Urea \(\left[\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\right]\) is used for fertilizer and many other things. Calculate the number of \(\mathrm{N}, \mathrm{C}, \mathrm{O},\) and \(\mathrm{H}\) atoms in \(1.68 \times 10^{4} \mathrm{~g}\) of urea.

Short Answer

Expert verified
The number of Nitrogen atoms is \(3.36 * 10^{27}\), the number of Hydrogen atoms is \(6.74 * 10^{27}\), and the number of each Carbon and Oxygen atoms is \(1.68 * 10^{27}\).

Step by step solution

01

Determine the molar mass of urea.

The molar mass of urea can be calculated by summing the molar masses of its constituent atoms. Each urea molecule consists of 2 nitrogen (N) atoms, 4 hydrogen (H) atoms, 1 carbon (C) atom, and 1 oxygen (O) atom. Using the molar masses from the periodic table: Nitrogen (14.007 g/mol) * 2 + Hydrogen (1.008 g/mol) * 4 + Carbon (12.01 g/mol) * 1 + Oxygen (15.999 g/mol) * 1 = 60.056 g/mol.
02

Calculate the number of moles of urea.

The number of moles of urea can be calculated using the formula: moles = mass / molar mass. Given that the mass of urea is 1.68 * 10^4 g and the molar mass is 60.056 g/mol, the number of moles of urea = 1.68 * 10^4 g / 60.056 g/mol = 279.67 moles.
03

Calculate the number of atoms.

We now have the number of moles of urea and the number of atoms of each type per molecule of urea. We can calculate the number of atoms using Avogadro's number (\(6.022 * 10^{23}\) atoms/mole). For nitrogen: 2 atoms/molecule * 279.67 moles * \(6.022 * 10^{23}\) atoms/mole = \(3.36 * 10^{27}\) atoms. Similarly, for hydrogen: 4 atoms/molecule * 279.67 moles * \(6.022 * 10^{23}\) atoms/mole = \(6.74 * 10^{27}\) atoms. For carbon and oxygen: 1 atom/molecule * 279.67 moles * \(6.022 * 10^{23}\) atoms/mole = \(1.68 * 10^{27}\) atoms.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
Understanding molar mass is crucial for performing stoichiometry calculations in chemistry. The molar mass of a substance is the mass of one mole of that substance, typically expressed in grams per mole (g/mol). It is calculated by summing the masses of all the atoms in a molecule of the substance. For instance, the molar mass of urea, \( \left[\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\right] \), is determined by adding the masses of nitrogen (\(2 \times 14.007\) g/mol), hydrogen (\(4 \times 1.008\) g/mol), carbon (\(12.01\) g/mol), and oxygen (\(15.999\) g/mol), resulting in \(60.056\) g/mol.

This value allows us to navigate from the macroscopic scale (grams of the substance) to the microscopic scale (moles of the substance), essential for understanding quantities in chemical reactions.
Avogadro's Number
Avogadro's number, \(6.022 \times 10^{23}\), is a fundamental constant in chemistry that represents the number of entities, such as atoms or molecules, in one mole of a substance. This constant allows chemists to perform moles to atoms conversion, bridging the gap between the macroscopic amounts of substances and the microscopic counting of individual atoms or molecules.

The importance of Avogadro's number in stoichiometry cannot be overstated—it converts theoretical chemical equations into practical quantities that can be measured and used in laboratory settings.
Moles to Atoms Conversion
Moles to atoms conversion is a simple yet vital step in stoichiometry that involves using Avogadro's number to determine the number of atoms in a given number of moles of a substance. For example, if we have \(279.67\) moles of urea, the calculation for the number of nitrogen atoms would be \(2 \times 279.67 \times 6.022 \times 10^{23}\), since each urea molecule contains two nitrogen atoms.

This approach allows us to understand the exact number of atoms or molecules involved in a chemical process and is foundational for predicting the outcomes of reactions.
Chemical Compound Composition
The composition of a chemical compound, such as urea, determines its chemical and physical properties. This composition is expressed in terms of the types and numbers of atoms that make up the molecule. In urea, each molecule consists of two nitrogen, four hydrogen, one carbon, and one oxygen atom.

By knowing the atomic composition and using stoichiometry calculations, we can determine the amounts of each element in a given sample of the compound. This is essential for applications such as creating fertilizers where a specific ratio of elements is necessary for optimum performance.

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Most popular questions from this chapter

Industrially, nitric acid is produced by the Ostwald process represented by the following equations: \(\begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow & \mathrm{HNO}_{3}(a q)+\mathrm{HNO}_{2}(a q) \end{aligned}\) What mass of \(\mathrm{NH}_{3}\) (in g) must be used to produce 1.00 ton of \(\mathrm{HNO}_{3}\) by the above procedure, assuming an 80 percent yield in each step? \((1\) ton \(=2000 \mathrm{lb}\); \(1 \mathrm{lb}=453.6 \mathrm{~g} .)\)

When heated, lithium reacts with nitrogen to form lithium nitride: $$ 6 \mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{Li}_{3} \mathrm{~N}(s) $$ What is the theoretical yield of \(\mathrm{Li}_{3} \mathrm{~N}\) in grams when \(12.3 \mathrm{~g}\) of \(\mathrm{Li}\) are heated with \(33.6 \mathrm{~g}\) of \(\mathrm{N}_{2} ?\) If the actual yield of \(\mathrm{Li}_{3} \mathrm{~N}\) is \(5.89 \mathrm{~g}\), what is the percent yield of the reaction?

Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition can be represented by $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9} \longrightarrow 6 \mathrm{~N}_{2}+12 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} $$ This reaction generates a large amount of heat and many gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. (a) What is the maximum amount of \(\mathrm{O}_{2}\) in grams that can be obtained from \(2.00 \times 10^{2} \mathrm{~g}\) of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of \(\mathrm{O}_{2}\) generated is found to be \(6.55 \mathrm{~g}\).

Tin (Sn) exists in Earth's crust as \(\mathrm{SnO}_{2}\). Calculate the percent composition by mass of \(\mathrm{Sn}\) and \(\mathrm{O}\) in \(\mathrm{SnO}_{2}\).

The anticaking agent added to Morton salt is calcium silicate, \(\mathrm{CaSiO}_{3}\). This compound can absorb up to 2.5 times its mass of water and still remains a freeflowing powder. Calculate the percent composition of \(\mathrm{CaSiO}_{3}\)

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