Question

orce Completion Once started, this test must be completed in one sitting Do not leave the test beforeiding Save and Submit Remaining Time: 49 minutes, 14 seconds. Question Completion Status: > Moving to another question will save this response Question 1 Combustion analysis of 0.600 g of an unknown compound containing carbon, hydrogen and oxygen produced 100% of CO2 and 0.5670 of 20. What is the empirical formula of the compound? O CaHgO₂ осно H502 C₂H₂O3 Moving to another question will save this response

Answer 1

Empirical formula : C₃H₈O₂

Explanation

mass CO₂ = 1.043 g

moles CO₂ = (mass CO₂) / (molar mass CO₂)

moles CO₂ = (1.043 g) / (44.01 g/mol)

moles CO₂ = 0.0237 mol

moles C = moles CO₂

moles C = 0.0237 mol

mass C = (moles C) * (molar mass C)

mass C = (0.0237 mol) * (12.01 g/mol)

mass C = 0.2846 g

Similarly, moles H = 0.06295 mol

mass H = 0.06345 g

mass O = (total mass) - (mass C + mass H)

mass O = 0.600 g - (0.2846 g + 0.06345 g)

mass O = 0.25195 g

moles O = 0.01575 mol

moles C / moles O = (0.0237 mol) / (0.01575 mol)

moles C / moles O = 1.5

moles H / moles O = (0.06295 mol) / (0.01575 mol)

moles H / moles O = 4

Empirical formula : C_1.5H₄O

Empirical formula : C₃H₈O₂