a) Calculate the molar mass of sucrose, C12H22O11.
(b) How many moles of sucrose are present in:
- (i) 1.00 gram;
- (ii) 1.00 pound;
- (iii) 1.00 kg of sucrose?
(c) How many sucrose molecules are present in a 4.536-g sample of sucrose?
(d) How many atoms of carbon, hydrogen, and oxygen, respectively, are present in 4.536 g of sucrose?
(1 lb = 453.6 g; Avogadro’s number = 6.022 x 1023/mol)
The correct answer and explanation is:
Answers:
(a) The molar mass of sucrose (C₁₂H₂₂O₁₁) is 342.296 g/mol.
(b) The number of moles of sucrose present in:
- (i) 1.00 g of sucrose = 0.00292 moles
- (ii) 1.00 lb (453.6 g) of sucrose = 1.325 moles
- (iii) 1.00 kg (1000 g) of sucrose = 2.921 moles
(c) The number of sucrose molecules in a 4.536 g sample:
- 7.98 × 10²¹ molecules
(d) The number of atoms in 4.536 g of sucrose:
- Carbon (C): 9.58 × 10²² atoms
- Hydrogen (H): 1.76 × 10²³ atoms
- Oxygen (O): 8.78 × 10²² atoms
Explanation:
Sucrose is composed of carbon (C), hydrogen (H), and oxygen (O) with a molecular formula of C₁₂H₂₂O₁₁. To determine its molar mass, we sum the atomic masses of its elements: (12×12.01)+(22×1.008)+(11×16.00)=342.296 g/mol(12 \times 12.01) + (22 \times 1.008) + (11 \times 16.00) = 342.296 \text{ g/mol}
For part (b), the number of moles is found using the formula: Moles=MassMolar Mass\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}
This calculation is applied to different masses (1 g, 1 lb, and 1 kg) of sucrose.
In part (c), we use Avogadro’s number (6.022 × 10²³ molecules/mol) to convert moles to molecules: Molecules=Moles×6.022×1023\text{Molecules} = \text{Moles} \times 6.022 \times 10^{23}
Finally, for part (d), we determine the individual atom counts by multiplying the number of molecules by the number of atoms per sucrose molecule.