Question

CHML 201 Data Pag Expt 055 Determination of the molar mass of a basic oxide, an example of a"back-titration." M Hel-300ma M N4OH=d00 Name A. Standardiration of NaOH Solution Na0H in burette Trial D Tral 25 Trial3 fnal burct reading initial buret reading Volume oh NAOH sed ml, of standand acid used molarity of HC molarity of NaOH Av. molality ofNaOH B. Determisatlon of Melar Mass: Sample No Trial I Trial 2 Trial 3 mass paper+ sample e erec mass paper mass sample final buret reading .O 21-5 M initial burct reading De Volume of NaOH used 915 20- mL of standand acid addedi mol of HCl added 00?nel mol of NaOH used Do-2ls 00262 mol of HCI rcacted oo 20/ Deil mol of HCI left Oo535 Oosy9 Molar mass, ef oside Av. Molar mgse Suggested formula for oxide (Show calculations on separate pages). Ne O.ege inale e Hel for neutraligat eole

Answer 1

The molar weight of the oxide in each sample is calculated, dividing the mass of the sample among the moles of HCl left:

MM = g / n =

0.104 g / 0.00535 mol = 19.44 g / mol

Trial 2: 20.49 g / mol

Trial 3: 30.09 g / mol

Average = 23.34 g / mol

Li2O and BeO are proposed as possible oxides, due to their molar weight similar to the calculated one.

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