Answer to Question #150703 in Inorganic Chemistry for No

Question

empirical formula of a compound with 1.20% H, 42.00% Cl, and 56.80% O

Answer 1

Solution:

1) We should take a hypothetical sample of exactly 100 grams of the compound:

$n_{(hydrogen)}= \dfrac{1.2g}{1.007947( \tfrac{g}{mole} )}=1.19054$ $moles$

$n_{(chlorine)} =\dfrac{42g}{35.4532( \tfrac{g}{mole} )}=1.184660$ $moles$

$n_{(oxygen)}= \dfrac{56.8g}{15.99943( \tfrac{g}{mole} )}=3.550126$ $moles$

2) We should divide by the smallest number of moles:

$\dfrac{1.19054moles}{1.184660moles}=1.00496$

$\dfrac{1.184660}{1.184660} =1$

$\dfrac{3.550126}{1.184660}=2.9967$

3) Finally, we should round to the nearest whole numbers to find the empirical formula:

HClO₃