Question

A sample of urea contains \(1.121 \mathrm{~g} \mathrm{~N}, 0.161 \mathrm{~g} \mathrm{H}, 0.480 \mathrm{~g} \mathrm{C}\), and \(0.640 \mathrm{~g} \mathrm{O}\). What is the empirical formula of urea?

Short answer

The empirical formula of urea is \(NH_2CO\), which consists of 2 nitrogen atoms, 4 hydrogen atoms, 1 carbon atom, and 1 oxygen atom in the simplest whole-number ratio.

Step-by-step solution

Convert masses to moles

To convert the mass of each element to moles, we'll need the atomic weights of nitrogen (N), hydrogen (H), carbon (C), and oxygen (O). Using the periodic table, we can find their atomic weights: - N: 14.01 g/mol - H: 1.01 g/mol - C: 12.01 g/mol - O: 16.00 g/mol Now, divide the given masses by their respective atomic weights to get the number of moles: moles of N = 1.121 g N / 14.01 g/mol = 0.0800 moles moles of H = 0.161 g H / 1.01 g/mol = 0.159 moles moles of C = 0.480 g C / 12.01 g/mol = 0.0400 moles moles of O = 0.640 g O / 16.00 g/mol = 0.0400 moles

Determine the mole ratio

Now, divide each number of moles by the smallest number (which is 0.0400 moles in this case) to get the simplest whole-number ratio: N: 0.0800 / 0.0400 = 2 H: 0.159 / 0.0400 = 3.98, which can be rounded to 4 C: 0.0400 / 0.0400 = 1 O: 0.0400 / 0.0400 = 1

Write the empirical formula

Based on the obtained mole ratio, the empirical formula is: \( NH_2 CO \) To clarify, there are 2 nitrogen atoms, 4 hydrogen atoms, 1 carbon atom, and 1 oxygen atom in the simplest whole-number ratio.