# Report On Determination Of A Composition Of A Mixture

Type of paper: Report

Pages: 2

Words: 550

Published: 2020/10/25

## Experimental Objectives

The laboratory assignment aims to test the water absorption by sodium carbonate and its capability to store water.
Experimental Procedure
The chemicals used for the assignment are sample that consist of 60% sodium hydrocarbonate and 40% sodium carbonate, and the sample with unknown composition.
During the experiment, the moisture from the sample is removed by drying and the sample is dried until the constant mass. Taking the mass difference, the quantity of evaporated water is calculated, and the composition of the unknown mixture is defined.

## Weight of the unknown sample

Weight of the sample after the constant drying.
Experimental Data
The raw experimental results are present in Tables 1 and 2.

## Analysis of Data

Heating induces water evaporation and carbon dioxide disengagement (Miessler and Tarr):
2NaHCO3 (heating) → Na2CO3 + H2O + CO2

## According to molar mass, the atom economy calculates as:

Atom economy=Mass of the desired product Mass of the reactant ∙100%=Na2CO3NaHCO3∙100%
Applying the molar mass, we obtain (the number of molecules of sodium hydrocarbonate should be taken into account): Atom economy=2∙23+12+3∙16 2∙(23+1+12+3∙16) ∙100%=10684∙100%=63.1%
Therefore, the substance percent loss is: 100 – 63.1 = 36.9%

## Trial 1

The average final mass calculates as:
The final mass =0.39 + 0.40 = 0.395 = 0.40 g

## The error of the experimental measurement:

% difference = (The trial value 1 – The trial value 2) / The average of the two trial values ∙ 100% = (0.40 + 0.39) / 0.395 ∙ 100% = 2.5 %.

## The mass of H2O and CO2 evaporated is:

Mass H2O and CO2 = The initial mass - The final mass = 0.50 – 0.40 = 0.1 g.

## The mass lost is attributed to H2O and CO2 . Thus, we can compose an equation:

0.1= m(NaHCO3) ∙ 36.9 %/ 100 %;
0.1 = m(NaHCO3) ∙ 0.369;
m(NaHCO3) = 0.1 / 0.369 = 0.27 g.

## The experimentally determined sample composition:

ω(NaHCO3) = m(NaHCO3)exp / The initial mass ∙ 100% = 0.27 / 0.5 ∙ 100% = 54.2%
% error = (The experimentally determined composition –The theoretical composition) / The theoretical composition ∙ 100% = (54.2 – 60) ∙ 100% = 5.15 % (the absolute value is taken)
% error = 5.15%.

## Trial 2

The average final mass calculates as:
The final mass =0.92 + 0.91 = 0.915 = 0.92 g

## The error of the experimental measurement:

% difference = (The trial value 1 – The trial value 2) / The average of the two trial values ∙ 100% = (0.92 + 0.91) / 0.915 ∙ 100% = 1.1 %.

## The mass of H2O and CO2 evaporated is:

Mass H2O and CO2 = The initial mass - The final mass = 1.0 – 0.92 = 0.08 g.

## The mass lost is attributed to H2O and CO2 . Thus, we can compose an equation:

0.08 = m(NaHCO3) ∙ 36.9 %/ 100 %;
0.08 = m(NaHCO3) ∙ 0.369;
m(NaHCO3) = 0.08 / 0.369 = 0.217 g.

## The experimentally determined sample composition:

ω(NaHCO3) = m(NaHCO3)exp / The initial mass ∙ 100% = 0.217 / 1 ∙ 100% = 21.7%

## Results

The experimental results and calculations are presented in Tables 3 and 4.

The loss of sample mass appears due to dehydration and carbon dioxide disengagement. This allows determining the unknown sample composition.

## Conclusions

The experiments on sodium hydrocarbonate dehydration were performed. The composition of the unknown mixture was calculated and the experimental errors were assessed.

## Works Cited

Miessler, G. L, and Tarr, D. A. Inorganic Chemistry. Upper Saddle River, N.J: Pearson Education, 2004. Print.

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Report On Determination Of A Composition Of A Mixture. Free Essay Examples - WePapers.com. https://www.wepapers.com/samples/report-on-determination-of-a-composition-of-a-mixture/. Published Oct 25, 2020. Accessed June 30, 2022.
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