Answer to Question #150703 in Inorganic Chemistry for No

Question #150703

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


1
Expert's answer
2020-12-14T14:41:38-0500

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}\t)}=1.19054" "moles"


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


"n_{(oxygen)}= \\dfrac{56.8g}{15.99943( \\tfrac{g}{mole}\t)}=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}\t=1"


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



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


HClO3

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS