Molecular Formulas

Moderators: Chem_Mod, Chem_Admin

kshalbi
Posts: 68
Joined: Fri Sep 25, 2015 3:00 am

Molecular Formulas

Postby kshalbi » Tue Sep 29, 2015 3:20 pm

Some problems, when asking you to find the molecular formula, tell you that the substance is composed of 14% C and certain percentages of other elements. Then you proceed by treating the 14% as 14 grams. My question is that if the problem says that a substance is made of 15 g of C, 2 g of H, and 3 g of O, why do you just covert these into moles of each element instead of first finding what percent of the substance is made of C and each other element. (e.x.) (15) / (15+2+3) = % of C in the substance.

Jennifer Cheng 2L
Posts: 13
Joined: Fri Sep 25, 2015 3:00 am

Re: Molecular Formulas

Postby Jennifer Cheng 2L » Tue Sep 29, 2015 3:47 pm

Not sure about this, but you could convert to percentages and then use the amount of grams of each element you would find in an 100g sample, but then you would have to convert to moles again afterwards anyways. I think it's just easier to go straight from grams to moles directly.

Crystal Eshraghi 2L
Posts: 24
Joined: Fri Sep 25, 2015 3:00 am

Re: Molecular Formulas

Postby Crystal Eshraghi 2L » Tue Sep 29, 2015 4:40 pm

If the problem directly tells you that the substance consists of 15 g of C, 2 g of H, and 3 g of O, and that the total mass of the substance is 20 g, then yes, you would just directly calculate the appropriate percentage compositions by dividing 15 g by 20 g for the percent composition of C, 2 g by 20 g for that of H, and 3 g by 20 g for that of O.

However, I think what you are referring to are those problems which give you the amount of C, H, or N PRODUCED in the chemical reaction. In these cases, you would have to convert the given masses of elements to moles, and then use the appropriate chemical reactions to find the appropriate molar ratios for each element in the compound and its corresponding molecule in the chemical reaction. You would then use those ratios to calculate the grams of each element in the substance itself. From there, you would calculate the particular percentage compositions for each element. Then, after finding all of the percentage compositions, you would use a 100 g sample as a model to convert those percentages into an amount in grams of each element in the substance. You would then use the usual methods to obtain the appropriate empirical and/or molecular formulas.

I hope this helped!

Chem_Mod
Posts: 19138
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 820 times

Re: Molecular Formulas

Postby Chem_Mod » Tue Sep 29, 2015 5:30 pm

To clarify, the question refers to different methods in finding the empirical formulas of an unknown compound. Methods are as follows:

1)Given the % percent composition of an unknown substance, we find the empirical formula by ASSUMING we have 100 grams of the unknown. Therefore, if the problem states that the %percent composition of the substance is "...14% carbon and certain percentages of other elements", then 14% of 100 grams is 14 grams. We convert these masses into their respective molar quantities and proceed to find their molar ratio. The molar ratios are normalized with respect to the element that has the fewest moles.

2)Given the mass composition of an unknown substance, we find the empirical formula by "...convert these masses into their respective molar quantities and proceed to find their molar ratio. The molar ratios are normalized with respect to the element with the fewest quantity."

3)Given the mass composition produced by an unknown substance due to some process, we find the empirical formula by first determining the elements that ONLY originate from the unknown. These types of problems usually involve combustion (adding O2 to the left side of the reaction) of the unknown substance. Since oxygen was introduced from another source, we cannot use the amount of oxygen from the right hand side to determine the oxygen composition of the unknown substance.

So, you find the molar amount of C, H, ... produced and if the unknown had oxygen, you would need to find that amount. You find it by knowing that the sum of the %composition of C,H,...O must equal 100% so any missing %percent composition is from the oxygen. This is when it would be necessary to change your masses into %percent composition.

You may solve types of problems like (2) by converting to %percent composition, but, in the end, you just want to get to moles. Once you have all of the elemental composition in moles of the unknown substance, then you would be able to find the empirical formula.


Return to “Empirical & Molecular Formulas”

Who is online

Users browsing this forum: No registered users and 1 guest