Question

How many atoms are in 73.9g of potassium oxide

Answer 1

Answer: $4.69(10)^{23} atoms$

Explanation:

Firstly, we have to find the Molecular mass of potassium oxide ($K_{2}0$):

$K$ atomic mass: 39 u

$O$ atomic mass: 16 u

$K_{2}O$ molecular mass: $39(2) g/mol+16g/mol=94 g/mol$

This means that in 1 mole of $K_{2}O$ there are $94 g$ and we need to find how many moles there are in $73.9 g$ $K_{2}O$:

1 mole of $K_{2}O$-----$94 g$ of $K_{2}O$

$X$-----$73.9 g$ of $K_{2}O$

$X=\frac{(73.9 g)(1 mole)}{94 g}$

$X=0.78 mole$ This is the quantity of moles in 73.9 g of potassium oxide

Now we can calculate the number of atoms in 73.9 g of potassium oxide by the following relation:

$N_{atoms}=(X)(N_{A})$

Where:

$N_{atoms}$ is the number of atoms in 73.9g of potassium oxide

$N_{A}=6.0221(10)^{23}/mol$ is the Avogadro's number, which is determined by the number of particles (or atoms) in a mole.

Then:

$N_{atoms}=(0.78 mole)(6.0221(10)^{23}/mol)$

$N_{atoms}=4.69(10)^{23} atoms$ This is the quantity of atoms in 73.9g of potassium oxide