calculate the percentage of copper [||] oxide in the impure sample

Question

0.5g of impure copper [||]oxide were reacted with 50cm³ of 0.1M HNO3. Calculate the percentage of the copper [||] oxide in the impure sample. (C=64.0,O=14.0,H=1.0)

Answer 1

2CuO(s) + 4HNO3(aq) --> 2CU(NO3)2(aq) + 2H2O(l)

Mole of HNO3 in 50cm³ = (50cm³ × 0.1)/1000cm³
  = 0.005moles


CuO : HNO3 = 2:4

Moles of CuO = 0.0025
RFM of CuO = 64 + 16= 80g/mole

Mass = 0.0025 × 80

Mass = 0.2g

0.2/0.5 × 100%

40%

Note: RAM of oxygen is 16 NOT '14'