By the mass percentage, [tex]100[/tex] grams of the sample would contain
Which corresponds to
Hence the approximate ratio [tex]n(\text{N}) : n(\text{C}) \approx 1 : 1[/tex] and the empirical formula [tex]\text{CN}[/tex].
Apply the idea gas formula to determine the number of moles of the gas molecules present in the [tex]462 \; \text{cm}^{3} = 0.462 \; \text{L} [/tex] sample:
[tex]\begin{array}{lll}n &= & (\text{P}\cdot \text{V}) / (\text{R} \cdot \text{T})\\ & = & 1.01 \times 10^2 \times 0.462 /(8.314 \times 273) \\ &= &0.0206\; \text{mol}\end{array}[/tex]
Thus
[tex]M = m / n = 1.048 / 0.0206 = 51.0 \; \text{g} \cdot \text{mol}^{-1}[/tex]
which is approximately twice of [tex]26.02\; \text{g} \cdot \text{mol}^{-1}[/tex], formula mass for the empirical formula [tex]\text{CN}[/tex]. Hence the molecular formula [tex]\text{C}_2\text{N}_2[/tex]
The molecular formula of the compound is C2N2.
From the ideal gas equation;
PV = nRT
P = 1.01 x 10^5 Pa or 1.01 atm
V = 462 cm^3 or 0.462 L
R = 0.082 atm Lmol-1K-1
T = 273 K
n = ?
n = PV/RT
n = 1.01 atm × 0.462 L/ 0.082 atm Lmol-1K-1 × 273 K
n =0.467 /22.386
n = 0.02086 moles
Now;
Molar mass of the compound is obtained from;
number of moles = mass/molar mass
molar mass = mass/number of moles
molar mass = 1.048g/0.02086 moles
molar mass = 50.24 g/mol
The empirical formula of the compound is;
C - 53.8%/12, N - 46.2%/14
C - 4.48, N - 3.3
Dividing through by the lowest ratio
C -4.48/3.3 N - 3.3/3.3
C - 1 N - 1
The empirical formula is CN
The molecular formula is obtained thus;
[12 + 14]n = 50.24
n = 50.24/26
n = 2
The molecular formula of the compound is C2N2
Learn more: https://brainly.com/question/14530520
