Question

What is the molecular formula for a compound having a molar mass of 90.04g/mol containing carbon, hydrogen, and oxygen? Combustion analysis reveals .501g of CO2 and .103 grams of water from burning .514 grams of the compound.

Answer 1

Solution :-

Lets first calculate the mass of the C and H using the percent of the C and H in water

mass of C = 0.501 g CO2 * 27.29 % C / 100% = 0.1367 g C

mass of H = 0.103 g H2O * 11.19 % H / 100 % = 0.01153 g H

now lets find mass of oxygen

mass of oxygen = mass of compound - ( mass of C+ mass H)

                         = 0.514 g - (0.1367 g C + 0.01153 g H)

                         = 0.3658 g O

now lets calculate moles of each element

moles = mass / molar mass

moles of C =0.1367 g / 12.01 g per mol = 0.01138 mol C

moles of H = 0.01153 g / 1.0079 g per mol = 0.01144 mol H

moles of O = 0.3658 g / 16 g per mol = 0.02286 mol O

now lets find the mole ratio by dividing the moles of each element by smallest mole value

C= 0.01138/0.01138 = 1

H = 0.01144 / 0.001138 = 1

O = 0.02286 / 0.001138 = 2

so the empirical formula = CHO2

now lets find the mass of empirical formula

mass of CHO2 = 12.01 g + 1.0079 g +(16*2) = 45.0 g

now lets find the ratio of the molar mass and empirical formula mass

molar mass / empirical formula mass = 90.04 g per mol / 45.0 g per mol = 2

so the molecular formula = 2*empirical formul

                                        = 2*CHO2

                                        = C2H2O4

So the molecualr formula = C2H2O4