You need to use the % information to determine the empirical formula of the compound first.
The empirical formula is the simplest ratio of atoms in the molecule.
Then use the rest of the data to determine moles of gas, and use this to determine molar mass of gas...
Empirical formula calculations:
Assume you have 100 g, calculate the moles of each atom in the 100 g
moles = mass / molar mass molar mass C = 12.01 g/mol molar mass H = 1.008 g/mol molar mass O = 16.00 g/mol
C = 64.9 % = 64.6 g H = 13.5 % = 13.5 g O = 21.6 % = 21.6 g
moles C = 64.6 g / 12.01 g/mol = 5.38 mol moles H = 13.5 g / 1.008 g/mol = 13.39 mol moles O = 21.6 g / 16.00 g/mol = 1.35 mol
So ratio of C : H : O is 5.38 mol : 13.39 mol : 1.35 mol
Divide each number in the ratio by the lowest number to get the simplest whole number ratio
(5.38 / 1.35) : (13.39 / 1.35) : (1.35 / 1.35)
4 : 10 : 1
empirical formula is C4H10O
Finding moles and molar mass calcs
Now, you know that at 120 deg C and 750 mmHg that 1.00L compound weighs 2.30 g.
We can use this information to determine the molar mass of the gas after first working out how many moles the are in the 1.00 L
PV = nRT P = pressure = 750 mmHg V = volume = 1.00 L n = moles (unknown) T = temp in Kelvin (120 deg C = (273.15 + 120) Kelvin) - T = 393.15 Kelvin R = gas constant, which is 62.363 mmHg L K^-1 mol^-1 (when your P is in mmHg and volume is in L)
n = PV / RT n = (750 mmHg x 1.00 L) / (62.363mmHg L K^-1 mol^-1 x 393.15 K) n = 0.03059 moles of gas
We know moles = 0.03509 and mass = 2.30 g So we can work out molar mass of the gas
moles = mass / molar mass Therefore molar mass = mass / moles molar mass = 2.30 g / 0.03059 mol = 75.19 g/mol
Determine molecular formula
So empirical formula is C4H10O molar mass = 75.19 g/mol
To find the molecular formula you divide the molar mass by the formula weight of the empirical formula... This tells you how many times the empirical formula fits into the molecular formula. Tou then multiply every atom in the empirical formula by this number
formula weight C4H10O = 74.12 g/mol
Divide molar mass by formula weight empirical 75.15 g/mol / 74.12 g/mol = 1 (It doesn't matter that the number don't quite match, they rarely do in this type of calc (although I could have made a slight error somewhere) but the numbers are very close, so we can say 1.)
The empirical formula only fits into the molar mass once,
molecular formula thus = empirical formula <span> C4H10O
Therefore, the </span>molecular formula of the compound is <span>C4H10O.
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