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Suppose a student adds 0.105 g of an unknown copper carbonate compound to a crucible and reacts it under a bunsen burner until all the copper has turned into cuo. she measures the mass of cuo and it is 0.068 g. a. what is the mass of copper, cu, in the sample? show your calculations and circle the final answer. b. what is the % mass of cu in the unknown copper carbonate compound? show your calculations and circle the final answer. c. using your answer to 3b, and the % mass of cu you did for question

User HackToHell
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Final answer:

The mass of copper in the sample is calculated by determining the moles of CuO formed and using molar masses to find the mass of Cu. Subsequently, the percent mass of Cu in the original compound is obtained using the ratio of the mass of Cu to the mass of the compound multiplied by 100%, resulting in a value of 51.43%.

Step-by-step explanation:

To calculate the mass of copper, Cu, in the sample, you subtract the mass of the copper oxide, CuO, from the initial mass of the copper carbonate compound. Since CuO consists of Cu and one atom of oxygen (O), and because the mass of the oxygen atom in the CuO can be calculated using the molar mass of oxygen (16.00 g/mol), you can establish the mass of Cu in CuO. Once you know the mass of Cu in CuO, you can find the percent mass of Cu in the original copper carbonate compound using the formula:

Percent mass of Cu = (Mass of Cu / Mass of the compound) x 100%

The calculations will look like this:

  1. Calculate the molar mass of CuO, which is 63.55 g/mol for Cu + 16.00 g/mol for O = 79.55 g/mol.
  2. Determine the moles of CuO formed: 0.068 g CuO × (1 mol CuO / 79.55 g CuO) = 8.55×10-4 mol CuO.
  3. Calculate the mass of Cu from the moles of CuO: 8.55×10-4 mol CuO × (63.55 g Cu / 1 mol Cu) = 0.054 g Cu.
  4. Finally, calculate the percent mass of Cu in the compound:Percent mass of Cu = (0.054 g Cu / 0.105 g compound) x 100% = 51.43%.

Therefore, the mass of Cu in the sample is 0.054 g and the percent mass of Cu in the unknown copper carbonate compound is 51.43%.

User Fpointbin
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