Formula of a Chemical Compound PDF

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4. Data Tables Table 1: Recorded Mass Mass of crucible (g)

11.4708 g

Mass of crucible and copper wire (g)

11.8740 g

Mass of crucible and copper wire after first heating (g)

11.9752 g

Mass of crucible and copper wire after second heating (g)

N/A

5. Results/Discussion: Sample calculations: Mass of copper = mass of crucible and copper wire - mass of crucible = 1.8740 g - 11.4708 g = 0.4032 g Moles of copper = mass of copper / atomic mass copper = 0.4032 g / 63.546 g/mol = 6.345*10-3  mol Mass of CuxSy = mass of crucible + copper wire after 1st heating - mass of crucible = 11.9752 g - 11.4708 g = 0.5044 g Mass of sulfur = 0.5044 g - 0.4032 g = 0.1012 g Moles of sulfur = 0.1012 g / 32.06 g/mol = 3.157*10-3  mol CuxSy = Cu0.006345S0.003157 Divide lowest ratio and round to find empirical formula = 0.006345 / 0.003157 ≅ 2 Empirical Formula = Cu2S

% Cu in product = (63.546*2) / 159.16 = 0.79852*100 = 79.852%

6. Discussion questions 1a. How would the empirical formula of copper sulfide be affected, if not all of the copper wire were converted to copper sulfide? The empirical formula would be Cu1-xS because there would be an excess in sulfur.

1b. What if all the excess sulfur was not burned off— how would the empirical formula of copper sulfide be affected? There would not be a change to the empirical formula of the copper sulfide.

2. Explain why the formula you determined was an empirical and not a molecular formula.

It is empirical and not molecular because what was calculated is the ratio of copper to sulfur. I did so by dividing it by the smallest ratio instead of providing the actual relationship between the atoms in the compound.

3. Suppose that the mass you determined for your final piece of data, the mass of crucible plus copper sulfide after second heating, was 0.002 g too low. What would be the percentage of copper calculated? Would this error change the empirical formula you would report? Explain. The percentage of copper calculated would still be 79.852%. This error would not change the empirical formula reported.

4. Calculate the uncertainty in the subscript in the formula you found in this experiment. Cu % uncertainty = (0.00005 g + 0.00005 g/ 0.4032 g Cu)+(0.0001 g/ 0.1012 g S)*100%

= 0.12% Cu S % uncertainty = (0.00005 g + 0.00005 g/ (0.1012 g S*2))*100% = 0.099% S

7. Conclusion Using the stoichiometric information, such as the masses of the different elements in the compound, the empirical formula of the compound could be determined to be Cu2S....