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Although carbon-12 is now used as the standard for atomic weights, this has not always been the case. Early attempts at classification used hydrogen as the standard, with the weight of hydrogen being set equal to 1.0000 u. Later attempts defined atomic weights using oxygen (with a weight of 16.0000 ). In each instance, the atomic weights of the other elements were defined relative to these masses. (To answer this question, you need more precise data on current atomic weights: \(\mathrm{H}\), \(1.00794 \mathrm{u} ; \mathrm{O}, 15.9994 \mathrm{u} .)\) (a) If \(\mathrm{H}=1.0000\) u was used as a standard for atomic weights, what would the atomic weight of oxygen be? What would be the value of Avogadro's number under these circumstances? (b) Assuming the standard is \(\mathrm{O}=16.0000,\) determine the value for the atomic weight of hydrogen and the value of Avogadro's number.

Short Answer

Expert verified
(a) Oxygen would be 15.915 u; Avogadro's number stays the same. (b) Hydrogen is 1.008 u; Avogadro's number remains unchanged.

Step by step solution

01

Adjusting Oxygen's Atomic Weight Based on Hydrogen Standard

Given that the current atomic weight of hydrogen is \(1.00794\, \mathrm{u}\) and hydrogen's weight is set to \(1.0000\, \mathrm{u}\) as a standard, we need to adjust the atomic weight of oxygen. The formula for this conversion is:\[ \mathrm{O}_{\text{new}} = \mathrm{O}_{\text{current}} \times \frac{\mathrm{H}_{\text{standard}}}{\mathrm{H}_{\text{current}}} \]Substituting the given values, we find:\[ \mathrm{O}_{\text{new}} = 15.9994 \, \mathrm{u} \times \frac{1.0000}{1.00794} \approx 15.915\, \mathrm{u} \]
02

Calculating Avogadro's Number with Hydrogen as Standard

Avogadro's number can be calculated using the standard molar volume or other fixed values; however, it remains constant as it is defined as the number of atoms in 12g of carbon-12. Thus, when \( \mathrm{H} = 1.0000 \, \text{u} \), Avogadro's number is not influenced directly by the conversion:
03

Adjusting Hydrogen's Atomic Weight Based on Oxygen Standard

Using oxygen's weight set to \(16.0000\, \mathrm{u}\), we convert the atomic weight of hydrogen using a similar formula:\[ \mathrm{H}_{\text{new}} = \mathrm{H}_{\text{current}} \times \frac{\mathrm{O}_{\text{standard}}}{\mathrm{O}_{\text{current}}} \]Substituting the values, we get:\[ \mathrm{H}_{\text{new}} = 1.00794 \, \mathrm{u} \times \frac{16.0000}{15.9994} \approx 1.008\, \mathrm{u} \]
04

Avogadro's Number with Oxygen Standard

Similarly, when \( \mathrm{O} = 16.0000 \, \text{u} \), Avogadro's number remains constant since it is defined independently of these changes in atomic weights. Therefore, Avogadro's number remains \(6.022 \times 10^{23}\, \text{mol}^{-1}\).

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Key Concepts

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

Hydrogen Atomic Weight
In chemistry, the atomic weight of an element is crucial for understanding how it behaves and reacts. Hydrogen, being the simplest and lightest element, often serves as a reference point in these calculations. Its atomic weight is defined according to the mass of its atoms. In current standards, hydrogen's atomic weight is approximately 1.00794 atomic mass units (u). This small value is derived from taking into account the natural isotopic composition of hydrogen.

Historically, when hydrogen's atomic weight was set to exactly 1.0000 u, it served as a standard for comparing the weights of other elements. This practice changed when more advanced methods allowed chemists to measure atomic weights with higher precision.

To see how the atomic weight of one element affects another, we can adjust the atomic weight of oxygen based on hydrogen being the standard. Using the conversion formula:
  • \[\mathrm{O}_{\text{new}} = \mathrm{O}_{\text{current}} \times \frac{\mathrm{H}_{\text{standard}}}{\mathrm{H}_{\text{current}}}\]
Substituting the known values gives us an adjusted atomic weight for oxygen, demonstrating hydrogen's critical role in setting atomic weight standards.
Oxygen Atomic Weight
Oxygen is a vital element, common in the Earth's crust, water, and atmosphere. Its atomic weight is a reflection of its atomic structure and isotopic abundance. The modern accepted atomic weight of oxygen is approximately 15.9994 u.

For a period, oxygen was used as a reference point, assigning it a standard atomic weight of exactly 16.0000. This decision made it easier to compare and classify other elements systematically.

Using oxygen as a standard allows us to calculate the adjusted atomic weight of hydrogen or any other element. The formula we use is:
  • \[\mathrm{H}_{\text{new}} = \mathrm{H}_{\text{current}} \times \frac{\mathrm{O}_{\text{standard}}}{\mathrm{O}_{\text{current}}}\]
By plugging in the values, we can determine the new atomic weight of hydrogen when using oxygen as the standard. This process shows how simply changing the basis of comparison can influence the numeric values of other elements' weights, offering a glimpse into the complex tapestry of atomic theory.
Avogadro's Number
Avogadro's number is a fundamental concept in chemistry that represents the number of atoms or molecules in one mole of a substance. It is a constant value, approximately equal to \(6.022 \times 10^{23}\), and provides a bridge between the atomic scale and the macroscopic world we observe.

Named after the Italian scientist Amedeo Avogadro, this number allows chemists to conduct calculations involving large quantities of atoms or molecules using moles rather than individual particles. For substances like water (\(\text{H}_2\text{O}\)), knowing Avogadro's number helps us predict how much of a substance we have in moles given its mass.

Regardless of which atomic weight standard is used—whether hydrogen or oxygen—as a basis, Avogadro's number remains unchanged. It is fixed because it is derived from carbon-12, where precisely 12 grams contain Avogadro's number of atoms. This stability provides consistency in chemical calculations, ensuring uniformity in scientific communication and research globally.

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Most popular questions from this chapter

Consider the two ionic compounds \(\mathrm{NaCl}\) and \(\mathrm{CaO} .\) In which compound are the cation-anion attractive forces stronger? Explain your answer.

Elemental sulfur \((1.256 \mathrm{g})\) is combined with fluorine, \(\mathrm{F}_{2},\) to give a compound with the formula \(\mathrm{SF}_{x},\) a very stable, colorless gas. If you have isolated \(5.722 \mathrm{g}\) of \(\mathrm{SF}_{x}\) what is the value of \(x ?\)

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