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Posted: Sun Sep 29, 2019 10:55 am
The homework question claims that a chemist measured out 8.61 g of CuCl2 * 4H2O.
Part (d) asks for what fraction of the total mass of the sample is due to oxygen?
I just want to double check that we only need to divide the molar mass of the sample 206.51 g by 64 g of oxygen (since there are 4 oxygen atoms).
Is this correct?
Re: Question E29D.
Posted: Sun Sep 29, 2019 11:15 am
Not quite. The question is asking for what fraction of the given total mass is due to oxygen. In such a case, you would first need to use the given mass to solve for the number of moles of the CuCl2*4H2O compound in 8.61 grams of that compound. Then, you can use the mole ratio between the compound and Oxygen to find the moles of oxygen in the sample. In this case, there is a mole ratio of 1:4 between the compound and oxygen, so every mole of the compound has 4 moles of oxygen. Once you find the moles of oxygen in 8.61 grams of the compound, you can use the molar mass of oxygen to find how many grams of oxygen are in the determined moles of oxygen. Finally, you would use the mass of oxygen that you find and divide it by the given mass to determine what fraction of the given mass is due to the oxygen in the sample. Hope this helps!
Edit: After double checking, your method provides the same answer, which makes sense since it finds the fraction of the molar mass of 4 moles of oxygen in the molar mass of the entire compound, it would give the same percentage. So yes that would be correct. Thanks for teaching me another method!
Re: Question E29D.
Posted: Sun Sep 29, 2019 11:26 am
I first calculated the total molar mass of the compound CuCL24H2O, which came out to be 206.51 grams/mol. Dividing the given mass by this molar mass gave me 0.042 moles of the compound. Since the molar ratio of the compound to the oxygen is 1:4, you know you have 0.167 moles of Oxygen in the compound. Thus, you can multiply this mole count with the molar mass 16 to get 2.67 g Oxygen. Divide this by the total mass of the compound and you should be able to obtain the percentage by mass of oxygen in the compound.