Answer:
C₁₀H₁₀
Explanation:
We can use the Ideal Gas Law to solve this problem:
pV = nRT
Since n = m/M, the equation becomes
pV = (m/M)RT
Data:
p = 0.740 atm
V = 92.7 mL
m = 0.212 g
R = 0.082 16 L·atm·K⁻¹mol⁻¹
T =200 °C
Calculations:
(a) Convert the volume to litres
V = 92.7 mL = 0.0927 L
(b) Convert the temperature to kelvins
T = (200 + 273.15) = 473.15 K
(c) Calculate the molecular mass
[tex]\begin{array}{rcl}\\pV&=&\dfrac{m}{M}RT\\\\\text{0.740 atm} \times \text{0.0972 L} & = &\dfrac{\text{0.242 g}}{M} \times 0.08206 \text{L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{473.15 K}\\\\0.07193 & = & \dfrac{9.396 \text{ g$\cdot$mol}^{-1}}{M}\\\\M& = & \dfrac{9.405\text{ g$\cdot$mol}^{-1}}{0.07193}\\\\& = & \text{131 g$\cdot$mol}^{-1}\\\end{array}[/tex]
The molar mass is 131 g/mol, so the molecular mass is 131 u.
(d) Calculate the empirical formula
EF Mass = (12.01 + 1.008) u = 13.02 u
The molecular formula is an integral multiple of the empirical formula.
MF = (EF)ₙ
[tex]n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{131 u}}{\text{13.02 u}} = 10.0 \approx 10[/tex]
MF = (CH)₁₀ = C₁₀H₁₀
