Qualitative analysis of an unknown acid found only carbon, hydrogen, and oxygen.
ID: 910272 • Letter: Q
Question
Qualitative analysis of an unknown acid found only carbon, hydrogen, and oxygen. In a quantitative analysis, a 10.46 mg sample was burned in oxygen and gave 15.332 mg CO2 and 6.28 mg H2O. The molecular mass was determined to be 60.05 g/mol. When a 0.1680 g sample of the acid was titrated with 0.156 M NaOH, the end point was reached after 17.93 mL of the base had been added.
a.) What is the molecular formula for the acid? Enter your answer in form CxHyOz
b.) Is the acid monoprotic, di-, or trioprotic?
Explanation / Answer
Molar mass of CO2 = 44 g/mole ; H2O = 18 g/mole
moles of C present in the sample = moles of CO2 = 0.015332/44 = 3.485*10-4
Mass of C present in the sample = moles*molar mass of C = 0.004182 g
moles of H present in the sample = 2*moles of H2 produced = 2*(0.00628/18) = 6.978*10-4
mass of H present in the sample = moles of H*molar mass of H = 0.0006978 g
Hence mass of O present in the sample = mass of sample - mass of H - mass of C = 0.00558 g
moles of O present = mass/molar mass = 3.488*10-4
Hence. C,H & O are present in the molar ratio of :- 0.3485:0.6978:0.3488 = 1:2:1
Hence the empirical formula of the compound is :- CH2O
Empirical mass of the compound is :- 30 g
Thus, molecular formula of the compound is :- (CH2O)n ; where n = molar mass/empirical mass = 60.05/30 = 2
Molecular formula of the acid = C2H4O2 = CH3COOH (acetic acid)
Now, moles of NaOH reacted = molarity*volume of solution in litres = 0.156*0.01793 = 2.797*10-3
moles of acid reacted = mass/molar mass = 0.168/60.05 = 2.797*10-3
Since, moles of acid recated = moles of NaOH reacted
Therepore, the acid is monoprotic acid