Question
Upon combustion, 4.30 grams sample of a compound containing only carbon, hydrogen, and oxygen produced 8.59 g of CO2 and 3.52g of H2o. what is empirical formula of the compound ?
Answers
Convert 8.59g CO2 to grams C.
8.59 x (atomic mass C/molar mass CO2) =?
Convert 3.52 g H2O to g H (atoms).
3.52 g H2O x (2*atomic mass H/molar mass H2O) =?
Then g O = 4.30-g C - g H = ?
Now convert grams each to mols.
mols C = grams C/atomic mass C
mols H = grams H/atomic mass H
mols O = grams O/atomic mass O
Now find the ratio to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself, then follow suit with the other numbers. Round to whole numbers.
Post your work if you get stuck.
8.59 x (atomic mass C/molar mass CO2) =?
Convert 3.52 g H2O to g H (atoms).
3.52 g H2O x (2*atomic mass H/molar mass H2O) =?
Then g O = 4.30-g C - g H = ?
Now convert grams each to mols.
mols C = grams C/atomic mass C
mols H = grams H/atomic mass H
mols O = grams O/atomic mass O
Now find the ratio to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself, then follow suit with the other numbers. Round to whole numbers.
Post your work if you get stuck.
I got a weird answer. C1.9H4O. Do u think this is right?
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