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Ch.22 - The Main Group Elements
Chapter 22, Problem 149a,b

Niobium reacts with fluorine at room temperature to give a solid binary compound that is 49.44% Nb by mass. (a) What is the empirical formula of the compound? (b) Write a balanced equation for the reaction.

Verified step by step guidance
1
Step 1: Determine the mass percentage of fluorine in the compound. Since the compound is 49.44% niobium (Nb) by mass, the percentage of fluorine (F) is 100% - 49.44% = 50.56%.
Step 2: Convert the mass percentages to moles. Assume you have 100 grams of the compound. This means you have 49.44 grams of Nb and 50.56 grams of F. Use the molar masses: Nb (92.91 g/mol) and F (19.00 g/mol) to convert grams to moles.
Step 3: Calculate the mole ratio of Nb to F. Divide the number of moles of each element by the smallest number of moles calculated in the previous step to find the simplest whole number ratio.
Step 4: Write the empirical formula based on the mole ratio. The empirical formula is the simplest whole number ratio of atoms in the compound.
Step 5: Write a balanced chemical equation for the reaction of niobium with fluorine. Use the empirical formula to determine the product and balance the equation by ensuring the number of atoms of each element is the same on both sides of the equation.