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The atomic mass of element \(\mathrm{X}\) is 33.42 amu. A \(27.22-\mathrm{g}\) sample of \(\mathrm{X}\) combines with \(84.10 \mathrm{~g}\) of another element \(\mathrm{Y}\) to form a compound XY. Calculate the atomic mass of Y.

Short Answer

Expert verified
The atomic mass of Y is approximately 103.34 amu.

Step by step solution

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01

Write the chemical equation

The compound formed is XY, which means one atom of element X combines with one atom of element Y.
02

Find the number of moles of X

The number of moles of element X in the sample can be found by dividing the mass of X by its atomic mass. \[ \text{Moles of X} = \frac{27.22 \text{ g}}{33.42 \text{ g/mol}} \approx 0.814 \text{ moles} \]
03

Use the mole ratio to find moles of Y

Since the compound is XY, the number of moles of Y will be the same as that of X. Therefore, the moles of Y are also 0.814.
04

Calculate the atomic mass of Y

The atomic mass of Y can be calculated by dividing the mass of Y by the moles of Y: \[ \text{Atomic mass of Y} = \frac{84.10 \text{ g}}{0.814 \text{ moles}} \approx 103.34 \text{ g/mol} \]
05

Conclusion

The atomic mass of element Y has been calculated to be approximately 103.34 amu.

Key Concepts

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

Chemical Equation
When studying chemistry, one of the fundamental concepts is the chemical equation. A chemical equation symbolizes a chemical reaction. It showcases the reactants (starting materials) and products (end materials) involved. Each substance in the equation is represented by its chemical formula. The chemical equation also indicates the proportions in which elements and compounds react or form. For example, consider the equation for our compound, XY. In this scenario, one atom of element X reacts with one atom of element Y to form the compound XY. This simple representation highlights the fixed ratio and type of atoms involved in the chemical change.
Moles of an Element
The mole is an essential concept in chemistry used to quantify the amount of substance. It's similar to how we use "dozen" to count eggs. One mole is equivalent to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities, like atoms or molecules. To find the number of moles, you can use the formula:
  • Moles = \( \frac{\text{Mass}}{\text{Atomic or Molar Mass}} \)
For element X, whose atomic mass is 33.42 amu, and given a mass of 27.22 grams, the number of moles can be calculated as follows:\[ \text{Moles of X} = \frac{27.22 \text{ g}}{33.42 \text{ g/mol}} \approx 0.814 \text{ moles} \]This calculation allows us to transition from the macroscopic world of grams to the atomic scale.
Mole Ratio
In a chemical equation, the mole ratio tells you how many moles of one substance are related to moles of another substance. The mole ratio is derived from the coefficients in the balanced chemical equation. For our compound XY, there are no coefficients in front of the elements, meaning the ratio is 1:1. This indicates that one mole of element X combines with one mole of element Y to form compound XY. The mole ratio plays a crucial role in converting between quantities of different substances. In our calculation, since the moles of X are 0.814, the moles of Y must also be 0.814 because of the 1:1 ratio. This equality allows us to understand the interaction between elements X and Y in producing the compound.
Compounds
A compound is a substance composed of two or more different elements that are chemically bonded. In compounds, elements combine in fixed ratios to form a substance with unique properties different from its constituent elements. In this exercise, the compound in question is XY. It is formed when an atom from element X combines with an atom from element Y. Compounds can be represented by their chemical formulas, which detail the elements involved and the ratio of atoms present. Knowing the total mass of a compound and how its components contribute to this mass enables us to assess atomic or molecular masses. Here, knowing the atomic mass of X and the total mass provided the means to calculate the atomic mass of Y, illustrating how elemental properties integrate within compound structures.

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Most popular questions from this chapter

Leaded gasoline contains an additive to prevent engine "knocking." On analysis, the additive compound is found to contain carbon, hydrogen, and lead (Pb) (hence, "leaded gasoline"). When \(51.36 \mathrm{~g}\) of this compound is burned in an apparatus such as that shown in Figure \(3.5,55.90 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(28.61 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are produced. Determine the empirical formula of the gasoline additive. Because of its detrimental effect on the environment, the original lead additive has been replaced in recent years by methyl tert-butyl ether (a compound of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) ) to enhance the performance of gasoline. (As of \(1999,\) this compound is also being phased out because of its contamination of drinking water.) When \(12.1 \mathrm{~g}\) of the compound is burned in an apparatus like the one shown in Figure \(3.5,30.2 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(14.8 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are formed. What is the empirical formula of this compound?

A sample of \(10.0 \mathrm{~g}\) of sodium reacts with oxygen to form \(13.83 \mathrm{~g}\) of sodium oxide \(\left(\mathrm{Na}_{2} \mathrm{O}\right)\) and sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right) .\) Calculate the percent composition of the product mixture.

Determine whether each of the following equations represents a combination reaction, a decomposition reaction, or a combustion reaction: (a) \(\mathrm{C}_{3} \mathrm{H}_{8}+\) \(5 \mathrm{O}_{2} \longrightarrow 3 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O},(\mathrm{b}) 2 \mathrm{NF}_{2} \longrightarrow \mathrm{N}_{2} \mathrm{~F}_{4}\) (c) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CuSO}_{4}+5 \mathrm{H}_{2} \mathrm{O} .\)

Determine whether each of the following equations represents a combination reaction, a decomposition reaction, or a combustion reaction: (a) \(2 \mathrm{NaHCO}_{3} \longrightarrow\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O},(\mathrm{b}) \mathrm{NH}_{3}+\mathrm{HCl} \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}\) (c) \(2 \mathrm{CH}_{3} \mathrm{OH}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O}\)

Which of the following has the greater mass: \(0.72 \mathrm{~g}\) of \(\mathrm{O}_{2}\) or \(0.0011 \mathrm{~mol}\) of chlorophyll \(\left(\mathrm{C}_{55} \mathrm{H}_{72} \mathrm{MgN}_{4} \mathrm{O}_{5}\right) ?\)

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