How many particles of CuCr2O7 are present in a 64.5 gram sample?

1.39 × 10²³ particles CuCr₂O₇

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] 64.5 g CuCr₂O₇

[Solve] particles CuCr₂O₇

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Cu - 63.55 g/mol

[PT] Molar Mass of Cr - 52.00 g/mol

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

Molar Mass of CuCr₂O₇ - 63.55 + 2(52.00) + 7(16.00) = 279.55 g/mol

Step 3: Convert

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

Step 4: Check

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

1.38944 × 10²³ particles CuCr₂O₇ ≈ 1.39 × 10²³ particles CuCr₂O₇

The correct answer are

Vectors have a size and a direction.
Vectors are indicated with arrows.
Vectors can have positive or negative values