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Chapter 3: Problem 27

Calculate the number of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) atoms in \(1.50 \mathrm{~g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\), a sugar.

Short Answer

Expert verified
The given amount of glucose contains approximately \(3.01 \times 10^{23}\) C atoms, \(6.01 \times 10^{23}\) H atoms, and \(3.01 \times 10^{23}\) O atoms.

Step by step solution

01

Calculate the molecular weight of glucose

The molecular weight of glucose is calculated by adding the weighted averages of the elements found in it, which are Carbon (C), Hydrogen (H), and Oxygen (O). Their molecular weights are approximately 12.01 g/mol, 1.008 g/mol, and 16.00 g/mol respectively. The molecule of glucose contains 6 carbon atoms(C), 12 hydrogen atoms(H), and 6 oxygen atoms(O). Therefore, the molecular weight of glucose is \( (6 \times 12.01) + (12 \times 1.008) + (6 \times 16.00) = 180.156 \, g/mol\)
02

Calculate the moles of glucose

Using the molecular weight of glucose, the moles of glucose can be calculated. Given that the moles are computed by dividing the mass by the molecular weight, 1.50 g (mass of glucose) divided by 180.156 g/mol (molecular weight of glucose) equals \( \approx 0.00832 \) moles of glucose
03

Calculate the number of atoms

The number of atoms in a mole is defined as Avogadro's number, which is approximately \( 6.022 \times 10^{23} \). To find the number of each type of atom in the sample of glucose, multiply the number of moles by Avogadro's number, then by the number of those atoms in the glucose molecule. This gives \( 0.00832 \times 6 \times 6.022 \times 10^{23} \approx 3.01 \times 10^{23} \) C atoms, \( 0.00832 \times 12 \times 6.022 \times 10^{23} \approx 6.01 \times 10^{23} \) H atoms, and \( 0.00832 \times 6 \times 6.022 \times 10^{23} \approx 3.01 \times 10^{23} \) O atoms.

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