Problem 48 - Chemistry The Central Science

Question

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

Step-by-Step Solution

Answer

The empirical formulas for the given compounds are: (a) KPO₄ (b) Na₂SiF₆ (c) C₅H₀.₄₀₇N¹.₀₂₂O₂.₀₃₃

Step 1: Convert percentage composition to mass

Assume that each compound has a total mass of 100 grams. This assumption allows us to easily convert the percentage composition directly into grams of each element. For example, For compound (a): - 55.3% K implies 55.3 grams of K - 14.6% P implies 14.6 grams of P - 30.1% O implies 30.1 grams of O We will make similar calculations for compounds (b) and (c).

Step 2: Determine moles of each element

Next, we need to determine the moles of each element in the compound. To do this, divide the mass (in grams) of each element by its respective molar mass. For compound (a): - Moles of K: \(\frac{55.3}{39.10}\) (molar mass of K is 39.10 g/mol) - Moles of P: \(\frac{14.6}{30.97}\) (molar mass of P is 30.97 g/mol) - Moles of O: \(\frac{30.1}{16.00}\) (molar mass of O is 16.00 g/mol) Calculate the moles for elements in compounds (b) and (c) similarly.

Step 3: Determine the mole ratio

To determine the mole ratio, divide all the moles calculated in Step 2 by the smallest moles value among the elements in that compound. For compound (a): - Calculate the smallest moles value: min(K, P, O) - Divide all the moles by the smallest moles value to get the mole ratio. Do this also for compounds (b) and (c).

Step 4: Determine the empirical formula

Use the mole ratios obtained in Step 3 to construct the empirical formula for each compound. For compound (a): - Combine the elements with their respective ratios, rounding to the nearest whole number if the ratios are close to integers. Perform this step for compounds (b) and (c), and the empirical formulas will be determined.

Key Concepts

Percentage Composition

When trying to find the empirical formula of a compound, the starting point is understanding the percentage composition. This refers to the percent by mass of each element in the compound. For instance, if a compound is said to contain 55.3% potassium (K), it means that in 100 grams of this compound, there are 55.3 grams of potassium.

This percentage composition is vital because it allows us to determine the relative amount of each element present. By assuming a total mass of 100 grams, converting percentages directly into grams becomes straightforward, simplifying your work in later steps.

• This conversion lays the foundation for quantitative analysis.
• All percentages should ideally add up to 100%. Any remainder often belongs to unspecified elements such as oxygen, as seen in problem (c) where the remaining percentage accounts for oxygen.

This method ensures accuracy in translating percentage into mass, which is crucial for calculations that follow.

Molar Mass

Molar mass is a key concept that connects the mass of an element to the number of atoms. In practice, it's defined as the mass of one mole of a substance, typically measured in grams per mole (g/mol).

Knowing the molar mass of an element is essential for converting grams to moles during empirical formula calculations. For example, potassium (K) has a molar mass of 39.10 g/mol. To find the moles of potassium from its mass, you'd divide the mass of potassium by its molar mass.

• Molar mass helps equate macroscopic measurements with the microscopic scale of atoms and molecules.
• This is a critical step because the number of moles is used to determine the mole ratio.

Remember, recorded atomic weights of elements—found in periodic tables—represent their molar masses and are used directly for these conversions.

Mole Ratio

Determining the mole ratio is crucial for deriving the empirical formula. After converting masses to moles using molar mass, the next step is comparing the mole quantities of the elements. This is done by dividing the moles of each element by the smallest number of moles found among them.

For instance, in compound (a), if the smallest moles value is that of phosphorus (P), you would divide the moles of potassium (K), phosphorus (P), and oxygen (O) by the moles of phosphorus to get their respective ratios.

• Mole ratios indicate the simplest whole number ratio of atoms in the compound.
• If the ratios aren't whole numbers, we'll sometimes need to multiply through by a factor to get them as close to whole numbers as possible.

This simplification step is pivotal in ultimately determining the empirical formula.

Stoichiometry

Stoichiometry is like the math of chemistry, helping you to balance relationships between quantities of reactants and products. When determining empirical formulas, stoichiometry enables conversion of mass data into meaningful atomic ratios, ensuring that chemical equations reflect real substance ratios.

The process involves using stoichiometric coefficients (like in mole ratios) to ensure calculations align with the laws of conservation of mass—indicating that the number and ratio of atoms are consistent before and after reactions.

• Stoichiometry ensures proportionality in chemical reactions.
• It's the backbone for accurately determining empirical formulas from experimental data.

By maintaining these balances and conversions, stoichiometry solidifies understanding the compound's basic formula from a macroscopic viewpoint.