Problem 11 - Chemistry and Chemical Reactivity

Question

Ethylene glycol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2},\) is an ingredient of automobile antifreeze. Its density is \(1.11 \mathrm{g} / \mathrm{cm}^{3}\) at \(20^{\circ} \mathrm{C}\). If you need exactly \(500 .\) mL of this liquid, what mass of the compound, in grams, is required?

Step-by-Step Solution

Answer

The mass of ethylene glycol needed is 555 grams.

Step 1: Understand the Given Data

We are given the chemical compound ethylene glycol, with a chemical formula of \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\). Its density is provided as \(1.11 \mathrm{g}/\mathrm{cm}^{3}\) at \(20^{\circ} \mathrm{C}\). The volume needed is \(500\) mL, which is equivalent to \(500\) cubic centimeters (since \(1 \mathrm{mL} = 1 \mathrm{cm}^3\)).

Step 2: Density Formula Introduction

The density formula is \(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\). To find the mass, rearrange the formula to \(\text{Mass} = \text{Density} \times \text{Volume}\).

Step 3: Substitute Given Values

Substitute the given values into the rearranged formula: \(\text{Mass} = 1.11 \, \mathrm{g/cm^3} \times 500 \, \mathrm{cm^3}\). This step involves multiplication of the density and volume.

Step 4: Perform the Calculation

Calculate \(1.11 \, \mathrm{g/cm^3} \times 500 \, \mathrm{cm^3} = 555 \, \mathrm{g}\). This value represents the mass of 500 mL of ethylene glycol.

Key Concepts

Chemical Formula

Chemical formulas are a shorthand way to represent chemical substances using symbols for the elements and subscripts for the number of atoms. For ethylene glycol, the chemical formula is \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\). This indicates that each molecule consists of two carbon atoms, six hydrogen atoms, and two oxygen atoms. Understanding the chemical formula is crucial for making calculations about the substance, such as determining its molar mass or predicting its chemical behavior.

• In \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\), "C" stands for carbon, "H" stands for hydrogen, and "O" stands for oxygen.
• The subscripts (2, 6, and 2) show the number of each type of atom in a molecule.

The chemical formula provides important insight into the structure and properties of ethylene glycol, which is useful not only in calculations but also in understanding how it acts as an antifreeze.

Volume Conversion

In chemistry, converting measurements into compatible units is essential. In this exercise, volume conversion from milliliters (mL) to cubic centimeters (cm³) is a simple yet important step. Understanding that 1 mL is equivalent to 1 cm³ helps ensure consistency across calculations. Since 500 mL was provided in the problem, you don't need complex conversions.

• 1 mL = 1 cm³
• This means that the volume of the liquid is 500 cm³ when given as 500 mL.

Converting measurements appropriately is important to accurately apply formulas and compute results in science.

Density Formula

The density formula is a critical concept in science, relating the mass of a substance to its volume. Density is expressed as \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). In this context, density allows for determining how much a given volume of material 'weighs,' or rather, its mass. For ethylene glycol, the density is given as \(1.11 \text{ g/cm}^3\).

• Density is a measure of how much mass a substance contains in a given volume.
• It is typically expressed in grams per cubic centimeter (g/cm³) in this context.

Rearranging the density formula to find mass from density and volume is straightforward: simply multiply the density by the volume. Understanding and using the density formula is key to solving many practical problems in chemistry, including finding the mass of substances given their volume and density.

Mass Calculation

Mass calculations often require the use of formulas involving density and volume. In the given problem, the objective was to find the mass of 500 mL of ethylene glycol. Using the rearranged density formula, \( \text{Mass} = \text{Density} \times \text{Volume} \), you can easily find the mass once the density and volume are known.

• With \( \text{Density} = 1.11 \text{ g/cm}^3 \) and \( \text{Volume} = 500 \text{ cm}^3 \), the calculation becomes:
• \( \text{Mass} = 1.11 \times 500 = 555 \text{ g} \).

This mass calculation provides the mass of ethylene glycol required for a specific volume of 500 mL. Understanding how to perform mass calculations using the density formula is useful for a variety of applications in chemistry and related fields.