Question

Two students are analyzing a compound containing copper and oxygen only. Fred’s sample weighs 3.12g and contains 2.50g of copper. Wilma’s sample weighs 1.62g and contained 1.44g of copper. Are the two students analyzing the same compound? Justify your answer.

Answer 1

The formula of the analyzed samples could be presented as follows

Cu_x1O_y1(Fred) and Cu_x2O_y2 (Wilma)

Calculations of the elements quantity in Fred's sample

n(Cu) = x1 = m(Cu)/M(Cu) = 2.50 g/64 g/mol = 0.039 mol

n(O) = y1 = m(O)/M(O) = (3.12 - 2.50) g/ 16 g/mol = 0.039 mol

Elements ratio in the sample is x1/y1 = 1/1 -> the formula of the analyzed substance is CuO

Calculations of the elements quantity in Wilma's sample

n(Cu) = x2 = m(Cu)/M(Cu) = 1.44 g/64 g/mol = 0.0225 mol

n(O) = y2 = m(O)/M(O) = (1.62 - 1.44) g/ 16 g/mol = 0.01125 mol

Elements ratio in the sample is x2/y2 = 2/1 -> the formula of the analyzed substance is Cu₂O

The answer is "No"