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

14900 g AgNO₃

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Chemistry

Atomic Structure

Reading a Periodic TableMolesAvogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional AnalysisExplanation:

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

[DA] Set up:                                                                                                        [DA] Multiply/Divide [Cancel out units]:                                                          

Step 4: Check

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

14923 g AgNO₃ ≈ 14900 g AgNO₃

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