This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Using Stoichiometry to Determine a Formula
Burn 0.115 g of a hydrocarbon, C xHy, and produce 0.379 g of CO 2 and 0.1035 g of and H2O. What is the empirical formula of CxHy? CxHy + some oxygen ---> some 0.379 g CO 2 + 0.1035 g H 2O 0.1035 Using Stoichiometry to Determine a Formula
CxHy + some oxygen ---> some 0.379 g CO2 + 0.1035 g H2O 0.1035 Using Stoichiometry to Determine a Formula
CxHy + some oxygen ---> some 0.379 g CO2 + 0.1035 g H2O 0.1035 First, recognize that all C in CO 2 and all H and in H2O is from CxHy. 1. 2. Calculate moles of C in CO 2 8.61 x 10-3 mol C mol Calculate moles of H in H 2O
-2 mol 1.149 x 10 -2 mol H Now find ratio of mol H/mol C to find H/mol values of x and y in C xHy.
-2 mol 1.149 x 10 -2 mol H/ 8.61 x 10 -3 mol C mol = 1.33 mol H / 1.00 mol C = 4 mol H / 3 mol C Empirical formula = C 3H4 Page 1 ...
View Full Document
This note was uploaded on 01/11/2011 for the course ENGINEERIN MAE 107 taught by Professor Pozikrizdis during the Fall '08 term at San Diego.
- Fall '08