Answer:
Question 1:
A) 0.289 moles.
B) 1.74 x 10²³ atoms.
Question 2:
A) 0.30 moles.
B) it contains 0.3 moles of both Na and Cl.
C) it contains 6.023 x 10²³ atoms of both Na and Cl.
Question 3:
A) The number of moles of sucrose (C₁₂H₂₂O₁₁) ≅ 0.0228 moles.
B) The number of moles of C atoms in sucrose (C₁₂H₂₂O₁₁) = 0.2763 mole of C atoms.
The number of moles of H atoms in sucrose (C₁₂H₂₂O₁₁) = 0.5016 mole of H atoms.
The number of moles of O atoms in sucrose (C₁₂H₂₂O₁₁) = 0.2508 mole of O atoms.
C) The number of C atoms = 1.65 x 10²³ atoms.
The number of H atoms = 3.02 x 10²³ atoms.
The number of O atoms = 1.51 x 10²³ atoms.
Step-by-step explanation:
Question 1:
A) The number of moles of Au in 57.01 g sample:
n = mass / molar mass,
mass = 57.01 g and molar mass = 196.966 g/mol e.
The number of moles of Au in the sample = (57.01 g) / (196.966 g/mole) = 0.289 moles.
B) The number of atoms of Au in the sample:
It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.
1.0 mole of Au → 6.023 x 10²³ atoms
0.289 mole of Au → ???? atoms
using cross multiplication:
The number of atoms of Au in the sample = (6.023 x 10²³ x 0.289 mole) / (1.0 mole) = 1.74 x 10²³ atoms.
Question 2:
A) The number of moles of 17.45 g of NaCl:
n = mass / molar mass,
mass = 17.45 g and molar mass = 58.44 g/mole.
The number of moles of NaCl = (17.45 g) / (58.44 g/mole) = 0.298 mole ≅ 0.30 moles.
B) The number of moles of each element in NaCl
NaCl → Na + Cl
Each mole of NaCl contains one mole of Na and one mole of Cl.
using cross multiplication:
1.0 mole NaCl → 1.0 mole Na
0.3 mole NaCl → ??? mole Na
The number of moles of Na atoms in NaCl = (1.0 mole Na x 0.3 mole NaCl) / (1.0 mole NaCl) = 0.3 mole of Na atoms.
by the same way; the number of moles of Cl atoms = (1.0 mole Cl x 0.3 mole NaCl) / (1.0 mole NaCl) = 0.3 mole of Cl atoms.
C) The number of atoms of each element in the sample:
It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.
1.0 mole of NaCl → 6.023 x 10²³ molecules
0.3 mole of NaCl → ???? molecules
using cross multiplication:
The number of molecules in 0.3 mole of NaCl = (6.023 x 10²³ x 0.3 mole) / (1.0 mole) = 1.8069 x 10²³ molecules.
Every molecule of NaCl contains one atom of Na and one atom of Cl.
So, it contains 6.023 x 10²³ atoms of both Na and Cl.
Question 3:
A) The number of moles of 7.801 g of sucrose (C₁₂H₂₂O₁₁):
n = mass / molar mass,
mass = 7.801 g and molar mass = 342.3 g/mole.
The number of moles of sucrose (C₁₂H₂₂O₁₁) = (7.801 g) / (342.3 g/mol) = 0.022789 mol ≅ 0.0228 moles.
B) The number of moles of each element in sucrose (C₁₂H₂₂O₁₁):
C₁₂H₂₂O₁₁ → 12C + 22H + 11O
Each mole of sucrose contains 12 moles of C, 22 moles of H, and 11 moles of O.
- using cross multiplication:
1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 12.0 moles C
0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles C
The number of moles of C atoms in sucrose (C₁₂H₂₂O₁₁) = (12.0 moles C x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.2763 mole of C atoms.
- By the same way; the number of moles of H atoms:
1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 22.0 moles H
0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles H
The number of moles of H atoms in sucrose (C₁₂H₂₂O₁₁) = (22.0 moles H x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.5016 mole of H atoms.
- Also; the number of moles of O atoms:
1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 11.0 moles O
0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles O
The number of moles of O atoms in sucrose (C₁₂H₂₂O₁₁) = (11.0 moles H x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.2508 mole of O atoms.
C) The number of atoms of each element in the sucrose (C₁₂H₂₂O₁₁) sample:
It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.
1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 6.023 x 10²³ molecules
0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ???? molecules
using cross multiplication:
The number of molecules in 0.0228 mole of sucrose (C₁₂H₂₂O₁₁) = (6.023 x 10²³ x 0.0228 mole) / (1.0 mole) = 1.273 x 10²² molecules.
Each molecule of sucrose contains 12 atoms of C, 22 atoms of H, and 11 atoms of O.
So, the number of each atom that the sucrose (C₁₂H₂₂O₁₁) sample contains are:
The number of C atoms = (12 x 1.273 x 10²² molecules) = 1.65 x 10²³ atoms.
The number of H atoms = (22 x 1.273 x 10²² molecules) = 3.02 x 10²³ atoms.
The number of O atoms = (11 x 1.273 x 10²² molecules) = 1.51 x 10²³ atoms.