IB DP Chemistry: SL復習筆記11.2.4 Percentage Error

Category: IB課程, 教材筆記, 福利干貨 Date: 2022年8月5日上午10:06

Percentage Error

Percentage error?is used to express the difference between a final calculated answer and an accepted or literature value
It is calculated using the following formula

11.2.4-Experimenal-Error-Formula

You should be able to comment on any differences between the experimental and literature values

Worked Example

1.023 g of propan-1-ol (M?= 60.11 g mol^-1) was burned in a spirit burner and used to heat 200 g of water in a copper calorimeter. The temperature of the water rose by 30?^oC.

Calculate the enthalpy of combustion of propan-1-ol using this data.
The literature value for this enthalpy change is -2021 kJ mol^-1. Calculate the percentage error and comment on your findings

Answer 1:

Step 1: Calculate q

q = m x c x ΔT

q = 200 g x 4.18 J g^-1?K^-1?x 30 K = – 25 080 J

Step 2: Calculate the amount of propan-1-ol burned

moles = mass ÷ molar mass = 1.023 g ÷? 60.11 g mol^-1?= 0.01702 mol

Step 3: Calculate ΔH

ΔH = q ÷ n =? -25 080 J ÷ 0.01702 mol = – 1 473 560 J = -1 474 kJ =?-1.5 x 10³ kJ

Answer 2:

Using the formula

11.2.4-Experimenal-Error-Formula 11.2.4-Worked-Example-Answer3 Heat losses are likely to be the largest source of error in this experiment