Empirical Formula of Compound: HClO | Solving for Empirical Formula

[annayn]

A sample of a compound of Cl and O reacts with an excess of H2 to give .233g of HCl and .403g of H2O Determine the empirical formula of the compound

Can anyone help explain how to solve this problem?

Thanks

 

[pizzasky]

ALL the Cl and O in the products come from the unknown compound. This fact can be used to determine the amount (in moles) of Cl and O present in that compound.

Please show some of your thoughts/ working if you require more assistance!

 

[Blueshift5]

for reaching out! To solve this problem, we first need to determine the moles of each element present in the compound. This can be done by using the given masses of HCl and H2O and their respective molar masses.

For HCl, we have:
0.233g HCl x (1 mol HCl/36.46 g HCl) = 0.00639 mol HCl

For H2O, we have:
0.403g H2O x (1 mol H2O/18.02 g H2O) = 0.0224 mol H2O

Next, we need to determine the ratio of moles of Cl to moles of O. Since we have 0.00639 mol HCl, we know that we have 0.00639 mol Cl. And since the amount of H2O in the compound is also the amount of O, we have 0.0224 mol O.

Therefore, the ratio of Cl to O is 0.00639 mol Cl : 0.0224 mol O, or approximately 1:3.5.

To get whole numbers for the ratio, we can multiply both values by 2, giving us a ratio of 2:7. This means that the empirical formula of the compound is HClO2.

I hope this helps! Let me know if you have any further questions.