Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Join our Q&A platform to get precise answers from experts in diverse fields and enhance your understanding. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Answer:
C2H4O3
Explanation:
Empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.
To solve this question we need to find the moles of carbon = Moles of CO2, the moles of hydrogen (Using moles of H2O) and the moles of oxygen (Finding the mass of the mass of each atom) as follows:
Moles Carbon -Molar mass CO2: 44.01g/mol-:
33.86g CO2 * (1mol/44.01g) = 0.769 moles CO2 = Moles C * (12g/mol) =
9.23g C
Moles Hydrogen -Molar mass H2O: 18.01g/mol-
13.86g H2O * (1mol/18.01g) = 0.770 moles H2O * (2mol H / 1mol H2O) = 1.54 moles H * (1g/mol) = 1.54g H
Moles Oxygen:
Mass: 29.26g - 9.23g C - 1.54g H = 18.49g O * (1mol/16g) = 1.156 moles O
Dividing each number of moles in the moles of C (Lowe number of moles):
C = 0.769 moles C / 0.769 moles C = 1
H = 1.54 moles H / 0.769 moles C = 2
O = 1.156 moles O / 0.769 moles C = 1.5
As the number must be a whole number each ratio twice:
C = 2
H = 4
O = 3
Empirical formula is:
C2H4O3
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.