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Chapter 3: Problem 66

Dimethylnitrosamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}_{2} \mathrm{O}\), is a carcinogenic (cancercausing) substance that may be formed in foods, beverages, or gastric juices from the reaction of nitrite ion (used as a food preservative) with other substances. a. What is the molar mass of dimethylnitrosamine? b. How many moles of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}_{2} \mathrm{O}\) molecules are present in \(250 \mathrm{mg}\) dimethylnitrosamine? c. What is the mass of \(0.050\) mol dimethylnitrosamine? d. How many atoms of hydrogen are in \(1.0 \mathrm{~mol}\) dimethylnitrosamine? e. What is the mass of \(1.0 \times 10^{6}\) molecules of dimethylnitrosamine? f. What is the mass in grams of one molecule of dimethylnitrosamine?

Short Answer

Expert verified
a. The molar mass of dimethylnitrosamine is 74.14 g/mol. b. There are 3.37 x 10^(-3) moles of dimethylnitrosamine in 250 mg. c. The mass of 0.050 mol dimethylnitrosamine is 3.71 g. d. There are 3.613 x 10^(24) hydrogen atoms in 1.0 mol of dimethylnitrosamine. e. The mass of 1.0 x 10^6 molecules of dimethylnitrosamine is 1.23 x 10^(-16) g. f. The mass of one molecule of dimethylnitrosamine is 1.23 x 10^(-22) g.

Step by step solution

01

Determine chemical formula and atomic masses

The molecular formula for dimethylnitrosamine is \((CH_3)_2N_2O\). The atomic masses (in g/mol) are: - Carbon (C): 12.01 - Hydrogen (H): 1.01 - Nitrogen (N): 14.01 - Oxygen (O): 16.00
02

Calculate molar mass

To find the molar mass of dimethylnitrosamine, we'll add the atomic masses of each atom considering the number of atoms: Molar mass = 2(12.01) + 6(1.01) + 2(14.01) + 1(16.00) = 74.14 g/mol. The molar mass of dimethylnitrosamine is 74.14 g/mol. #b. Calculate moles of dimethylnitrosamine in 250 mg#
03

Convert mass to grams

First, convert the mass of dimethylnitrosamine to grams: 250 mg = 0.250 g.
04

Calculate moles from mass and molar mass

To calculate the moles: Moles = (Mass) / (Molar mass) = 0.250 g / 74.14 g/mol = 3.37 x 10^(-3) moles. There are 3.37 x 10^(-3) moles of dimethylnitrosamine in 250 mg. #c. Calculate the mass of 0.050 mol dimethylnitrosamine#
05

Calculate mass from moles and molar mass

To calculate the mass: Mass = (Moles) x (Molar mass) = 0.050 mol x 74.14 g/mol = 3.71 g. The mass of 0.050 mol dimethylnitrosamine is 3.71 g. #d. Calculate the number of hydrogen atoms in 1.0 mol dimethylnitrosamine#
06

Calculate the hydrogen atoms in one molecule

In one molecule of dimethylnitrosamine, there are six hydrogen atoms (two methyl groups, each having three hydrogens).
07

Calculate the total hydrogen atoms

Since there are 6 hydrogen atoms per molecule and 1.0 mol of dimethylnitrosamine: Total hydrogen atoms = 6 (hydrogen atoms) x 1.0 (moles) x \(6.022 \times 10^{23}\) (Avogadro's number). Total hydrogen atoms = 3.613 x 10^(24) hydrogen atoms. There are 3.613 x 10^(24) hydrogen atoms in 1.0 mol of dimethylnitrosamine. #e. Calculate the mass of 1.0 x 10^6 molecules of dimethylnitrosamine#
08

Calculate moles from number of molecules

To calculate moles from number of molecules, use Avogadro's number: Moles = \(1.0 \times 10^{6}\) molecules / \(6.022 \times 10^{23}\) molecules/mol = 1.66 x 10^(-18) moles.
09

Calculate mass from moles and molar mass

To calculate the mass: Mass = (Moles) x (Molar mass) = 1.66 x 10^(-18) moles x 74.14 g/mol = 1.23 x 10^(-16) g. The mass of 1.0 x 10^6 molecules of dimethylnitrosamine is 1.23 x 10^(-16) g. #f. Calculate the mass in grams of one molecule of dimethylnitrosamine#
10

Calculate moles from number of molecules

To calculate moles for one molecule, use Avogadro's number: Moles = 1 molecule / \(6.022 \times 10^{23}\) molecules/mol = 1.66 x 10^(-24) moles.
11

Calculate mass from moles and molar mass

To calculate the mass: Mass = (Moles) x (Molar mass) = 1.66 x 10^(-24) moles x 74.14 g/mol = 1.23 x 10^(-22) g. The mass of one molecule of dimethylnitrosamine is 1.23 x 10^(-22) g.

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Key Concepts

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

Understanding the Mole Concept
To fully grasp chemistry, it's essential to understand the mole concept, which is a fundamental unit in chemistry representing a specific number of particles, atoms, or molecules. Essentially, the mole is a bridge between the microscopic world of atoms and molecules and the macroscopic world of grams and liters that we can measure in the lab.

The mole concept is integral for quantifying substances in chemical reactions. One mole of any substance contains exactly the same number of particles as there are atoms in exactly 12 grams of carbon-12. This number is Avogadro's number, which is approximately \(6.022 \times 10^{23}\).

When asked about the mass of moles or the number of particles in a substance, you'll be using the mole concept to convert between mass, moles, and particle count. In the textbook exercise, for example, you calculated the number of moles in a given mass of dimethylnitrosamine (\

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Most popular questions from this chapter

Reference Section \(3.2\) to find the atomic masses of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\), the relative abundance of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) in natural carbon, and the average mass (in amu) of a carbon atom. If you had a sample of natural carbon containing exactly 10,000 atoms, determine the number of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) atoms present. What would be the average mass (in amu) and the total mass (in amu) of the carbon atoms in this 10,000 -atom sample? If you had a sample of natural carbon containing \(6.0221 \times 10^{25}\) atoms, determine the number of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) atoms present. What would be the average mass (in amu) and the total mass (in amu) of this \(6.0221 \times 10^{23}\) atom sample? Given that \(1 \mathrm{~g}=6.0221 \times 10^{23} \mathrm{amu}\), what is the total mass of \(1 \mathrm{~mol}\) of natural carbon in units of grams?

Adipic acid is an organic compound composed of \(49.31 \% \mathrm{C}\). \(43.79 \% \mathrm{O}\), and the rest hydrogen. If the molar mass of adipic acid is \(146.1 \mathrm{~g} / \mathrm{mol}\), what are the empirical and molecular formulas for adipic acid?

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: $$ 2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If \(5.00 \times 10^{3} \mathrm{~kg}\) each of \(\mathrm{NH}_{3}, \mathrm{O}_{2}\), and \(\mathrm{CH}_{4}\) are reacted, what mass of \(\mathrm{HCN}\) and of \(\mathrm{H}_{2} \mathrm{O}\) will be produced, assuming \(100 \%\) yield?

Phosphorus can be prepared from calcium phosphate by the following reaction: \(2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{SiO}_{2}(s)+10 \mathrm{C}(s) \longrightarrow\) $$ 6 \mathrm{CaSiO}_{3}(s)+\mathrm{P}_{4}(s)+10 \mathrm{CO}(g) $$ Phosphorite is a mineral that contains \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) plus other nonphosphorus- containing compounds. What is the maximum amount of \(\mathrm{P}_{4}\) that can be produced from \(1.0 \mathrm{~kg}\) of phosphorite if the phorphorite sample is \(75 \% \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) by mass? Assume an excess of the other reactants.

The Freons are a class of compounds containing carbon, chlorine, and fluorine. While they have many valuable uses, they have been shown to be responsible for depletion of the ozone in the upper atmosphere. In 1991 , two replacement compounds for Freons went into production: \(\mathrm{HFC}-134 \mathrm{a}\left(\mathrm{CH}_{2} \mathrm{FCF}_{3}\right)\) and \(\mathrm{HCFC}-124\) (CHCIFCF \(_{3}\) ). Calculate the molar masses of these two compounds.
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