What is the chemical formula of the compound containing: \(6.02 \times 10^{23}\) atoms of hydrogen, \(35 g\) of chlorine, and 4 moles of oxygen atoms?
A. \(HCl _{4} O\) B. \(HClO\) C. \(HClO _{4}\) D. \(HCl _{2} O _{4}\)
Correct Answer: C
Explanation
We calculate the no of moles of the elements For \(H\), No of mole \(=\frac{\text { no of atoms }}{6.02 \times 10^{23}}\)or \(Cl\), \(=1\) mole of \(H\) \(\begin{aligned} \text { No of mole } &=\frac{\text { mass of } c l}{\text { Relative atomic mass }} \\ &=\frac{35 g }{35.5 g / mol } \end{aligned}\) \(=0.9859 mol \approx 1\) mole of \(cl\)nd 4 mole of \(O\) \(\therefore\) the formular is \(HClO _{4}\)