Question

A compound whose empirical formula is \(\mathrm{XF}_{3}\) consists of \(65 \%\) \(\mathrm{F}\) by mass. What is the atomic mass of \(\mathrm{X} ?\)

Short answer

The atomic mass of element X in the compound XF3, which is 65% fluorine by mass, is approximately \(30.69 g/mol\).

Step-by-step solution

Determine the atomic mass of fluorine and calculate the mass of 3 fluorine atoms in the compound

The atomic mass of fluorine (F) is approximately 19 g/mol. Since there are 3 fluorine atoms in the formula XF3, we can calculate the mass of fluorine in the compound: Mass of 3 F atoms = 3 × atomic mass of F = 3 × 19 g/mol = 57 g/mol

Calculate the mass of element X in the compound

Since the compound is 65% fluorine by mass, the remaining percentage must be the mass of element X in the compound. Therefore, the mass percentage of element X in the compound is: 100% - 65% = 35% Now, we can determine the mass of element X in the compound by finding the mass equivalent of 35%. Since the mass of 3 F atoms in the compound equals 57 g/mol (calculated in Step 1), we can set up the equation: (35 g X) / (57 g F) = 35% Solve for the mass of X: 35 g X = 0.35 × 57 g F Mass of X = 19.95 g/mol Since the mass percentage of a single atom of X should be the same as the mass percentage in the compound, we can set up the equation: Mass of 1 X atom / (57 g F + Mass of 1 X atom) = 35%

Calculate the atomic mass of element X

Now, we will solve the equation we set up in Step 2 to find the atomic mass of element X: Mass of 1 X atom / (57 g F + Mass of 1 X atom) = 35% Let the atomic mass of X be 'm'. m / (57 + m) = 0.35 Solving for 'm', we get: m = 0.35 × (57 + m) m = 19.95 + 0.35m Now, subtract 0.35m from both sides to get: 0.65m = 19.95 Finally, we can find the atomic mass of X by dividing both sides by 0.65: m = 19.95 / 0.65 m ≈ 30.69 g/mol Thus, the atomic mass of element X is approximately 30.69 g/mol.

Key concepts

Empirical Formula

The empirical formula is a crucial concept in chemistry and represents the simplest ratio of elements in a compound. For instance, the empirical formula of the compound in the exercise is \(\mathrm{XF}_{3}\). This signifies that for every atom of element \(X\), there are three atoms of fluorine.

The empirical formula does not necessarily convey the exact number of each type of atom in the compound. It only shows the lowest possible ratio. However, it provides foundational information necessary to determine the compound's molecular formula, if the molar mass is known.

• Empirical formulas simplify comparison between different compounds.
• Helps in calculating relative molar masses.
• Often used as a starting point for finding molecular formulas.

Fluorine

Fluorine is an essential element in chemistry, especially in organofluorine compounds. In the problem provided, fluorine's atomic mass is noted as 19 g/mol, and it plays a significant role in determining the mass percentage composition of the compound \(\mathrm{XF}_{3}\).

Fluorine is highly electronegative, affecting the properties of compounds it forms. Its presence in a compound can influence both its chemical reactivity and stability. Calculating the mass of 3 fluorine atoms involved a multiplication of its atomic mass by 3, which helped in setting the groundwork for further calculations in this problem.

• Commonly used in halogenation reactions.
• A major component in non-stick coatings and refrigerants.
• Presents in high percentages, influences compound's properties significantly.

Mass Percentage Composition

Understanding mass percentage composition is vital for determining the relative amounts of components in a compound. In our exercise, around 65% of the compound is made up of fluorine. This means that when you calculate the mass of component elements, you attribute 65% of the total mass to fluorine.

Mass percentage is calculated by dividing the mass of the element of interest by the total mass of the compound, and then multiplying the result by 100. In this problem:

• Knowing fluorine's contribution helped determine the remaining mass percentage for element X.
• Essential in calculating unknown component’s proportional weight in a compound.
• Useful in various applications, including formulation of medicines and materials.

By understanding and calculating mass percentage composition, we can clearly see the distribution of each element in the compound, which plays a crucial role in determining molecular and empirical formulas.