Empirical Formula calculation
Favorites|Homepage
Subscriptions | sitemap
HOME > > Empirical Formula calculation

Empirical Formula calculation

[From: ] [author: ] [Date: 12-06-22] [Hit: ]
What is the empirical formula of the compound?Please give a detailed answer thanks.-Mass of C in 0.372g CO2 = 12/44*0.372 = 0.Mass of H in 0.......
Elemental analysis of 0.2511g of a compound produced 0.372g of CO2, 0.178g of H2O and 0.039g of N2. What is the empirical formula of the compound?

Please give a detailed answer thanks.

-
Mass of C in 0.372g CO2 = 12/44*0.372 = 0.1015g
Mass of H in 0.178g H2O = 2/18*0.178 = 0.0198g
Mass of N = 0.039g

Divide each by atomic mass

C = 0.1015/12 = 0.008458
H = 0.0198/1= 0.0198
N = 0.039/14 = 0.002786

divide by smallest:
C = 3.036
H = 7.10
N = 1

The closest I can make this is C3H5NH2 = cyclopropylamine.

But there is something not quite correct with this - the masses of the 3 elements making up the compound do not add up to the mass of the original compound. I am particularly suspicious of the statement "and 0.039g of N2"are you certain that this is correct? .

-
You might want to check the numbers. If you have a 0.2511g compound, you can't have 0.372g of CO2 as part of the compound. Part of the compound can't exceed the mass of the entire compound. Is it 2.511g by chance?
1
keywords: Empirical,calculation,Formula,Empirical Formula calculation
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .