Combustion analysis of toluene, a common organic solvent, gives 5.27 mg of co2 and 1.23 mg of h2o. if the compound contains only carbon and hydrogen, what is its empirical formula?

C7H8 First, determine the number of relative moles of each element we have and the molar masses of the products. atomic mass of carbon = 12.0107 atomic mass of hydrogen = 1.00794 atomic mass of oxygen = 15.999 Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087 Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 We have 5.27 mg of CO2, so 5.27 / 44.0087 = 0.119749 milli moles of CO2 And we have 1.23 mg of H2O, so 1.23 / 18.01488 = 0.068277 milli moles of H2O Since there's 1 carbon atom per CO2 molecule, we have 0.119749 milli moles of carbon. Since there's 2 hydrogen atoms per H2O molecules, we have 2 * 0.068277 = 0.136554 milli moles of hydrogen atoms. Now we need to find a simple integer ratio that's close to 0.119749 / 0.136554 = 0.876937 Looking at all fractions n/m where n ranges from 1 to 10 and m ranges from 1 to 10, I find a closest match at 7/8 = 0.875 with an error of only 0.001937, the next closest match has an error over 6 times larger. So let's go with the 7/8 ratio. The numerator in the ratio was for carbon atoms, and the denominator was for hydrogen. So the empirical formula for toluene is C7H8.

It has to be b i hope this

Show an image of the data (˵¯͒ བ¯͒˵)