Question

Calculate the number of cations and anions in each of the following compounds: (a) \(8.38 \mathrm{~g}\) of \(\mathrm{KBr}\), (b) \(5.40 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{SO}_{4},(\mathrm{c}) 7.45 \mathrm{~g}\) of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

Short answer

a) 4.24x10^22 ions each; b) 4.58x10^22 Na⁺, 2.29x10^22 SO₄²⁻; c) 4.34x10^22 Ca²⁺, 2.89x10^22 PO₄³⁻.

Step-by-step solution

Calculate Molar Mass

First, calculate the molar mass of each compound by summing up the atomic masses of each constituent element.(a) For \( \mathrm{KBr} \): - Potassium (K) has an atomic mass of approx 39.1 g/mol. - Bromine (Br) has an atomic mass of approx 79.9 g/mol. Molar mass of \( \mathrm{KBr} \) = \( 39.1 + 79.9 = 119.0 \) g/mol.(b) For \( \mathrm{Na}_{2}\mathrm{SO}_{4} \): - Sodium (Na) has an atomic mass of approx 23.0 g/mol. - Sulfate (SO4) has a combined atomic mass of approx 96.1 g/mol. Molar mass of \( \mathrm{Na}_{2}\mathrm{SO}_{4} \) = \( 2 \times 23.0 + 96.1 = 142.1 \) g/mol.(c) For \( \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \): - Calcium (Ca) has an atomic mass of approx 40.1 g/mol. - Phosphate (PO4) has a combined atomic mass of approx 94.1 g/mol. Molar mass of \( \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \) = \( 3 \times 40.1 + 2 \times 94.1 = 310.3 \) g/mol.

Find Moles of Each Compound

Convert grams of each compound to moles by dividing by the molar mass calculated in step 1.(a) Moles of \( \mathrm{KBr} \): \[ \frac{8.38 \text{ g}}{119.0 \text{ g/mol}} = 0.0704 \text{ moles} \](b) Moles of \( \mathrm{Na}_{2}\mathrm{SO}_{4} \): \[ \frac{5.40 \text{ g}}{142.1 \text{ g/mol}} = 0.0380 \text{ moles} \](c) Moles of \( \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \): \[ \frac{7.45 \text{ g}}{310.3 \text{ g/mol}} = 0.0240 \text{ moles} \]

Calculate Number of Cations and Anions

Use Avogadro's number to find the number of cations and anions per mole for each compound. Multiplication with the number of ions per formula unit is necessary.(a) \( \mathrm{KBr} \): Contains 1 K⁺ and 1 Br⁻ per formula unit. - Number of K⁺ and Br⁻ ions: \( 0.0704 \times 6.022 \times 10^{23} = 4.24 \times 10^{22} \) ions.(b) \( \mathrm{Na}_{2}\mathrm{SO}_{4} \): Contains 2 Na⁺ and 1 SO₄²⁻ per formula unit. - Na⁺ ions: \( 2 \times 0.0380 \times 6.022 \times 10^{23} = 4.58 \times 10^{22} \) ions. - SO₄²⁻ ions: \( 0.0380 \times 6.022 \times 10^{23} = 2.29 \times 10^{22} \) ions.(c) \( \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \): Contains 3 Ca²⁺ and 2 PO₄³⁻ per formula unit. - Ca²⁺ ions: \( 3 \times 0.0240 \times 6.022 \times 10^{23} = 4.34 \times 10^{22} \) ions. - PO₄³⁻ ions: \( 2 \times 0.0240 \times 6.022 \times 10^{23} = 2.89 \times 10^{22} \) ions.

Key concepts

Molar Mass Calculation

The concept of molar mass is crucial in chemistry as it relates to the mass of a given substance divided by the amount of substance (in moles). It essentially acts as a bridge between the microscopic world of atoms and the macroscopic quantities we measure. Molar mass is expressed in grams per mole (g/mol), and it helps in converting grams to moles and vice versa. To calculate the molar mass of a compound, you add up the atomic masses of all the atoms in a single molecule of that compound. For instance, the molar mass of potassium bromide (KBr) is calculated by adding the atomic mass of potassium (K), approximately 39.1 g/mol, and that of bromine (Br), approximately 79.9 g/mol, resulting in a total of 119.0 g/mol.
It's essential to be accurate when determining molar mass, as even small inaccuracies can lead to significant errors in subsequent calculations such as stoichiometry and determining the amount of substances in a reaction.

Avogadro's Number

Avogadro's number is a fundamental constant in chemistry that describes the number of constituent particles, usually atoms or molecules, in one mole of a substance. Numerically, it is equal to approximately 6.022 x 10^23. This large number is a bit daunting, but it is crucial for converting between the amount of substance in moles and the number of individual particles.
Using Avogadro's number, chemists can easily determine how many of a specific type of ion or molecule are present in a given sample. For example, once the amount of substance in moles is known, multiplying by Avogadro's number gives the total number of particles. In the case of KBr, calculating with Avogadro's number illustrated how many K⁺ and Br⁻ ions are present in a sample of known mass.

Cations and Anions

Cations and anions are types of ions that differ by their charge. Cations are positively charged ions, while anions are negatively charged. The charge disparity arises from a difference between the number of protons and electrons in an atom or molecule. Understanding the roles of cations and anions helps in deciphering the properties of ionic compounds and predicting the behavior of compounds during a chemical reaction.
For example, in ionic compounds such as KBr, ions bond together due to opposite charges attracting. Potassium ions (K⁺) are cations, while bromide ions (Br⁻) are anions. This complementary charge attraction is what keeps the ionic compound stable. Knowing which ions are cations or anions, and in what ratio they exist, is vital for calculating how many ions are in any given compound.

Stoichiometry

Stoichiometry is the field of chemistry that studies the dimensions of quantities in chemical reactions. This includes how reactants combine and products form in specific ratios. These ratios are derived from molecular or empirical formulas of reactants and products, which means stoichiometry relies on precise molar mass and balancing chemical equations.
Stoichiometry allows chemists to predict the outcomes of reactions and to calculate the amounts of products and reactants needed or produced. For instance, when calculating the number of ionic species in a reaction, you must consider stoichiometric coefficients. These coefficients show the balance of ions, demonstrating how compounds dissociate to form the ions mentioned in the original exercise, such as Na⁺, SO₄²⁻, Ca²⁺, and PO₄³⁻ in the sample compounds.