What will be the weight of CO having the same number of oxygen atoms as present in \(22 \mathrm{~g}\) of \(\mathrm{CO}_{2} ?\) (a) \(28 \mathrm{~g}\) (b) \(22 \mathrm{~g}\) (c) \(44 \mathrm{~g} \quad\) (d) \(72 \mathrm{~g}\)

Understanding the molar mass of a substance is essential in the field of stoichiometry chemistry problems. Molar mass refers to the mass of one mole of a substance and it is expressed in grams per mole (g/mol). It is a physical property that is calculated by summing the atomic masses of all atoms present in the chemical formula of a substance.

For instance, in the given exercise, the molar mass of carbon dioxide (CO2) is calculated by adding the atomic mass of carbon (12 g/mol) to twice the atomic mass of oxygen, since there are two oxygen atoms in each molecule of CO2. The atomic mass of oxygen is 16 g/mol, thus giving us a molar mass of CO2 as: \[12 \text{ g/mol (C)} + 2 \times 16 \text{ g/mol (O)} = 44 \text{ g/mol}\].

Similarly, calculating the molar mass of carbon monoxide (CO) involves adding the atomic mass of carbon to the atomic mass of oxygen, resulting in:\[12 \text{ g/mol (C)} + 16 \text{ g/mol (O)} = 28 \text{ g/mol}\].

Knowing how to find molar mass accurately is the first step in solving many stoichiometry problems, as it allows you to connect the mass of a substance with the number of its molecules or atoms.

After determining the molar mass, stoichiometry often requires converting moles to grams or vice versa. This conversion is pivotal since it links the weight of a substance with the number of particles it contains, which is a fundamental concept in chemistry.

To convert moles to grams, we multiply the number of moles by the molar mass of the substance: \[\text{Mass in grams} = \text{Moles} \times \text{Molar Mass}\].

In the example exercise, the weight of carbon dioxide is given, and through the calculated moler mass, we can find the number of moles. And conversely, knowing the number of oxygen atoms (in moles) allows us to calculate the mass of carbon monoxide (CO) in grams. This process is instrumental in quantifying substances during chemical reactions and is often used to scale reactions to practical quantities in lab settings or industrial processes.

Finally, chemical formula stoichiometry involves using the coefficients of a balanced chemical equation to understand the quantitative relationships between reactants and products. These stoichiometric calculations rely on the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction.

For example, if you have a reaction where one molecule of CO2 yields one molecule of CO and half a molecule of O2 (not considering practical chemical equations, but just for the stoichiometric relationship), you can deduce that one mole of CO2 will produce one mole of CO, assuming complete conversion.

Using the stoichiometric principles, one can calculate the required amount of reactants or predict the amount of products formed. In the context of the exercise, this concept allows us to understand that the weight of CO is directly related to the number of oxygen atoms initially present in CO2 based on their stoichiometric relationship in the respective molecular formulas. This comprehension is crucial for solving more complex chemistry problems involving multiple reactions and compounds.