Answer:
3.2 g CaCO₃
General Formulas and Concepts:
Math
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
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
- Set up:

- Multiply/Divide:

Step 4: Check
Follow sig fig rules and round. We are given 2 sig figs.
3.15794 g CaCO₃ ≈ 3.2 g CaCO₃