Chemistry, 07.07.2019 20:30 VamPL

Atoxic gas. a, consists of 53.8% nitrogen and 46.2% carbon by mass. at 273 k and 1.01 x 10^5 pa, 1.048g of a occupies 462 cm^3. determine the empirical formula of a. calculate the moral mass of the compound and determine its molar mass

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Answer from: kcarstensen59070

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

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Answer from: Yungnaudie

By the mass percentage, 100 grams of the sample would contain

$53.8 \; \text{g}$ of nitrogen, and $46.2 \; \text{g}$ of carbon.

Which corresponds to

$53.8 / 14.01 \approx 3.84 \; \text{mol}$ of nitrogen atoms, and $46.2 / 12.01 \approx 3.85 \; \text{mol}$ of carbon atoms.

Hence the approximate ratio $n(\text{N}) : n(\text{C}) \approx 1 : 1$ and the empirical formula $\text{CN}$ .

Apply the idea gas formula to determine the number of moles of the gas molecules present in the $462 \; \text{cm}^{3} = 0.462 \; \text{L}$ sample:

$\begin{array}{lll}n &= & (\text{P}\cdot \text{V}) / (\text{R} \cdot \text{T})\\ & = & 1.01 \times 10^2 \times 0.462 /(8.314 \times 273) \\ &= &0.0206\; \text{mol}\end{array}$

Thus

$M = m / n = 1.048 / 0.0206 = 51.0 \; \text{g} \cdot \text{mol}^{-1}$

which is approximately twice of $26.02\; \text{g} \cdot \text{mol}^{-1}$ , formula mass for the empirical formula $\text{CN}$ . Hence the molecular formula $\text{C}_2\text{N}_2$

Answer from: Quest

The answer is b i think

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.