Question

A certain anesthetic contains 64.9 percent C. 13.5 percent H. and 21.6 percent O by mass. At 116 degree C and 8- mmHg. 1.70 L of the gaseous compound weighs 3.71 g, What is the molecular formula of the compound?

Answer 1

A certain anesthetic contains 64.9 percent C, 13.5 percent H, and 21.6 percent O by mass. At 116°C and 785 mmHg, 1.70 L of the gaseous compound weighs 3.71 g. What is the molecular formula of the compound?

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 % = 64.9 g
H = 13.5 % = 13.5 g
O = 21.6 % = 21.6 g

moles C = 64.9 g / 12.01 g/mol = 5.40 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

C	H	O
5.40 mol	13.39 mol	1.35 mol
5.40 mol/1.35	13.39 mol/1.35	1.35 mol /1.35
4	10	1

empirical formula =C₄H₁₀O

PV = nRT
P = pressure = 785 mmHg
V = volume = 1.70 L
n = moles (unknown)
T = temp in Kelvin (116 deg C = (273.15 + 116) Kelvin)
- T = 389.15Kelvin
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 = (785 mmHg x 1.70 L) / (62.363mmHg L K^-1 mol^-1 x 389.15 K)
n = 0.054988moles of gas

moles = mass / molar mass
Therefore molar mass = mass / moles
molar mass = 3.71 g / 0.054988moles
= 67.46 g/mol ------ANSWER