Answered

Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

Combustion of 29.26 gg of a compound containing only carbon, hydrogen, and oxygen produces 33.86 gCO2gCO2 and 13.86 gH2OgH2O. Part A What is the empirical formula of the compound

Sagot :

jufsanabriasa

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

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.