Stoichiometry, Molar Quantities, and Chemical Reactions of Carbon and Limestone

Concepts of Molar Mass and Atomic Weights

The molar mass, represented by the symbol $M$ , is defined as the sum of the atomic masses of all atoms present in a molecule. In the context of chemical calculations, the mass of an atom is expressed in terms of the amount of substance ( $n$ ). For example, the atomic mass of hydrogen corresponds to $1$ gram per mole ( $1\,g/mol$ ) per hydrogen atom. Carbon has an atomic mass that corresponds to $12\,g/mol$ per carbon atom, and oxygen corresponds to $16\,g/mol$ per oxygen atom. These values are used to determine the total mass of a substance portion based on its chemical formula.

The fundamental formula relating mass ( $m$ ), amount of substance ( $n$ ), and molar mass ( $M$ ) is given by: $M = \frac{m}{n}$

If a substance has a molar mass of $100\,g/mol$ and an amount of substance of $2\,mol$ , the total mass can be calculated as: $m = M \times n = 100\,g/mol \times 2\,mol = 200\,g$

Chemical Reactions and the Functional Principles of Airbags

The functioning of an airbag relies on specific chemical principles and rapid gas-forming reactions. The process can be broken down into four essential points:

A solid mixture is ignited, which produces a large volume of gaseous substance that expands rapidly to inflate the airbag.
The reaction products generated must be non-toxic, and while the reaction is explosive in nature to ensure speed, it must be controlled so that it does not destroy the airbag material or harm the occupant.
A common reaction used is the decomposition of sodium azide ( $NaN_3$ ): $NaN_3 \rightarrow Na + 3N_2$
Another example of decomposition mentioned is the breakdown of limestone (calcium carbonate): $CaCO_3 \rightarrow CaO + CO_2$

Systematic Procedure for Stoichiometric Calculations

To solve problems involving chemical quantities and transformations, a four-step (or five-step) systematic approach is employed to ensure accuracy. These steps are as follows:

Identify the given and sought substances and write the balanced chemical reaction equation.
Determine the molar mass ( $M$ ) of the given and sought substances.
Calculate the amount of substance ( $n$ ) for the given substance using its known mass ( $m$ ) or volume.
Determine the amount of substance ( $n$ ) for the sought substance based on the stoichiometric ratios provided by the reaction equation coefficients.
Calculate the desired size or quantity ( $m, V$ ) of the sought substance.

Molar Volume of Gaseous Substances

The molar volume ( $V_m$ ) indicates the volume occupied by exactly one mole of a gaseous substance. This value is relatively constant for all gases under specific conditions. At a temperature of $18^{\circ}C$ and standard air pressure, the molar volume is approximately: $V_m = 24\,dm^3/mol$

Applying this to the decomposition of limestone ( $CaCO_3 \rightarrow CaO + CO_2$ ), if we are given an amount of substance $n(CaCO_3) = 2\,mol$ , we can find the volume of the resulting gas ( $CO_2$ ). Given the $1:1$ ratio, the amount of $CO_2$ is also $2\,mol$ . The volume is calculated as: $V = V_m \times n = 24\,dm^3/mol \times 2\,mol = 48\,dm^3$

To calculate the mass of the byproduct calcium oxide ( $CaO$ ): Given: $n = 2\,mol$ and $M(CaO) = 40 + 16 = 56\,g/mol$ $m(CaO) = n \times M = 2\,mol \times 56\,g/mol = 112\,g$

In the specific example of an airbag calculation where the mass of reactant starts at $65\,g$ : Given: $m = 65\,g$ , $M(NaN_3) = 65\,g/mol$ $n = \frac{m}{M} = \frac{65}{65} = 1\,mol$ Following another specific data point provided: $V = V_m \times n = 24 \times 0.231 = 5.544\,L$ .

The Amount of Substance and the Avogadro Constant

When calculating quantities and conversions, the physical number of particles (atoms, molecules, ions, or electrons) in a macroscopic sample is exceedingly large, typically on the order of trillions. To handle these quantities, the concept of the "amount of substance" ( $n$ ) is used. It indicates how many particles are in a defined portion of matter as a multiple of the unit "mol."

One mole is defined as the number of particles found in exactly $12\,g$ of Carbon-12 isotopes ( $^{12}C$ ), which is approximately $6 \times 10^{23}$ particles. The number of particles ( $N$ ) is a dimensionless count (it has no unit). The relationship between the number of particles ( $N$ ) and the amount of substance ( $n$ ) is mediated by the Avogadro constant ( $N_A$ ), defined as: $N_A = 6.022 \times 10^{23}\,mol^{-1}$

The formulas used are: $n = \frac{N}{N_A}$ $N = n \times N_A$

Example calculations:

If given $N = 1.2 \times 10^{24}$ particles, find $n$ : $n = \frac{1.2 \times 10^{24}}{6.022 \times 10^{23}\,mol^{-1}} \approx 2\,mol$
If given $n = 3\,mol$ , find $N$ : $N = 3\,mol \times 6 \times 10^{23}\,mol^{-1} = 1.8 \times 10^{24}$

Stoichiometry and Relationship Ratios in Chemical Equations

The coefficients used in chemical equations, often represented by the Greek letter $\nu$ (nu), indicate the stoichiometric ratio. These numbers show how many particles (or moles) of a substance are necessary for a complete chemical reaction to take place. They establish the ratio between the starting materials (reactants) and the reaction products.

Examples of reaction equations showing these ratios: Burning charcoal (carbon): $C + O_2 \rightarrow CO_2$

Sulfur burning with oxygen to form sulfur trioxide: $2S + 3O_2 \rightarrow 2SO_3$

In the sulfur trioxide reaction, the stoichiometric coefficients indicate that $2$ units of sulfur react with $3$ units of oxygen gas to produce $2$ units of sulfur trioxide. This translates directly to the molar ratios used in larger scale calculations.