1
$\begingroup$

A sample of copper(II) sulfate pentahydrate (CuSO4·5H2O) contains 0.360 g of water. What is the total number of atoms in the compound

what's the idea of this question?

what i know is that atoms number is found by multiplying the Avogadro number by the mol's number

??

$\endgroup$
1
$\begingroup$

You have to calculate the amount of substance with the mass of water.

$n[H_2O]=\frac{m}{M}=\frac{0.360\,g}{18\,g} \cdot mol = 0.02\, mol$

$1~ n[CuSO_4\cdot 5 H_2O] = 5~n[H_2O]$

$1~ n[CuSO_4\cdot 5 H_2O] = 0.004\, mol$

In 0.004 mol copper sulfate pentahydrate you have 0.36 g water.

1 molecule of copper sulfate pentahydrate is composed of 21 atoms and the Avogadro constant is $N_A = 6.022 \cdot 10^{23}\, mol^{-1}$.

Now you can simply multiply all values.

$\text{#}Atoms = 0.004\, mol \cdot 6.022 \cdot 10^{23}\, mol^{-1} \cdot 21 = \underline{5.06 \cdot 10^{22}}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.