Problem 46 - Chemistry: The Central Science

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.

Step-by-Step Solution

Answer

The empirical formula for each compound is as follows: (a) K₂CO₃ (b) SnF₄ (c) NH₂

Step 1: Write down the given moles

We are given 0.104 mol K, 0.052 mol C, and 0.156 mol O.

Step 2: Find the smallest number of moles

In this case, the smallest number of moles is for C, which is 0.052 mol.

Step 3: Divide all moles by the smallest mole value

Divide the moles of each element by the smallest mole value (0.052 mol): K: \(0.104/0.052 = 2\) C: \(0.052/0.052 = 1\) O: \(0.156/0.052 = 3\)

Step 4: Write the empirical formula

Now we have the simplest ratio of the elements: K₂CO₃ , so the empirical formula is K₂CO₃. Part (b):

Step 1: Convert grams to moles for each element

Using the molar masses for Sn and F: Sn: \(5.28g ÷ 118.71g/mol ≈ 0.0445mol\) F: \(3.37g ÷ 19.00g/mol ≈ 0.177mol\)

Step 2: Find the smallest number of moles

The smallest number of moles is for Sn: 0.0445 mol.

Step 3: Divide all moles by the smallest mole value

Divide the moles of each element by the smallest mole value (0.0445 mol): Sn: \(0.0445/0.0445 = 1\) F: \(0.177/0.0445 ≈ 4\)

Step 4: Write the empirical formula

Now we have the simplest ratio of the elements: SnF₄, so the empirical formula is SnF₄. Part (c):

Step 1: Assume 100g of compound

Since we have the percentage composition, assume we have 100g of the compound. So we have 87.5g of N and 12.5g of H.

Step 2: Convert grams to moles for each element

Using the molar masses for N and H: N: \(87.5g ÷ 14.01g/mol ≈ 6.25mol\) H: \(12.5g ÷ 1.008g/mol ≈ 12.4mol\)

Step 3: Find the smallest number of moles

The smallest number of moles is for N: 6.25 mol.

Step 4: Divide all moles by the smallest mole value

Divide the moles of each element by the smallest mole value (6.25 mol): N: \(6.25/6.25 = 1\) H: \(12.4/6.25 ≈ 2\)

Step 5: Write the empirical formula

Now we have the simplest ratio of the elements: NH₂, so the empirical formula is NH₂.

Key Concepts

Mole Ratio

Understanding the concept of mole ratio is essential in chemistry workshops and assignments when working with chemical formulas. The mole ratio represents the relative quantities of elements in a compound and is crucial in determining the empirical formula. In our scenario, the mole ratio is found by dividing the number of moles of each element by the smallest number of moles present among the elements in the compound.
For instance, when calculating for potassium carbonate (K₂CO₃), we have:

• Moles of K = 0.104
• Moles of C = 0.052
• Moles of O = 0.156

Here, the smallest mole is 0.052 mol (C), so we divide all molar quantities by this value. This process helps to simplify the numbers to the smallest whole-number ratio for the empirical formula:

• K: 0.104/0.052 = 2
• C: 0.052/0.052 = 1
• O: 0.156/0.052 ≈ 3

This gives us the ratio of K:C:O = 2:1:3, leading to the empirical formula of K₂CO₃.

Percentage Composition

Percentage composition refers to the percentage by mass of each element in a compound. It is pivotal for deducing the empirical formula, especially when the composition is given instead of direct mole values. In essence, by knowing the percentage composition, you can estimate the actual mass of each element in a certain sample of the compound.
For example, let's assume we have a compound made of 87.5% nitrogen and 12.5% hydrogen by mass. If we consider a 100g sample for simplicity, we derive the mass of nitrogen to be 87.5g and that of hydrogen 12.5g. From here, converting these masses to moles gives us a clearer picture of the substance composition:

• Nitrogen: 87.5g ÷ 14.01g/mol = 6.25 mol
• Hydrogen: 12.5g ÷ 1.008g/mol = 12.4 mol

By calculating the mole ratios, we comprehensively identify the elements' ratios in forms considered during chemical reactions, hence constructing the empirical formula. Here, this ratio helps to determine that the simplest form is NH₂.

Molar Mass

Molar mass is another fundamental concept when working with chemical compounds and understanding their formulas. It is defined as the mass of one mole of a substance (element or compound) and is usually expressed in grams per mole (g/mol). You can determine it by summing up the atomic masses of all the atoms in a molecule.
In practical exercise problems, knowing the molar mass aids in converting grams of elements into moles, a necessary step in formulating reactions and studying compounds' properties.
For instance, take tin fluoride (SnF₄), where conversion of grams to moles entails:

• Finding the molar mass of tin (Sn), which is about 118.71 g/mol
• Finding the molar mass of fluorine (F), around 19.00 g/mol

Using these values helps in converting a 5.28g mass of Sn to moles via this calculation: 5.28g ÷ 118.71g/mol = 0.0445 mol, and similarly for F. The conversion of mass to moles makes it possible to determine the mole ratio, leading to the empirical formula SnF₄. Calculating molar mass not only assists in these conversions but is pivotal to understanding the quantitative relationships in chemical reactions.