You need to use the % information to determine the empirical formula of the compound first.Â
The empirical formula is the simplest ratio of atoms in the molecule.Â
Then use the rest of the data to determine moles of gas, and use this to determine molar mass of gas...Â
Empirical formula calculations:
Assume you have 100 g, calculate the moles of each atom in the 100 gÂ
moles = mass / molar massÂ
molar mass C = 12.01 g/molÂ
molar mass H = 1.008 g/molÂ
molar mass O = 16.00 g/molÂ
C = 64.9 % = 64.6 gÂ
H = 13.5 % = 13.5 gÂ
O = 21.6 % = 21.6 gÂ
moles C = 64.6 g / 12.01 g/mol = 5.38 molÂ
moles H = 13.5 g / 1.008 g/mol = 13.39 molÂ
moles O = 21.6 g / 16.00 g/mol = 1.35 molÂ
So ratio of C : H : OÂ
is 5.38 mol : 13.39 mol : 1.35 molÂ
Divide each number in the ratio by the lowest number to get the simplest whole number ratioÂ
(5.38 / 1.35) : (13.39 / 1.35) : (1.35 / 1.35)Â
4 : 10 : 1Â
empirical formula isÂ
C4H10OÂ
Finding moles and molar mass calcsÂ
Now, you know that at 120 deg C and 750 mmHg that 1.00L compound weighs 2.30 g.Â
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 LÂ
PV = nRTÂ
P = pressure = 750 mmHgÂ
V = volume = 1.00 LÂ
n = moles (unknown)Â
T = temp in Kelvin (120 deg C = (273.15 + 120) Kelvin)Â
- T = 393.15 KelvinÂ
R = gas constant, which is 62.363 mmHg L K^-1 mol^-1 (when your P is in mmHg and volume is in L)Â
n = PV / RTÂ
n = (750 mmHg x 1.00 L) / (62.363mmHg L K^-1 mol^-1 x 393.15 K)Â
n = 0.03059 moles of gasÂ
We know moles = 0.03509 and mass = 2.30 gÂ
So we can work out molar mass of the gasÂ
moles = mass / molar massÂ
Therefore molar mass = mass / molesÂ
molar mass = 2.30 g / 0.03059 molÂ
= 75.19 g/molÂ
Determine molecular formulaÂ
So empirical formula is C4H10OÂ
molar mass = 75.19 g/molÂ
To find the molecular formula you divide the molar mass by the formula weight of the empirical formula...Â
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 numberÂ
formula weight C4H10O = 74.12 g/molÂ
Divide molar mass by formula weight empiricalÂ
75.15 g/mol / 74.12 g/molÂ
= 1Â
(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.)Â
The empirical formula only fits into the molar mass once,Â
molecular formula thus = empirical formulaÂ
C4H10O
Therefore, the molecular formula of the compound is C4H10O.
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