Question

What is the mass of 1.9 x 10^22 molecules of CaCO3?

Answer 1

3.2 g CaCO₃

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

1.9 × 10²² molecules CaCO₃

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Ca - 40.08 g/mol

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

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

Molar Mass of CaCO₃ - 40.08 + 12.01 + 3(16.00) = 100.09 g/mol

Step 3: Convert

1. Set up:
   `\displaystyle 1.9 \cdot 10^(22) \ molecules \ CaCO_3((1 \ mol \ CaCO_3)/(6.022 \cdot 10^(23) \ molecules \ CaCO_3))((100.09 \ g \ CaCO_3)/(1 \ mol \ CaCO_3))`
2. Multiply/Divide:
   `\displaystyle 3.15794 \ g \ CaCO_3`

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

3.15794 g CaCO₃ ≈ 3.2 g CaCO₃