A student dissolved a piece of copper wire, then converted it to CuO, filtered t

Question

Explanation / Answer

% yield = mass recovered / starting mass x 100%

let's start with the reactions.....

2Cu + 1O2 ------> 2CuO + 2e-

and

2CuO + 2e- ------> 2Cu + O2

where the 2e-'s indicates a redox reaction....

anyway,

you have have 2 moles of Cu ----> 2 moles of CuO
and 2 moles of CuO going to 2 moles of Cu

so calculate starting moles of Cu, then moles of CuO, then mass of CuO. That's your theoretical starting mass of CuO. then from your filter paper info, calculate actual mass of CuO recovered and actual mass of Cu recovered. then calculate yield. Ready?

1) moles of Cu

moles = mass / mw = .260g / (63.5 g/mole) = 4.09 x 10^-3 moles Cu

2) theoretical moles of CuO

4.09 x 10^-3 moles Cu x ( 2 moles CuO / 2 moles Cu) = 4.09 x 10^-3 moles CuO

3) theoretical starting mass of CuO

moles = wt / mw
wt = moles x mw = 4.09 x 10^-3 moles CuO x (79.5 g / moles) = 0.326 g CuO

4) mass of CuO recovered

mass of filter paper + CuO - mass of filter paper = mass CuO recovered right?

actual mass CuO = 0.834 g - 0.525 g = 0.309 g CuO

5) mass of Cu recovered

0.732 g - 0.490 g = 0.242 g Cu

6) % yield

% yield CuO = mass recovered / theoretical starting mass x 100%
= 0.309 g CuO / 0.326 g CuO x 100% = 94.8%

% yield Cu = mass recovered / starting mass x 100%
= 0.242 g Cu / 0.260g = 93.1%

fyi, the percents need to be reported to 3 sig figs since all of your data is 3 sig figs.....

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