Problem 188 - Introductory Chemistry Atoms First

Question

What mass of each substance occupies a volume of \(50.0 \mathrm{~mL} ?\) (Densities are shown in parentheses.) (a) Lead \((11.4 \mathrm{~g} / \mathrm{mL})\) (b) Ethanol \((0.785 \mathrm{~g} / \mathrm{mL})\) (c) Oxygen gas \(\left(1.4 \times 10^{-3} \mathrm{~g} / \mathrm{mL}\right)\) (d) Hydrogen gas \(\left(8.4 \times 10^{-5} \mathrm{~g} / \mathrm{mL}\right)\) (e) Mercury (13.6 g/mL) (f) Gold \((19.3 \mathrm{~g} / \mathrm{mL})\)

Step-by-Step Solution

Answer

The mass of each substance occupying a volume of 50.0 mL is as follows: (a) Lead: 570 g (b) Ethanol: 39.25 g (c) Oxygen gas: 0.07 g (d) Hydrogen gas: 0.0042 g (e) Mercury: 680 g (f) Gold: 965 g

Step 1: Convert volume to liters

Since the given volume is in mL, we need to convert it to liters, as we will work with units of g / L. Remember that 1 L = 1000 mL. So, 50.0 mL is equivalent to 0.050 L.

Step 2: Calculate mass for each substance

We will now use the formula Mass = Density × Volume to calculate the mass of each substance: (a) Lead: Mass = Density × Volume Mass = \(11.4 \frac{g}{mL} × 50.0 mL\) Mass = 570 g (b) Ethanol: Mass = Density × Volume Mass = \(0.785 \frac{g}{mL} × 50.0 mL\) Mass = 39.25 g (c) Oxygen gas: Mass = Density × Volume Mass = \(1.4 × 10^{-3} \frac{g}{mL} × 50.0 mL\) Mass = 0.07 g (d) Hydrogen gas: Mass = Density × Volume Mass = \(8.4 × 10^{-5} \frac{g}{mL} × 50.0 mL\) Mass = 0.0042 g (e) Mercury: Mass = Density × Volume Mass = \(13.6 \frac{g}{mL} × 50.0 mL\) Mass = 680 g (f) Gold: Mass = Density × Volume Mass = \(19.3 \frac{g}{mL} × 50.0 mL\) Mass = 965 g

Step 3: Present the masses of each substance

The mass of each substance occupying a volume of 50.0 mL is as follows: (a) Lead: 570 g (b) Ethanol: 39.25 g (c) Oxygen gas: 0.07 g (d) Hydrogen gas: 0.0042 g (e) Mercury: 680 g (f) Gold: 965 g

Key Concepts

Understanding Mass-Volume Relationship

The mass-volume relationship is essential in understanding how much space a given amount of a chemical substance will occupy or conversely, how much of that substance is needed to fill up a certain volume. This relationship is ultimately governed by density, which is a fundamental property representing the mass of a substance per unit volume. In essence, density is a measure of how “tightly packed” or “concentrated” a substance is.

When given the density of a substance, and a specified volume, one can calculate the mass using the simple formula:
\[ \text{Mass} = \text{Density} \times \text{Volume} \]
For instance, if you have a density of ethanol at \(0.785 \frac{g}{mL}\) and you are interested in finding out the mass of ethanol that would occupy a 50.0 mL container, you would simply multiply the two values:
\[ \text{Mass} = 0.785 \frac{g}{mL} \times 50.0 \text{mL} = 39.25 \text{g} \]
This formula is crucial in many scientific calculations, especially ones involving solutions and mixtures in chemistry.

The Process of Unit Conversion

Unit conversion is a critical skill in scientific disciplines as it enables one to express measurements in different units of volume or mass, making comparisons and calculations possible. It's somewhat like translating languages but with numbers. For example, in the context of the mass-volume relationship, it is often necessary to convert volume from milliliters (mL) to liters (L), as some formulas may require specific units.

The standard conversion is that \(1 \text{L} = 1000 \text{mL}\). Hence, converting 50.0 mL to liters would involve dividing by 1000, yielding 0.050 L. This conversion step is paramount before you apply the density formula if the density is provided in different volume units (like g/L). Such conversions ensure consistency in the units across a calculation, preventing errors and confusion.

It is important for students to remember unit conversions, as forgetting to do so can result in incorrect answers and misunderstandings in both homework and real-world applications.

Chemical Substance Properties

Chemical substances each possess unique properties that dictate their behavior and interactions. Density, mentioned earlier, is one such property that describes the mass per unit volume of a substance. Understanding the properties of substances like melting and boiling points, solubility, reactivity, and specific heat capacity can be important when conducting experiments or solving problems.

Density itself can vary based on conditions like temperature and pressure, especially for gases. In the exercise provided, densities for each substance are given under standard conditions to calculate mass. For example, Oxygen gas is much less dense compared to metals like Gold or Lead, which are significantly more “heavy” or dense. The implications are extensive: lower density substances will occupy more volume than higher density substances if their masses are equal.

Knowledge of such properties is not only vital in solving textbook problems but is essential for practical applications, such as designing materials, medication dosage, or engineering processes in a wide variety of fields.