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Magazine Chapter 22: Problem 151
Niobium reacts with fluorine at room temperature to give a solid binary compound that is \(49.44 \% \mathrm{Nb}\) by mass. (a) What is the empirical formula of the compound? (b) Write a balanced equation for the reaction. (c) The compound reacts with hydrogen to regenerate metallic niobium. Write a balanced equation for the reaction.
Short Answer
Expert verified
(a) \( \text{NbF}_5 \); (b) \( \text{Nb} + \frac{5}{2} \text{F}_2 \rightarrow \text{NbF}_5 \); (c) \( \text{NbF}_5 + \frac{5}{2} \text{H}_2 \rightarrow \text{Nb} + 5 \text{HF} \).
Step by step solution
01
Determine Moles of Niobium and Fluorine
To find the empirical formula, we need to calculate the moles of niobium (Nb) and fluorine (F) present in the compound. Start by assuming a 100 g sample of the compound. This means there are 49.44 g of niobium in the compound. The molar mass of niobium is 92.91 g/mol, so the moles of niobium are \( \frac{49.44 \text{ g}}{92.91 \text{ g/mol}} \approx 0.532 \text{ moles} \). The rest of the compound is fluorine, which is 100 g - 49.44 g = 50.56 g. The molar mass of fluorine is 19.00 g/mol, so the moles of fluorine are \( \frac{50.56 \text{ g}}{19.00 \text{ g/mol}} \approx 2.66 \text{ moles} \).
02
Determine the Simplest Mole Ratio
Next, we need to find the simplest whole number ratio of moles of niobium to moles of fluorine. Divide each by the smallest number of moles calculated in Step 1: Nb: \( \frac{0.532}{0.532} = 1 \) and F: \( \frac{2.66}{0.532} \approx 5 \). The simplest ratio of niobium to fluorine is approximately 1:5.
03
Write the Empirical Formula
Using the simplest whole number ratio of elements derived in Step 2, we can write the empirical formula of the compound as \( \text{NbF}_5 \).
04
Write a Balanced Equation for the Reaction with Fluorine
Niobium reacts with fluorine gas to form niobium pentafluoride. The balanced chemical equation for this reaction is \( \text{Nb}_{(s)} + \frac{5}{2} \text{F}_{2(g)} \rightarrow \text{NbF}_{5(s)} \).
05
Write a Balanced Equation for the Reaction with Hydrogen
In the reaction with hydrogen, niobium pentafluoride reacts to produce niobium metal and hydrogen fluoride. The balanced chemical equation is \( \text{NbF}_{5(s)} + \frac{5}{2} \text{H}_{2(g)} \rightarrow \text{Nb}_{(s)} + 5 \text{HF}_{(g)} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Empirical Formula
To understand the empirical formula, let's start with a basic definition. The empirical formula of a compound represents the simplest whole number ratio of the elements present in the compound. It does not provide information about the actual number of atoms but rather the relative proportions of each element.
In the original exercise, we determined the empirical formula of a niobium-fluorine compound. Here's how we did it:
In the original exercise, we determined the empirical formula of a niobium-fluorine compound. Here's how we did it:
- Assumed a 100 g sample of the compound, which has 49.44 g of niobium and 50.56 g of fluorine.
- Converted these masses to moles using their respective molar masses: 92.91 g/mol for niobium and 19.00 g/mol for fluorine.
- Calculated the moles of niobium as approximately 0.532 and fluorine as approximately 2.66.
- Divided the moles of each element by the smallest number of moles to find the simplest ratio: nb = 1, F = 5.
Molar Mass
Molar mass is an essential concept in chemistry when you calculate quantitative relationships. It refers to the mass of one mole of a substance and is expressed in grams per mole (g/mol). Molar mass enables us to convert between grams and moles, offering a bridge to predict and analyze chemical reactions.
In the niobium-fluorine compound exercise, we used molar mass for both elements to find the ratios necessary for deducing the empirical formula. A few key steps include:
In the niobium-fluorine compound exercise, we used molar mass for both elements to find the ratios necessary for deducing the empirical formula. A few key steps include:
- The molar mass of niobium is 92.91 g/mol, while that of fluorine is 19.00 g/mol.
- These values allow us to convert from grams to moles efficiently as they express how much one mole of each element will weigh.
- For example, with 49.44 g of niobium, calculating the moles involved dividing by its molar mass, giving us approximately 0.532 moles.
Balancing Chemical Equations
Balancing chemical equations is a foundational skill in chemistry. It ensures that the same amount of each element is present on both sides of the equation, reflecting the law of conservation of mass. Each chemical reaction must be balanced because matter is not created or destroyed in chemical processes.
In the given exercise, two reactions required balancing:
In the given exercise, two reactions required balancing:
- The reaction of niobium with fluorine to form niobium pentafluoride is expressed as \( \text{Nb}_{(s)} + \frac{5}{2} \text{F}_{2(g)} \rightarrow \text{NbF}_{5(s)} \). Although it can appear balanced, to adhere to conventional equations, you can multiply through by 2 to avoid fractional coefficients, resulting in \( 2\text{Nb}_{(s)} + 5\text{F}_{2(g)} \rightarrow 2\text{NbF}_{5(s)} \).
- In the reaction of niobium pentafluoride with hydrogen, the equation is \( \text{NbF}_{5(s)} + \frac{5}{2} \text{H}_{2(g)} \rightarrow \text{Nb}_{(s)} + 5 \text{HF}_{(g)} \). Similarly, multiplying through by 2 would yield integer coefficients.
Stoichiometry
Stoichiometry deals with the quantitative relationship between reactants and products in a chemical reaction. It uses balanced equations to understand how much of each reactant is needed and what will be produced. It combines concepts like molar mass, mole ratios, and balanced equations to solve practical problems in chemistry.
The two balanced reactions discussed in the exercise illustrate stoichiometry in action. Consider these aspects:
The two balanced reactions discussed in the exercise illustrate stoichiometry in action. Consider these aspects:
- The empirical formula and balanced equations allow calculation of the molar ratios: for instance, in producing \( \text{NbF}_5 \), 1 mole of niobium reacts with 2.5 moles of \( \text{F}_2 \) according to the fractional equation.
- Conversions between masses and moles, critical in stoichiometry, align with the molar mass calculations we covered. For instance, knowing how much \( \text{NbF}_5 \) will form when a certain mass of niobium is reacted requires these stoichiometric relationships.
