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. 4.25 mol of oxygen gas, \(\mathrm{O}_{2}\) b. 1.27 millimol of platinum (1 millimol \(=1 / 1000\) mol) c. 0.00101 mol of iron(II) sulfate, \(\mathrm{FeSO}_{4}\) d. 75.1 mol of calcium carbonate, \(\mathrm{CaCO}_{3}\) e. \(1.35 \times 10^{-4}\) mol of gold f. 1.29 mol of hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\) g. 6.14 mol of copper(II) sulfide, Cus

Short answer

The masses of each sample are as follows: a. 136 g of \(\mathrm{O}_2\) b. 0.2476 g of Pt c. 0.15348 g of \(\mathrm{FeSO}_4\) d. 7516.76 g of \(\mathrm{CaCO}_3\) e. 0.0266 g of Au f. 43.86 g of \(\mathrm{H}_2 \mathrm{O}_2\) g. 587.04 g of \(\mathrm{CuS}\)

Step-by-step solution

1. Finding the molecular weight of each compound

Calculate the molecular weight (in g/mol) of each compound by adding the average atomic masses of the involved elements. For instance, for \(\mathrm{CaCO}_3\), multiply the atomic mass of Calcium by 1, Carbon by 1, and Oxygen by 3, and then add all the products together. Now, we find the molecular weights of each compound using their atomic masses: a. \(\mathrm{O}_2\): (16 * 2) = 32 g/mol b. Pt: 195.08 g/mol c. \(\mathrm{FeSO}_4\): (55.85 + 32.06 + 4 * 16) = 151.97 g/mol d. \(\mathrm{CaCO}_3\): (40.08 + 12.01 + 3 * 16) = 100.09 g/mol e. Au: 197.0 g/mol f. \(\mathrm{H}_2 \mathrm{O}_2\): (2 * 1.01 + 2 * 16) = 34.02 g/mol g. \(\mathrm{CuS}\): (63.55 + 32.06) = 95.61 g/mol

2. Converting moles to grams

Now, perform the conversion of moles to grams for each compound/sample by multiplying its moles with its molecular weight: a. 4.25 mol of \(\mathrm{O}_2\): (4.25 * 32) = 136 g b. 1.27 millimol of Pt: (1.27 * 195.08) ÷ 1000 = 0.2476 g c. 0.00101 mol of \(\mathrm{FeSO}_4\): (0.00101 * 151.97) = 0.15348 g d. 75.1 mol of \(\mathrm{CaCO}_3\): (75.1 * 100.09) = 7516.76 g e. \(1.35 \times 10^{-4}\) mol of Au: \((1.35 × 10^{-4} * 197.0)\) = 0.0266 g f. 1.29 mol of \(\mathrm{H}_2 \mathrm{O}_2\): (1.29 * 34.02) = 43.86 g g. 6.14 mol of \(\mathrm{CuS}\): (6.14 * 95.61) = 587.04 g So, the masses of each sample are as follows: a. 136 g of \(\mathrm{O}_2\) b. 0.2476 g of Pt c. 0.15348 g of \(\mathrm{FeSO}_4\) d. 7516.76 g of \(\mathrm{CaCO}_3\) e. 0.0266 g of Au f. 43.86 g of \(\mathrm{H}_2 \mathrm{O}_2\) g. 587.04 g of \(\mathrm{CuS}\)

Key concepts

Molecular Weight Calculation

Understanding the molecular weight calculation is crucial for any chemistry student. Molecular weight, also known as molecular mass, is the sum of the atomic weights of all atoms in a molecule. To calculate it, you refer to the periodic table to find each element's atomic mass and multiply it by the number of times the element occurs in the compound. For example, water (H₂O) has a molecular weight of about 18 g/mol because it is composed of two hydrogen atoms with an atomic mass approximately of 1 g/mol each, plus one oxygen atom with an atomic mass of about 16 g/mol.

Stoichiometry

The term stoichiometry originates from the Greek words 'stoicheion' meaning 'element' and 'metron' meaning 'measure'. In chemistry, stoichiometry relates quantities of reactants and products in a chemical reaction. It is based on the conservation of mass, where the mass of the reactants equals the mass of the products. Stoichiometry uses the mole concept, Avogadro's number, and the balanced chemical equation to predict the amounts of substances consumed and produced in a reaction. For instance, if a reaction between substance A and B yields compound C, stoichiometry helps determine how much of C can be produced from a known quantity of A and B.

Molar Mass

The molar mass is the mass of one mole of substance, expressed in grams per mole (g/mol). It is numerically equivalent to the molecular weight but includes units, making it useful for conversion between moles and grams. To find the molar mass, you calculate the molecular weight and then use it to convert moles to grams or vice versa. This becomes especially handy when dealing with chemical equations where you need to convert moles of a reactant to grams. For instance, carbon dioxide (CO₂) has a molar mass of about 44 g/mol, being the sum of the molar masses of one carbon atom and two oxygen atoms.

Chemistry Mole Concept

The chemistry mole concept is a fundamental principle that allows chemists to count particles by weighing. One mole is Avogadro's number (approximately 6.022 x 10²³) of particles. These particles could be atoms, molecules, ions, or electrons. The mole bridges the microscopic and macroscopic worlds, making it possible to work with amounts of substances that can be physically measured. Through the mole concept, chemists can calculate the number of particles in a given mass of a substance by dividing the mass in grams by the substance's molar mass. This concept is foundational for the quantitative understanding of chemical compounds and reactions.