Problem 117 - Chemistry and Chemical Reactivity

Question

A drop of water has a volume of about \(0.050 \mathrm{mL}\). How many molecules of water are in a drop of water? (Assume water has a density of \(1.00 \mathrm{g} / \mathrm{cm}^{3} .\) )

Step-by-Step Solution

Answer

There are approximately \(1.67 \times 10^{21}\) molecules in a drop of water.

Step 1: Convert Volume to Mass

Given the density of water is \(1.00 \text{ g/cm}^3\), and the volume is \(0.050 \text{ mL}\) or \(0.050 \text{ cm}^3\), the mass of the water drop is calculated using the formula: \(\text{mass} = \text{density} \times \text{volume}\). Therefore, the mass is: \(1.00 \text{ g/cm}^3 \times 0.050 \text{ cm}^3 = 0.050 \text{ g}\). This means the mass of the water drop is \(0.050 \text{ g}\).

Step 2: Convert Mass to Moles

Use the molar mass of water, which is approximately \(18.015 \text{ g/mol}\), to convert grams to moles using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). Thus, \(\text{moles} = \frac{0.050 \text{ g}}{18.015 \text{ g/mol}} \approx 0.002776 \text{ moles}\).

Step 3: Convert Moles to Molecules

Using Avogadro's number \(6.022 \times 10^{23}\, \text{molecules/mol}\), calculate the number of molecules from the moles of water using the formula: \(\text{molecules} = \text{moles} \times \text{Avogadro's number}\). Therefore, \(0.002776 \text{ moles} \times 6.022 \times 10^{23}\, \text{molecules/mol} \approx 1.67 \times 10^{21} \text{ molecules}\).

Key Concepts

Avogadro's Number

Avogadro's Number is a fundamental concept in chemistry that helps us bridge microscopic and macroscopic worlds. It plays a crucial role in understanding chemical reactions at the molecular level.

Simply put, Avogadro's Number is the number of particles, usually atoms or molecules, present in one mole of a substance. The value of Avogadro's Number is about \(6.022 \times 10^{23}\) particles per mole.

Think of it as a giant counting unit like a dozen, but for an enormous number of molecules. Just as a dozen means 12 of something, a mole signifies \(6.022 \times 10^{23}\) of something. This massive number allows chemists to count molecules in practical amounts of material. Without Avogadro's Number, understanding the amount of substances we deal with in everyday chemistry would be incredibly difficult.

In this exercise, Avogadro's Number helps to convert the amount of water in moles to the actual number of water molecules in a drop.

Molar Mass

Molar Mass represents the mass of one mole of a particular substance. This is usually expressed in grams per mole (\( ext{g/mol}\)) and is a key concept in stoichiometry, which is the science of measuring the quantitative aspects of chemical reactions.

To find the molar mass of a compound, you sum up the atomic masses of the constituent elements. For water \( ext{H}_2 ext{O}\), this calculation involves the atomic masses of hydrogen (about 1.008 \( ext{g/mol}\)) and oxygen (about 16.00 \( ext{g/mol}\)). So, the molar mass of water is \(2 \times 1.008 + 16.00 = 18.015 \text{g/mol}\).

This value is incredibly useful when converting between mass and moles, allowing chemists to determine how many molecules are in a given amount of substance. In the case provided, the molar mass of water helps convert a water drop's mass into a mole count, which in turn gets converted into the number of molecules using Avogadro’s Number.

Density of Water

Density is a measure of how much mass is contained in a given volume and it can differ from substance to substance. It's typically expressed in grams per cubic centimeter (\( ext{g/cm}^3\)) for solids and liquids.

Water, with its density of \(1.00 \text{ g/cm}^3\) at standard temperature and pressure, serves as a convenient baseline for calculating the mass of a volume of water because 1 mL of water weighs exactly 1 gram.

This unit density simplifies many problems. For example, measuring a certain volume of water immediately gives you its mass. In this exercise, the density of water allows for direct conversion of the drop's volume into mass without complex calculations. Understanding and using density efficiently is a fundamental skill in solving various chemistry problems, especially those involving conversions between mass and volume.