Question

You are given 1.0 -mol amounts of \(\mathrm{He}, \mathrm{Fe}, \mathrm{Li}, \mathrm{Si},\) and \(\mathrm{C}\) Which sample has the largest mass?

Short answer

Iron (Fe) has the largest mass.

Step-by-step solution

Identify the Problem

We need to determine which of the given 1.0-mol samples has the largest mass among \( \mathrm{He}, \mathrm{Fe}, \mathrm{Li}, \mathrm{Si},\) and \(\mathrm{C}\). Since the number of moles is the same for each, we can compare their molar masses.

Find Molar Mass

To find the mass of each sample, we need to know the molar mass of each element. The molar mass (in grams per mole) is equal to the atomic mass found on the periodic table. - \( \mathrm{He} \): 4.00 \, \text{g/mol}- \( \mathrm{Fe} \): 55.85 \, \text{g/mol}- \( \mathrm{Li} \): 6.94 \, \text{g/mol}- \( \mathrm{Si} \): 28.09 \, \text{g/mol}- \( \mathrm{C} \): 12.01 \, \text{g/mol}

Calculate Mass of Each Sample

Since we have 1.0 mol of each element, the mass of each sample is equal to its molar mass:\[ \text{Mass of } \mathrm{He} = 1.0 \text{ mol} \times 4.00 \, \text{g/mol} = 4.00 \text{ g} \\text{Mass of } \mathrm{Fe} = 1.0 \text{ mol} \times 55.85 \, \text{g/mol} = 55.85 \text{ g} \\text{Mass of } \mathrm{Li} = 1.0 \text{ mol} \times 6.94 \, \text{g/mol} = 6.94 \text{ g} \\text{Mass of } \mathrm{Si} = 1.0 \text{ mol} \times 28.09 \, \text{g/mol} = 28.09 \text{ g} \\text{Mass of } \mathrm{C} = 1.0 \text{ mol} \times 12.01 \, \text{g/mol} = 12.01 \text{ g}\]

Compare Masses

Now compare the calculated masses: - \( \mathrm{He}\): 4.00 \, \text{g}- \( \mathrm{Fe}\): 55.85 \, \text{g}- \( \mathrm{Li}\): 6.94 \, \text{g}- \( \mathrm{Si}\): 28.09 \, \text{g}- \( \mathrm{C}\): 12.01 \, \text{g}The largest mass is that of \(\mathrm{Fe}\).

Key concepts

Understanding Atomic Mass

Atomic mass is a critical concept in chemistry, representing the average mass of an atom of an element, measured in atomic mass units (amu). This value is generally found on the periodic table and is often similar to the element's molar mass expressed in grams per mole.

Atomic mass helps us identify how heavy an element's atom is relative to another element's atom. For example, iron (Fe) has an atomic mass of 55.85 amu, which is considerably heavier than helium (He) with an atomic mass of 4.00 amu.

This atomic mass difference is the core reason why a mole of iron atoms weighs significantly more than a mole of helium atoms. Understanding this concept is the baseline of more involved calculations in chemistry.

Essential Mole Calculations

Mole calculations are foundational in chemistry as they relate to the amount of substance. One mole of a substance is Avogadro's number of particles, typically atoms or molecules, which is approximately 6.022 × 10^{23} particles.

The beauty of mole calculations is their ability to simplify the complex world of atomic measures into understandable quantities. When dealing with calculations involving different substances, the number of moles helps balance chemical equations and predict how substances interact.

For instance, in our exercise, because each element is given in 1.0 moles, the calculations are straightforward. The mass of each 1.0 mol sample equals the element's molar mass, derived from the atomic mass multiplied by 1 mole.

Comparing Elements By Molar Mass

When comparing elements, molar mass is a handy tool. It tells you the mass of one mole of any element, making it easier to predict which element will be heavier in a given amount. This calculation is essential especially in situations where elements are compared quantitatively.

In our exercise, although each element sample is 1.0 mol which implies equal number of particles, their weights significantly vary. Calculating based on molar mass, iron (Fe) with a molar mass of 55.85 g/mol is much heavier than helium (He) at just 4.00 g/mol per mole.

This comparison ensures we can quickly deduce that iron would have the largest mass among the given elements, when quantities are the same in moles. It underscores how molar mass provides insight into the bulk of materials in chemical experiments and day-to-day applications.