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# Empirical Formula Lab Report: CuSO4.5H2O and CuO

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1. Data
(a) mass of copper powder used (g): 25g
(b) mass of copper powder and beaker (g): 72.420g
(c) mass of copper oxide and beaker (g): 78.714g

2. Calculate the following for the copper:

(a) The number of moles of copper powder, given a molecular weight of copper equal to 63.55 g/mole. (1 pt) {Convert the value from 1(a) to moles}

3. Calculate the following for the oxygen:

(a) The mass of the beaker (g) {1(b) - 1(a)}47.42g

(b) The final mass of copper oxide (g) {1(c) - 3(a)}31.3g

(c) The gain in mass in the beaker which is equal to the added mass of oxygen (g) {3(b) - 1(a)}6.3g

(d) The number of moles of oxygen in the copper oxide, given a molecular weight of oxygen (O) equal to 16.00 g/mole. {convert 3(c) to moles}

4. Calculate the molar ratio of copper to oxygen from:

(moles copper powder) / (moles of oxygen in the copper oxide)

5. According to this molar ratio, what is the empirical formula of copper oxide?

6. Write a balanced equation for the reaction occurring between copper and oxygen.

Procedure 2

1. Data:
(a) Mass of CuSO4*5H2O (g): 2g
(b) Mass of empty test tube (g): 18.200g
(c) Mass of test tube with copper oxide (g): 18.837g

2. Calculate the following about the copper sulfate pentahydrate:

(a) Molecular weight of CuSO4*5H2O

(b) Moles of CuSO4*5H2O: {convert 1(a) to moles}

(c) Moles of Cu in the copper sulfate:

3. Calculate the following for the oxygen:

(a) The initial mass of copper in the moles of copper from the copper sulfate salt (MW of Cu = 63.55 g/mole) (g): {convert 2(c) to grams}

(b) The mass of copper oxide obtained (g): {1(c) - 1(b)}0.637g

(c) The gain in mass, equal to the added mass of oxygen (g): {3(b) - 3(a)}

(d) The number of moles of oxygen in the copper oxide, given a molecular weight of oxygen (O) equal to 16.00 g/mole: {convert 3(c) to moles}

4. Calculate the molar ratio of copper to oxygen from:

(moles copper) / (moles of oxygen in the copper oxide)

5. According to this molar ratio, what is the empirical formula of copper oxide?

6. Calculate the yield of copper oxide:

(a) Expected number of moles of copper oxide (using the empirical formula of copper oxide and the starting moles of Cu):

(b) Actual mass of copper oxide (g):

(c) Actual moles of copper oxide (dividing the mass by the MW of the copper oxide):

7. Calculate the percent difference between the expected and actual values for the moles of copper oxide, according to the following equation:

% difference = |expected moles - actual moles| / expected moles * 100%

8. Can you find an explanation for the difference?

Procedure 2
A 0.750 g sample of tin metal reacts with 0.201 g of oxygen gas to form tin oxide. Calculate the empirical formula of tin oxide.
(a) moles of tin
(b) moles of oxygen
(c) molar ratio between tin and oxygen (moles tin/moles oxygen)
(d) empirical formula of tin oxide

A 0.626 g sample of copper oxide was reduced to 0.500 g of copper metal by heating in a stream of hydrogen gas (oxygen was lost). Calculate the empirical formula of the copper oxide.
(a) moles of copper
(b) mass of oxygen lost
(c) moles of oxygen
(d) molar ratio between oxygen and copper
(e) empirical formula