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Chapter 3: Problem 84

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

Expert verified
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

01

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
02

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
03

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.

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Most popular questions from this chapter

A given sample of a xenon fluoride compound contains molecules of the type \(\mathrm{XeF}_{n}\), where \(n\) is some whole number. Given that \(9.03 \times 10^{20}\) molecules of \(\mathrm{XeF}_{\text {a }}\) weigh \(0.368 \mathrm{~g}\), determine the value for \(n\) in the formula.

Arrange the following substances in order of increasing mass percent of carbon. a. caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\) b. sucrose, \(\bar{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) c. ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)

Calculate the percent composition by mass of the following compounds that are important starting materials for synthetic polymers: a. \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{2}\) (acrylic acid, from which acrylic plastics are made) b. \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{2}\) (methyl acrylate, from which Plexiglas is made) c. \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}\) (acrylonitrile, from which Orlon is made)

A compound contains only carbon, hydrogen, and oxygen. Combustion of \(10.68 \mathrm{mg}\) of the compound yields \(16.01 \mathrm{mg}\) \(\mathrm{CO}_{2}\) and \(4.37 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\). The molar mass of the compound is \(176.1 \mathrm{~g} / \mathrm{mol}\). What are the empirical and molecular formulas of the compound?

A binary compound between an unknown element \(\mathrm{E}\) and hydrogen contains \(91.27 \% \mathrm{E}\) and \(8.73 \% \mathrm{H}\) by mass. If the formula of the compound is \(\mathrm{E}_{3} \mathrm{H}_{8}\), calculate the atomic mass of \(\mathrm{E}\).
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