Question

Using the average atomic masses given inside the front cover of this book, calculate the mass in grams of each of the following samples. a. 0.341 mole of potassium nitride b. \(2.62 \mathrm{mmol}\) of neon \((1 \mathrm{mmol}=1 / 1000 \mathrm{~mol})\) c. 0.00449 mole of manganese(II) oxide d. \(7.18 \times 10^{5}\) moles of silicon dioxide e. 0.000121 mole of iron( III) phosphate

Short answer

a. 44.79 g of potassium nitride b. 0.05287 g of neon c. 0.31867 g of manganese(II) oxide d. \(4.315 \times 10^{7}\) g of silicon dioxide e. 0.01825 g of iron(III) phosphate

Step-by-step solution

Find the molar mass of potassium nitride

Determine the molar mass of potassium nitride (K₃N). This can be found by adding the molar masses of the constituent elements. Molar mass of K = 39.10 g/mol Molar mass of N = 14.01 g/mol Molar mass of K₃N = 3 × Molar mass of K + Molar mass of N = 3 × 39.10 g/mol + 14.01 g/mol = 131.31 g/mol

Calculate the mass of the potassium nitride

Now multiply the given moles (0.341) by the molar mass (131.31 g/mol) to find the mass in grams. Mass = 0.341 mol × 131.31 g/mol = 44.79 g The mass of 0.341 mole of potassium nitride is 44.79 grams. b. 2.62 mmol of neon (\(1 mmol = 1/1000 mol\))

Convert mmol to mol

We are given the number of moles in mmol, so we need to convert it to mol first: 2.62 mmol × (1 mol / 1000 mmol) = 0.00262 mol

Find the molar mass of neon

The molar mass of neon (Ne) is found to be 20.18 g/mol.

Calculate the mass of the neon sample

Now multiply the number of moles (0.00262 mol) by the molar mass (20.18 g/mol) to find the mass in grams. Mass = 0.00262 mol × 20.18 g/mol = 0.05287 g The mass of 2.62 mmol of neon is 0.05287 grams. c. 0.00449 mole of manganese(II) oxide

Find the molar mass of manganese(II) oxide

We need to find the molar mass of manganese(II) oxide (MnO). This can be found by adding the molar masses of the constituent elements. Molar mass of Mn = 54.94 g/mol Molar mass of O = 16.00 g/mol Molar mass of MnO = Molar mass of Mn + Molar mass of O = 54.94 g/mol + 16.00 g/mol = 70.94 g/mol

Calculate the mass of the manganese(II) oxide sample

Now multiply the given moles (0.00449) by the molar mass (70.94 g/mol) to find the mass in grams. Mass = 0.00449 mol × 70.94 g/mol = 0.31867 g The mass of 0.00449 mole of manganese(II) oxide is 0.31867 grams. d. \(7.18 \times 10^{5}\) moles of silicon dioxide

Find the molar mass of silicon dioxide

We need to find the molar mass of silicon dioxide (SiO₂). This can be found by adding the molar masses of the constituent elements. Molar mass of Si = 28.09 g/mol Molar mass of O = 16.00 g/mol Molar mass of SiO₂ = Molar mass of Si + 2 × Molar mass of O = 28.09 g/mol + 2 × 16.00 g/mol = 60.09 g/mol

Calculate the mass of the silicon dioxide sample

Now multiply the given moles (\(7.18 \times 10^{5}\)) by the molar mass (60.09 g/mol) to find the mass in grams. Mass = \(7.18 \times 10^{5}\) mol × 60.09 g/mol = \(4.315 \times 10^{7}\) g The mass of \(7.18 \times 10^{5}\) moles of silicon dioxide is \(4.315 \times 10^{7}\) grams. e. 0.000121 mole of iron(III) phosphate

Find the molar mass of iron(III) phosphate

We need to find the molar mass of iron(III) phosphate (FePO₄). This can be found by adding the molar masses of the constituent elements. Molar mass of Fe = 55.85 g/mol Molar mass of P = 30.97 g/mol Molar mass of O = 16.00 g/mol Molar mass of FePO₄ = Molar mass of Fe + Molar mass of P + 4 × Molar mass of O = 55.85 g/mol + 30.97 g/mol + 4 × 16.00 g/mol = 150.82 g/mol

Calculate the mass of the iron(III) phosphate sample

Now multiply the given moles (0.000121) by the molar mass (150.82 g/mol) to find the mass in grams. Mass = 0.000121 mol × 150.82 g/mol = 0.01825 g The mass of 0.000121 mole of iron(III) phosphate is 0.01825 grams.

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships among reactants and products in chemical reactions. In essence, it provides the calculations for predicting how much of a product will form from a given amount of reactant or how much reactant is needed to create a certain amount of product.

For students to master stoichiometry, understanding the mole concept and molar masses is essential, as these are the foundational tools used in stoichiometric calculations. As shown in the exercise, stoichiometry is not just about balancing equations but also about converting quantities, such as the number of moles, to measurable weights in grams using the molar mass of the substances involved.

Mole Concept

The mole concept is central to understanding stoichiometry and chemical equations. One mole is defined as the amount of any substance that contains the same number of entities, such as atoms, molecules, or ions, as there are atoms in 12 grams of carbon-12. This number is known as Avogadro's constant and is approximately equal to \(6.022 \times 10^{23}\).

Whether dealing with moles of a pure element or a compound, the mole concept allows chemists to count atoms and molecules in a practical way by weighing them. For example, in part a of the given exercise, the problem starts by providing the quantity of potassium nitride in moles, which must then be converted to grams. Such conversions hinge on the mole concept and utilize the molar mass of the substance in question.

Atomic Masses

Atomic masses represent the mass of individual atoms and are expressed in atomic mass units (amu). However, for bulk chemical quantities, it's more practical to use grams. Molar mass bridges this gap, converting atomic mass units to grams per mole. The molar mass is numerically equal to the atomic mass but scaled up to grams for one mole of atoms. In other words, the molar mass of an element in grams per mole is numerically equivalent to its atomic mass in amu.

Understanding how to calculate molar mass is crucial, as illustrated in the step-by-step solutions to our problem. Students should remember that the molar mass of a compound is the sum of the molar masses of its constituent elements, each proportionate to the number of atoms of that element in the formula unit. By accurately determining the molar mass, as done for substances like silicon dioxide and iron(III) phosphate in the exercise, you can then convert moles to grams in order to solve stoichiometric problems.