Question

How many molecules are present in 98.2g of NaNO3

Answer 1

6.96 × 10²³ molecules NaNO₃

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

98.2 g NaNO₃

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of Na - 22.99 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of NaNO₃ - 22.99 + 14.01 + 3(16.00) = 85.00 g/mol

Step 3: Convert

1. Set up:
   `\displaystyle 98.2 \ g \ NaNO_3((1 \ mol \ NaNO_3)/(85.00 \ g \ NaNO_3))((6.022 \cdot 10^(23) \ molecules \ NaNO_3)/(1 \ mol \ NaNO_3))`
2. Multiply:
   `\displaystyle 6.95718 \cdot 10^(23) \ molecules \ NaNO_3`

Step 4: Check

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

6.95718 × 10²³ molecules NaNO₃ ≈ 6.96 × 10²³ molecules NaNO₃