1. If Ar for Cl is 35.45 , what is the ratio of $$ 35_{\mathrm{Cl}}: 37_{\mathrm{Cl}} $$ present in a sample of Cl atoms containing naturally occurring Cl ? 2. Calculate the value of Ar for naturally occurring Cu if the distribution of isotopes is $$ 69.2 \% 63_{\mathrm{Cu}} $$ and $$ 30.8 \% 65_{\mathrm{Cu}} $$; accurate masses are 62.93 and 64.93 . 3. Why in question 2 is it adequate to write $$ 63_{\mathrm{Cu}} $$ rather than $$ { }_{29}^{63} \mathrm{Cu} $$ ? 4. Calculate Ar for naturally occurring Mg if the isotope distribution is $$ 78.99 \% 24_{\mathrm{Mg}}, 10.00 \% 25_{\mathrm{Mg}} $$ and $$ 11.01 \% 26_{M g} $$; accurate masses are $$ 23.99,24.99 $$ and 25.98 .

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---

### **1. Ratio of $$ ^{35}\mathrm{Cl} : ^{37}\mathrm{Cl} $$ in Naturally Occurring Chlorine**

**Given:**
- Atomic mass of Cl, $$ \text{Ar}_{\text{Cl}} = 35.45 $$
- Isotopes: $$ ^{35}\mathrm{Cl} $$ and $$ ^{37}\mathrm{Cl} $$

**Let:**
- Abundance of $$ ^{35}\mathrm{Cl} $$ = $$ x $$
- Abundance of $$ ^{37}\mathrm{Cl} $$ = $$ 1 - x $$

**Calculation:**
The atomic mass is the weighted average:
$$
35x + 37(1 - x) = 35.45
$$
$$
35x + 37 - 37x = 35.45
$$
$$
-2x = -1.55
$$
$$
x = 0.775 \quad \text{or} \quad 77.5\%
$$
$$
\text{Abundance of } ^{37}\mathrm{Cl} = 22.5\%
$$

**Ratio:**
$$
\frac{^{35}\mathrm{Cl}}{^{37}\mathrm{Cl}} = \frac{77.5}{22.5} = \frac{31}{9} \approx 3.44 : 1
$$

**Final Answer:**
The natural abundance ratio of $$ ^{35}\mathrm{Cl} : ^{37}\mathrm{Cl} $$ is approximately **3.44 : 1**.

---

### **2. Calculating the Atomic Mass ($$ \text{Ar} $$) for Naturally Occurring Copper (Cu)**

**Given:**
- Isotope distribution:
  - $$ ^{63}\mathrm{Cu} $$: 69.2%
  - $$ ^{65}\mathrm{Cu} $$: 30.8%
- Accurate masses:
  - $$ ^{63}\mathrm{Cu} $$: 62.93 u
  - $$ ^{65}\mathrm{Cu} $$: 64.93 u

**Calculation:**
$$
\text{Ar}_{\text{Cu}} = (0.692 \times 62.93) + (0.308 \times 64.93)
$$
$$
= 43.57 + 19.99
$$
$$
= 63.56 \, \text{u}
$$

**Final Answer:**
The atomic mass of naturally occurring copper is **63.56 u**.

---

### **3. Notation of Copper Isotopes: Why $$ ^{63}\mathrm{Cu} $$ Suffices**

When performing calculations related to atomic masses and isotope abundances, it's sufficient to denote isotopes by their mass number alone (e.g., $$ ^{63}\mathrm{Cu} $$) because the atomic number (which defines the element) is already indicated by the chemical symbol "Cu". Including the full nuclear notation (e.g., $$ _{29}^{63}\mathrm{Cu} $$) is unnecessary for these purposes since the chemistry and mass considerations don't require information about the number of protons explicitly.

**Final Answer:**
Including only the mass number (e.g., $$ ^{63}\mathrm{Cu} $$) is sufficient because the element’s identity and mass number uniquely specify the isotope, making full nuclear notation unnecessary for atomic mass calculations.

---

### **4. Calculating the Atomic Mass ($$ \text{Ar} $$) for Naturally Occurring Magnesium (Mg)**

**Given:**
- Isotope distribution:
  - $$ ^{24}\mathrm{Mg} $$: 78.99%
  - $$ ^{25}\mathrm{Mg} $$: 10.00%
  - $$ ^{26}\mathrm{Mg} $$: 11.01%
- Accurate masses:
  - $$ ^{24}\mathrm{Mg} $$: 23.99 u
  - $$ ^{25}\mathrm{Mg} $$: 24.99 u
  - $$ ^{26}\mathrm{Mg} $$: 25.98 u

**Calculation:**
$$
\text{Ar}_{\text{Mg}} = (0.7899 \times 23.99) + (0.1000 \times 24.99) + (0.1101 \times 25.98)
$$
$$
= 18.948 \, \text{u} + 2.499 \, \text{u} + 2.859 \, \text{u}
$$
$$
= 24.306 \, \text{u}
$$

**Final Answer:**
The atomic mass of naturally occurring magnesium is **24.31 u**.

---
1. The ratio of $$ ^{35}\mathrm{Cl} : ^{37}\mathrm{Cl} $$ is approximately **3.44 : 1**. 2. The atomic mass of naturally occurring copper is **63.56 u**. 3. It's sufficient to use $$ ^{63}\mathrm{Cu} $$ instead of $$ _{29}^{63}\mathrm{Cu} $$ because the element's identity is already clear from the chemical symbol. 4. The atomic mass of naturally occurring magnesium is **24.31 u**.

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