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 \(\mathrm{S}\)

Short answer

The empirical formulas of the given compounds are: (a) C5H7N (b) NaOH (c) N2H8S

Step-by-step solution

(a) Conversion to Moles

For the first compound with the given mass percentages, determine the moles of each element by assuming 100 grams of compound. 74.0 g of C: \(\frac{74.0~g}{12.01~g/mol} \approx 6.16~mol\) of C 8.7 g of H: \(\frac{8.7~g}{1.01~g/mol} \approx 8.61~mol\) of H 17.3 g of N: \(\frac{17.3~g}{14.01~g/mol} \approx 1.235~mol\) of N

(a) Calculate the Mole Ratios

Divide all the mole values obtained above by the smallest value to get the mole ratios for the elements. C: \(\frac{6.16}{1.235} \approx 5\) H: \(\frac{8.61}{1.235} \approx 7\) N: \(\frac{1.235}{1.235} \approx 1\)

(a) Final Empirical Formula

Write the empirical formula for the first compound by noting the mole ratios. Empirical formula of compound (a): C5H7N

(b) Conversion to Moles

For the second compound, we do the same process as in (a) to get the moles of each element: 57.5 g of Na: \(\frac{57.5~g}{22.99~g/mol} \approx 2.50~mol\) of Na 40.0 g of O: \(\frac{40.0~g}{16.00~g/mol} \approx 2.50~mol\) of O 2.5 g of H: \(\frac{2.5~g}{1.01~g/mol} \approx 2.475~mol\) of H

(b) Calculate the Mole Ratios

Divide the mole values obtained above by the smallest value. Na: \(\frac{2.50}{2.475} \approx 1\) O: \(\frac{2.50}{2.475} \approx 1\) H: \(\frac{2.475}{2.475} \approx 1\)

(b) Final Empirical Formula

Write the empirical formula for the second compound by considering the mole ratios: Empirical formula of compound (b): NaOH

(c) Determining the Percentage of S

Since the remainder of the compound is sulfur, we need to calculate its mass percentage. 100% - 41.1 (N) - 11.8 (H) = 47.1% S

(c) Conversion to Moles

For the third compound, similarly, obtain the moles of each element and S: 41.1 g of N: \(\frac{41.1~g}{14.01~g/mol} \approx 2.934~mol\) of N 11.8 g of H: \(\frac{11.8~g}{1.01~g/mol} \approx 11.68~mol\) of H 47.1 g of S: \(\frac{47.1~g}{32.06~g/mol} \approx 1.468~mol\) of S

(c) Calculate the Mole Ratios

Divide the mole values obtained above by the smallest value. N: \(\frac{2.934}{1.468} \approx 2\) H: \(\frac{11.68}{1.468} \approx 8\) S: \(\frac{1.468}{1.468} \approx 1\)

(c) Final Empirical Formula

Write the empirical formula for the third compound by considering the mole ratios: Empirical formula of compound (c): N2H8S So, the empirical formulas of the given compounds are: (a) C5H7N (b) NaOH (c) N2H8S

Key concepts

Mole Ratios

To determine the empirical formula of a compound, scientists often start by finding the mole ratios of the elements present. This involves calculating the number of moles of each element, based on a known mass, typically using the elements' respective molar masses.

Let's break it down with a simple approach:

• First, assume you have a 100-gram sample of the compound. This makes it easy to convert percentages directly into grams.
• Convert the mass of each element into moles by dividing the mass by the atomic mass (from the periodic table) of the element. For instance, if we have 74 grams of carbon (C) in our composition, we would divide by its atomic mass of approximately 12.01 g/mol to find the moles of carbon.
• Once you have the moles of each element, divide each one by the smallest number of moles calculated to determine mole ratios.

For example in compound (a) with elements C, H, and N:

• Carbon: approximately 6.16 moles leads us to a ratio compared to Nitrogen's approximately 1.235 moles resulting in a 5:1 ratio.
• Hydrogen: with its moles yielding a roughly 7:1 ratio compared to nitrogen.

Mole ratios provide the simplest whole number ratio of elements in a compound, essential for writing correct empirical formulas. Understanding how to calculate and interpret these ratios is a crucial skill in chemistry.

Mass Percentage

Mass percentage plays a critical role in determining the empirical formula of a compound by providing a way to express the proportion of each element in the compound. It represents how much of an element's mass contributes to the total mass of the compound. Typically, mass percent is given in percentage, making it intuitive and easily understandable.

The calculation process is straightforward:

• The mass percentage of an element is calculated by dividing the mass of that element in the compound by the total mass of the compound, then multiplying by 100%.
• For example, consider a compound that is 57.5% sodium by mass. This means that in 100 grams of this compound, 57.5 grams are sodium.
• Knowing this percentage helps us convert to moles when determining the empirical formula, as we can assume a simple 100-gram sample for convenience in calculations.

For compound (c), using mass percentages like 41.1% for Nitrogen, 11.8% for Hydrogen, and calculating the remainder for Sulfur is essential to getting an accurate count of the elements' contributions. These percentages form the baseline from which all other empirical formula calculations are made. It's how chemists keep track of the elemental make-up of complex mixtures.

Elemental Composition

Elemental composition is a quantification of the distribution of elements within a compound. It's the detailed percentage makeup of each element within a molecule, often expressed in mass or atom count, providing clarity in how compounds are constructed.

• Knowing the elemental composition allows chemists to write empirical formulas, which show the simplest integer ratio of the atoms in a compound.
• For instance, if a compound has an elemental composition by mass of 41.1% nitrogen, 11.8% hydrogen, and the remainder sulfur, it provides clear instruction on what elements are present and in what relative quantities.
• This information can also be used to infer physical and chemical properties of the compound since elemental composition dictates bonding and interactions.

Elemental composition not only simplifies the empirical formula process but also enables more profound insights into compound characteristics and reactivity. By understanding which elements and how much of each are present, scientists and students can make informed predictions about a compound's behavior and use this knowledge practically in chemical analysis and synthesis.