Question

Complete combustion of 4.20 g of a hydrocarbon produced 12.9 g of CO2 and 6.15 g of H20. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary. empirical formula:

Answer 1

1) Hydoo Carton C and H only. mass of mass & agar tid goo cę is 12-9 g Hao is 6.15g molas mass mass nolat nolaio nolat un g/mol mass of He is 18 g/mol mass of c is 12 hmot mass of his 1.01 g/mol hans mol ob co = = - nolas mase mass nolas mass = 12.9 g mass 12.9 g un Ilmul soola -=0.293 To ) ic and zo's. I molt Imolco, moes ob c = 0.293 but to x mol ( 7 = 0.293 molc IH to - 2t's mols of 2ts and so and 10 – Hoo) 2 molt Imol Ho. mols ob 6.15g hor =0.342 18 Ilmol hool ho.

mols of H = 0.34 2 molt. x 2 molta methox Limotto = o.684 molt. ad from above mol values divide values with I mol least least mol Valul. so with divide 0.293 0.293 mol 0.295 mol H = 0.684 but 0.29 ind = 2:33 = 21/3 we reed whole number Coeffients, so multiply them with least possible number to get whole number. - B cu 1x3=3 H. - 2 X3=7 .. Empirical formala" C3 H7 ..An [Ca ty