Question

The weight of \(1 \times 10^{22}\) molecules of \(\mathrm{CuSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) is (a) \(42.42 \mathrm{~g}\) (b) \(41.42 \mathrm{~g}\) (c) \(44.44 \mathrm{~g}\) (d) \(48.94 \mathrm{~g}\)

Short answer

The calculated weight is 2.72 g, none of the options match.

Step-by-step solution

Determine Molar Mass of Compound

First, calculate the molar mass of \( \mathrm{CuSO}_{4} \cdot 2\mathrm{H}_{2} \mathrm{O} \). The atomic masses are as follows: Cu = 63.55 g/mol, S = 32.07 g/mol, O = 16.00 g/mol, H = 1.01 g/mol. Thus, the molar mass is calculated as: \[ 63.55 + 32.07 + 4(16.00) + 2(2(1.01) + 16.00) = 63.55 + 32.07 + 64.00 + 4.04 + 32.00 = 163.66 \text{ g/mol} \].

Use Avogadro's Number

Avogadro's number tells us that one mole of a substance contains approximately \(6.022 \times 10^{23}\) molecules. We want to find out how many moles \(1 \times 10^{22}\) molecules correspond to. Calculate the moles of \(\mathrm{CuSO}_{4} \cdot 2\mathrm{H}_{2} \mathrm{O}\) using the formula: \[ \text{moles} = \frac{1 \times 10^{22}}{6.022 \times 10^{23}} \].

Calculate Moles of Molecules

Compute the moles: \[ \text{moles} = \frac{1 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.0166 \text{ moles} \].

Calculate the Weight of the Compound

The weight of the \(0.0166\) moles of \(\mathrm{CuSO}_{4} \cdot 2\mathrm{H}_{2} \mathrm{O}\) is given by multiplying the moles by the molar mass: \[ \text{weight} = 0.0166 \times 163.66 \approx 2.72 \text{ g} \]. Therefore, it seems there is an error or perhaps the option for this question isn't accurately listed as per the calculations.

Key concepts

Molar Mass Calculation

Understanding the molar mass of a compound is key in stoichiometry, as it helps us convert between grams and moles. Molar mass is the sum of the atomic masses of all the atoms in a molecule. For

• Copper (Cu): 63.55 g/mol
• Sulfur (S): 32.07 g/mol
• Oxygen (O): 16.00 g/mol
• Hydrogen (H): 1.01 g/mol

For the compound \[\mathrm{CuSO}_{4} \cdot 2\mathrm{H}_{2} \mathrm{O}\]Calculate the molar mass step-by-step:- Copper contributes = 63.55 g/mol- Sulfur contributes = 32.07 g/mol- Oxygen in the sulfate adds up as 4 Oxygen atoms = 4 \times 16.00 = 64.00 g/mol- Water of crystallization gives 2 \times (2 \times 1.01 + 16.00) = 4.04 + 32.00 = 36.04 g/molAdding these together, the total molar mass is:\[63.55 + 32.07 + 64.00 + 36.04 = 195.66 \text{ g/mol}\]This is crucial for converting moles to grams or vice versa in calculations.

Avogadro's Number

Avogadro's Number is a cornerstone concept in chemistry that connects the atomic scale with the macroscopic scale. It allows us to count particles by weighing them. Avogadro's Number, \[6.022 \times 10^{23}\] represents the number of atoms, ions, or molecules in one mole of a substance. It's much like a universal constant for chemistry.To find the number of moles from a given number of molecules, you use the formula: \[\text{moles} = \frac{\text{Given number of molecules}}{\text{Avogadro's Number}}\]For example, if you want to find out how many moles are in \[1 \times 10^{22} \text{ molecules}, \]use:\[\text{moles} = \frac{1 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.0166\text{ moles}\]This conversion is essential in determining how many particles you have in a quantifiable mass.

Mole Concept

The mole concept is a fundamental chemistry concept that allows chemists to work at the particle level with measurable amounts of a substance. A mole is a unit that measures the amount of a substance. One mole corresponds to

• Avogadro's number: 6.022 × 10^{23} particles
• Can refer to molecules, atoms, ions, or electrons, depending on the context.

Using the mole concept simplifies chemical calculations and reactions. If we have \[0.0166\text{ moles}\] of \[\mathrm{CuSO}_{4} \cdot 2\mathrm{H}_{2} \mathrm{O}\], we can multiply this by the molar mass:\[\text{mass} = \text{moles} \times \text{molar mass} = 0.0166 \times 195.66 \approx 3.25 \text{ g}\]This shows how much this number of moles would weigh, helping translate microscopic chemical laws into practical, tangible terms. This conceptual framework is pivotal for performing stoichiometric calculations effectively in chemistry.