Question

A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of the compound yields 16.01 mg and 4.37 mg . The molar mass of the compound is 176.1 g/mol. What are the empirical and molecular formulas of the compound

Answer 1

Answer: See below

Step-by-step explanation:

n of CO2​ = 0.364mmol

Mass of C = 0.364*12 = 4.368 mg

n of H2​O = 184.37​ = 0.243 mol

The compound has 2*0.243mmol of H

Mass of H = 0.486 mg

Mass of O = 10.68 − (4.368+0.486) = 5.826mg

Moles of O = 0.364

C:H:O Ratios

0.364 : 0.486 : 0.364

= 1 : 1.34 : 1

= 3 : 4 : 3

So the empirical formula is C3​H4​O3​,

Empirical formula mass

= 88= 2 × Molar mass

And the molecular formula is C6​H8​O6