Calculating Atomic Weight: Solving a Chemistry Problem

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SUMMARY

The discussion focuses on calculating the atomic weight of element Y in the compound XY, given the atomic mass of element X as 46.80 g/mol and a sample mass of 35.78 grams. The total mass of the compound XY formed is 119.88 grams (35.78 g of X + 84.10 g of Y). The number of moles of X is calculated using the formula: moles = mass/atomic mass, resulting in 0.764 moles of X. Since the reaction occurs in a 1:1 mole ratio, the atomic weight of Y can be determined by rearranging the equation to find the mass of Y per mole.

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  • Understanding of atomic mass and atomic weight concepts
  • Basic knowledge of stoichiometry
  • Ability to perform mole calculations
  • Familiarity with chemical reaction equations
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  • Study stoichiometry in chemical reactions
  • Learn how to calculate moles from mass and molar mass
  • Explore the differences between atomic mass and atomic weight
  • Practice problems involving mole ratios in chemical reactions
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Big Question!

In my chem class the teacher doesn't explain very much so I decided to see if this site may help. Ok, so my question is: The atomic mass of an element X is 46.80. A 35.78 gram sample of X combines with 84.10 grams of another element Y, to form the compound XY. Calculate the atomic weight of Y.
Explanations are greatly appreciated :)
 
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atomic weight and atomic mass can be used to mean the same thing for purposes of this problem. let y = unknown formula weight, or 'atomic weight' of Y. The X and Y react in a 1:1 mole ratio. How many moles of X was combined in the reaction to form XY? Remember, you were given 35.78 grams of X to react. Enough now to find your set-up and result?
 
A 35.78 gram sample of X combines with 84.10 grams of another element Y, to form the compound XY. Calculate the atomic weight of Y.
Note you made (35.78+84.10) grams of XY. Number of moles = (35.78g)/(46.80g/mole) of XY because the number of moles of X in the reaction is equivalent to the number of moles of XY formed, according to the quantities used in the reaction.
 

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