Question F.13 [ENDORSED]
Moderators: Chem_Mod, Chem_Admin
Question F.13
In this question is 4.14g also its percent composition? Would you assume that 4.14g translates to 4.14%?
-
- Posts: 102
- Joined: Fri Aug 02, 2019 12:15 am
Re: Question F.13 [ENDORSED]
In this question, the 4.14g does not translate into 4.14%. Since 4.14g of phosphorus combined with chlorine to form a 27.8g compound, we know that therefore the mass of chlorine must be 27.8 - 4.14 = 23.66g. Now that we know the mass of the chlorine used, 23.66g, and the mass of the phosphorus, 4.14g, you can now convert them into moles and find the empirical formula.
Re: Question F.13
No, in this case you first subtract 4.14 g of P by 27.8g of the solid compound to find the mass of Cl.
27.8g compound - 4.14g P + 23.66 g Cl
then you divide both elements by their molar masses
4.14g P(1/30.97g) = 0.1337 mol P
23.66g Cl(1/35.45g)=0.6674 mol Cl
finally you divide both moles by the smaller element (0.1337 mol P) to get 5 moles of Cl and 1 mol of P, giving you the empirical formula PCl5
27.8g compound - 4.14g P + 23.66 g Cl
then you divide both elements by their molar masses
4.14g P(1/30.97g) = 0.1337 mol P
23.66g Cl(1/35.45g)=0.6674 mol Cl
finally you divide both moles by the smaller element (0.1337 mol P) to get 5 moles of Cl and 1 mol of P, giving you the empirical formula PCl5
-
- Posts: 105
- Joined: Wed Sep 18, 2019 12:19 am
Re: Question F.13
4.14 g would not be translated as the percent composition. If the problem said there is 4.14 g of phosphorous and the total mass of the white compound is 100g, then you can say that the percent composition of phosphorous is 4.14%. Instead, it just tells you the mass of phosphorous which you would then convert to moles. Next you would convert the mass of chlorine to moles. Then you find the ratio which will help you determine the empirical formula
Last edited by Amy Luu 2G on Wed Oct 02, 2019 11:33 pm, edited 1 time in total.
-
- Posts: 105
- Joined: Sat Jul 20, 2019 12:16 am
Re: Question F.13
Hi!
Here is the problem for anyone wondering:
"4.14 g of phosphorus combined with chlorine to produce 27.8 g of a white solid compound. (a.) What is the empirical formula of the compound? (b.) Assuming that the empirical and molecular formulas of the compound are the same, what is its name?"
For this problem, 4.14 g would not be the percentage mass. If, for whatever reason, you wanted to find the percentage mass of phosphorus (not necessary for this problem), you would d o it this way:
4.14 g P/27.8 g compound x100%= 14.9%.
As I said before, you don't actually need that to solve the problem. It actually makes it easier for you because rather than having to change percentages into grams, you could move straight into converting grams into moles. You can do this by dividing 4.14 g P by the molar mass of P (30.97 g/mol-1).
Hope this helps!
Here is the problem for anyone wondering:
"4.14 g of phosphorus combined with chlorine to produce 27.8 g of a white solid compound. (a.) What is the empirical formula of the compound? (b.) Assuming that the empirical and molecular formulas of the compound are the same, what is its name?"
For this problem, 4.14 g would not be the percentage mass. If, for whatever reason, you wanted to find the percentage mass of phosphorus (not necessary for this problem), you would d o it this way:
4.14 g P/27.8 g compound x100%= 14.9%.
As I said before, you don't actually need that to solve the problem. It actually makes it easier for you because rather than having to change percentages into grams, you could move straight into converting grams into moles. You can do this by dividing 4.14 g P by the molar mass of P (30.97 g/mol-1).
Hope this helps!
Return to “Empirical & Molecular Formulas”
Who is online
Users browsing this forum: No registered users and 1 guest