Chem1A Fall 2021 Quiz 1 Notes

Quiz 1 Notes

Part A: Molarity

• Children's Tylenol oral suspension contains 160 mg of acetaminophen per 5 mL.
• Molar mass of acetaminophen is 151.163 g/mol.

1.1 Dosage Calculation

• If a doctor recommends 3.75 mL of the suspension, the calculation to find the milligrams of acetaminophen is as follows:
  $\frac{160 \text{mg}}{5 \text{mL}} \times 3.75 \text{mL} = 120 \text{mg}$

1.2 Molarity Calculation

• To calculate the molarity of children’s Tylenol:
  • First, convert mg to grams: $160 \text{ mg} = 0.16 \text{ g}$.
  • Then, calculate moles of acetaminophen: $\frac{0.16 \text{ g}}{151.163 \text{ g/mol}} = 1.058 \times 10^{-3} \text{ mol}$.
  • Finally, calculate molarity: $\frac{1.058 \times 10^{-3} \text{ mol}}{0.005 \text{ L}} = 0.21 \text{ M}$.

Part B: Isotopes and Average Atomic Mass

2.1 Chromium Isotope Abundance

• The most abundant isotope of Chromium is the one closest to the mass on the periodic table.
• Cr52 is the most abundant because its mass is closest to the value listed on the periodic table for Chromium.

2.2 Gallium Average Atomic Mass

• Gallium has two isotopes: 69Ga (68.9256 amu, 60.11%) and 71Ga (70.9247 amu, 39.89%).
• The average atomic mass is calculated as:
  $(68.9256 \times 0.6011) + (70.9247 \times 0.3989) = 69.723 \text{ amu}$

Part C: Converting Reactant to Product

3.1 Balanced Equation for Rust Formation

• The balanced equation for the reaction of solid iron with oxygen gas to produce solid iron(III) oxide (rust) is:
  $4Fe(s) + 3O<em>2(g) \rightarrow 2Fe</em>2O_3(s)$

3.2 Mass of Iron(III) Oxide Formed

• Starting with 5.0 g of iron:
  • Convert grams of Fe to moles: $\frac{5.0 \text{ g Fe}}{55.845 \text{ g/mol}} = 0.0895 \text{ mol Fe}$.
  • Use stoichiometry to find moles of $Fe<em>2O</em>3$: $\frac{0.0895 \text{ mol Fe}}{1} \times \frac{2 \text{ mol } Fe<em>2O</em>3}{4 \text{ mol Fe}} = 0.04475 \text{ mol } Fe<em>2O</em>3$
  • Convert moles of $Fe<em>2O</em>3$ to grams: $0.04475 \text{ mol } Fe<em>2O</em>3 \times 159.69 \text{ g/mol} = 7.146 \text{ g } Fe<em>2O</em>3$

3.3 Moles of Oxygen Gas Required

• To find moles of oxygen gas required to react with 5.0 g of iron:
  • Use stoichiometry: $\frac{0.0895 \text{ mol Fe}}{1} \times \frac{3 \text{ mol } O<em>2}{4 \text{ mol Fe}} = 0.0671 \text{ mol } O</em>2$

Part D: Limiting Reactant

4.1 Balanced Reaction for Barium Sulfate Formation

• The balanced reaction between barium chloride and sodium sulfate to produce barium sulfate and sodium chloride is:
  $BaCl<em>2(aq) + Na</em>2SO<em>4(aq) \rightarrow BaSO</em>4(s) + 2NaCl(aq)$

4.2 Limiting Reactant and Moles Calculation

• Starting with 50.0 g of barium chloride (MM: 208.23 g/mol) and 50.0 g of sodium sulfate (MM: 142.04 g/mol):
  • Calculate moles of $BaCl<em>2$: $\frac{50.0 \text{ g}}{208.23 \text{ g/mol}} = 0.240 \text{ mol } BaCl</em>2$.
  • Calculate moles of $Na<em>2SO</em>4$: $\frac{50.0 \text{ g}}{142.04 \text{ g/mol}} = 0.352 \text{ mol } Na<em>2SO</em>4$.
  • $BaCl<em>2$ is the limiting reactant because it produces fewer moles of $BaSO</em>4$.
  • Moles of $BaSO_4$ formed = 0.240 mol.
  • Moles of excess reactant ($Na<em>2SO</em>4$) leftover: 0.352 mol - 0.240 mol = 0.112 mol.

4.3 Molarity of Barium Chloride and Ions

• 50.0 g of $BaCl_2$ in 250 mL solution:
  • Molarity of $BaCl_2$: $\frac{0.240 \text{ mol}}{0.250 \text{ L}} = 0.960 \text{ M}$.
  • Molarity of $Ba^{2+}$ ion: 0.960 M.
  • Molarity of $Cl^-$ ion: $2 \times 0.960 \text{ M} = 1.92 \text{ M}$.

4.4 Molarity of Sodium Sulfate and Ions

• 50.0 g of $Na<em>2SO</em>4$ in 250 mL solution:
  • Molarity of $Na<em>2SO</em>4$: $\frac{0.352 \text{ mol}}{0.250 \text{ L}} = 1.408 \text{ M}$.
  • Molarity of $Na^+$ ion: $2 \times 1.408 \text{ M} = 2.816 \text{ M}$.
  • Molarity of $SO_4^{2-}$ ion: 1.408 M.

Part E: Limiting Reactant and Theoretical Yield

5. Theoretical Yield of Carbon Dioxide

• Gaseous methane ($CH<em>4$) reacts with gaseous oxygen ($O</em>2$) to produce gaseous carbon dioxide ($CO<em>2$) and gaseous water ($H</em>2O$).
• The balanced equation is: $CH<em>4 + 2O</em>2 \rightarrow CO<em>2 + 2H</em>2O$
• Starting with 0.64 g of $CH<em>4$ (MM: 16.04 g/mol) and 1.5 g of $O</em>2$ (MM: 32.00 g/mol):
  • Convert grams of $CH<em>4$ to moles: $\frac{0.64 \text{ g}}{16.04 \text{ g/mol}} = 0.0399 \text{ mol } CH</em>4$.
  • Convert grams of $O<em>2$ to moles: $\frac{1.5 \text{ g}}{32.00 \text{ g/mol}} = 0.0469 \text{ mol } O</em>2$.
  • Determine the limiting reactant:
    • For $CH<em>4$: 0.0399 mol $CH</em>4$ would require $2 \times 0.0399 = 0.0798 \text{ mol } O_2$.
    • Since we only have 0.0469 mol $O_2$, oxygen is the limiting reactant.
  • Calculate the moles of $CO<em>2$ produced from $O</em>2$: $\frac{0.0469 \text{ mol } O<em>2}{1} \times \frac{1 \text{ mol } CO</em>2}{2 \text{ mol } O<em>2} = 0.02345 \text{ mol } CO</em>2$.
  • Convert moles of $CO<em>2$ to grams: $0.02345 \text{ mol } CO</em>2 \times 44.01 \text{ g/mol} = 1.03 \text{ g } CO_2$.