Problem 46 - Chemistry: The Central Science

Question

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(55.3 \% \mathrm{~K}, 14.6 \% \mathrm{P}\), and \(30.1 \% \mathrm{O}\) (b) \(24.5 \% \mathrm{Na}, 14.9 \% \mathrm{Si}\), and \(60.6 \% \mathrm{~F}\) (c) \(62.1 \% \mathrm{C}, 5.21 \% \mathrm{H}, 12.1 \% \mathrm{~N}\), and \(20.7 \% \mathrm{O}\)

Step-by-Step Solution

Answer

The empirical formulas for the given compounds are: (a) K3PO4 (b) Na2SiF6 (c) C12H12N2O3

Step 1: Assume a 100g sample and convert mass percentages to grams

Assume that we have a 100g sample. The given percentages will represent the mass of each element in the 100g sample. - 55.3g K - 14.6g P - 30.1g O

Step 2: Convert grams to moles

Use the molar masses to convert the grams of each element to moles: Moles of K = \(\frac{55.3g}{39.1g/mol}\) = 1.413 moles Moles of P = \(\frac{14.6g}{30.97g/mol}\) = 0.471 moles Moles of O = \(\frac{30.1g}{16g/mol}\) = 1.881 moles

Step 3: Find the simplest mole ratio

Divide all mole values by the smallest mole value (0.471 moles, which corresponds to P) to obtain the simplest whole number ratio: K: \(\frac{1.413}{0.471}\) ≈ 3 P: \(\frac{0.471}{0.471}\) = 1 O: \(\frac{1.881}{0.471}\) ≈ 4

Step 4: Write the empirical formula

The empirical formula for this compound is K3P1O4 or simply K3PO4. (b) Determine the empirical formula for compound with 24.5% Na, 14.9% Si, and 60.6% F

Step 1: Convert mass percentages to grams

Assuming a 100g sample, we have: - 24.5g Na - 14.9g Si - 60.6g F

Step 2: Convert grams to moles

Moles of Na = \(\frac{24.5g}{22.99g/mol}\) = 1.065 moles Moles of Si = \(\frac{14.9g}{28.09g/mol}\) = 0.531 moles Moles of F = \(\frac{60.6g}{19g/mol}\) = 3.190 moles

Step 3: Find the simplest mole ratio

Divide all mole values by the smallest mole value (0.531 moles, which corresponds to Si) to obtain the simplest whole number ratio: Na: \(\frac{1.065}{0.531}\) ≈ 2 Si: \(\frac{0.531}{0.531}\) = 1 F: \(\frac{3.190}{0.531}\) ≈ 6

Step 4: Write the empirical formula

The empirical formula for this compound is Na2Si1F6 or simply Na2SiF6. (c) Determine the empirical formula for compound with 62.1% C, 5.21% H, 12.1% N, and 20.7% O

Step 1: Convert mass percentages to grams

Assuming a 100g sample, we have: - 62.1g C - 5.21g H - 12.1g N - 20.7g O

Step 2: Convert grams to moles

Moles of C = \(\frac{62.1g}{12.01g/mol}\) = 5.17 moles Moles of H = \(\frac{5.21g}{1.01g/mol}\) = 5.16 moles Moles of N = \(\frac{12.1g}{14.01g/mol}\) = 0.865 moles Moles of O = \(\frac{20.7g}{16g/mol}\) = 1.294 moles

Step 3: Find the simplest mole ratio

Divide all mole values by the smallest mole value (0.865 moles, which corresponds to N) to obtain the simplest whole number ratio: C: \(\frac{5.17}{0.865}\) ≈ 6 H: \(\frac{5.16}{0.865}\) ≈ 6 N: \(\frac{0.865}{0.865}\) = 1 O: \(\frac{1.294}{0.865}\) ≈ 1.5 Since we obtained a ratio that is not a whole number for O (1.5), we can multiply all ratios by 2 to obtain whole numbers. C: 6 * 2 = 12 H: 6 * 2 = 12 N: 1 * 2 = 2 O: 1.5 * 2 = 3

Step 4: Write the empirical formula

The empirical formula for this compound is C12H12N2O3.

Key Concepts

Percentage Composition

Understanding percentage composition is crucial for determining an empirical formula, as it represents the fraction of each element's mass in a compound. To start, imagine you have a 100g sample of a compound—this simplifies calculations because the percentage then directly translates to grams.

For instance, if a compound is composed of 55.3% potassium (K), that simply means there are 55.3 grams of potassium in every 100 grams of the compound. By converting these percentages to actual masses, students find it easier to visualize the constituent elements and take the first step towards determining the compound's empirical formula.

After establishing the mass of each element, next steps will include converting these masses into moles, which is how chemists compare quantities of substances based on the number of particles, rather than their weight.

Molar Mass

Molar mass is the bridge between the mass of a substance and the amount of moles it represents. It is denoted in grams per mole (g/mol) and can be found on the periodic table for each element. For instance, potassium has a molar mass of 39.1 g/mol, so to convert grams to moles, divide the element's mass by its molar mass.

In our exercise, dividing 55.3 grams of potassium by its molar mass (39.1 g/mol) gives us the moles of potassium in the compound. It's like converting currency: if you have dollars and need to know how many euros they equate to, you divide by the exchange rate. Similarly, to 'exchange' grams for moles, you divide by the molar mass.

Mole Ratio

Determining the mole ratio is akin to finding the simplest recipe for a compound. Once you've converted the masses of elements to moles, the next step is to divide all mole amounts by the smallest number of moles obtained for any single element. This 'divisor' helps you achieve the simplest whole number ratio, which corresponds to the empirical formula of the compound.

For example, if the smallest number of moles is 0.471 for phosphorus, dividing all elements' moles by this number normalizes the ratio. If the result is not already a whole number, you may need to multiply all ratios by the smallest number to achieve whole numbers. The ability to identify an appropriate multiplier to obtain whole numbers is a critical step and ensures the ratios accurately represent the number of atoms in the simplest form of the compound.

Chemical Formula

The chemical formula of a compound is a symbolic representation showing the types and numbers of atoms involved. The empirical formula is the simplest form of the chemical formula, giving the lowest whole number ratio of atoms in the compound. Completing the previous steps—converting percentage composition to grams, then to moles, and finally finding the mole ratio—sets the stage to write out the empirical formula.

For the exercise's compound (c), after adjusting for the non-whole number ratio, we derive an empirical formula of C12H12N2O3. This formula tells us that the simplest version of this compound contains 12 carbon atoms, 12 hydrogen atoms, 2 nitrogen atoms, and 3 oxygen atoms. Understanding how to interpret these ratios and symbols is essential for students delving into chemical compositions and reactions.