An experiment requires that enough C3H6O3 be used to yield 15.0 g of oxygen. How much C3H6O3 must be weighed out?

Answer 1

Hello, Jaden,

We technically need a balanced equation for this reaction to be certain of the clculations. I'm unfamiliar with this reaction, but will assume it is a decomposition that realeases all the O atoms as O₂ molecules.

This assumption would allow us to simply calculate how many grams of C₃H₆O₃ we need to have 15 grams of oxygen.

Let's calculate the percentage of the total mass that is due to oxygen. For example, the molar mass of C₃H₆O₃ is 90.03 g/mole. It has 3 O atoms per molecule, which add to (3 x 16.0 =) 48.0 AMU. The percentage of O in C₃H₆O₃ is therefore (48.0 AMU/90.03 AMU), or 53.31%.

To find the mass required to contain 15.0 grams of O, use:

X*(0.5331) = 15 g [What mass times the O percentage produced 15g of O].

X = 37.48 grams of C₃H₆O₃ contains 15.0 g of O.

Bob