Question

How many grams does 5.29 x 10^25 molecules of AgNO3 weigh?

Answer 1

14900 g AgNO₃

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

Stoichiometry

• Using Dimensional Analysis

Step-by-step explanation:

Step 1: Define

[Given] 5.29 × 10²⁵ molecules AgNO₃

[Solve] grams AgNO₃

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Ag - 107.87 g/mol

[PT] Molar Mass of N - 14.01 g/mol

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

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

Step 3: Convert

1. [DA] Set up:
   `\displaystyle 5.29 \cdot 10^(25) \ molecules \ AgNO_3((1 \ mol \ AgNO_3)/(6.022 \cdot 10^(23) \ molecules \ AgNO_3))((169.88 \ g \ AgNO_3)/(1 \ mol \ AgNO_3))`
2. [DA] Multiply/Divide [Cancel out units]:
   `\displaystyle 14923 \ g \ AgNO_3`

Step 4: Check

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

14923 g AgNO₃ ≈ 14900 g AgNO₃