Calculations: (Show all work and provide answers with correct significant figures.)1. Determine the molecular weight of the unknown acid from:a. Trial 10.12M * 18.55mL0.47g / 2.226 x10-3 mol= 2.226 mmol= 2.226 x10-3 mol= 211.1 g/molb. Trial 20.12M * 17.6mL0.46g / 2,112 x10-3 mol=2.112 x10-3 mol= 213.8 g/molc. Trial 30.48g / 2.268 x10-3 mol0.12M * 18.90mL= 2.268 x10-3 mol =211.6 g/mol2. Calculate the average molecular weight of the unknown monoprotic acid from Trials 1-3.212.16 g/molThe actual identity of the monoprotic acid is potassium hydrogen phthalate (KHCaHOe), which has a molecularweight of 204 g/mol. Calculate the percent error using the average molecular weight from Calculation #2.

The calculated molecular weight of monoprotic acid is 212.16 g/mol (from question no. 2)and actual…

The actual identity of the monoprotic acid is potassium hydrogen phthalate (KHCaH.O4), which has a molecular weight of 204 g/mol. Calculate the percent error using the average molecular weight from Calculation #2.

The actual identity of the monoprotic acid is potassium hydrogen phthalate (KHCaH.O4), which has a molecular weight of 204 g/mol. Calculate the percent error using the average molecular weight from Calculation #2.

Appl Of Ms Excel In Analytical Chemistry

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ISBN:9781285686691

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Chapter4: Least-squares And Calibration Methods

Section: Chapter Questions

Problem 1P

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Concept explainers

Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can be used in many calculations related to mole calculation, molar amount, mole ratio, and so on, which are significant in chemistry.

Question

Calculations: (Show all work and provide answers with correct significant figures.)
1. Determine the molecular weight of the unknown acid from:
a. Trial 1
0.12M * 18.55mL
0.47g / 2.226 x10-3 mol
= 2.226 mmol
= 2.226 x10-3 mol
= 211.1 g/mol
b. Trial 2
0.12M * 17.6mL
0.46g / 2,112 x10-3 mol
=2.112 x10-3 mol
= 213.8 g/mol
c. Trial 3
0.48g / 2.268 x10-3 mol
0.12M * 18.90mL
= 2.268 x10-3 mol =211.6 g/mol
2. Calculate the average molecular weight of the unknown monoprotic acid from Trials 1-3.
212.16 g/mol
The actual identity of the monoprotic acid is potassium hydrogen phthalate (KHCaHOe), which has a molecular
weight of 204 g/mol. Calculate the percent error using the average molecular weight from Calculation #2.

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