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Determining the Formula of a Hydrocarbon by Combustion

Determining the Formula of a Hydrocarbon by Combustion. CCR, page 138. Using Stoichiometry to Determine a Formula. Burn 0.115 g of a hydrocarbon, C x H y , and produce 0.379 g of CO 2 and 0.1035 g of H 2 O . C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O

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Determining the Formula of a Hydrocarbon by Combustion

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  1. Determining the Formula of a Hydrocarbon by Combustion CCR, page 138

  2. Using Stoichiometry to Determine a Formula Burn 0.115 g of a hydrocarbon, CxHy, and produce 0.379 g of CO2 and 0.1035 g of H2O. CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O What is the empirical formula of CxHy?

  3. +O2 Puddle of CxHy +O2 0.115 g Using Stoichiometry to Determine a Formula CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O First, recognize that all C in CO2 and all H in H2O is from CxHy. 0.379 g CO2 1 CO2 molecule forms for each C atom in CxHy 0.1035 g H2O 1 H2O molecule forms for each 2 H atoms in CxHy

  4. Using Stoichiometry to Determine a Formula CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O First, recognize that all C in CO2 and all H in H2O is from CxHy. 1. Calculate amount of C in CO2 8.61 x 10-3 mol CO2 --> 8.61 x 10-3 mol C 2. Calculate amount of H in H2O 5.744 x 10-3 mol H2O -- >1.149 x 10-2 mol H

  5. Using Stoichiometry to Determine a Formula CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O Now find ratio of mol H/mol C to find values of x and y in CxHy. 1.149 x 10 -2 mol H/ 8.61 x 10-3 mol C = 1.33 mol H / 1.00 mol C = 4 mol H / 3 mol C Empirical formula = C3H4

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