Determine the empirical formula of a compound containing 47 37 grams of carbongrams of hydrogen, and of oxygen In an expenmen the me compound was determined to de 228 276 gis the molecular of the compound? For both questions your work or explain how you detemined the formulas by giving specille values used in calculations

Molecular formula :  C₉H₂₄O₆

Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen. In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound? For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations.

mol of C :

mol of H :

mol of O :

divide by smallest (2.6275)

C : H : O ⇒

Empirical formula(EF) : C₃H₈O₂

(EF)n=MF(molecular formula)

( C₃H₈O₂)n=228.276

(3x12.0107+8x1,00784+2x15,999)n=228.276

(76.09282)n=228.276⇒n≈3

Molecular formula : C₉H₂₄O₆

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation:

2b2t

explanation:

2b2t