Question

How many grams are in 3.93 x 10^24 molecules of CCl4?

Answer 1

1000 g CCl₄

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
• Using Dimensional Analysis
• Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Step-by-step explanation:

Step 1: Define

3.93 × 10²⁴ molecules CCl₄

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of CCl₄ - 12.01 + 4(35.45) = 153.81 g/mol

Step 3: Convert

1. Set up:
   `\displaystyle 3.93 \cdot 10^(24) \ molecules \ CCl_4((1 \ mol CCl_4)/(6.022 \cdot 10^(23) \ molecules \ CCl_4))((153.81 \ g \ CCl_4)/(1 \ mol \ CCl_4))`
2. Multiply:
   `\displaystyle 1003.77 \ g \ CCl_4`

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1003.77 g CCl₄ ≈ 1000 g CCl₄