Determination of the molecular formula by the percentage of elements and the molar mass
Exercise 5
The analysis of a substance $S$ gives following results:
- presence of $C$, $H$, $O$ and $N$.
- 0.654 g produce 0.977 g $CO_2$ and 0.499 g $H_2O$ by combustion.
- The Dumas method applied to $0.710\;g$ of this substance consists in eliminating all of the combustion products. excluding nitrogen $N_2$ which is collected by water displacement: $153.4\; mL$ of gas are so collected on sea level on a water tank at $20\;^oC$ under atmospheric pressure of $1.00665\; 10^5 \frac{N}{m^2}$.
The water level in the cylinder is $68\; mm$ above the level of the tank. Knowing that the saturation vapor pressure of water at $ 20\;^oC$ is $0.025\;10^5\frac{N}{m^2}$ and the volumic mass of water at $20\;^oC $ is $0.998\frac{g}{mL}$. give the percentage composition of $C$, $H$ , $O$ and $N$.
-$1.5\; g$ of this substance. dissolved in $100\; g$ water lower the freezing point by $0.47\;^oC$. The cryoscopic constant of water is $1.850\;^oC$.
Find the molar mass of the unknown substance. and write all possible developed formulas .
1) %N by Dumas:
According to the basic principle of hydrostatic we have:
$p_A =$ $p_B$ $= $ $1.00665\; 10^5 \frac{N}{m^2}$
$p_B - p_C$ $=$ $\rho_{H_2O}\cdot g \cdot h $ =
$998 \frac{kg}{m^3}$ $\cdot \; 9.81 \frac{N}{kg} $ $\; \cdot 0.068 m$ =
$0.00665\; 10^5 \frac{N}{m^2}$.
so:
$p_C $ $=$ $1.00000\cdot 10^5 \frac{N}{m^2} $
Pressure of nitrogen alone = $p_N$
$p_C-p_{water\;vapour}$ =
$0.975\cdot 10^5 \frac{N}{m^2}$
According to the perfect gas law we have:
$n_{N_2}$=
$ \frac{0.975\cdot 10^5\cdot 153.4\cdot 10^{-6}}{8.314\cdot 293.15}$ =
$0.00614$
$m_N$ $=$ $0.00614 \cdot 28$ $=$ $0.172\; g$
$\%_N$ $=$ $\frac{0.172\cdot 100}{0.710}$
=$24.2\; \%$
2) %others by the combustion products:
$m_C$ $=$ $n_{CO_2}\cdot 12$ =
$\frac{0.977}{44}\cdot 12$ $=$ $0.266 \; g$
$\%_C$ $=$ $\frac{0.266\cdot 100}{0.654}\; g$ =
$40.7\; \%$
$m_H$ $=$ $n_{H_2O}\cdot 2$ =
$\frac{0.499}{18}\cdot 2$ $=$ $0.0554 \; g$
$\%_H$ $=$ $\frac{0.0554\cdot 100}{0.654}\; g$ =
$8.5\; \%$
$\%_O$ $=$ $100-40.7-8.5-24.2$ =
$26.6\; \%$
3) Molar mass by cryometry:
$\mu_{S}$ $=$ $\frac{\Delta T}{K_f}$ =
$ \frac{0.47}{1.850}$ =
$0.254$
$n_S$ $=$ $\mu_{S}\cdot 0.100$ $=$ $0.0254$
$M_S=\frac{m_S}{n_S}$ =
$\frac{1.5}{0.0254}$ $=$ $59.1\frac{g}{mol}$
5) Molecular formula:
One mole S contains
$\frac{59.1\cdot 0.242}{14}$ $=$ $1\; mole\; N$
$\frac{59.1\cdot 0.407}{12}$ $=$ $2\; mole\; C$
$\frac{59.1\cdot 0.085}{1}$ $=$ $5 \;mole\; H$
$\frac{59.1\cdot 0.266}{16}$ $=$ $1\; mole\; O $
Molecular formula:
$C_2H_5NO$
The water level in the cylinder is $68\; mm$ above the level of the tank. Knowing that the saturation vapor pressure of water at $ 20\;^oC$ is $0.025\;10^5\frac{N}{m^2}$ and the volumic mass of water at $20\;^oC $ is $0.998\frac{g}{mL}$. give the percentage composition of $C$, $H$ , $O$ and $N$.
-$1.5\; g$ of this substance. dissolved in $100\; g$ water lower the freezing point by $0.47\;^oC$. The cryoscopic constant of water is $1.850\;^oC$.
Find the molar mass of the unknown substance. and write all possible developed formulas .
