Question

1.)) An unknown compound containing carbon, hydrogen and oxygen is combusted. If 25.000 g of the compound produced 61.024 g of CO2 and 24.981 g of H2O, what is the empirical formula of the compound?

2.)) If the molar mass of this compound is 72.1 g/mol, what is the molecular formula?

Answer 1

Solution=

solution it since the formula compound CH and o. will be CxHy oz. so the molecular No, ok moles of con in the initial product is - - 61.024 g x 1 mol 44 g/mol = 1.3869 mol of coz so moles ok 6 is equal to 1.3869 mol. ." No. ok H is 24,981 g x 1 mal x2 18 g/mol = 2.7756 mol so moles ok n = 2.7756 mol. NOW convert mola ok C and H to greams, c = 12g x 1.3869 mol = 16.6428 H = 1 g x 2.7756 mol = 2.7756 g/mol g/mol. from the compound Now substract the amount of c H to give the amount ok Or. = 259 - (16.64289 +2.77563) = 5.5816 g no. ok moles ok 0 = 0.34 885 mol and so = 0.34885 mo = 1.3869 mol , H = 2.7756 mol divide all 3 with smallest mole - NOW Co to Rounding = 8.97 gives H = 7.95, an empirical 0 - 1 formula NOW C4H8 0 and Molecalove ormula is C4H8O.