A mixture of Cu2 and CuO of mass 8.828g is reduced to copper metal with hydrogen:
Cu2O + H2 --> 2Cu + H2O
CuO + H2 --> Cu + H2O
If the mass of pure copper isolated was 7.214g, determine the percent by mass of CuO in the original sample
Let x = grams of CuO in the original sample.
y = grams of Cu2O in the original sample.
Eq. #1 x + y = 8.828 grams
Molar mass of CuO = 63.5 + 16 = 79.5 grams
Moles of CuO = x ÷ 79.5
Molar mass of Cu2O = 63.546 + 32 = 95.5 grams
Moles of Cu2O = y ÷ 95.5
According to the 2nd balanced equation, CuO + H2 --> Cu + H2O ,
1 mole of CuO produces 1 mole of Cu.
So, x ÷ 79.5 moles of CuO will produce x ÷ 79.5 moles of Cu
According to the 1st balanced equation, Cu2O + H2 --> 2Cu + H2O,
1 mole of Cu2O produces 2 moles of Cu
So, (y ÷ 95.5) moles of Cu2O will produce 2 * (y ÷ 95.5) moles of Cu
Since, the mass of pure copper isolated was 7.214 grams
Moles of Cu = (7.214 ÷ 63.5)
Moles of Cu from Cu2O + moles of Cu from CuO = total moles of Cu!!
2 * (y ÷ 95.5) + (x ÷ 79.5) = (7.214 ÷ 63.5)
Multiply by both sides by 95.5 * 79.5 * 63.5 to get rid of denominators
(2 * 79.5 * 63.5) y + (95.5 * 63.5) x = (7.214 * 95.5 * 79.5)
10,096.5 y + 6,064.25 x = 36,418.0755
Divide both sides by 6,064.25
x + 1.665 y = 6
Eq.#2 x = 6 – 1.665 y
Eq. #1 x + y = 8.828
x = 8.828 – y
8.828 – y = 6 – 1.665 y
0.665 y = 2.828
y = 4.25 grams of Cu2O
x = 8.828 – 4.25 = 4.58 grams of CuO
% CuO = (4.58 ÷ 8.828) * 100 = 51.88% CuO
No units are provided in the data table, though one would usually assume that 0.244 is molarity of NaOH, and buret data of 14.7 is probably the final volume of NaOH in mL. No identification is given for the 38, but maybe this is the initial buret volume in mL.
If the buret was rinsed with water, it should have been left to dry for some time, or else the added water may slightly dilute the NaOH solution placed in it.