Question

In a 3.7- gram sample, there are______ atoms of gold

Answer 1

1.1 × 10²² atoms Au

General Formulas and Concepts:

Chemistry

Atomic Structure

• Reading a Periodic Table
• Using Dimensional Analysis
• Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Math

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Step-by-step explanation:

Step 1: Define

3.7 g Au

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of Au - 196.97 g/mol

Step 3: Convert

`3.7 \ g \ Au((1 \ mol \ Au)/(196.97 \ g \ Au) )((6.022 \cdot 10^(23) \ atoms \ Au)/(1 \ mol \ Au) )` = 1.13121 × 10²² atoms Au

Step 4: Check

We are given 2 sig figs. Follow sig fig rules and round.

1.13121 × 10²² atoms Au ≈ 1.1 × 10²² atoms Au