Looks OK except that you do not need to calculate the weight percent of oxygen. Starting from the point where you have 86.74 g of CaCO3 (the weight of the cube), do the following:
1) convert to moles of CaCO3 by dividing by 100.0869 g/mol
2) moles of O = moles of CaCO3 X 3 mol O / mol CaCO3
3) multiply moles of O by 6.022 X 1023 atoms of O / mol O
The percentage route gives the same answer, just adds an extra step or two! 
Steve
>> 8. Consider a sample of calcium carbonate in the form of a cube measuring 1.25 in on each edge. If the sample has a density of 2.71 g/cm3, how many oxygen atoms does it contain? 1 in = 2.54 cm. See attached - does it look ok? <<
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