Question

When 2.050 grams of a hydrocarbon, C_xH_y, were burned in a combustion analysis apparatus, 6.280 grams of CO₂ and 3.000grams of H₂O were produced.

In a separate experiment, the molar mass of the compound was found to be 86.18 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

empirical formula =?

molecular formula =?

Answer 1

Empirical formula is the least coefficient formula, that is

CxHyOz

we must find x,y,z via gravimetry.

Typically we do this relating to moles of C,H,O

mol of CO2 = mass of CO2/MW of CO2 = (6.28)/(44) = 0.142727 mol of CO2

1 mol of CO2 = 1 mol of C ---> 0.142727 mol of CO2 = 0.142727 mol of C

mol of H2O = mass of H2O/MW of H2O= (3)/(18) = 0.16666mol of H2O

1 mol of H2O= 2 mol of H ---> 0.16666 mol of H2O= 0.16666*2 = 0.33332 mol of H

Ratios:

H:C =0.33332 /0.142727 = 2.33 = 2 + 1/3 = (6+1)/3 = 7/3

or 7 H per 3 C

C3H7 empirical

MW of C3H7 -> 12*3 + 7*1 = 43 g/mol, then

MW ratio --> 86.18/43 = 2x

C3H7 --> 2x --> C6H14