Asked by martina
a compound contains 5.2% by mass Nitrogen. It also contains carbon, hydrogen and oxygen. Combustion of 0.085g of the compound produced 0.224g of carbon dioxide and 0.0372g of water. Calculate the empirical formula of the compound.
Answers
Answered by
DrBob222
g N = 0.0850*0.052 = ?
g C = 0.224 g CO2 x (atomic mass C /molar mass CO2)= ?
g H = 0.0372 g H x (2*atomic mass H/molar mass H2O) = ?
g O = 0.085 - gC - gH - gN = ?
Now convert g to mols.
mols C = g C/atomic mass C = ?
mols H = g H/atomic mass H = ?
mols O = g O/atomic mass O = ?
mols N = g N/atomic mass N = ?
Now find the ration of these elements to each other with the smallest number no less than 1.00. The easy way to do that is to divide the smallest number by itself (which makes it 1.00), then divide the other numbers by the same small number. I've looked at these numbers and they don't come out to be whole numbers so you do this. Multiply each of the numbers, in succession, by 2, 3, 4, 5 etc until you get numbers that are close enough to whole numbers to round to a whole number for each. That will give you the empirical formula. Post your work if you run into trouble and I can help you thorough it.
g C = 0.224 g CO2 x (atomic mass C /molar mass CO2)= ?
g H = 0.0372 g H x (2*atomic mass H/molar mass H2O) = ?
g O = 0.085 - gC - gH - gN = ?
Now convert g to mols.
mols C = g C/atomic mass C = ?
mols H = g H/atomic mass H = ?
mols O = g O/atomic mass O = ?
mols N = g N/atomic mass N = ?
Now find the ration of these elements to each other with the smallest number no less than 1.00. The easy way to do that is to divide the smallest number by itself (which makes it 1.00), then divide the other numbers by the same small number. I've looked at these numbers and they don't come out to be whole numbers so you do this. Multiply each of the numbers, in succession, by 2, 3, 4, 5 etc until you get numbers that are close enough to whole numbers to round to a whole number for each. That will give you the empirical formula. Post your work if you run into trouble and I can help you thorough it.
Answered by
Shruti
C^20H^16O
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