Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
Answer:
C2H3O3
Explanation:
Empirical formula is the simplest whole number ratio of moles of atoms that you can find in a molecule.
In combustion analysis all Carbon reacts producing CO2 and all hydrogen reacts producing H2O. With the differences in masses we can find the mass of oxygen and their moles:
Moles CO2 = Moles C:
14.08g * (1mol/44.01g) = 0.3199 moles C * (12.01g/mol) = 3.8423g C
Moles H2O:
4.32g H2O * (1mol/18.01g) = 0.2399 moles H2O * (2mol H / 1molH2O) = 0.4797moles H = 0.4797g H
Mass O:
12.01g = Mass O + 3.8423g C + 0.4797g H
Mass O = 7.688g O
Moles O:
7.688g O * (1mol/16g) = 0.48 moles O
The ratio of atoms (Dividing in the moles of C that are the lower number of moles):
O: 0.48moles O / 0.3199 moles C = 1.50
C: 0.3199 moles C / 0.3199 moles C = 1
H: 0.4797 moles H / 0.3199 moles C = 1.50
As empirical formula requires whole numbers:
O: 1.50* 2 = 3
C: 1*2 = 2
H: 1.50*2 = 3
The empirical formula is:
C2H3O3
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.