Empirical formula, PLEASE HELP!!!
Favorites|Homepage
Subscriptions | sitemap
HOME > > Empirical formula, PLEASE HELP!!!

Empirical formula, PLEASE HELP!!!

[From: ] [author: ] [Date: 12-01-27] [Hit: ]
254 g sample of an organic compound that contains only carbon, hydrogen, and oxygen reacts with a stream of chlorine gas, Cl2. After the reaction, 4.......
1) A compound has the formulas X2O5, where X is an unknown element. The compound is 44.0% oxygen by mass. What is the identity of element X?

2) A 1.254 g sample of an organic compound that contains only carbon, hydrogen, and oxygen reacts with a stream of chlorine gas, Cl2. After the reaction, 4.730 g of HCI, and 9.977 g of CCl4 are obtained. Determine the empirical formula of the organic compound.

How would I solve these? I tried and failed a lot, and am sure that they are actually quite easy T_T

-
1)
We know that 5 moles of O = 44 % so 1 mole of O in the compound = 44 / 5 % = 8.8%
if 8.8% = 16 g/ mole the mass of 100 % = 16 x 100 / 8.8 = 181.5 g / mole
the 5 O have mass of 5 x 16 or 80 g so the rest of the mass = 181.5- 80 = 101.5g
this is the mass of 2 moles of X so molar masx of X = 101.5 / 2 = 50.75 g / mole .. vanadium


2)
mass of H in 4.730g of HCl = 4.730 x 1/ 36.5 = .130g of H or .130 moles of H
mass of C in 9.977g of CCl4 = 9.977 x 12/ 153.82 = 0.7783 g or .7783 / 12 moles = 0.06486
mass of O = 1.254 - ( .130+.7783) = 0.3457g or 0.3457 / 16 moles = 0.02161 moles

molar ratio of H :C:O = 0.130:0.064861:0.02161
divide by the smallest to simplify the ratio and we have 6:3:1

empirical formula is C3H6O
1
keywords: Empirical,HELP,PLEASE,formula,Empirical formula, PLEASE HELP!!!
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .