Runner Posted September 29, 2003 Share Posted September 29, 2003 Hello all, I have some problems with a question and was wondering if there is a brilliant mind out there that could help. It would be much appreciated, thank you. The question states: A chloride of phosphorus, A, contains 22.5% P by mass. Reaction of this compound with more chlorine gives another chloride, B, containing 14.9% Calculate the empirical formulae of the two chlorides. Link to comment Share on other sites More sharing options...
JaKiri Posted September 29, 2003 Share Posted September 29, 2003 Divide the % by the RAM. Divide by the lower value. Cancel any common factors. Tadaa. Link to comment Share on other sites More sharing options...
Runner Posted September 29, 2003 Author Share Posted September 29, 2003 MrL_JaKiri said in post #2 :Divide the % by the RAM. Divide by the lower value. Cancel any common factors. Tadaa. I know how to calculate the empirical formula given the percentage masses; I’m just not sure what the hell is going on in the question. *Very confused* Thanks for your help anyway. Link to comment Share on other sites More sharing options...
JaKiri Posted September 30, 2003 Share Posted September 30, 2003 You have Phosphorus. You have Chlorine. Phosphorus makes up 22.5% by mass. How much by mass do you think chlorine makes up? Link to comment Share on other sites More sharing options...
YT2095 Posted September 30, 2003 Share Posted September 30, 2003 it sounds like one is Dimer and the other Trimer. but either way, like Mr_L says, you can still work it`s molarity out quite easily Link to comment Share on other sites More sharing options...
Runner Posted September 30, 2003 Author Share Posted September 30, 2003 I think this is how it goes.. P Cl 22.5 100-22.5= 77.5% so. 22.5/31=0.726 77.5/35.5=2.18 0.726/0.726=1 2.18/0.726=3 PCL3 Second part P Cl 14.9 100-14.9=85.1 so 14.9/31=0.480 85.1/35.5=2.39 PCL5 Link to comment Share on other sites More sharing options...
JaKiri Posted September 30, 2003 Share Posted September 30, 2003 Yep. except the L is lower case Link to comment Share on other sites More sharing options...
Runner Posted September 30, 2003 Author Share Posted September 30, 2003 Thank you for your help. Link to comment Share on other sites More sharing options...
JaKiri Posted September 30, 2003 Share Posted September 30, 2003 np Link to comment Share on other sites More sharing options...
Runner Posted October 13, 2003 Author Share Posted October 13, 2003 Could someone please help, these calculations are driving me insane. BTW, this has nothing to do with calculating the empirical formula, I thought it would have been a waste of space if I had started a new thread. Here goes... Magnesium oxide is not very soluble in water, and is difficult to titrate directly. Its purity can be determined by use of a 'back titration' method. 4.06g of impure magnesium oxide was completely dissolved in 100cm^3 of hydrochloric acid, of concentration 2.0mol/dm^3 (in excess). The acid required 19.7cm^3 of sodium hydroxide (0.20mol/dm^3) for neutralisation. This 2nd titration is called a 'back-titration', and is used to determine the unreacted acid. 1.)Calculate the moles of HCl reacting with MgO. 2.)Calculate the mass of MgO that reacted with the initial HCl, hence the % purity of the MgO Link to comment Share on other sites More sharing options...
neo_maya Posted October 16, 2003 Share Posted October 16, 2003 Runner said in post #10 :Could someone please help, these calculations are driving me insane. BTW, this has nothing to do with calculating the empirical formula, I thought it would have been a waste of space if I had started a new thread. Here goes... Magnesium oxide is not very soluble in water, and is difficult to titrate directly. Its purity can be determined by use of a 'back titration' method. 4.06g of impure magnesium oxide was completely dissolved in 100cm^3 of hydrochloric acid, of concentration 2.0mol/dm^3 (in excess). The acid required 19.7cm^3 of sodium hydroxide (0.20mol/dm^3) for neutralisation. This 2nd titration is called a 'back-titration', and is used to determine the unreacted acid. 1.)Calculate the moles of HCl reacting with MgO. 2.)Calculate the mass of MgO that reacted with the initial HCl, hence the % purity of the MgO I am not good in chemistry ( I am not good in anything, but here goes a small effort) ________________________________________________ Given, Used NaOH = .2M 19.7 cm3 = .2M 19.7 cm3 HCl = 1M 3.94 cm3 HCl = 2M 7.88 cm3 HCl ____________________________________ So, Total HCl = 2M 100 cm3 Used with NaOH = 2M 7.88 cm3 Used with MgO = 2M 92.12 cm3 ____________________________________ So, Used with MgO Mole of HCl = (92.12/1000) * 2 mole HCl mass = (92.12/1000) *36.5 * 2 = 6.72476 gm ____________________________________ MgO + 2HCl = MgCl2 + H2O 40gm 73gm So, 73 gm HCl reacts with MgO of = 40 gm 6.72476 gm HCl reacts with MgO of = 3.6848 gm ____________________________________ Purity of MgO = (3.6848/4.06) * 100 = 90.75 % ____________________________________ Answer : 1. 6.72476 gm 2. 3.6848 gm 3. 90.75 gm Link to comment Share on other sites More sharing options...
neo_maya Posted October 16, 2003 Share Posted October 16, 2003 Sorry, Couldn't find the equivalence sign. And couldn't find the subscript or superscript options either. So, please don't mind. Someone please check it. It's sure to have something wrong somewhere. Or, there is a huge chance that the entire procedure is wrong. Oops, sorry about the third answer. It should have been 90.75 % Link to comment Share on other sites More sharing options...
Runner Posted October 16, 2003 Author Share Posted October 16, 2003 neo_maya said in post #12 :Sorry, Couldn't find the equivalence sign. And couldn't find the subscript or superscript options either. So, please don't mind. Someone please check it. It's sure to have something wrong somewhere. Or, there is a huge chance that the entire procedure is wrong. Oops, sorry about the third answer. It should have been 90.75 % Good try. I manage to work it out after spending almost 10 hours on it. (i) MgO + 2HCl ==> MgCl2 + 2H2O (ii) NaOH + HCl ==> NaCl + H2O moles of hydrochloric acid added to the magnesium oxide = 100 ÷ 1000 x 2 = 0.20 mol HCl moles of excess hydrochloric acid titrated = 19.7 ÷ 1000 x 0.2 = 0.00394 mol HCl (mole ratio NaOH:HCl is 1:1 from equation (ii)) moles of hydrochloric acid reacting with the magnesium oxide = 0.20 - 0.00394 = 0.196 mole MgO reacted = 0.196 ÷ 2 = 0.098 (1: 2 in equation (i)), the formula mass of MgO = 40.3, so mass of MgO reacting with acid = 0.098 x 40.3 = 3.95g, % purity = 3.95 ÷ 4.06 x 100 = 97.3% Link to comment Share on other sites More sharing options...
neo_maya Posted October 16, 2003 Share Posted October 16, 2003 neo_maya said in post #11 : I am not good in chemistry ( I am not good in anything, but here goes a small effort) ________________________________________________ Given, Used NaOH = .2M 19.7 cm3 = .2M 19.7 cm3 HCl = 1M 3.94 cm3 HCl = 2M 7.88 cm3 HCl ____________________________________ So, Total HCl = 2M 100 cm3 Used with NaOH = 2M 7.88 cm3 Used with MgO = 2M 92.12 cm3 ____________________________________ So, Used with MgO Mole of HCl = (92.12/1000) * 2 mole HCl mass = (92.12/1000) *36.5 * 2 = 6.72476 gm ____________________________________ MgO + 2HCl = MgCl2 + H2O 40gm 73gm So, 73 gm HCl reacts with MgO of = 40 gm 6.72476 gm HCl reacts with MgO of = 3.6848 gm ____________________________________ Purity of MgO = (3.6848/4.06) * 100 = 90.75 % ____________________________________ Answer : 1. 6.72476 gm 2. 3.6848 gm 3. 90.75 gm ____________________________________________________ ____________________________________________________ ________________________________________________________________________________________________________ I knew it - I had to make a mistake and be the greatest jerk of all time. Just a calculation error. Sorry. Instead of dividing, I multiplied by mistake. I think I have marked the line in blue. Sorry . ________________________________________________ Given, Used NaOH = .2M 19.7 cm3 = .2M 19.7 cm3 HCl = 1M 3.94 cm3 HCl = 2M 7.88 cm3 HCl The right line is - = 2 M 1.97 cm3 HCl ____________________________________ So, Total HCl = 2M 100 cm3 Used with NaOH = 2M 1.97 cm3 Used with MgO = 2M 98.03 cm3 ____________________________________ So, Used with MgO Mole of HCl = (98.03/1000) * 2 mole HCl mass = (98.03/1000) *36.5 * 2 = 7.15619 gm ____________________________________ MgO + 2HCl = MgCl2 + H2O 40.3 gm 73gm So, 73 gm HCl reacts with MgO of = 40.3 gm 7.15619 gm HCl reacts with MgO of = 3.95 gm ____________________________________ Purity of MgO = (3.95/4.06) * 100 = 97.305 % ____________________________________ Answer : 1. 7.15619gm 2. 3.95 gm 3. 97.305 % Link to comment Share on other sites More sharing options...
neo_maya Posted October 16, 2003 Share Posted October 16, 2003 I just wish that someday - ( God knows when ) - I will become a lesser jerk. Link to comment Share on other sites More sharing options...
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