Moderators: Chem_Mod, Chem_Admin

Jose Hernandez 1D
Posts: 16
Joined: Tue Nov 14, 2017 3:02 am


Postby Jose Hernandez 1D » Thu Apr 05, 2018 10:18 pm

For this problem on the homework do you divide the percentage of each element by the mass or by the molar mass?

Elizabeth Parker 1K
Posts: 33
Joined: Wed Nov 15, 2017 3:00 am

Re: F19

Postby Elizabeth Parker 1K » Fri Apr 06, 2018 9:36 am

Hi! For this problem you convert the percentages to grams and then divide by the mass of each element. For example 5.19 g H * 1 mol H/1.008 g H.

Jesus A Cuevas - 1E
Posts: 29
Joined: Fri Apr 06, 2018 11:02 am

Re: F19

Postby Jesus A Cuevas - 1E » Fri Apr 06, 2018 12:53 pm

1) Start with converting the percentage of the element into grams (16.48% O = 16.48g O).
(2) Divide the mass of the elements from the compound by the actual molar mass from the periodic table for each element (O = 16.48g/15.999g = 1.03).
(3) Then divide the result for each element by the smallest resultant from the previous step (O = 1.03 /1.03 = 1 mol)
(4) Write empirical formula using the mols from previous step as subscript for each element (O1)
(5) Calculate the molarity of the compound based on the empirical formula from the previous step, by multiplying the number of mols by the atomic mass for each element (O= 1*15.999) and adding them all together.
(6) Then, divide the found molarity of the compound by the molarity given in the question.
(7) Multiply the subscript of empirical formula found in step 4 by the result from the previous step, and that gives you the molecular formula for that problem.

Posts: 18693
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 616 times

Re: F19

Postby Chem_Mod » Sun Apr 08, 2018 11:28 am

Jesus's post (see above) is very well explained but I want to clarify the assumption in step 1. With regards to step 1, the assumption you have there is that you assume that you now have some amount of the sample with that composition. The example in parenthesis assumes that you have 100.00 grams of the sample. You are welcome to assume any amount, but 100.00 g is just a nice number to convert % to grams readily.

Return to “Molarity, Solutions, Dilutions”

Who is online

Users browsing this forum: No registered users and 2 guests