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Chapter 13: Problem 87

A certain acid has a molecular mass of 63 u. By mass, it consists of \(1.6 \%\) hydrogen, \(22.2 \%\) nitrogen, and \(76.2 \%\) oxygen. What is the chemical formula for this acid?

Short Answer

Expert verified
Answer: The chemical formula of the acid is HNO3.

Step by step solution

01

Determine the moles of each element

Assume that we have 100 g of the acid, as this will make it easier to work with the given percentages. In this 100 g of the acid, we have: - \(1.6 \%\) hydrogen, which is equivalent to 1.6 g of hydrogen - \(22.2 \%\) nitrogen, which is equivalent to 22.2 g of nitrogen - \(76.2 \%\) oxygen, which is equivalent to 76.2 g of oxygen To determine the moles of each element, we will use the molar mass of each element: (1 g/mol for hydrogen, 14 g/mol for nitrogen, and 16 g/mol for oxygen) and divide the mass by the molar mass: - Moles of hydrogen: \(\frac{1.6}{1} = 1.6\) mol - Moles of nitrogen: \(\frac{22.2}{14} = 1.586\) mol - Moles of oxygen: \(\frac{76.2}{16} = 4.763\) mol
02

Find the empirical formula of the acid

To determine the empirical formula, we need to find the mole ratio of each element. First, we will divide all the mole values by the smallest mole value (1.586 mol of nitrogen in this case) to obtain the mole ratios. - Hydrogen: \(\frac{1.6}{1.586} = 1.009 \approx 1\) - Nitrogen: \(\frac{1.586}{1.586} = 1 \) - Oxygen: \(\frac{4.763}{1.586} = 3.002 \approx 3\) The mole ratios of the elements yield an empirical formula of HNO\(_3\).
03

Calculate the molecular formula of the acid

To calculate the molecular formula, we need to determine the multiple of the empirical formula that is within the given molecular mass. First, calculate the molar mass of the empirical formula. Molar mass of HNO\(_3\): \((1) + (14) + (3 \times 16) = 63\) g/mol Since the molar mass of the empirical formula (HNO\(_3\), 63 g/mol) matches the given molecular mass (63 u), there is no need to find a whole number multiple for the empirical formula. Therefore, the chemical formula for this acid is HNO\(_3\).

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