Chemical Concepts and Calculations

Composition of one molecule of Citric Acid:
- $6$ Carbon atoms
- $8$ Hydrogen atoms
- $7$ Oxygen atoms
Composition of one mole of Citric Acid:
- $6$ moles of Carbon atoms
- $8$ moles of Hydrogen atoms
- $7$ moles of Oxygen atoms
Calculation of Molecular Weight:
- Carbon ( $C$ ): $6 \times 12.011 \text{ g/mol} = 72.066 \text{ g/mol}$
- Hydrogen ( $H$ ): $8 \times 1.008 \text{ g/mol} = 8.064 \text{ g/mol}$
- Oxygen ( $O$ ): $7 \times 15.999 \text{ g/mol} = 111.993 \text{ g/mol}$
- Total Molecular Weight of Citric Acid: $72.066 + 8.064 + 111.993 = 192.123 \text{ g/mol}$
- This means $1$ mole of Citric Acid weighs $192.123 \text{ g}$ and contains $6.022 \times 10^{23}$ molecules.

Question: How many grams of Carbon atoms are in $44.4$ grams of Citric Acid ( $C6H8O_7$ )?
Solution Steps:
1. Convert grams of Citric Acid to moles of Citric Acid.
2. Use the molar ratio to convert moles of Citric Acid to moles of Carbon.
3. Convert moles of Carbon to grams of Carbon.
Calculation:
$44.4 \text{ g Citric Acid} \times \frac{1 \text{ mol Citric Acid}}{192.123 \text{ g Citric Acid}} \times \frac{6 \text{ mol C}}{1 \text{ mol Citric Acid}} \times \frac{12.011 \text{ g C}}{1 \text{ mol C}} = 16.63 \text{ g Carbon}$
Note: The transcript shows a slight calculation error leading to $116.7$ grams, but the corrected calculation is $16.63$ grams based on the provided values.

Question: What is the percent of Carbon in $44.4$ grams of Citric Acid?
Using Mass of Carbon in Sample:
- $\text{Percent Carbon} = \frac{\text{Mass of Carbon}}{\text{Total Mass of Citric Acid}} \times 100$ (using value from previous calculation)
- $\text{Percent Carbon} = \frac{16.63 \text{ g}}{44.4 \text{ g}} \times 100 = 37.45\%$ (approx. $37.5\%$ )
Using Molar Mass for Elemental Percent Composition in Citric Acid ( $C6H8O_7$ ):
- This method calculates the percentage of each element in a compound based on its chemical formula and atomic weights, providing a constant percentage for the compound.
- Carbon Percentage:
 - $\frac{(\text{Moles of C})(\text{Atomic Mass of C})}{\text{Molecular Weight of Citric Acid}} \times 100$
 - $\frac{(6)(12.011 \text{ g/mol})}{192.123 \text{ g/mol}} \times 100 = 37.51\%$ (approx. $37.5\%$ )
- Hydrogen Percentage:
 - $\frac{(\text{Moles of H})(\text{Atomic Mass of H})}{\text{Molecular Weight of Citric Acid}} \times 100$
 - $\frac{(8)(1.008 \text{ g/mol})}{192.123 \text{ g/mol}} \times 100 = 4.19\%$ (approx. $4.23\%$ )
- Oxygen Percentage:
 - $\frac{(\text{Moles of O})(\text{Atomic Mass of O})}{\text{Molecular Weight of Citric Acid}} \times 100$
 - $\frac{(7)(15.999 \text{ g/mol})}{192.123 \text{ g/mol}} \times 100 = 58.30\%$ (approx. $58.3\%$ )
- Note: The sum of these percentages should be approx. $100\%$ ( $37.51 + 4.19 + 58.30 = 100.00\%$ )

Signs that a Chemical Reaction has Occurred:
- Temperature Change:
  - If the temperature of the reaction mixture decreases, the reaction is Endothermic (absorbs heat from surroundings).
  - If the temperature of the reaction mixture increases, the reaction is Exothermic (releases heat to surroundings).
- Color Changes: Indicates a change in the chemical identity of substances.
- Changes in Phase (e.g., precipitate forms): Formation of a new solid (precipitate) from a solution, or gas formation (bubbles).

Basic Structure:
- Reactants: The starting materials, shown on the left side of the arrow.
- Products: The substances formed at the end of the reaction, shown on the right side of the arrow.
- Arrow: Represents the transformation from reactants to products.

Context: After performing combustion analysis, an empirical formula (the simplest whole-number ratio of atoms in a compound) can be determined. To find the molecular formula (the actual number of atoms of each element in a molecule), the molecular weight is also needed.
Technique for Molecular Weight: Mass Spectrometry is a technique that can provide the molecular weight of a compound.
Example:
- Given: Empirical formula $CHO$
- Given: Molecular weight is determined to be $116 \text{ g/mol}$
- Objective: Determine the molecular formula.
- Steps:
 1. Calculate the empirical formula mass (EFM) of $CHO$ :
 - $1 \times 12.011 \text{ (C)} + 1 \times 1.008 \text{ (H)} + 1 \times 15.999 \text{ (O)} = 29.018 \text{ g/unit}$
 2. Divide the molecular weight (MW) by the empirical formula mass (EFM) to find the scaling factor ( $n$ ):
 - $n = \frac{\text{MW}}{\text{EFM}} = \frac{116 \text{ g/mol}}{29.018 \text{ g/unit}} \approx 4$
 3. Multiply the subscripts in the empirical formula by $n$ to get the molecular formula:
 - $(CHO)4 = C4H4O4$

Definition: A solution where water is the solvent.
Solvent: The component of a solution present in the largest number of moles (in aqueous solutions, this is water).
Solute: The substance that is dissolved in the aqueous solution.

Molarity (M): A measure of the concentration of a solute in a solution.
Formula:
$Molarity (M) = \frac{\text{Moles of Solute}}{\text{Liters of Solution}}$
Example Calculation:
- Question: What is the molarity of a $155 \text{ mL}$ solution that contains $0.200$ moles of glucose?
- Given:
  - Moles of solute (glucose) = $0.200 \text{ moles}$
  - Volume of solution = $155 \text{ mL}$
- Steps:
  1. Convert the volume of the solution from milliliters ( $mL$ ) to liters ( $L$ ):
    - $155 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.155 \text{ L}$
  2. Apply the molarity formula:
    - $M = \frac{0.200 \text{ moles}}{0.155 \text{ L}} = 1.2903 \text{ M}$
    - Result (rounded): $1.29 \text{ M}$ (or $1.30 \text{ M}$ as shown in transcript, likely due to rounding to two decimal places).