Acertain anesthetic contains 64.9% c, 13.5% h, and 21.6% o by mass. at 120 deg celsius & 750 mmhg, 1.00 l of the gaseous compound has a mass of 2.30g. what is the molecular formula of the compound? ! !

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 

C4H10O

Therefore, the molecular formula of the compound is C4H10O.

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Because it was a dramatically big guy

The answer is c. 320 g