Problem 46 - Chemistry The Central Science

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the moles of each element

In this case, we are given the moles directly: 0.104 mol K, 0.052 mol C, and 0.156 mol O.

Step 2: Divide by the smallest mole value

The smallest mole value is 0.052 mol (C). So, divide each mole value by 0.052. - K: 0.104/0.052 = 2 - C: 0.052/0.052 = 1 - O: 0.156/0.052 = 3

Step 3: Write the empirical formula

The empirical formula for part a is K2C1O3 or K2CO3. ##Part b##

Step 1: Convert mass to moles using the molar mass

We have 5.28 g Sn and 3.37 g F. Convert these masses to moles using the molar mass of the elements. - Sn: 5.28 g / 118.71 g/mol (Sn molar mass) ≈ 0.0445 mol - F: 3.37 g / 19.00 g/mol (F molar mass) ≈ 0.177 mol

Step 2: Divide by the smallest mole value

The smallest mole value is 0.0445 mol (Sn). So, divide each mole value by 0.0445. - Sn: 0.0445/0.0445 = 1 - F: 0.177/0.0445 ≈ 4

Step 3: Write the empirical formula

The empirical formula for part b is Sn1F4 or SnF4. ##Part c##

Step 1: Convert mass percentages to moles

Assume a 100 g sample, then we have 87.5 g N and 12.5 g H. Convert these masses to moles using the molar mass of the elements. - N: 87.5 g / 14.01 g/mol (N molar mass) ≈ 6.245 mol - H: 12.5 g / 1.01 g/mol (H molar mass) ≈ 12.375 mol

Step 2: Divide by the smallest mole value

The smallest mole value is 6.245 mol (N). So, divide each mole value by 6.245. - N: 6.245/6.245 = 1 - H: 12.375/6.245 ≈ 2

Step 3: Write the empirical formula

The empirical formula for part c is N1H2 or NH2.

Key Concepts

Mole Concept

The mole concept is a critical piece in understanding chemical calculations. A mole is a convenient unit that measures the amount of a substance. It is similar to the concept of a dozen, but instead of 12, a mole corresponds to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities (such as atoms, molecules, or ions).
Understanding moles helps you count the vast number of tiny entities in a lump of a substance, using grams and molar masses.

• Moles connect macroscopic amounts of substances to their molecular-scale quantities.
• They allow scientists to measure how much of each substance participates in chemical reactions.
• Conveniently, equating measurements in grams to molar masses lets you convert between mass and moles.

In the exercise, you determined moles from given data by either directly having the mole quantities or converting from grams using molar masses.

Stoichiometry

Stoichiometry is the calculation process aimed at understanding and predicting the quantities of reactants and products involved in chemical reactions. It's central to chemical equations and involves using relationships from balanced chemical reactions to calculate unknown quantities.

• Start with a balanced chemical equation.
• Convert quantities from masses to moles (using the mole concept).
• Use the mole ratios from the balanced equation to find other quantities.
• Convert from moles back to masses if needed.

This concept plays an essential role in everyday chemical applications, ensuring correct proportions for reactions. In the exercise, stoichiometry helps us find the proportional relationships needed to establish empirical formulas.

Molar Mass

Molar mass is a physical property defining how much one mole of a substance weighs. It is quantitative, expressed in grams per mole, and based on the atomic masses of elements found on the periodic table.

• Calculate molar mass by summing the atomic masses of all atoms in a molecule.
• It's crucial for converting between grams and moles.
• A compound's molar mass influences stoichiometric calculations.

In the example exercise, molar mass conversion is a key step. For instance, calculating the moles of Sn or F involved assessing their respective molar masses.

Chemical Composition

Chemical composition defines the types and proportions of elements in a compound. The empirical formula represents these as the simplest whole-number ratio of atoms in a compound. It highlights how fundamental these relationships are to chemistry.

• The empirical formula helps identify the basic building blocks of a substance.
• It differs from the molecular formula, which shows actual numbers of atoms.
• Weights and mole ratios derived from it assist in many chemical calculations.

In solving for the empirical formula, you rely on mole ratios and percentage compositions to decide the formula that best represents the compound's structure. Throughout the exercise, distinguishing these compositions emphasizes how substances are characterized and identified.