Question

Calculate the number of moles of the indicated substance present in each of the following samples. a. \(1.28 \mathrm{~g}\) of iron(II) sulfate b. \(5.14 \mathrm{mg}\) of mercury \((\mathrm{II})\) iodide c. \(9.21 \mu \mathrm{g}\) of tin(IV) oxide d. 1.26 lb of cobalt(II) chloride e. \(4.25 \mathrm{~g}\) of copper(II) nitrate

Short answer

The number of moles present in each sample is as follows: a. \(0.0084 \ moles\) of iron(II) sulfate b. \(1.13 \times 10^{-5} \ moles\) of mercury(II) iodide c. \(6.11 \times 10^{-8} \ moles\) of tin(IV) oxide d. \(4.41 \ moles\) of cobalt(II) chloride e. \(0.0226 \ moles\) of copper(II) nitrate

Step-by-step solution

a. Finding moles of iron(II) sulfate

Iron(II) sulfate has the formula FeSO4. To calculate its molar mass, we sum the molar mass of each element in the compound: Molar mass of FeSO4 = (1 * 55.8) + (1 * 32.1) + (4 * 16.0) = 55.8 + 32.1 + 64.0 = 151.9 g/mol Given mass of iron(II) sulfate = 1.28 g Number of moles of iron(II) sulfate = \( \frac{1.28 \ g}{151.9 \ g/mol} = 0.0084 \ moles \)

b. Finding moles of mercury (II) iodide

Mercury(II) iodide has the formula HgI2. To calculate its molar mass, we sum the molar mass of each element in the compound: Molar mass of HgI2 = (1 * 200.6) + (2 * 126.9) = 200.6 + 253.8 = 454.4 g/mol Given mass of mercury(II) iodide = 5.14 mg = 0.00514 g Number of moles of mercury(II) iodide = \( \frac{0.00514 \ g}{454.4 \ g/mol} =1.13 \times 10^{-5} \ moles \)

c. Finding moles of tin(IV) oxide

Tin(IV) oxide has the formula SnO2. To calculate its molar mass, we sum the molar mass of each element in the compound: Molar mass of SnO2 = (1 * 118.7) + (2 * 16.0) = 118.7 + 32.0 = 150.7 g/mol Given mass of tin(IV) oxide = 9.21 μg = 0.00000921 g Number of moles of tin(IV) oxide = \( \frac{0.00000921 \ g}{150.7 \ g/mol} =6.11 \times 10^{-8} \ moles \)

d. Finding moles of cobalt(II) chloride

Cobalt(II) chloride has the formula CoCl2. To calculate its molar mass, we sum the molar mass of each element in the compound: Molar mass of CoCl2 = (1 * 58.93) + (2 * 35.45) = 58.93 + 70.90 = 129.83 g/mol Given mass of cobalt(II) chloride = 1.26 lb = 1.26 * 453.6 g = 572.54 g Number of moles of cobalt(II) chloride = \( \frac{572.54 \ g}{129.83 \ g/mol} = 4.41 \ moles \)

e. Finding moles of copper(II) nitrate

Copper(II) nitrate has the formula Cu(NO3)2. To calculate its molar mass, we sum the molar mass of each element in the compound: Molar mass of Cu(NO3)2 = (1 * 63.55) + (2 * (14.01 + (3*16.0)))= 63.55 + (2 * (14.01 + 48.0))= 63.55 + 124.02 = 187.57 g/mol Given mass of copper(II) nitrate = 4.25 g Number of moles of copper(II) nitrate = \( \frac{4.25 \ g}{187.57 \ g/mol} = 0.0226 \ moles \)

Key concepts

Understanding Molar Mass

Molar mass is a foundational concept in chemistry that represents the mass of one mole of any substance, usually in grams per mole (g/mol). This quantity is calculated by summing the atomic masses of each element in the compound, and these atomic masses can be found on the periodic table of elements.

For example, in the case of iron(II) sulfate, with the chemical formula FeSO4, the molar mass is composed of the mass of one atom of iron (Fe), one atom of sulfur (S), and four atoms of oxygen (O). By multiplying the atomic mass of each element by the number of atoms in the formula and adding them together, we arrive at the molar mass of the compound.

To improve understanding, imagine you have a collection of small Lego blocks, each representing a different element with its own unique weight. Building a model (a molecule) involves joining these blocks together in a specific configuration (the chemical formula). The total weight of the model is analogous to the molar mass of the compound.

Demystifying Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of the mole. The stoichiometric coefficients in a balanced equation tell us the proportion of molecules or moles required for the reaction.

In a balanced chemical equation, the ratio of the substances involved can be used to calculate how much of a reactant is needed to produce a certain amount of product, or vice versa. When applying stoichiometry to solve problems, as done in the provided exercise, the concept allows us to predict how many moles of a substance we can obtain from a given mass, or determine the mass from a given number of moles, connecting to the concept of molar mass.

Real-Life Analogy of Stoichiometry

To better grasp this concept, consider a recipe that requires two eggs for every cup of flour. If you know how many eggs you have, you can determine how much flour is needed. Similarly, with stoichiometry, knowing the amount of one reactant can help you calculate the amount of another reactant or product.

Deciphering Chemical Formulas

Chemical formulas provide a lot of information in a very compact form. They tell us the types of atoms in a molecule and the number of each type present. These formulas serve as the blueprint for building the structure of the compound and play a critical role in the calculations of stoichiometry and molar mass.

For example, the chemical formula of copper(II) nitrate, Cu(NO3)2, shows that each molecule is made up of one copper atom, two nitrogen atoms, and six oxygen atoms. The subscripts in the formula indicate the number of atoms of each type in the molecule, and if there's no subscript, it's implied that there's only one atom of that element.

Visualizing Chemical Structures

A helpful way to internalize chemical formulas is to visualize or draw the structure they represent, showing the connections between atoms. This helps in understanding the composition and proportions of different elements in the compound, which is crucial when calculating molar mass and performing stoichiometric calculations.