Question

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

Answer 1

Answer:

1.39 × 10²³ particles CuCr₂O₇

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

Explanation:

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

1. [DA] Set up:                                                                                                     $\displaystyle 64.5 \ g \ CuCr_2O_7(\frac{1 \ mol \ CuCr_2O_7}{279.55 \ g \ CuCr_2O_7})(\frac{6.022 \cdot 10^{23} \ particles \ CuCr_2O_7}{1 \ mol \ CuCr_2O_7})$
2. [DA] Divide/Multiply [Cancel out units]:                                                         $\displaystyle 1.38944 \cdot 10^{23} \ particles \ CuCr_2O_7$

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