Problem: Combustion analysis of toluene, a common organic solvent, gives 6.45 mg of CO2 and 1.51 mg of H2O. If the compound contains only carbon and hydrogen, what is its empirical formula?

Recall that in combustion analysis, a compound reacts with excess O₂ to form products. 

mole-to-mole comparison:

• 1 mmole of C in 1 mmole of CO₂ (molar mass = 44.01 g/mol or mg/mmol)
• 2 mmoles of H in 1 mmole of H₂O (molar mass = 18.02 mg/mmol)

$\mathbf{mmoles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{45}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}}{\mathbf{44}\mathbf{.}\mathbf{01}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mmol}\mathbf{ }\mathbf{C}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}}$= 0.147 mmol C

$\mathbf{mmoles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{51}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }\mathbf{H}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}{\mathbf{18}\mathbf{.}\mathbf{02}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}\mathbf{\times }\frac{\mathbf{2}\mathbf{ }\mathbf{mmol}\mathbf{ }\mathbf{H}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}$

= 0.168 mmol H