Question

Naphthalene, a hydrocarbon, has an approximate molar mass of 128 g/mole. If the combustion of 64 produces 0.3599 g H20 and 2.1977 g CO2, what is the molecular formula of this compound? 3. A mass of 0.4113 g of an unknown acid, HA, is titrated with NaOH. If the acid reacts with 28.10 mL of 0.1055 M NaOH(aq), what is the molar mass of the acid? 4. The amount of calcium in a 15.0-g sample was determined by converting the calcium to calcium oxalate, CaC204. The CaC2O4 weighed 11.0 g. What is the percent of calcium in the original sample?

Answer 1

2. C₁₀H₈ (Naphthalene)

Males of_Napthalene combusted : 0.6400_j 128 g/mol = 0.005 male Males of H2O produce = 0.3599 Moly of CO2 produce = 0.02 mole = 2.1977 44 0.05 moles of (02 + O₂(excess) Napthalene -Todos miele)___ .CO2 + H2O 1otos mol) Co.o2 mul). 0.05 male CO2 colain -0.02 mole H2o contain because H₂O contain 2 0.05 mole of 2x0.02 mole hydrogen in carbon of Hydrogen. 1 molecule .. So, Male hatio of c and H is = formula of 0.05:0.04 5:4 napthalene ce Emperical molecular

wen atoms. contains 5 carbon atom and 4 Hydrogen ato so, Emperical formula - C5 H4 Now, calculate Emperical mass of C5 H4 = 5x12 + 4x1_ - = 60 + 4 = 649- But Malar mass of hap thalene is 128 g/mol hatio of Malar mass and Emperical mass. - 128:64 = 2:1 ie emperical mass is half of molar mass. so multiply the emperical formula by 2. (CSM) x 2 = Co H& Now, check molar mass_of C10H8 = 12X10 + 8x1__ = 120+ 8 = 128 g/mol . Co H& is correct formula of Napthalene.
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