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Chapter 4: Problem 131

If the atomic weight of oxygen was 50, what would its equivalent weight be?

Short Answer

Expert verified
If the atomic weight of oxygen was 50 and its valency is 2, its equivalent weight would be \( \frac{50}{2} = 25 \).

Step by step solution

01

Recall the formula to calculate equivalent weight

The formula to calculate the equivalent weight of an element is as follows: Equivalent Weight = (Atomic Weight) / (Valency)
02

Gather the given data

We are given: - Atomic weight of oxygen = 50 - Valency of oxygen = 2
03

Calculate the equivalent weight of oxygen

Now we can use the formula to calculate the equivalent weight: Equivalent Weight of Oxygen = (50) / (2) = 25 So if the atomic weight of oxygen was 50, its equivalent weight would be 25.

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Key Concepts

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

Atomic Weight
Atomic weight is one of the fundamental properties of an element. It's like the mass of an atom, which we can compare with others. By using a standard element, typically carbon-12, scientists have created a unified scale for measuring atomic weight.
  • Atomic weight is essentially the average mass of an element's atoms, considering all its isotopes and their abundances.
  • It's important to note that atomic weight can sometimes be referred to as atomic mass. However, these concepts serve different purposes. Atomic weight is dimensionless, which means it's a pure number without units, whereas atomic mass comes with units of atomic mass units (amu).
  • In the case of oxygen, its atomic weight is typically around 16, not 50. However, this problem shows an interesting change to help us calculate equivalent weight.
Understanding atomic weight helps in calculating equivalent weights and understanding chemical reactions.
Valency
Valency plays a crucial role in chemistry and helps us understand how atoms connect. It's the combining power of an element, represented by the number of hydrogen atoms an element can combine or displace in a reaction.
  • Valency provides the number of electrons an atom can lose, gain, or share to form chemical bonds.
  • For oxygen, the valency is often 2. This is because oxygen can take on two additional electrons to complete its outer shell, which consists of six electrons naturally.
  • Using valency in calculations helps determine equivalent weights, a necessary step for many chemical reactions and processes.
By recognizing the valency, we can predict how elements like oxygen will react and form compounds.
Oxygen
Oxygen is a life-sustaining element. It's essential for respiration in the animal kingdom and necessary for combustion processes.
  • It is a non-metal found in abundance in the Earth's crust and atmosphere, usually existing as a diatomic molecule, meaning two oxygen atoms bonded together, written as \( ext{O}_2\).
  • Oxygen has a standard atomic weight of 16 but for calculation purposes here, it is portrayed with an atomic weight of 50. This is to demonstrate the concept of equivalent weight.
  • Oxygen is essential for various chemical reactions, including those crucial in industrial processes and the survival of living organisms.
Recognizing these basic properties shows why oxygen's equivalent weight is valuable in chemistry.

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Most popular questions from this chapter

When a piece of magnesium ribbon weighing \(0.32 \mathrm{~g}\) is burned in oxygen, the resultant oxide weighs \(0.53 \mathrm{~g}\). What is the percentage composition of the oxide?

Two thirds of the atoms in a molecule of water \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are hydrogen. What percentage of the weight of a water molecule is the weight of the two hydrogen atoms? The atomic weight of hydrogen is \(1.008 \mathrm{~g} /\) mole and of oxygen is \(16.00 \mathrm{~g} / \mathrm{mole} .\)

A sample of the poisonous compound nicotine extracted from cigarette smoke was found to contain \(74.0 \%\) by weight of carbon (C, atomic weight \(=12.0 \mathrm{~g} / \mathrm{mole}), 8.65 \%\) by weight of hydrogen \((\mathrm{H}\), atomic weight \(=1.01 \mathrm{~g} / \mathrm{mole})\), and \(17.3 \%\) by weight of nitrogen \((\mathrm{N}\), atomic weight \(=14.0\) \(\mathrm{g} /\) mole). What is the empirical formula of nicotine?

The density of a \(25.0 \%\) sugar solution is \(1.208 \mathrm{~g} / \mathrm{ml}\). What weight of sugar would be contained in \(1.00\) liter of this solution ?

If the density of ethylene is \(1.25 \mathrm{~g}\) /liter at S.T.P. and the ratio of carbon to hydrogen atoms is \(1: 2\), what is molecular weight and formula of ethylene?
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