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Given a pair of elements and their mass relation, answer the following questions. (a) The mass of 4 atoms of \(A=\) the mass of 6 atoms of \(B\). Which element has the smaller molar mass? (b) The mass of 6 atoms of \(C\) is less than the mass of 3 atoms of the element D. Which element has more atoms/gram? (c) Six atoms of \(\mathrm{E}\) have larger mass than six atoms of \(\mathrm{F}\). Which has more atoms/gram? (d) Six atoms of \(F\) have the same mass as 8 atoms of \(G .\) Which has more atoms/mole?

Short Answer

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Answer: (a) B has a smaller molar mass, (b) D has more atoms per gram, (c) F has more atoms per gram, and (d) G has more atoms per mole.

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01

Identify the given information

We are given the mass relations of A and B as: \(4 \times \text{Mass of A} = 6 \times \text{Mass of B}\)
02

Find the molar mass ratio

We can find the molar mass ratio of A to B by dividing both sides of the equation by 4 and 6 (the respective number of atoms) respectively: \(\frac{\text{Mass of A}}{\text{Mass of B}} = \frac{6}{4}\) This simplifies to: \(\frac{\text{Mass of A}}{\text{Mass of B}} = \frac{3}{2}\)
03

Determine the element with the smaller molar mass

Since the ratio of the mass of A to B is \(\frac{3}{2}\), this means that A is more massive than B. Therefore, element B has a smaller molar mass. (b) The mass of 6 atoms of C is less than the mass of 3 atoms of the element D.
04

Identify the given information

We are given the inequality for the mass relations of C and D as: \(6 \times \text{Mass of C} < 3 \times \text{Mass of D}\)
05

Compare the mass ratio

Dividing both sides of the inequality by 6 and 3 (the respective number of atoms) respectively, we have: \(\frac{\text{Mass of C}}{\text{Mass of D}} < \frac{3}{6}\) This simplifies to: \(\frac{\text{Mass of C}}{\text{Mass of D}} < \frac{1}{2}\)
06

Determine which element has more atoms/gram

The inequality shows that the mass of C is less than half the mass of D. As the mass of a substance is directly related to its number of atoms, we can conclude that element D has more atoms per gram. (c) Six atoms of E have larger mass than six atoms of F.
07

Identify the given information

We are given the inequality for the mass relations of E and F as: \(6 \times \text{Mass of E} > 6 \times \text{Mass of F}\)
08

Compare the mass ratio

Dividing both sides of the inequality by 6 (the number of atoms), we have: \(\frac{\text{Mass of E}}{\text{Mass of F}} > 1\)
09

Determine which element has more atoms/gram

Since the mass of E is greater than the mass of F, we can conclude that element F has more atoms per gram. This is because a smaller mass per atom means there are more atoms in a fixed mass (or gram). (d) Six atoms of F have the same mass as 8 atoms of G.
10

Identify the given information

We are given the mass relations of F and G as: \(6 \times \text{Mass of F} = 8 \times \text{Mass of G}\)
11

Find the molar mass ratio

We can find the molar mass ratio of F to G by dividing both sides of the equation by 6 and 8 (the respective number of atoms) respectively: \(\frac{\text{Mass of F}}{\text{Mass of G}} = \frac{8}{6}\) This simplifies to: \(\frac{\text{Mass of F}}{\text{Mass of G}} = \frac{4}{3}\)
12

Determine which element has more atoms/mole

Since the ratio of the mass of F to G is \(\frac{4}{3}\), this means that F is more massive than G. Considering Avogadro's number, if an element has a smaller molar mass, it will have more atoms per mole. Therefore, element G has more atoms per mole.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Ratio
When comparing elements, understanding their **mass ratio** is essential. A mass ratio is essentially a comparison of the masses of two different elements based on a defined relationship between their quantities. It can help determine which element has a smaller molar mass (the mass of one mole of an element) when equal numbers of atoms are compared.
For instance, if 4 atoms of element A have the same mass as 6 atoms of element B, we can say the mass of A to B is in the ratio of 3:2 after simplifying the equation: \[ \frac{\text{Mass of A}}{\text{Mass of B}} = \frac{3}{2} \]
  • If A is more massive, the ratio indicates that A has a higher molar mass than B.
  • By simplifying complex mass relationships, you can identify which element has a smaller molar mass.
Mass ratio comparisons are a powerful tool in chemistry, helping in evaluating the properties of elements and understanding their behavior in reactions.
Atomic Mass
The **atomic mass** provides insight into the mass of a single atom of an element, often expressed in atomic mass units (amu). Understanding atomic mass is crucial for comparing different elements, as it plays a key role in determining relative mass and calculating molar masses.
Atomic mass can easily be compared using inequalities between different elements, as was demonstrated with elements C and D:
  • If 6 atoms of C are lighter than 3 atoms of D, the inequality shows \( \frac{\text{Mass of C}}{\text{Mass of D}} < \frac{1}{2} \).
  • Such comparisons tell us that despite having fewer atoms, each atom of C is relatively lighter compared to each atom of D.
This concept confirms which element is heavier when compared atom by atom and helps in determining which element can provide more atoms per gram or more atoms in any given weight.
Atoms Per Gram
**Atoms per gram** describe how many atoms of an element exist in one gram of that element. This concept is inversely linked to the concept of molar mass - the lower the molar mass of an element, the more atoms it can have per gram.
Consider elements like E and F, when comparing six atoms of each:
  • If the mass of six atoms of E is greater than that of F, it implies element F is lighter, meaning it has more atoms per gram than E.
  • Conclusively, the number of atoms per gram is not just about atomic weight but also how compactly atoms can "pack" into a given weight.
This gives students insights into how different substances might weigh the same, yet one might contain many more particles (atoms) than the other, affecting how we utilize these elements in reactions and calculations.

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