Analysis of an unknown oxide of copper Mass of beaker (g) = 48.451g Mass of beak

Question

Analysis of an unknown oxide of copper

Mass of beaker (g) = 48.451g
Mass of beaker plus copper oxide (g) = 48.606g
Mass of copper oxide (g) = .151g
Mass of beacker plus copper (g) = 48.571g
Mass of copper obtained (g) = .035g
Mass of oxygen in the copper oxide (g) = .116g

Questions:
1) Calculate the observed percentage composition of your copper oxide. Show your calculations. From this result, deduce which pure oxide you were given.

2) Calculate how many milliliters of 6M hydrochloric acid were actually required to react with your sample of copper oxide (Show your calculations). Approximately what percentage excess did you use? (2.5 mL was used)

3)Calculate the observed empirical formula of your copper oxide, i.e., the formula based exactly on your experimental results (note: you will likely have fractional stoichiometric subscripts here)

Explanation / Answer

Firstly, the eqn: CuxOy = xCu + yO

Moles of O: 0.036gO x (1 mole/16 g) = 0.00225 moles O

Moles of Cu: 0.151g Cu x(1 mole/63.5 g)= 0.002377 moles Cu

Cu O ratio : Cu/O= 0.002377/0.00225=1.056 -> AS THE RATIO IS CLOSE TO 1, THEN THE VALUES OF X AND Y WOULD BE 1, THEN WE HAVE CuO

Then as there are 0.00225 moles of oxigen, there must be 0,00225 moles of CuO, the weight of pure CuO is

0.00225 moles CuO x (79.54 g/ mole)=0.17 g CuO

%CuO=100 x(0.1789/0.187)= 95.7%

Thre reaction between HCl and CuO is as follows:

CuO + 2 HCl -> CuCl2 + H2O

0.00225 moles CuO x (2 moles HCl/1 mole CuO)=0.0045 moles HCl, then we need 0.0045 moles of HCl

0.0045 moles HCl x (1L/ 6 moles) = 0.00075 L=0.75 ml, then 0.75 ml of HCl solution are needed to react.

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