Problem 116 - Chemistry Matter and Change

Question

A sample of a compound contains 3.86 g of sulfur and 4.08 g of vanadium. How many atoms of sulfur and vanadium does the compound contain?

Step-by-Step Solution

Answer

The compound contains approximately \(7.26 \times 10^{22}\) atoms of sulfur and \(4.82 \times 10^{22}\) atoms of vanadium.

Step 1: Find the molar mass of sulfur and vanadium.

To calculate the number of moles of each element, we need to know the molar mass of each. Using the periodic table, we find the molar masses (rounded to the nearest whole number): Sulfur (S): 32 g/mole Vanadium (V): 51 g/mole

Step 2: Calculate the number of moles of sulfur and vanadium.

Now that we have the molar masses, we can calculate the number of moles of each element: Moles of sulfur = (mass of sulfur) / (molar mass of sulfur) = (3.86 g) / (32 g/mole) = 0.120625 moles Moles of vanadium = (mass of vanadium) / (molar mass of vanadium) = (4.08 g) / (51 g/mole) = 0.08 moles

Step 3: Use Avogadro's number to convert moles to atoms.

Avogadro's number (N_A) is the number of atoms, ions, or particles in one mole of a substance: \(N_A = 6.022 \times 10^{23}\) per mole. Using Avogadro's number, we can convert from moles to atoms: Atoms of sulfur = (moles of sulfur) × \(N_A\) = (0.120625 moles) × \(6.022 \times 10^{23}\) atoms/mole ≈ \(7.26 \times 10^{22}\) atoms Atoms of vanadium = (moles of vanadium) × \(N_A\) = (0.08 moles) × \(6.022 \times 10^{23}\) atoms/mole ≈ \(4.82 \times 10^{22}\) atoms

Step 4: Write the final answer.

The compound contains approximately \(7.26 \times 10^{22}\) atoms of sulfur and \(4.82 \times 10^{22}\) atoms of vanadium.

Key Concepts

Molar Mass Calculation

Understanding molar mass is crucial for mastering chemistry concepts. It's essentially the weight of one mole of a substance, typically expressed in grams per mole (g/mol). Think of it as the 'chemical weight ticket' that tells you how much one mole of an element or compound weighs.

To calculate the molar mass of an element, just look at the periodic table. The molar mass is the number listed, usually below the chemical symbol. For compounds, you'll need to do a bit more math - add up the molar masses of all the individual elements according to the compound's formula. For example, water (H₂O) has a molar mass of two times hydrogen's molar mass plus one time oxygen's molar mass.

Avogadro's Number

Have you ever wondered how many items are in a dozen? That's simple, it's 12. Now, what if you wanted to count the basic units in chemistry, like atoms or molecules, in a bulk amount? That's where Avogadro's number comes in handy. A mole, which is like a chemical dozen, is defined as having exactly Avogadro's number of units. As intimidating as the name might sound, Avogadro's number, which is approximately \(6.022 \times 10^{23}\) units per mole, is just the chemistry world's version of a dozen, but for an inconceivably huge number of particles.

This number isn't arbitrary; it's chosen so that the mass of one mole of a substance in grams is approximately equal to the substance’s atomic or molecular mass. Fancy, right? Avogadro's number allows chemists to convert between atomic scale measurements and human scale measurements.

Moles to Atoms Conversion

When you have the hang of working with moles and you're familiar with Avogadro’s number, converting between moles and atoms (or molecules) becomes almost second nature. It's just like converting dozens into individual items.

Here's how you do it: if you have the number of moles of a substance and you want to find out the number of atoms or molecules, multiply the moles by Avogadro’s number. It’s like how you would multiply by 12 to find out how many eggs are in several cartons if you knew how many dozen cartons you had.

For instance, let’s take \(0.08\) moles of substance. To find out how many atoms there are, you'd multiply \(0.08\) by \(6.022 \times 10^{23}\) to get about \(4.82 \times 10^{22}\) atoms. This simple multiplication allows chemists to switch from the very small scale of atom counts, which are useful for reactions at the particle level, to the more practical macro scale of grams and moles usable in the laboratory.