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How many oxygen atoms are present in 90.1 g of glucose (C6H12O6)?

a. 18.1 x 10²³
b. 3.01 x 10²³
c. 6.02 x 10²³
d. 9.81 x 10²³

1 Answer

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Final answer:

The number of oxygen atoms in 90.1 g of glucose is calculated by dividing the mass of glucose by its molar mass to find the moles, then multiplying by 6 (the number of oxygen atoms in one molecule of glucose) and Avogadro's number, giving a total of 3.01 × 10²³ oxygen atoms.The correct answer is therefore (b) 3.01 × 10²³.

Step-by-step explanation:

To determine how many oxygen atoms are present in 90.1 g of glucose (C6H12O6), we must first find out how many moles of glucose this mass represents. Using the molecular mass of glucose, which is 180.16 g/mol, we can perform this calculation:

Number of moles = mass of substance / molar mass = 90.1 g / 180.16 g/mol = 0.50 mol.

Since one molecule of glucose contains six oxygen atoms, one mole of glucose will contain 6 times Avogadro's number of oxygen atoms. Therefore, we multiply the number of moles of glucose by 6 and then by Avogadro's number (6.022 × 10²³) to find the total number of oxygen atoms:

Total oxygen atoms = moles of glucose × 6 × Avogadro's number = 0.50 mol × 6 × 6.022 × 10²³ = 3.01 × 10²³ oxygen atoms.

The correct answer is therefore (b) 3.01 × 10²³.

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