Let's solve **Question 1(a)** in two clear steps as requested. --- ## 1(a) Calculate the monoisotopic mass of C₇H₇OH in kg and unified atomic mass unit (u). ### **Step 1: Calculate the monoisotopic mass in unified atomic mass units (u)** **Monoisotopic mass** uses the most abundant (lightest) isotope for each element. - **C₇H₇OH** can be rewritten as **C₇H₈O** (Benzyl alcohol| Element | No. of atoms | Isotope Used | Mass (u) | Contribution (u) | |---------|--------------|--------------|--------------|---------------------| | Carbon | 7 | ¹²C | 12.000 | 7 × 12.000 = 84.000 | | Hydrogen| 8 | ¹H | 1.008 | 8 × 1.008 = 8.064 | | Oxygen | 1 | ¹⁶O | 15.995 | 1 × 15.995 = 15.995 | **Total monoisotopic mass (u):** \[ 84.000 + 8.064 + 15.995 = 108.059 \, \text{u} \] **Explanation:** We summed the mass of the most common isotope for each atom in the formula. --- ### **Step 2: Convert the monoisotopic mass to kilograms (kg)** **1 unified atomic mass unit (u) = \( 1.66053906660 \times 10^{-27} \) kg** So, \[ \text{Mass in kg} = 108.059 \times 1.66053906660 \times 10^{-27} \text{ kg} \] \[ = 1.794 \times 10^{-25} \text{ kg} \] **Explanation:** We multiply the mass in atomic mass units by the conversion factor to get the mass in kilograms. --- ### **Final Answers** - **Monoisotopic mass in u:** \( \boxed{108.059 \, \text{u}} \) - **Monoisotopic mass in kg:** \( \boxed{1.79 \times 10^{-25} \, \text{kg}} \) (rounded to 3 significant figures) --- Let me know if you need a step-by-step for part (b) or the others!

## Question 1 ### 1(a) Calculate the monoisotopic mass of C₇H₈O in kg and unified atomic mass unit (u). #### **Step 1: Calculate the monoisotopic mass in unified atomic mass units (u)** - **Molecular formula:** C₇H₈O - **Element Contributions:** | Element | No. of Atoms | Isotope Used | Atomic Mass (u) | Contribution (u) | |-----------|--------------|--------------|------------------|---------------------| | Carbon | 7 | ¹²C | 12.000 | \( 7 \times 12.000 = 84.000 \) | | Hydrogen | 8 | ¹H | 1.008 | \( 8 \times 1.008 = 8.064 \) | | Oxygen | 1 | ¹⁶O | 15.995 | \( 1 \times 15.995 = 15.995 \) | **Total monoisotopic mass (u):** \[ 84.000 + 8.064 + 15.995 = 108.059 \, \text{u} \] **Explanation:** The contributions of each element are calculated using the most abundant isotopes. --- #### **Step 2: Convert the monoisotopic mass to kilograms (kg)** **Conversion factor:** 1 unified atomic mass unit (u) = \( 1.66053906660 \times 10^{-27} \) kg **Calculation:** \[ \text{Mass in kg} = 108.059 \times 1.66053906660 \times 10^{-27} \text{ kg} \] \[ = 1.794 \times 10^{-25} \text{ kg} \] **Explanation:** The mass in atomic mass units is converted to kilograms using the known conversion factor. --- ### **Final Answers for 1(a)** - **Monoisotopic mass in u:** \( \boxed{108.059 \, \text{u}} \) - **Monoisotopic mass in kg:** \( \boxed{1.79 \times 10^{-25} \, \text{kg}} \) (rounded to 3 significant figures) --- ### 1(b) Determine which reagent gases will ionize Benzyl alcohol C₇H₈O and propose reagent ions. #### **Step 1: Compare the proton affinity (PA)** - **Benzyl alcohol PA = 186 kcal/mol** - **Reagent gases:** - **H₂** (PA = 101 kcal/mol) - **CH₄** (PA = 130 kcal/mol) - **NH₃** (PA = 204 kcal/mol) **Reagent selection:** - Only NH₃ (PA = 204 kcal/mol) has a higher PA than Benzyl alcohol (186 kcal/mol). **Explanation:** A reagent gas can ionize the alcohol only if its proton affinity is greater than that of the alcohol. --- #### **Step 2: Propose the reagent ions** **Reaction with NH₃:** \[ \text{C₇H₈O} + \text{NH}_3 \rightarrow \text{C₇H₉O}^+ + \text{NH}_2^- \] **Reagents involved:** - **Reagent ion:** \(\text{C₇H₉O}^+\) (protonated Benzyl alcohol) - **By-product:** \(\text{NH}_2^-\) (deprotonated ammonia) **Explanation:** The reaction shows how Benzyl alcohol can be ionized in the presence of NH₃ by forming a protonated species. --- ### **Final Answer for 1(b)** - **Reagent gas that will ionize Benzyl alcohol:** **NH₃** - **Proposed reaction:** \(\text{C₇H₈O} + \text{NH}_3 \rightarrow \text{C₇H₉O}^+ + \text{NH}_2^-\) --- ### 1(c) Sketch the isotopic distributions of triply charged cations of C₇H₈O and C₇H₅Cl (Benzyl chloride). #### **Step 1: Determine isotopic patterns for C₇H₈O** - **Major isotopes:** - **C:** ¹²C, ¹³C (approximately 1.1% abundance) - **H:** ¹H, ²H (approximately 0.015% abundance) - **O:** ¹⁶O (dominant) **Isotopic distribution:** - **C₇H₈O:** Major peak for \((C_7H_8O)^{3+}\), minor peaks for \((C_7H_7^{13}C H_8O)^{3+}, (C_6^{13}C H_8O)^{3+}\), etc. #### **Step 2: Determine isotopic patterns for C₇H₅Cl** - **Chlorine isotopes:** - \(^3\text{Cl}\) (75% abundance) and \(^35\text{Cl}\) (25% abundance) **Isotopic distribution:** - **C₇H₅Cl:** Peaks for \((C_7H_5^{35}Cl)^{3+}\) and \((C_7H_5^{37}Cl)^{3+}\). **Sketch:** - The x-axis represents mass/charge ratio, and the y-axis shows intensity. - Major peaks at \(108.059/3\) and minor peaks for isotopes. **Explanation:** The distribution reflects the natural abundance of isotopes contributing to the molecular ion peaks. --- ### **Final Answer for 1(c)** - **Isotopic distributions:** - **C₇H₈O:** Peaks at \(36.02, 36.03, ...\) (triply charged) - **C₇H₅Cl:** Peaks at \(36.02, 36.03, ...\) based on Cl isotopes. --- ### 1(d) Features required in a MALDI matrix #### **Step 1: Key features** 1. **Absorption of laser energy:** - Should efficiently absorb the laser wavelength used for ionization. 2. **Low molecular weight:** - Minimizes interference with the analyte mass. 3. **Solubility in solvent:** - Should dissolve analyte well for uniform sample preparation. 4. **Stability:** - Must remain stable under laser irradiation without decomposing. **Explanation:** These features ensure efficient ionization and accurate mass detection of the analyte in MALDI-TOF mass spectrometry. --- ### **Final Answer for 1(d)** - **Required features in a MALDI matrix:** 1. Absorption of laser energy 2. Low molecular weight 3. Solubility in solvent 4. Stability under laser irradiation --- This concludes the solutions for parts a, b, c, and d of the question. Let me know if you need further assistance!