Empirical Formula of Compound: HClO | Solving for Empirical Formula

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To determine the empirical formula of the compound HClO, the masses of HCl and H2O produced from the reaction are used to calculate the moles of chlorine (Cl) and oxygen (O). The total mass of Cl and O in the products is derived from the stoichiometry of the reaction, allowing for the calculation of their respective moles. The mole ratio of Cl to O can then be established, leading to the determination of the empirical formula. The process emphasizes the importance of understanding the contributions of each element from the original compound. This method effectively illustrates how to derive empirical formulas from reaction products.
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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
 
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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!
 
Thread 'Confusion regarding a chemical kinetics problem'

Thread 'Confusion regarding a chemical kinetics problem'

TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
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