OK, you have 12.0 g sample.
On combustion, you obtained 1.20 g H2O and 5.87 g CO2, plus some oxygen (which may or may not have found its way into the H2O and CO2; however, that doesn't matter. We will find O another way).
1. Convert 1.20 g H2O to g H and convert 5.87 g CO2 to grams C. Now, grams oxygen = 12.0 - g H - g C = g oxygen. I think there is an error in the problem somewhere. These numbers don't look right to me.
2. Convert g C to moles. moles = g/molar mass.
Do the same for g H.
Do the same for g O.
3. Find the empirical formula by finding the ratio of these three elements to each other in small whole numbers. The easy way to do that is to divide the smallest number by itself, thereby assuring you a 1.000 for that element. Then divide the other numbers by the same small number. Round to whole numbers.
Calculate the empirical formula and molecular formula of a compound that contains only carbon, oxygen and hydrogen and has a molecular mass of approximately 90. g/mol. Upon combustion of 12.0g of the compund with an excess of oxygen, the yield was 1.2 g of water and 5.87 of carbon dioxide.
I tried to set the equation like this
12.0 --> 1.2 H2O + 5.87 CO2
but it said "with an excess of oxygen" and I was confuse on that...
I know how to find the molecular mass but in here it says 90. g/mol not just regular 90 grams, I don't know what to do with it. Do i have to change it to just grams or something? or just leave it.. also there was another problem similar to this. the molecular mass is "approximately 150 g/ml."
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