has 3 moles of H and 1 mol of N (NH₃)

Then, 1 mol of N₂O has 2 moles of oxygen and 1 mol of O

The molar mass for our compound is 44 g/mol

So in 44 g of compound (1 mol) we have 2 moles of N

In 4.19 g of compound we would have ( 4.19  .2 ) /  44  = 0.190 moles of N

Now, we know that 1 mol of anything has NA particles so:

1 mol of N has  6.02×10²³ atoms

0.190 moles of N, would have (0.190  .NA)  = 1.14×10²³

Let's verify the amount of molecules, in our compound.

First of all, let's convert the mass to moles (mass / molar mass)

4.19 g / 44 g/m = 0.095 moles

moles . NA = amount of atoms → 5.71×10²² ≅

Answer 2

Answer:

We have 1.15 *10^23 nitrogen atoms

Explanation:

Step 1: Data given

Mass of N2O = 4.19 grams

Number of N2O molecules = 5.73 *10^22 molecules

Molar mass = 44.01 g/mol

Step 2: Calculate moles N2O

Moles N2O = mass N2O / molar mass N2O

Moles N2O = 4.19 grams / 44.01 g/mol

Moles N2O = 0.0952 moles

Step 3: Calculate moles nitrogen

For each mol N2O we have 2 moles of Nitrogen

For 0.0952 moles N2O we have 2*0.0952 = 0.1904 moles nitrogen

Step 4: Calculate atoms nitrogen

Number of nitrogen atoms = moles * number of Avogadro

Number of nitrogen atoms = 0.1904 moles * 6.022*10^23

Number of nitrogen atoms = 1.15 *10^23 atoms

We have 1.15 *10^23 nitrogen atoms