Question

If \(3.02 \times 10^{19}\) molecules are removed from \(98 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\), then the number of moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) left are: (a) \(0.1 \times 10^{-3}\) (b) \(5 \times 10^{-4}\) (c) \(1.2 \times 10^{-4}\) (d) \(1.5 \times 10^{-3}\)

Short answer

Option (d), approximately 1.5 x 10^{-3} moles, matches best.

Step-by-step solution

Calculate Moles in Original Mass

First, calculate the number of moles in 98 mg of \( \mathrm{H}_2 \mathrm{SO}_4 \). The molar mass of \( \mathrm{H}_2 \mathrm{SO}_4 \) is approximately 98 g/mol.For 98 mg, convert mg to grams: \( 98 \mathrm{mg} = 0.098 \mathrm{g} \).Moles in 0.098 g is given by: \( \text{Moles} = \frac{0.098 \text{g}}{98 \text{g/mol}} = 0.001 \text{mol} \).

Convert Molecules Removed to Moles

Given \(3.02 \times 10^{19}\) molecules were removed. One mole of any substance contains \( 6.022 \times 10^{23} \) molecules. The moles removed are:\[ \text{Moles removed} = \frac{3.02 \times 10^{19} \text{ molecules}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 5.016 \times 10^{-5} \text{mol} \]

Calculate Remaining Moles

Subtract the moles removed from the initial moles:\[ 0.001 \text{ mol} - 5.016 \times 10^{-5} \text{ mol} = 0.00094984 \text{ mol} \]

Express Remaining Moles in Exponential Form

Express the result as \( \approx 9.4984 \times 10^{-4} \text{ mol} \) and identify the closest option. Given the choices,\( \approx 1 \times 10^{-3} \text{ mol} \) aligns closely to option (d).

Key concepts

Molar Mass Calculation

To calculate the number of moles in a given mass of a compound, we first need to know the molar mass of the compound. The molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). For example, the molar mass of sulfuric acid, \( \mathrm{H}_2\mathrm{SO}_4 \), is approximately 98 g/mol.

To convert mass to moles, we use the formula:

• \( \text{Moles} = \frac{\text{Given mass in grams}}{\text{Molar mass}} \)

For the given problem, if we have 98 mg of \( \mathrm{H}_2\mathrm{SO}_4 \), we first convert it to grams: 98 mg = 0.098 g. Then using the molar mass:

• \( \text{Moles} = \frac{0.098 \text{ g}}{98 \text{ g/mol}} = 0.001 \text{ mol} \)

Understanding this calculation allows us to determine how many moles of a substance are present in any given mass. This foundational concept is key in many areas of chemistry.

Moles to Molecules Conversion

The conversion from moles to molecules involves Avogadro's number, which is approximately \( 6.022 \times 10^{23} \) molecules/mol. This number tells us how many individual atoms or molecules are present in one mole of a substance.

To convert a given number of molecules to moles, you use the formula:

• \( \text{Moles} = \frac{\text{Number of molecules}}{6.022 \times 10^{23}} \)

For example, if we remove \( 3.02 \times 10^{19} \) molecules of \( \mathrm{H}_2\mathrm{SO}_4 \), we calculate the moles as:

• \( \text{Moles removed} = \frac{3.02 \times 10^{19} \text{ molecules}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 5.016 \times 10^{-5} \text{ mol} \)

This conversion is essential for understanding quantities at the molecular level, helping to bridge the gap between macroscopic and microscopic perspectives in chemistry.

Stoichiometric Calculations

Stoichiometry involves the calculation of reactants and products in chemical reactions. It's based on the conservation of mass and the concept of moles.

In the context of a problem where molecules are removed from a substance, like the removal of \( 3.02 \times 10^{19} \) molecules from \( \mathrm{H}_2\mathrm{SO}_4 \), stoichiometry allows us to calculate what remains. We already know:

• Initial moles in 0.098 g of \( \mathrm{H}_2\mathrm{SO}_4 \) is \( 0.001 \text{ mol} \)
• Moles removed is \( 5.016 \times 10^{-5} \text{ mol} \)

To find the remaining moles:

• \( 0.001 \text{ mol} - 5.016 \times 10^{-5} \text{ mol} = 0.00094984 \text{ mol} \)

Expressing this result as \( \approx 9.4984 \times 10^{-4} \text{ mol} \), it is close to \( 1 \times 10^{-3} \text{ mol} \), corresponding to option (d) as the closest match.

Stoichiometric calculations help ensure the quantities of all substances involved in a reaction are determined precisely, a crucial skill for predicting the outcomes of chemical reactions.