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How many total atoms are in 181 g of sucrose (C\(_{12}\)H\(_{22}\)O\(_{11}\))?

Answer :

Final answer:

The 181 g of sucrose contains approximately 1.44 x 10^25 atoms. This has been calculated using the molar mass and Avogadro's number.

Explanation:

First, let's understand that mol stands for mole, which is a unit representing 6.02 x 10^23 entities, be it atoms or molecules. In chemistry, it is commonly used to calculate the number of atoms or molecules in a compound. The molecular formula for sucrose is C12H22O11. This means that for each molecule of sucrose, it contains 12 carbon (C) atoms, 22 hydrogen (H) atoms and 11 oxygen (O) atoms, making a total of 45 atoms per molecule of sucrose.

The molar mass of sucrose (C12H22O11) is about 342.30 g/mol. So, the number of moles in 181 g of sucrose will be 181 g / 342.30 g/mol = 0.53 mol. This represents the amount of sucrose molecules present in 181g of sucrose.

Then, if we want to get the total number of atoms in this 181g of sucrose, we have to multiply the total number of atoms in one molecule of sucrose (which is 45) by the number of moles we have (which is 0.53 mol) and also by Avogadro's number (6.02 x 10^23), as 1 mol contains that number of entities. Hence, the total number of atoms will be 45 x 0.53 x 6.02 x 10^23 = 1.44 x 10^25 atoms.

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