Fun Sheet to Accompany Chapter 3
31 Atomic Mass
1) Chlorine exists as a mixture of 2 major isotopes, Chlorine-35 and Chlorine-37. Calculate the average atontic mass for Chlorine. See Problem 3.5
Isotope
35
17
37C1
Explain why the atomic masses on the periodic table are decimals.
3.2 Molar Mass of an Element and Avogadro’s Number
2) Setting up using the conversion factor between the number of atoms in a mole & Avogadro’s number, convert 6.00 x 10″ atoms of Cobalt (Co) to moles of cobalt atom. See Problem 3.14
3) Setting up using the conversion factarx of molar mass and/or Avogadro’s number:
a) A modern penny weighs 2.5 g, but contains only 0.63 g of copper(Cu). How many copper atoms are present in a modern penny? See Problem 3.20
b) Determine the number of atoms of sulfur in 5.10 moles of sulfur? See Problem 3.13
4) Which sample has more atoms: 1.10 g of hydrogen atoma, or 14.7 g of chromium atoms7See Problem 3:21
33 Molecular Mass
5) Calculate the molecular mass (in amu) and molar mass(in grams) of each of the following CHA
substances: See Problem 3.23-24
b) K-50
c)
Cas(PO)
Abundance(%)
75.53
24.47
34.968
36.956
The Correct Answer and Explanation is :
Question 1: Atomic Mass of Chlorine (Cl)
Chlorine exists as a mixture of two isotopes, Chlorine-35 and Chlorine-37, with the following natural abundances:
- Chlorine-35 has an abundance of 75.53%.
- Chlorine-37 has an abundance of 24.47%.
To calculate the average atomic mass, we use the formula for the weighted average: Average Atomic Mass=(m1×f1)+(m2×f2)\text{Average Atomic Mass} = (m_1 \times f_1) + (m_2 \times f_2)
Where:
- m1m_1 and m2m_2 are the atomic masses of Chlorine-35 (34.968 amu) and Chlorine-37 (36.956 amu), respectively.
- f1f_1 and f2f_2 are the fractional abundances of Chlorine-35 and Chlorine-37.
Substituting the values: Average Atomic Mass=(34.968 amu×0.7553)+(36.956 amu×0.2447)\text{Average Atomic Mass} = (34.968 \, \text{amu} \times 0.7553) + (36.956 \, \text{amu} \times 0.2447) Average Atomic Mass=26.409 amu+9.038 amu=35.447 amu\text{Average Atomic Mass} = 26.409 \, \text{amu} + 9.038 \, \text{amu} = 35.447 \, \text{amu}
So, the average atomic mass of Chlorine is 35.447 amu.
Explanation for Decimals on the Periodic Table: Atomic masses on the periodic table are often decimal numbers because they represent weighted averages of the atomic masses of all naturally occurring isotopes of an element. Since elements can have multiple isotopes with different abundances, the atomic mass reflects these variations.
Question 2: Moles of Cobalt (Co)
Given:
- Atoms of Cobalt (Co) = 6.00×10246.00 \times 10^{24}
- Avogadro’s Number = 6.022×10236.022 \times 10^{23} atoms per mole.
To convert atoms to moles, we use the conversion factor: Moles of Co=6.00×1024 atoms6.022×1023 atoms/mol=9.97 moles\text{Moles of Co} = \frac{6.00 \times 10^{24} \, \text{atoms}}{6.022 \times 10^{23} \, \text{atoms/mol}} = 9.97 \, \text{moles}
So, the number of moles of cobalt is 9.97 moles.
Question 3a: Copper Atoms in a Penny
Given:
- Mass of copper in the penny = 0.63 g
- Molar mass of copper (Cu) = 63.55 g/mol
- Avogadro’s Number = 6.022×10236.022 \times 10^{23} atoms/mol.
First, convert grams of copper to moles using molar mass: Moles of Cu=0.63 g63.55 g/mol=0.00991 mol\text{Moles of Cu} = \frac{0.63 \, \text{g}}{63.55 \, \text{g/mol}} = 0.00991 \, \text{mol}
Next, convert moles to atoms: Atoms of Cu=0.00991 mol×6.022×1023 atoms/mol=5.96×1021 atoms\text{Atoms of Cu} = 0.00991 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 5.96 \times 10^{21} \, \text{atoms}
So, there are approximately 5.96 × 10²¹ atoms of copper in the penny.
Question 3b: Atoms of Sulfur in 5.10 Moles
Given:
- Moles of sulfur = 5.10 mol
- Avogadro’s Number = 6.022×10236.022 \times 10^{23} atoms/mol.
To find the number of atoms: Atoms of S=5.10 mol×6.022×1023 atoms/mol=3.07×1024 atoms\text{Atoms of S} = 5.10 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 3.07 \times 10^{24} \, \text{atoms}
So, there are 3.07 × 10²⁴ atoms of sulfur.
Question 4: More Atoms of Hydrogen or Chromium?
Given:
- 1.10 g of hydrogen
- 14.7 g of chromium
For hydrogen (H):
- Molar mass of hydrogen = 1.008 g/mol
- Moles of hydrogen = 1.10 g1.008 g/mol=1.092 mol\frac{1.10 \, \text{g}}{1.008 \, \text{g/mol}} = 1.092 \, \text{mol}
- Atoms of hydrogen = 1.092 mol×6.022×1023 atoms/mol=6.57×1023 atoms1.092 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 6.57 \times 10^{23} \, \text{atoms}
For chromium (Cr):
- Molar mass of chromium = 52.00 g/mol
- Moles of chromium = 14.7 g52.00 g/mol=0.2827 mol\frac{14.7 \, \text{g}}{52.00 \, \text{g/mol}} = 0.2827 \, \text{mol}
- Atoms of chromium = 0.2827 mol×6.022×1023 atoms/mol=1.70×1023 atoms0.2827 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 1.70 \times 10^{23} \, \text{atoms}
Conclusion: There are more atoms in 1.10 g of hydrogen (6.57×10236.57 \times 10^{23}) than in 14.7 g of chromium (1.70×10231.70 \times 10^{23}).
Question 5: Molecular Mass
For molecular mass calculations, sum the atomic masses of all atoms in the formula. For example:
For C2H6\text{C}_2\text{H}_6 (ethane):
- Carbon = 12.01 g/mol, Hydrogen = 1.008 g/mol.
- Molecular mass = 2×12.01+6×1.008=24.02+6.048=30.07 g/mol2 \times 12.01 + 6 \times 1.008 = 24.02 + 6.048 = 30.07 \, \text{g/mol}.
This is the process for determining molecular mass.