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An unknown compound contans only C, H, and O. Combustion of 5.70 g of this compound produced 11.4g of CO2 and 4.66g of H2O. Wha...Asked by Casey
An unknown compound contains only C, H, and O. Combustion of 4.50 g of this compound produced 10.2 g of CO2 and 4.19 g of H2O. What is this empirical formula of the unknown compound?
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Answered by
DrBob222
First, determine grams C, H, and O.
g CO2 = 10.2 x (atomic mass C/molar mass CO2) = about 2.8 but you need to be more accurate than that. The other numbers will be estimates, also; therefore you need to go through the problem and get the right numbers.
g H = 4.19 x (2*atomic mass H/molar mass H2O) = about 0.47
g O = 4.50 - g H - g C = about 1.3.
Next, convert g C, H, O to mols.
mols C = about 2.7/12 = about 0.23
mols H = about .47/1 = about 0.47
mols O = about 1.3/16 = about 0.08
Now find the ratio of the elements to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself (which makes it 1.0000 every time), then divide the other two numbers by the same small number. Round to whole numbers and that will be the empirical formula.
Post your work if you get stuck and I can help you through it.
g CO2 = 10.2 x (atomic mass C/molar mass CO2) = about 2.8 but you need to be more accurate than that. The other numbers will be estimates, also; therefore you need to go through the problem and get the right numbers.
g H = 4.19 x (2*atomic mass H/molar mass H2O) = about 0.47
g O = 4.50 - g H - g C = about 1.3.
Next, convert g C, H, O to mols.
mols C = about 2.7/12 = about 0.23
mols H = about .47/1 = about 0.47
mols O = about 1.3/16 = about 0.08
Now find the ratio of the elements to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself (which makes it 1.0000 every time), then divide the other two numbers by the same small number. Round to whole numbers and that will be the empirical formula.
Post your work if you get stuck and I can help you through it.
Answered by
Cam
C2H20
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