Foundation and Fundamentals

Gay-Lussac's law of gaseous volumes states that when gases react, they combine in volumes that bear a simple whole-number ratio to one another and to the volumes of the gaseous products, all measured at the same temperature and pressure.

For example, 2 volumes of hydrogen combine with 1 volume of oxygen to give 2 volumes of steam.

Relation. Vapour density is the ratio of the mass of a certain volume of gas to the mass of the same volume of hydrogen at the same temperature and pressure:

$\text{V.D.} = \frac{\text{mass of } V \text{ of gas}}{\text{mass of } V \text{ of H}_2}$

By Avogadro's hypothesis, equal volumes contain equal numbers of molecules ($n$), so

$\text{V.D.} = \frac{n\times\text{mass of 1 molecule of gas}}{n\times\text{mass of 1 molecule of H}_2} = \frac{M}{2}$

since H$_2$ has molecular mass 2. Hence

$M = 2\times \text{V.D.}$

Numerical. Molecular mass of the oxide $= 2\times15 = 30$.

Let the oxide be N$_x$O$_y$. 1 volume of oxide contains $\tfrac12$ volume of nitrogen (as N$_2$), i.e. 1 molecule of oxide contains $\tfrac12$ molecule of N$_2$ = 1 atom of N. So $x = 1$.

Mass of oxygen $= 30 - 14 = 16$, so $y = 1$.

Molecular formula: NO (nitric oxide).

According to Dalton, matter is made of indivisible atoms, and atoms can neither be created nor destroyed. In a chemical reaction atoms are only rearranged: bonds between reactant atoms break and new combinations form. Since the number and type of atoms (each with fixed mass) is the same before and after the reaction, the total mass of products equals the total mass of reactants. This explains the law of conservation of mass.

Moles of CO in 0.28 g $= \frac{0.28}{28} = 0.01$ mol

Moles removed $= \frac{2\times10^{21}}{6.022\times10^{23}} = 3.32\times10^{-3}$ mol

Moles left $= 0.01 - 0.00332 = 6.68\times10^{-3}$ mol

Moles of CO left $\approx 6.7\times10^{-3}$ mol

Relation between vapour density and molecular mass:

Vapour density is the ratio of the mass of a certain volume of gas to the mass of the same volume of hydrogen under the same conditions:

$VD = \frac{\text{mass of } V \text{ of gas}}{\text{mass of } V \text{ of } H_2}$

By Avogadro's hypothesis, equal volumes contain equal numbers of molecules, say $n$. Then

$VD = \frac{n \times \text{mass of 1 molecule of gas}}{n \times \text{mass of 1 molecule of } H_2} = \frac{M}{2}$

since the molecular mass of $\mathrm{H_2}$ is 2. Hence

$M = 2 \times VD$

Numerical:

Molecular mass of the oxide $= 2 \times 54 = 108$.

The oxide contains its own volume of nitrogen, i.e. 1 volume of the oxide gives 1 volume of $\mathrm{N_2}$. By Avogadro's hypothesis, 1 molecule of the oxide gives 1 molecule of $\mathrm{N_2}$, so each molecule has 2 atoms of N.

Let the formula be $\mathrm{N_2O_x}$:

$28 + 16x = 108 \Rightarrow x = 5$

Molecular formula: $\mathrm{N_2O_5}$

These data illustrate the law of multiple proportions. In $\mathrm{N_2O_3}$ and $\mathrm{N_2O_5}$, the masses of oxygen combining with a fixed mass (28 g) of nitrogen are 48 g and 80 g, in the simple ratio $3:5$.

Statement (Dalton): when two elements combine to form more than one compound, the different masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio.

$2\times10^{21}$ atoms weigh 0.4 g, so the mass of one mole ($6.022\times10^{23}$ atoms) is

$\frac{0.4\times6.022\times10^{23}}{2\times10^{21}} = 120.4\ \text{g mol}^{-1}$

Mass of 0.5 mole $= 0.5\times120.4 = 60.2\ \text{g}$.

(b) 4

Leading zeros are not significant. In $0.005230$, the significant digits are $5, 2, 3,$ and the final $0$, so there are 4 significant figures.

Given:
Mass of $\text{CO}_2 = 22\ \text{g}$
Molar mass of $\text{CO}_2 = 12 + 2(16) = 44\ \text{g mol}^{-1}$
$N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$

1. Number of moles of $\text{CO}_2$:

2. Number of molecules of $\text{CO}_2$:

So, $22\ \text{g}$ of $\text{CO}_2$ contains $3.011 \times 10^{23}$ molecules.

3. Number of atoms of each element:

One molecule of $\text{CO}_2$ contains:

• $1$ atom of carbon
• $2$ atoms of oxygen

Carbon atoms:

Oxygen atoms:

Also, in terms of moles of atoms:

• Moles of carbon atoms $= 0.5\ \text{mol}$
• Moles of oxygen atoms $= 2 \times 0.5 = 1.0\ \text{mol}$

Final answer:

Given:

• Percentage composition: C = 40%, H = 6.67%, O = 53.33%
• Molar mass = 180 g mol⁻¹
• Atomic masses: C = 12, H = 1, O = 16

Step 1: Consider 100 g of the compound.

Step 2: Divide each value by the smallest number of moles, i.e. 3.33.

So, the simplest whole-number ratio is:

Empirical formula = $\text{CH}_2\text{O}$

Step 3: Calculate empirical formula mass.

Step 4: Find $n$.

Step 5: Molecular formula.

Answer:

• Empirical formula = $\text{CH}_2\text{O}$
• Molecular formula = $\text{C}_6\text{H}_{12}\text{O}_6$

Given reaction:

$\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$

Molar masses:

• $\text{N}_2 = 2 \times 14 = 28\ \text{g mol}^{-1}$
• $\text{H}_2 = 2 \times 1 = 2\ \text{g mol}^{-1}$
• $\text{NH}_3 = 14 + 3(1) = 17\ \text{g mol}^{-1}$

Moles of reactants:

$n_{\text{N}_2}=\frac{28}{28}=1\ \text{mol}$

$n_{\text{H}_2}=\frac{6}{2}=3\ \text{mol}$

From the balanced equation, 1 mol $\text{N}_2$ requires 3 mol $\text{H}_2$.

Here, available $\text{H}_2$ is exactly 3 mol, so both reactants are present in the exact stoichiometric ratio.

(i) Limiting reagent

There is no limiting reagent. Both $\text{N}_2$ and $\text{H}_2$ are completely consumed.

(ii) Mass of ammonia produced

From the equation:

$1\ \text{mol N}_2 \rightarrow 2\ \text{mol NH}_3$

So,

$n_{\text{NH}_3}=2 \times 1 = 2\ \text{mol}$

Mass of $\text{NH}_3$:

$m = 2 \times 17 = 34\ \text{g}$

$\boxed{34\ \text{g}}$

(iii) Volume of ammonia at STP

At STP, 1 mol gas occupies $22.4\ \text{L}$.

$V = 2 \times 22.4 = 44.8\ \text{L}$

$\boxed{44.8\ \text{L}}$

(iv) Mass of excess reagent left unreacted

Since the reactants are in exact stoichiometric amounts, neither reactant is in excess.

$\boxed{0\ \text{g}}$

No excess reagent remains unreacted.