Question

A metal (M) forms an oxide with the formula \(\mathrm{M}_{2} \mathrm{O}\) . If the oxide con- tains 16.99\(\% \mathrm{O}\) by mass, what is the identity of the metal?

Short answer

The metal (M) is potassium (K) with an approximate molar mass of \(39.08\ g/mol\).

Step-by-step solution

Determine the mass percentage of metal (M)

Calculate the mass percentage of the metal (M) in the compound \(\mathrm{M}_2\mathrm{O}\) by subtracting the mass percentage of oxygen from 100\%. So, the mass percentage of metal (M) is 100\% - 16.99\% = 83.01\%.

Calculate the molar mass of \(\mathrm{M}_2\mathrm{O}\) based on composition

Let the molar mass of the metal be \(M_m\). The molar mass of \(\mathrm{M}_2\mathrm{O}\) is then \(2M_m + 16.00\ g/mol\) (since the molar mass of oxygen is 16.00 g/mol).

Set up a proportion for the metal's mass percentage

According to the mass percentage, the mass of the metal in \(100\ g\) of the compound is \(83.01\ g\). Therefore, we have \(83.01 = \frac{2M_m}{2M_m + 16} \times 100\).

Solve for the molar mass of the metal (M)

Solve the equation for \(M_m\): \(83.01(2M_m + 16) = 2M_m \times 100\). This simplifies to \(166.02M_m + 1328.16 = 200M_m\), which simplifies further to \(33.98M_m = 1328.16\). Divide both sides by 33.98 to get \(M_m \approx 39.08\ g/mol\).

Identify the metal

Compare the calculated molar mass to the periodic table to determine that the metal with a molar mass closest to \(39.08\ g/mol\) is potassium (K), which has an atomic mass of approximately \(39.10\ g/mol\).

Key concepts

Molar Mass Calculation

Understanding molar mass is essential for students embarking on a journey through chemistry. Molar mass is defined as the mass of one mole of a substance, and it is usually expressed in grams per mole (g/mol). The calculation of molar mass is a fundamental skill, as it not only helps in solving problems linked to chemical composition but also sets the base for stoichiometry.

To calculate the molar mass, you need to know the formula of the compound and the atomic masses of the elements present in the compound. For a compound like \(\mathrm{M}_{2} \mathrm{O}\), which was given in the exercise, the molar mass is inferred by multiplying the atomic mass of the metal (M) by two (since there are two moles of M in the compound) and adding the atomic mass of oxygen. As a practical lesson, when given the mass percent of oxygen, you can compute the mass percentage of the other elements by subtracting the known percentage from 100%, then use this information to derive the molar mass of unknown components, as beautifully illustrated in the problem provided.

Empirical Formula Determination

Another foundational block of chemistry is discerning an empirical formula from given data. An empirical formula represents the simplest whole-number ratio of elements in a compound. Detecting the empirical formula requires an analytical approach where you convert percentage composition to masses, use the molar mass to find moles, and then determine the simplest ratio of these moles.

When applied to the metallic oxide \(\mathrm{M}_{2} \mathrm{O}\) in the exercise, the calculation begins from the given percentage of oxygen. With oxygen's percentage and assuming a sample size of 100 g for ease of calculation, you can calculate the grams of each element, convert those to moles, and then simplify the ratio. This ratio translates to the empirical formula, pivotal in identifying unknown substances in a chemical query such as a metal with an unfamiliar molar mass.

Stoichiometry

The study of stoichiometry is tantamount to understanding the quantitative relationships in chemical reactions. Stoichiometry hinges on the law of conservation of mass and the concept of the mole, allowing chemists to predict the amounts of substances consumed and produced in a given reaction.

In exercises like the one in question, stoichiometry comes into play when you have determined the molar mass of each element in a compound. After establishing the empirical formula, students can use stoichiometry to deduce how much of each element is present in a sample of any size, or to predict how much of a compound can form from a given amount of reactant. The proportional reasoning applied to solve for the molar mass of metal (M) mirrors the stoichiometric calculations needed to balance chemical equations and perform related quantitative analyses.