# Calculate the number of atoms of oxygen present in 1.3 mol of sulfuric acid

Calculate the number of atoms of oxygen present in $\pu{1.3 mol}$ of $\ce{H_2SO_4}$

1 mol has $6.02\times 10^{23}$ atoms

So $\pu{1.3 mol}$ must have $1.3 \times 6.02 \times 10^{23} = 7.826 \times 10^{23}$ atoms

Since there are 4 oxygen atoms out of 7 atoms in total in $\ce{H_2SO_4}$,

${4 \over 7} \times 7.826 \times10^{23}$

So there are $4.472 \times 10^{23}$ oxygen atoms present.

However the answer key says that $3.1 \times 10^{24}$ atoms is the answer.

What went wrong?

## 2 Answers

The key about the unit mole is that it does not reference which type of entity is specified. Using the mole only makes sense if you define what you are using the mole for. This could be:

• protons
• atoms
• ions
• molecules
• formula units of an ionic or metallic solid
• aggregates
• apples
• stars

The mole is basically just a large number that you can use instead of e.g. the number 1,000,000 or the number 0.5.

So in your question, $\pu{1.3 mol}$ of $\ce{H2SO4}$ means that you have $1.3 \times 6.022 \times 10^{23}$ molecules of $\ce{H2SO4}$, each of which contains one sulphur atom, two hydrogen atoms and four oxygen atoms. Thus, in one mole of $\ce{H2SO4}$ you can find:

• $\pu{2mol}$ of hydrogen atoms
• $\pu{1mol}$ of sulphur atoms
• $\pu{50 mol}$ of protons
• $\pu{50mol}$ of electrons
• and $\pu{4mol}$ of oxygen atoms.

The correct solution is thus:

$$n(\ce{O}) = 4\times n(\ce{H2SO4}) = 4 \times 1.3 \times 6.022\times 10^{23} = 3.13 \times 10^{24}$$

• Does 50 moles proton means 50 x 6,022 10^23 units of protons? – Lifeforbetter Jan 7 '20 at 16:17
• @Lifeforbetter Not units of protons, individual protons. Protons are countable and a mole is essentially just a shorthand for big number. – Jan Jan 8 '20 at 8:22

There are 1.3 mol of the entire acid molecule, so there are four as many moles of oxygen as there are of the acid, because one acid contains four oxygens.

$4×1.3=5.2$ and 5.2 times Avogadro's constant is (about) $3.1×10^{24}$ atoms.

Your error is in dividing by seven, which is neither conceptually nor mathematically correct.