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Introduction to Analytical Chemistry — Chemistry STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceChemistryIntroduction to Analytical Chemistry2 Marks
Question
In laboratory experiment, 10 g potassium chlorate sample on decomposition gives following data: The sample contains 3.8 g of oxygen and the actual mass of oxygen in the quantity of potassium chlorate is 3.92 g. Calculate absolute error and relative error.
✓

Answer

The observed value is 3.8 g and accepted (true) value is 3.92 g.
i. Absolute error = Observed value – True value
= 3.8 – 3.92
= -0.12 g
$\begin{aligned} & \text { ii. } \quad \text { Relative error }=\frac{\text { Absolute error }}{\text { True value }} \times 100 \% \\ & =\frac{-0.12}{3.92} \times 100 \% \\ & \end{aligned}$
$=-3.06 \%$
[Note: The negative sign indicates that experimental result is lower than the true value.]
Ans: i. Absolute error = -0.12 g
ii. Relative error = -3.06%

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