Question

Quantitative Analysis a. A 1.861g sample of a certain monoprotic acid was dissolved in water and titrated with a 0.1984 M sodium hydroxide solution. The titration endpoint was reached after 45.9 mL of the titrant was added. What is the molar mass of the acid?

b. Combustion analysis of a hydrocarbon released 321 mg carbon dioxide and 164 mg water. What is the empirical formula of the compound?

Answer 1

m = 1.861 g of monoprotic acid

[NaOH] = 0.1984 M

V = 45.9 mL of titrant used...

mmol of base = MV = 0.1984 *45.9 = 9.10656 mmol of NaOH

ratio is 1:1 since monoprotic acid so

mmol of acid = 9.10656 mmol = 9.10656*10^-3 mol

MW = mass of acid / moles of acid = 1.861/(9.10656*10^-3) = 204.358 g /mol

b)

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 = (0.321)/(44) = 0.00729 mol of CO2

1 mol of CO2 = 1 mol of C ---> 0.00729 mol of CO2 = 0.00729 mol of C

mol of H2O = mass of H2O/MW of H2O= (0.164)/(18) = 0.009111 mol of H2O

1 mol of H2O= 2 mol of H ---> 0.009111 mol of H2O= 0.009111*2 = 0.018222 mol of H

Now... mol of Oxygen:

Mass of O = Mass of sample - Mass of C - Mass of H

Mass of C = mol of C * MW of C = 0.00729 *12 = 0.08748 g

Mass of H = mol of H * MW of H = 0.018222 *1= 0.018222 g

Ratios:

H:C =0.018222 /0.08748 = 0.2 = 1

C:H =  0.08748 /0.018222 = 4.8 = 5

CH5

empiricla formula C1H5