Problem 47 - Chemistry The Central Science In Si Units

Question

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(74.0 \% \mathrm{C}, 8.7 \% \mathrm{H},\) and \(17.3 \% \mathrm{~N}\) (b) \(57.5 \% \mathrm{Na}, 40.0 \% \mathrm{O},\) and \(2.5 \% \mathrm{H}\) (c) \(41.1 \% \mathrm{~N}, 11.8 \% \mathrm{H},\) and the remainder \(\mathrm{S}\)

Step-by-Step Solution

Answer

Empirical formulas are (a) \( C_5H_7N \), (b) \( NaOH \), and (c) \( N_2H_8S \).

Step 1: Convert Percentages to Grams

We assume we have a 100 g sample of each compound. Thus, the percentage values can be directly read as grams. For example, for compound (a), we have 74.0 g of C, 8.7 g of H, and 17.3 g of N.

Step 2: Convert Grams to Moles

Use the molar masses to convert grams to moles:(a) C: \( \frac{74.0 \text{ g}}{12.01 \text{ g/mol}} = 6.16 \text{ mol} \), H: \( \frac{8.7 \text{ g}}{1.008 \text{ g/mol}} = 8.63 \text{ mol} \), N: \( \frac{17.3 \text{ g}}{14.01 \text{ g/mol}} = 1.24 \text{ mol} \).(b) Na: \( \frac{57.5 \text{ g}}{22.99 \text{ g/mol}} = 2.50 \text{ mol} \), O: \( \frac{40.0 \text{ g}}{16.00 \text{ g/mol}} = 2.50 \text{ mol} \), H: \( \frac{2.5 \text{ g}}{1.008 \text{ g/mol}} = 2.48 \text{ mol} \).(c) N: \( \frac{41.1 \text{ g}}{14.01 \text{ g/mol}} = 2.93 \text{ mol} \), H: \( \frac{11.8 \text{ g}}{1.008 \text{ g/mol}} = 11.71 \text{ mol} \), S remainder: \( \frac{47.1 \text{ g}}{32.07 \text{ g/mol}} = 1.47 \text{ mol} \).

Step 3: Determine Simplest Molar Ratio

Divide each number of moles by the smallest number of moles in the compound:(a) C: \( \frac{6.16}{1.24} \approx 5.0 \), H: \( \frac{8.63}{1.24} \approx 7.0 \), N: \( \frac{1.24}{1.24} = 1.0 \), giving \( C_5H_7N \).(b) Na: \( \frac{2.50}{2.48} \approx 1.0 \), O: \( \frac{2.50}{2.48} \approx 1.0 \), H: \( \frac{2.48}{2.48} = 1.0 \), giving \( NaOH \).(c) N: \( \frac{2.93}{1.47} \approx 2.0 \), H: \( \frac{11.71}{1.47} \approx 8.0 \), S: \( \frac{1.47}{1.47} = 1.0 \), giving \( N_2H_8S \).

Step 4: Write the Empirical Formula for Each Compound

Based on previous calculations, the empirical formulas are:(a) \( C_5H_7N \)(b) \( NaOH \)(c) \( N_2H_8S \).

Key Concepts

Molar Mass Conversion

When working with chemical elements, it's essential to translate amounts from grams to moles. This is a crucial step because chemical reactions depend on moles rather than grams. To do this conversion, you need the element's molar mass, which tells you the mass of one mole of that element, usually found on the periodic table. For instance, the molar mass of carbon is 12.01 g/mol, hydrogen is 1.008 g/mol, and nitrogen is 14.01 g/mol. By dividing the mass (in grams) of the element by its molar mass, you find the number of moles. This allows you to compare and utilize these quantities in chemical reactions. For example, if you have 74.0 grams of carbon, dividing by its molar mass, you get approximately 6.16 moles of carbon. This conversion is crucial for calculating the correct proportions needed to determine the empirical formula of a compound.

Mole Ratio

Determining the simplest mole ratio is essential for finding a compound's empirical formula. Once you've converted the mass of each element in the compound to moles, the next step is to find their simplest ratio. This is done by dividing the moles of each element by the smallest number of moles in the sample. This step effectively normalizes the number of moles so that you can express the composition in terms of whole numbers. For instance, if a compound contains 6.16 moles of carbon, 8.63 moles of hydrogen, and 1.24 moles of nitrogen, you would divide each by the smallest value, 1.24. This simplification results in a mole ratio of C:H:N = 5:7:1, accurately representing the simplest whole number ratio of the elements in the compound.

Chemical Composition

The chemical composition of a substance indicates the types and amounts of atoms that make up the substance. In finding an empirical formula, you start with the compound's mass composition, which tells you what percentage of the compound's mass is made up by each element. This is the precursor to determining the chemical composition on an atomic level. With the mass percentages at hand, you assume a 100-gram sample to easily translate percentages directly to grams. Once in grams, the mass can be converted to moles, reflecting the number of atoms of each type in a representative quantity of the compound. This process simplifies our understanding of how atoms are combined in simple ratios to form compounds, facilitating the production of an empirical formula.

Percent Composition

Percent composition reveals the proportion by mass of each element in a compound. This is the initial information you need when determining an empirical formula. It provides the percentage that each element contributes to the total mass of the compound. For instance, if a compound contains 74% carbon, 8.7% hydrogen, and 17.3% nitrogen, this information is directly converted to a mass for each element in a hypothetical 100-gram sample. Consequently, 74 grams of carbon, 8.7 grams of hydrogen, and 17.3 grams of nitrogen can be used to determine the number of moles of each element. Calculating the percent composition helps break down the compound into its elemental components, setting the stage for further calculations needed to find the empirical formula.