Question

Calculate the following quantities: (a) mass, in grams, of 0.105 mol sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) (b) moles of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) in \(143.50 \mathrm{~g}\) of this substance (c) number of molecules in \(1.0 \times 10^{-6} \mathrm{~mol} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (d) number of \(\mathrm{N}\) atoms in \(0.410 \mathrm{~mol} \mathrm{NH}_{3}\)

Short answer

(a) The mass of 0.105 mol of sucrose (C12H22O11) is 35.95 g. (b) There are 0.7574 moles of Zn(NO3)2 in 143.50 g of the substance. (c) There are 6.022 x 10^17 molecules in 1.0 x 10^-6 mol of CH3CH2OH. (d) There are 2.47 x 10^23 N atoms in 0.410 mol of NH3.

Step-by-step solution

a) Mass of 0.105 mol of sucrose in grams

First, we need to find the molar mass of sucrose (C12H22O11). The molar mass of each element can be found on the periodic table: C = 12.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol. Molar mass of sucrose: C12H22O11 = 12 * 12.01 g/mol + 22 * 1.01 g/mol + 11 * 16.00 g/mol = 144.12 g/mol + 22.22 g/mol + 176.00 g/mol = 342.34 g/mol Now, we will use the relationship between moles, mass, and molar mass: Mass = Moles × Molar mass Mass of 0.105 mol of sucrose = 0.105 mol × 342.34 g/mol = 35.95 g So, the mass of 0.105 mol of sucrose is 35.95 g.

b) Moles of Zn(NO3)2 in 143.50 g of this substance

First, we need to find the molar mass of Zn(NO3)2: Molar mass of Zn(NO3)2 = (Zn) + 2 * (N + 3 * O) Molar masses: Zn = 65.38 g/mol, N = 14.01 g/mol, O = 16.00 g/mol Molar mass of Zn(NO3)2 = 65.38 g/mol + 2 * (14.01 g/mol + 3 * 16.00 g/mol) = 65.38 g/mol + 2 * (62.01 g/mol) = 189.40 g/mol Now, we will use the relationship between moles, mass, and molar mass, with the mass of Zn(NO3)2 given: Moles = Mass ÷ Molar mass Moles of Zn(NO3)2 = 143.50 g ÷ 189.40 g/mol = 0.7574 mol So, there are 0.7574 moles of Zn(NO3)2 in 143.50 g of the substance.

c) Number of molecules in 1.0 x 10^-6 mol of CH3CH2OH

To find the number of molecules, we will use Avogadro's number, where 1 mole of any substance contains 6.022 x 10^23 particles (atoms, ions, or molecules). Number of molecules = Moles × Avogadro's number Number of molecules in 1.0 x 10^-6 mol of CH3CH2OH = 1.0 x 10^-6 mol × 6.022 x 10^23 molecules/mol = 6.022 x 10^17 molecules So, there are 6.022 x 10^17 molecules in 1.0 x 10^-6 mol of CH3CH2OH.

d) Number of N atoms in 0.410 mol of NH3

First, note that there is 1 N atom in each NH3 molecule. So, to find the number of N atoms, we can directly use Avogadro's number. Number of N atoms = Moles of NH3 × Avogadro's number Number of N atoms in 0.410 mol of NH3 = 0.410 mol × 6.022 x 10^23 atoms/mol = 2.47 x 10^23 N atoms So, there are 2.47 x 10^23 N atoms in 0.410 mol of NH3.

Key concepts

Molar Mass

Understanding molar mass is crucial for converting moles of a substance to grams, which is often needed in chemical calculations. Molar mass is the mass of one mole of a substance (measured in grams per mole) and corresponds to the average weight of the atoms within a chemical element or compound. This value is derived from the sum of the atomic masses of all atoms in a molecular formula.

For example, to find the molar mass of sucrose (C\(_{12}\)H\(_{22}\)O\(_{11}\)), you refer to the periodic table for atomic masses: carbon (C) is 12.01 g/mol, hydrogen (H) is 1.01 g/mol, and oxygen (O) is 16.00 g/mol. The calculation is as follows: 12*(12.01 g/mol) + 22*(1.01 g/mol) + 11*(16.00 g/mol) = 342.34 g/mol. This molar mass allows us to determine the mass of any number of moles of sucrose, making it an essential component of stoichiometry.

Simply put, without knowing the molar mass, we cannot accurately convert between the mass and number of moles of a substance.

Avogadro's Number

Avogadro's number is a fundamental constant in chemistry, representing the number of atoms, ions, or molecules in one mole of a substance. The value of Avogadro's number is 6.022 \(\times 10^{23}\) particles per mole. This number makes it possible to count particles in moles, thereby bridging the gap between the atomic and macroscopic world.

For instance, in exercise part (c) involving ethanol (CH\(_3\)CH\(_2\)OH), when given moles of the substance, Avogadro's number helps calculate the actual count of molecules present. If you have 1.0 \(\times 10^{-6}\) mol of ethanol, multiplying this by Avogadro's number (6.022 \(\times 10^{23}\) molecules/mol) gives the total number of ethanol molecules: 6.022 \(\times 10^{17}\).

This method is inherently valuable in chemical equations and reactions, allowing chemists to predict the outcomes of reactions based on the number of molecules or atoms involved.

Moles and Molecules

The concept of moles is fundamental in chemistry, serving as a bridge between atomic-scale and real-world measurements. One mole is defined as an amount of any substance that contains as many particles (atoms, molecules, ions) as there are atoms in exactly 12 grams of carbon-12. This correlates to Avogadro's number, which is 6.022 \(\times 10^{23}\) particles.

When dealing with molecules, such as in the calculation of ethanol (CH\(_3\)CH\(_2\)OH) molecules in the previous exercise, chemists often need to convert the mass of a substance to moles before finding the number of molecules. In the exercise, by knowing we have 1.0 \(\times 10^{-6}\) moles of ethanol, and using Avogadro's number, we can find the exact number of molecules present.

This framework for conversion is essential for practical lab work, ensuring that reactions are carried out with precise amounts of each reactant, and helps prevent waste or dangerous chemical reactions by using incorrect ratios.

Chemical Calculations

Chemical calculations are crucial for understanding and predicting the outcomes of chemical reactions. They often involve conversions between mass, moles, and the number of molecules or atoms. These calculations rely heavily on concepts like molar mass and Avogadro's number.

Take for instance calculating the mass from moles as demonstrated in part (a) of the exercise. With a known amount of moles and molar mass, like 0.105 moles of sucrose, multiplying by the molar mass (342.34 g/mol) gives the mass in grams. Alternatively, converting grams to moles, as in example (b) with Zn(NO\(_3\)\(_2\)), involves dividing the given mass by the molar mass.

This systematic approach simplifies the assessment of quantities involved in a reaction and allows for scaling of reactions by adjusting the initial amounts of substances present. Mastery of these calculations enables a deeper understanding of reaction mechanics and the stoichiometry that dictates the relationship among reactants and products.