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The first sulfur-nitrogen compound was prepared in 1835 and has been used to synthesize many others. In the early \(1980 \mathrm{~s}\) researchers made another such compound that conducts electricity like a metal. Mass spectrometry of the compound shows a molar mass of \(184.27 \mathrm{~g} / \mathrm{mol}\), and analysis shows it to contain 2.288 g of \(\mathrm{S}\) for every \(1.000 \mathrm{~g}\) of \(\mathrm{N}\). What is its molecular formula?

Short Answer

Expert verified
The molecular formula is \( \text{S}_4 \text{N}_4 \).

Step by step solution

01

Determine the mass of each element

According to the problem, 2.288 g of sulfur (S) and 1.000 g of nitrogen (N) are given. Start by calculating the molar amounts of each.
02

Calculate moles of sulfur

The atomic mass of sulfur (S) is approximately 32.06 g/mol.Calculate the moles of sulfur: \[ \text{Moles of S} = \frac{2.288 \text{ g}}{32.06 \text{ g/mol}} \ \text{Moles of S} = 0.0714 \text{ mol} \ \text{(4 significant figures)} \]
03

Calculate moles of nitrogen

The atomic mass of nitrogen (N) is approximately 14.01 g/mol.Calculate the moles of nitrogen: \[ \text{Moles of N} = \frac{1.000 \text{ g}}{14.01 \text{ g/mol}} \ \text{Moles of N} = 0.0714 \text{ mol} \ \text{(4 significant figures)} \]
04

Determine the empirical formula

Since the moles of sulfur and nitrogen are in a 1:1 ratio:The empirical formula is \( \text{SN} \).
05

Determine the molecular formula

Calculate the molar mass of the empirical formula (S = 32.06 g/mol, N = 14.01 g/mol):\[ \text{Empirical Formula Molar Mass} = 32.06 + 14.01 = 46.07 \text{ g/mol} \]Now, use the formula to determine the molecular formula:\[ \text{Molecular Formula} = n \times \text{Empirical Formula} \ n = \frac{Molar Mass (given)}{Empirical Formula Molar Mass} \ n = \frac{184.27 \text{ g/mol}}{46.07 \text{ g/mol}} \ n = 4 \]Finally, multiply the subscripts in the empirical formula by this value:\[ \text{Molecular Formula} = (\text{SN})_4 = \text{S}_4 \text{N}_4 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

empirical formula
The empirical formula represents the simplest whole-number ratio of the atoms in a compound. Despite its simplicity, it provides crucial information about how elements are combined within a molecule. To find the empirical formula, start with the mass of each element given, convert these masses to moles, and then establish the simplest ratio. In our exercise, we had 2.288 grams of sulfur and 1.000 grams of nitrogen.
By using their atomic masses, we can convert these masses to moles: calculating 2.288 grams of sulfur converts to approximately 0.0714 moles, and 1.000 grams of nitrogen also converts to approximately 0.0714 moles.
Because these values are equal, the ratio of sulfur to nitrogen is 1:1. Thus, the empirical formula for the compound is SN.
molar mass
Molar mass is the mass of one mole of a substance, often measured in grams per mole (g/mol). It is a vital concept for converting between the mass of a substance and the number of moles. In our exercise, the molar mass of the chemical compound was found using mass spectrometry and given as 184.27 g/mol.
To verify the molecular formula using molar mass, take the empirical formula mass (46.07 g/mol for SN) and compare it to the molecular mass. You do this by dividing the given molar mass by the empirical formula mass: \(\frac{184.27 \text{g/mol}}{46.07 \text{g/mol}} = 4\).
This result signifies that the molecular formula contains four times the number of atoms in the empirical formula, leading us to \( \text{S}_4 \text{N}_4 \).
mass spectrometry
Mass spectrometry is an analytical technique used to measure the mass-to-charge ratio of ions. This technique is beneficial in determining the molecular mass of a compound, as it was in our exercise where the molecular mass was found to be 184.27 g/mol.
Mass spectrometry works by ionizing chemical compounds to generate charged molecules or molecule fragments and then measuring these ions with their respective mass-to-charge ratios. The method can precisely pinpoint the molar mass of compounds, crucial for identifying molecular formulas.
For students studying chemical compounds, understanding mass spectrometry can be incredibly beneficial.

Key points include:
  • How compounds ionize and break apart
  • How to read spectrometry data
  • Relating spectrometry data to molecular and empirical formulas
All these make mass spectrometry a powerful tool in modern chemistry.
sulfur-nitrogen compounds
Sulfur-nitrogen compounds are interesting because they exhibit unique chemical and physical properties. They were first synthesized in 1835, and since then, various forms have been discovered. Some of these compounds, like the one highlighted in our exercise, have conductivity properties akin to metals.
The compound in question, with a molecular formula of \(\text{S}_4\text{N}_4 \), illustrates the versatility of sulfur-nitrogen compounds. They serve as synthons in chemical synthesis and have relevance in materials science and electrical engineering.
Studying these compounds involves:
  • Understanding their synthesis and properties
  • Exploring their uses in different fields
  • Appreciating their role in historical and modern chemistry
Whether used for academic purposes or practical applications, sulfur-nitrogen compounds offer a fascinating glimpse into the world of advanced chemical engineering.

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Most popular questions from this chapter

Which of the following sets of information allows you to obtain the molecular formula of a covalent compound? In each case that allows it, explain how you would proceed (draw a road map and write a plan for a solution). (a) Number of moles of each type of atom in a given sample of the compound (b) Mass \% of each element and the total number of atoms in a molecule of the compound (c) Mass \% of each element and the number of atoms of one element in a molecule of the compound (d) Empirical formula and mass \(\%\) of each element (e) Structural formula

Butane gas is compressed and used as a liquid fuel in disposable cigarette lighters and lightweight camping stoves. Suppose a lighter contains \(5.50 \mathrm{~mL}\) of butane \((d=0.579 \mathrm{~g} / \mathrm{mL})\) (a) How many grams of oxygen are needed to burn the butane completely? (b) How many moles of \(\mathrm{H}_{2} \mathrm{O}\) form when all the butane burns? (c) How many total molecules of gas form when the butane burns completely?

Various nitrogen oxides, as well as sulfur oxides, contribute to acidic rainfall through complex reaction sequences. Nitrogen and oxygen combine during the high-temperature combustion of fuels in air to form nitrogen monoxide gas, which reacts with more oxygen to form nitrogen dioxide gas. In contact with water vapor, nitrogen dioxide forms aqueous nitric acid and more nitrogen monoxide. (a) Write balanced equations for these reactions. (b) Use the equations to write one overall balanced equation that does not include nitrogen monoxide and nitrogen dioxide. (c) How many metric tons (t) of nitric acid form when \(1350 \mathrm{t}\) of atmospheric nitrogen is consumed \((1 \mathrm{t}=1000 \mathrm{~kg}) ?\)

Two successive reactions, \(\mathrm{D} \longrightarrow \mathrm{E}\) and \(\mathrm{E} \longrightarrow \mathrm{F},\) have yields of \(48 \%\) and \(73 \%\), respectively. What is the overall percent yield for conversion of \(\mathrm{D}\) to \(\mathrm{F} ?\)

Isobutylene is a hydrocarbon used in the manufacture of synthetic rubber. When \(0.847 \mathrm{~g}\) of isobutylene was subjected to combustion analysis, the gain in mass of the \(\mathrm{CO}_{2}\) absorber was \(2.657 \mathrm{~g}\) and that of the \(\mathrm{H}_{2} \mathrm{O}\) absorber was \(1.089 \mathrm{~g} .\) What is the empirical formula of isobutylene?

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