A student wants to check the purity of a solid organic compound by measuring its melting point.
The literature melting point of the pure compound is 115 ∘C115\ ^\circ\text{C}115 ∘C.
Which equation represents the percentage (%) difference between the student's experimental value and the literature value?
% difference=100×(student’s value in ∘C)−115 ∘C115 ∘C\% \text{ difference} = 100 \times \frac{(\text{student's value in } ^\circ\text{C}) - 115\ ^\circ\text{C}}{115\ ^\circ\text{C}}% difference=100×115 ∘C(student’s value in ∘C)−115 ∘C
% difference=100×115 ∘C−(student’s value in ∘C)115 ∘C\% \text{ difference} = 100 \times \frac{115\ ^\circ\text{C} - (\text{student's value in } ^\circ\text{C})}{115\ ^\circ\text{C}}% difference=100×115 ∘C115 ∘C−(student’s value in ∘C)
% difference=(student’s value in ∘C)−115 ∘C115 ∘C\% \text{ difference} = \frac{(\text{student's value in } ^\circ\text{C}) - 115\ ^\circ\text{C}}{115\ ^\circ\text{C}}% difference=115 ∘C(student’s value in ∘C)−115 ∘C
% difference=115 ∘C−(student’s value in ∘C)115 ∘C\% \text{ difference} = \frac{115\ ^\circ\text{C} - (\text{student's value in } ^\circ\text{C})}{115\ ^\circ\text{C}}% difference=115 ∘C115 ∘C−(student’s value in ∘C)
46 exam-style questions on OCR GCSE Chemistry Purity and separating mixtures. Each one has a worked solution and a mark scheme showing where the marks go.