Question

Determine the empirical formula of each of the following compounds if a sample contains (a) \(0.104 \mathrm{~mol} \mathrm{~K}\). \(0.052 \mathrm{~mol} \mathrm{C}\), and \(0.156 \mathrm{~mol} \mathrm{O} ;\) (b) \(5.28 \mathrm{~g} \mathrm{Sn}\) and \(3.37 \mathrm{~g} \mathrm{~F}\); (c) \(87.5 \% \mathrm{~N}\) and \(12.5 \% \mathrm{H}\) by mass.

Short answer

The empirical formulas for each of the given compounds are: (a) K₂CO₃ (b) SnF₄ (c) NH₂

Step-by-step solution

(a) Calculate the moles of each element

Given moles are: - 0.104 mol K - 0.052 mol C - 0.156 mol O

(a) Determine smallest mole ratio

Divide the moles of each element by the smallest moles among them (0.052 mol C): - K: \( \frac{0.104}{0.052} \) = 2 - C: \( \frac{0.052}{0.052} \) = 1 - O: \( \frac{0.156}{0.052} \) = 3

(a) Write the empirical formula

The empirical formula of the compound is K₂CO₃.

(b) Calculate moles of Sn and F

First, we need to calculate moles of Sn and F from their given masses. For Sn: Moles = \(\frac{mass}{molar~mass} = \frac{5.28}{118.71} \approx 0.0445\) mol For F: Moles = \(\frac{mass}{molar~mass} = \frac{3.37}{19.00} \approx 0.177\) mol

(b) Determine smallest mole ratio

Divide the moles of each element by the smallest moles among them (0.0445 mol Sn): - Sn: \( \frac{0.0445}{0.0445} \) = 1 - F: \( \frac{0.177}{0.0445} \) ≈ 4

(b) Write the empirical formula

The empirical formula of the compound is SnF₄.

(c) Calculate moles of N and H from percentages

Let's assume we have a 100g sample. Then, the masses of N and H are 87.5g and 12.5g, respectively. For N: Moles = \(\frac{mass}{molar~mass} = \frac{87.5}{14.007} \approx 6.25\) mol For H: Moles = \(\frac{mass}{molar~mass} = \frac{12.5}{1.008} \approx 12.4\) mol

(c) Determine smallest mole ratio

Divide the moles of each element by the smallest moles among them (6.25 mol N): - N: \( \frac{6.25}{6.25} \) = 1 - H: \( \frac{12.4}{6.25} \) ≈ 2

(c) Write the empirical formula

The empirical formula of the compound is NH₂.

Key concepts

Mole Ratio

A mole ratio is a concept used to determine the proportion of different elements in a chemical compound. By comparing the moles of elements present, we can identify how many atoms of each type are present relative to each other. The process involves dividing the amount of each substance by the smallest quantity present. This gives a synthesized ratio that helps in determining the simplest form of the compound, known as the empirical formula. For example, in the given exercise, we had 0.104 mol of K, 0.052 mol of C, and 0.156 mol of O. Dividing each by the smallest mole value (0.052 mol) reveals a ratio of K: 2, C: 1, O: 3, leading to the empirical formula of K₂CO₃. By understanding mole ratios, students can transform experimental data into informative chemical formulas.

Percentage Composition

Percentage composition refers to the percentage mass of each element in a compound. It is calculated using the mass of each element divided by the total mass of the compound, multiplied by 100. This information helps in identifying the proportions of elements and is useful for determining empirical formulas. For instance, if a compound is composed of elements where N makes up 87.5% and H comprises 12.5% of its mass, we can assume a 100g sample for simplicity. This allocation allows us to directly use these mass percentages as the mass values of the elements. When converted into moles, these values then facilitate the calculation of the empirical formula by determining which mole quantity is smallest and setting the mole ratio.

Chemical Compound Analysis

Chemical compound analysis involves breaking down a substance to define its components and their respective proportions. This detailed analysis is fundamental in chemistry to understand the composition and characteristics of compounds. It typically involves steps such as weighing, calculating moles, and ultimately determining empirical formulas. In exercise cases, like determining the makeup of a compound from given elements, the analysis helps in converting weights or percentages into useful chemical data. The method includes forming a logical approach where the empirical formula acts as a simple representation of the compound's elemental distribution. This comprehensive approach enables chemists and students alike to predict reactions and properties of compounds more accurately.

Molar Mass Calculation

The molar mass is a key factor in converting between grams and moles of a chemical substance. It is defined as the mass of one mole of a substance, usually in grams per mole, and it is calculated by summing the atomic masses of all atoms in the element or compound. The ability to calculate molar mass is crucial as it helps in determining the number of moles for a given mass, a necessary step in empirical formula calculations. Looking at our example with Sn and F, knowing that tin has a molar mass of 118.71 g/mol and fluorine 19.00 g/mol, we first convert the given grams into moles by dividing the mass by the respective molar mass. This gives us a basis to compare and find mole ratios, translating quantitative data into actionable chemical information.