Question

1.204 x 10^24 molecules of CH3COOH is how many grams?

Answer 1

120.1 g CH₃COOH

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Chemistry

Atomic Structure

• Reading a Periodic Table
• Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

• Using Dimensional Analysis

Step-by-step explanation:

Step 1: Define

1.204 × 10²⁴ molecules CH₃COOH

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of CH₃COOH - 12.01 + 3(1.01) + 12.01 + 16.00 + 16.00 + 1.01 = 60.06 g/mol

Step 3: Convert

1. Set up:
   `\displaystyle 1.204 \cdot 10^(24) \ molecules \ CH_3COOH((1 \ mol \ CH_3COOH)/(6.022 \cdot 10^(23) \ molecules \ CH_3COOH))((60.06 \ g \ CH_3COOH)/(1 \ mol \ CH_3COOH))`
2. Multiply/Divide:
   `\displaystyle 120.08 \ g \ CH_3COOH`

Step 4: Check

Follow sig fig rules and round. We are given 4 sig figs.

120.08 g CH₃COOH ≈ 120.1 g CH₃COOH