Combustion analysis of toluene, a common organic solvent, gives 8.79 mg of co2 and 2.06 mg of h2o. if the compound contains only carbon and hydrogen, what is its empirical formula?

             C₇H₈

Solution:

Step 1: Convert mg into g as;

                            8.79 mg CO₂  =  0.00879 g CO₂

                            2.06 mg H₂O  =  0.00206 h H₂O

Step 2: Moles of CO₂ and H₂O as;

                            Mole of CO₂  =  mass of CO₂ ÷ M.mass of CO₂

                            Mole of CO₂  =  0.00879 g ÷ 44 g.mol⁻¹

                            Mole of CO₂  =  0.0002 mol

Also,

                            Mole of H₂O  =  mass of H₂O ÷ M.mass of H₂O

                            Mole of H₂O  =  0.00206 g ÷ 18 g.mol⁻¹

                            Mole of H₂O  =  0.0001144 mol

Step 3: Finding out moles of C and H as;

As 1 mole of CO₂ has 1 mole of C so, 0.0002 mol CO₂ will contain 0.0002 mol of C.

Aslo, 1 mole of H₂O has 2 moles of H so, 0.0001144 mol H₂O will contain (0.0001144 × 2) 0.0002288 moles of H.

So,

                            Moles of C  =  0.0002 mol

                            Moles of H  =  0.0002288 mol

Step 4: Find the simplest ratio;

Divide both moles with the smallest mole value as;

                            C  =  0.0002 ÷ 0.0002  =  1

                            H  =  0.0002288 ÷ 0.0002  =  1.14

Step 5: Get the whole number by multiplying the ratio by 7,

                            C  =  1 ×7  =  7

                           H =  1.14 × 7  =  8

Result:

          The Empirical formula (which is also the molecular formula) of TOLUENE is C₇H₈

i think slash and

the answer is 0.42 is that is an option. this question provides no info